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Golf superstar Inbee Park comes Down Under
South Korean great Inbee Park has entered next month's Vic Open and Women's Australian Open.
13-Jan-2020 By Australian Associated Press
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Golf superstar Inbee Park will return to Australia for the first time in six years to contest next month's Vic Open and Women's Australian Open.
The winner of seven major titles and the Rio Olympic gold medal, South Korean Park has not played Down Under since the 2014 Australian Ladies Masters in Queensland.
Recently voted the player of the decade, 31-year-old Park has committed to play the two LPGA Tour co-sanctioned events - the Vic Open at 13th Beach (Feb 6-9) and the Australian Open at Royal Adelaide (Feb 13-16).
She enters the 2020 LPGA Tour season ranked fourth on the career money list with $US15,356,126 in career earnings, third-most among active LPGA Tour players.
"I am so excited to be going back to Australia for these two tournaments in February," said Park.
"This is my first time in the country for a few years and I have heard so much about the Vic Open with its men's and women's fields, as well as good memories from my previous visits for the Australian Open."
Park's seven majors put her alongside legends like Juli Inkster and Australia's Karrie Webb, with only Patty Berg (15), Mickey Wright (13), Louise Suggs (11), Babe Zaharias (10), Annika Sorenstam (10) and Betsy Rawls (eight) ahead of them.
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2016 Weingut Von Schubert-Maximin Grünhauser, Riesling Kabinett, Abtsberg, Mosel, Germany.
The beautiful Kabinett Abtsberg 2016 is a crystal clean and mineral driven Riesling that while off dry and generous has a dry feel and complexity of flavors that make it far more serious than you'd expect. Brilliant layers of classic green apple, lime and peach fruits along with hints of earth, wet slate and a touch of tropical essences all come through on the vivid light/medium palate, this wine really impresses for it's detail and quality in the glass. The Maximin Grünhaus estate lies at the foot of a long steep south-facing slope on the left bank of the Ruwer river, about two kilometres before it joins the Mosel. The history of this place shines through in the wines and I am always intrigued to hear about it's past, the estate belongs to the Von Schubert family and is divided into three separate but contiguous vineyards: the Abtsberg, the Herrenberg, and the Bruderberg. According to the winery, each of these vineyards has its own distinct micro-climate, gradient and soil composition, differences in terroir which explain the unique character of the wines made at Grünhaus. The Abtsberg wines harvested from this vineyard were originally destined for the table of the Abbot (or Abt) of the Abbey of St Maximin. The site covers 14 hectares, parts of which have been planted with vines for over a thousand years, set on blue Devonian slate and the hillside runs south-east to south-west, achieving a gradient (slope angel) of nearly 70%. These Abtsberg wines, again according to the estate, are characterized by a finely structured subtle minerality, a racy acidity, generous fruit and great delicacy, of which I totally agree, especially with this Kabinett by Maximin Grünhaus, and they are also amongst the longest lived Rieslings of the region. Stefan Kraml took charge of viticulture and viniculture at Grünhaus. Since that time, the wines of the estate have benefited from his uncompromisingly high standards, the Grünhaus cuverie is sited conveniently close to the vineyards so that harvested grapes can be delivered to the press within a matter of minutes. Grünhaus employs a natural clarifying overnight via sedimentation, then the musts are brought to an optimal fermentation temperature, and fermented with naturally occurring wild yeasts (Sponti) in classic large oak casks (Fuder casks) or in some cases using small stainless steel tanks. These techniques promote optimal development of mineral intensity and fruit purity from the Abtsberg slate soils and ensures the longevity of their Rieslings. Neither the must nor the finished wine are fined and uniquely in the last two years most all the the oak for the barrels has come from the estate's own forests, and the barrels are crafted by a local cooper, everything done here is to highlight their terroir. I love these wines, they are not flashy or overt, but oh so pleasing and stylish, they are like drinking in history, enjoy the Abtsberg Kabinett over the next 5 to 10 years.


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Q: sending JSON with go I'm trying to send a JSON message with Go.
This is the server code:
func (network *Network) Join(
    w http.ResponseWriter,
    r *http.Request) {
    //the request is not interesting
    //the response will be a message with just the clientId value set
    log.Println("client wants to join")
    message := Message{-1, -1, -1, ClientId(len(network.Clients)), -1, -1}
    var buffer bytes.Buffer
    enc := json.NewEncoder(&buffer)

    err := enc.Encode(message)
    if err != nil {
        fmt.Println("error encoding the response to a join request")
        log.Fatal(err)
    }

    fmt.Printf("the json: %s\n", buffer.Bytes())
    fmt.Fprint(w, buffer.Bytes())
}

Network is a custom struct. In the main function, I'm creating a network object and registering it's methods as callbacks to http.HandleFunc(...)
func main() {
    runtime.GOMAXPROCS(2)
    var network = new(Network)
    var clients = make([]Client, 0, 10)
    network.Clients = clients

    log.Println("starting the server")
    http.HandleFunc("/request", network.Request)
    http.HandleFunc("/update", network.GetNews)
    http.HandleFunc("/join", network.Join)
    log.Fatal(http.ListenAndServe("localhost:5000", nil))
}

Message is a struct, too. It has six fields all of a type alias for int.
When a client sends an http GET request to the url "localhost:5000/join", this should happen

*

*The method Join on the network object is called

*A new Message object with an Id for the client is created

*This Message is encoded as JSON

*To check if the encoding is correct, the encoded message is printed on the cmd

*The message is written to the ResponseWriter

The client is rather simple. It has the exact same code for the Message struct. In the main function it just sends a GET request to "localhost:5000/join" and tries to decode the response. Here's the code
func main() {

    // try to join
    var clientId ClientId
    start := time.Now()
    var message Message
    resp, err := http.Get("http://localhost:5000/join")
    if err != nil {
        log.Fatal(err)
    }

    fmt.Println(resp.Status)
    dec := json.NewDecoder(resp.Body)
    err = dec.Decode(&message)
    if err != nil {
        fmt.Println("error decoding the response to the join request")
        log.Fatal(err)
    }

    fmt.Println(message)
    duration := time.Since(start)
    fmt.Println("connected after: ", duration)
    fmt.Println("with clientId", message.ClientId)
}

I've started the server, waited a few seconds and then ran the client. This is the result

*

*The server prints "client wants to join"

*The server prints "the json: {"What":-1,"Tag":-1,"Id":-1,"ClientId":0,"X":-1,"Y":-1}"

*The client prints "200 OK"

*The client crashes "error decoding the response to the join request"

*The error is "invalid character "3" after array element"

This error message really confused me. After all, nowhere in my json, there's the number 3. So I imported io/ioutil on the client and just printed the response with this code
b, _ := ioutil.ReadAll(resp.Body)
fmt.Printf("the json: %s\n", b)

Please note that the print statement is the same as on the server. I expected to see my encoded JSON. Instead I got this

*

*"200 OK"

*"the json: [123 34 87 104 97 116 ....]"  the list went on for a long time

I'm new to go and don't know if i did this correctly. But it seems as if the above code just printed the slice of bytes. Strange, on the server the output was converted to a string.
My guess is that somehow I'm reading the wrong data or that the message was corrupted on the way between server and client. But honestly these are just wild guesses.

A: In your server, instead of
fmt.Fprint(w, buffer.Bytes())

you need to use:
w.Write(buffer.Bytes())

The fmt package will format the Bytes() into a human-readable slice with the bytes represented as integers, like so:
[123 34 87 104 97 116 ... etc

A: You don't want to use fmt.Print to write stuff to the response.  Eg
package main

import (
    "fmt"
    "os"
)

func main() {
    bs := []byte("Hello, playground")
    fmt.Fprint(os.Stdout, bs)
}

(playground link)
Produces
[72 101 108 108 111 44 32 112 108 97 121 103 114 111 117 110 100]

Use the Write() method of the ResponseWriter instead
You could have found this out by telneting to your server as an experiment - always a good idea when you aren't sure what is going on!

redpajama_set_name: RedPajamaStackExchange


text:
28Matthew McHugh
28John Edward Porter
19Bob Dornan
19David Ross Obey
19James A. Traficant Jr.
18Bob Walker
16Helen Delich Bentley
15Benjamin Arthur Gilman
14Dan Burton
14Mary Rose Oakar
Matthew McHugh John Edward Porter
Showing 1 - 20 of 28 Show 28
House Session
H.R. 5368: A bill to make appropriations for foreign operations, export financing, and related programs. H.R. 5504: A bill to make appropriations for the Department of Defense. H.C.Res. 376: A resolution to provide for the preparation of official duplicates of certain legislative papers. S. 776: A bill to provide for improved energy effeciency. H.R. 5013:…
View in Congressional Chronicle503 views
August 11, 1992 Last Aired August 11, 1992
HR 5487: A bill to make appropriations for Agriculture, Rural Development, Food and the Drug Administration. HR 549: A resolution to elect Representative Mfume of Maryland to the Committee on Standards of…
View in Congressional Chronicle97 views
June 30, 1992 Last Aired June 30, 1992
HR 5487: A bill to make appropriations for agriculture, rural development, and the Food and Drug Administration for the fiscal year ending Semptember 30, 1993. S 2905: A bill to provide a…
HR 5368: A bill to make appropriations for foreign operations, export financing, and related programs for the fiscal year ending September 30, 1993. HR 429: A bill to amend certain federal…
HR 4364: A bill to authorize appropriations to the National Aeronautics and Space Administration for research and development, space flight, control and data communications, construction of…
The House approved the Journal of Tuesday, June 18, 1991. The House continued consideration of H.R. 2508, to amend the Foreign Assistance Act of 1961. The House agreed to the Roth (R-WI)…
The House continued consideration of H.R. 2508, to amend the Foreign Assistance Act of 1961. The Hall (D-OH) amendment en bloc that earmarks $335 million in fiscal year 1992, $135 million…
The House agreed to S. 3266, to reauthorize the Small Business Act of 1958. The House also agreed to H.Res. 538, amending the Immigration and Nationality Act, and H.Con.Res. 399, providing…
The House agreed to the conference report on S. 1430, to enhance national and community service. A letter from Rep. Donald E. Lukens, of Ohio, was read, wherein he resigned as a member of…
The House passed H.R. 5268 , making appropriations for Rural Development, Agriculture, and Related Agencies programs, and miscellaneous amendments. The House passed H.R. 5258, to require…
The House passed H.R. 5114, making appropriations for foreign operations for the fiscal year ending September 30, 1991. The Smith(R-NJ) amendment to allow the Agnecy for International…
March 15, 1990 Last Aired March 18, 1990
Promoting Political Stability, Part 1
The House Appropriations Subcommittee on Foreign Operations heard testimony concerning programs that promote economic freedom and political stability abroad.
March 1, 1990 Last Aired March 4, 1990
Foreign Aid Appropriations: Fiscal Year 1991
The subcommittee heard testimony from Secretary of State Baker concerning the fiscal year 1991 foreign operations request.
Under suspension of the rules the House passed several measures. Of special interest was H.R. 1602, the Trauma Care Systems Planning and Development Act to amend the Public Health Service…
The House debated H.R. 2916 appropriations bill for the Departments of Veterans' Affairs, Housing and Urban Development, and other independent agencies, boards, and commissions for FY90.…
May 1, 1989 Last Aired May 22, 1989
Democratic Revolution in Asia
Foreign Aid for Fiscal Year 1990
The subcommittee held a hearing to authorize funding for the Department of State.
The House agreed to the conference report on H.R. 3471, to establish the Veterans' Administration as an executive department. The House agreed to S.Con.Res. 154, to correct additional…
July 28, 1988 Last Aired September 28, 1988
The House agreed to S. Con. Res. 144, correcting the enrollment of S.J. Res. 317, commemorating the bicentennial of the French Revolution and the Declaration of the Rights of Man and of the…
September 14, 1988 Last Aired September 14, 1988
The House agreed to the Coelho motion to table the Walker motion to discharge the Committee on Armed Services from further consideration of H.R. 4264, National Defense Authorization Act for…

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First Professed Conference in Freeland.
First Professed Brothers met in Freeland from Friday 17th until Monday 21st April. It was a wonderfully relaxing time and a delight, as always, to spend time with the Second Order Sisters. Pictured are all of the First Professed Brothers (with the exception of Br Joseph Emmanuel who was taking the photograph) and Br Amos (currently the Guardian of the First Professed).
From Monday 22nd September until Friday 26th September six SSF Brothers joined with other Novices at Whitby to study the Enneagram. Although not a definitive statement on one's personality the Enneagram can shed some interesting insights into the way one reacts and where points of stress may occur in community life; a useful tool for the Religious Life!
First Professed Conference at Wantage.
The First Professed Brothers and Sisters.
until Friday 19th September. Over the three days we discussed using art as a form of prayer; Meister Eckhart and the Camino de Santiago. In addition there was plenty of time for reflection and we received a warm and generous welcome from the Sisters of the Community.
A fond farewell in Newcastle.
A service of thanksgiving was held in St Anthony's Church for the ministries of Brs David, Micael Cristoffer and Malcolm as they move on to new assignments. The three Brothers each chose a piece of Scripture or prose to reflect upon and then a favourite hymn. Canon Stephen Herbert (co-ordinator of MINE) officiated at the service along with Br Damian SSF. Afterwards the congregation – who consisted of parishioners and clergy of the MINE Parishes, members of the Third Order of the Society of St Francis and Brothers of the First Order – were treated to a lovely tea.


Risky business: when will the threat of natural disaster hit property prices?
Recently, Mexico City's civil society illustrated how an empowered community naturally coalesces and mitigates a disaster's destruction– ultimately finding itself better prepared and quicker to recover.
While building resilience to major shocks entails strengthening infrastructure, creating effective early warning systems and ensuring efficient government emergency procedures, addressing social stresses is also critical. In case after case, the cities and communities that fare best when disaster strikes, are those where social cohesion is already well-developed, and integrated into the other systems of the city.
With the growing frequency and severity of natural disasters, understanding this becomes even more crucial.
This entry was posted on 06/12/2017 by Jelenko Dragisic in Uncategorized and tagged children and natural disasters, children resilience, Mexico earthquake, Mexico earthquake recovery, property market and natural disasters.


For the better part of the last century, the jurisprudential focus in the U.S. has been on cognizable harms, or damages, resulting from statutory or common law privacy invasions. Courts almost invariably require a showing of damages — mostly financial — before a victim may be due some remedy. Because lawyers are trained in the law, and the bulk of privacy practitioners are lawyers, cognizable harms remain the common misconception about how to frame privacy risk.
But from a privacy-by-design perspective, harm prevention is not and should not be the primary goal. Rather, the goal should be a reduction in the incidences of privacy violations. Isn't that synonymous with harm reduction, you may be asking?
Certainly, the recent proposal from Intel frames privacy risk (see 3(h)) in terms of adverse consequences, with a fairly expansive definition of them. But consequences are not risks and privacy risk needn't involve harms. You might be familiar with the formulation of risk as a combination of likelihood and impact. In this formula, most people equate impact as harms, but as I illustrate below, these are not synonyms.
Consider this example: I have decided, without your consent, to put a video camera in your home. Most people would agree that this is a violation of your privacy, an invasion into your sanctum. Under the Solove Taxonomy this would be a surveillance violation. If you were to discover or be made aware of this camera, you might alter your behavior to avoid being seen. You might be embarrassed or upset at what the camera previously revealed. A hypersensitive individual might recoil in horror and might even commit suicide.
But what if you never found out? What if I didn't use the video in any way that might objectively harm you. I'm not selling it or disclosing it to another entity. I'm not using it to blackmail you or causing you to lose an opportunity, such as a loan or job. The surveillance is a violation, regardless of harms or damages to you. Even if you never found out or I never used it in a way that is disadvantageous to you, my installation of a camera in your home is not something society is willing to accept as appropriate behavior, as it violates our social norms in your personal and private space.
Now some will argue that there is a potential that you could uncover the camera causing embarrassment or a change of behavior. There is a potential the video might get leaked, leading to a lost job opportunity or friends. Yes, these possibilities exist. But now we're faced with a funny conundrum: What could I do to reduce these harms from taking place?
I could make the camera more obscure. I could encrypt the video feed. I could implement all sorts of controls that reduce the potential for the "harm" materializing. In other words, I could make it nearly impossible for you or others to find the camera or the video.
But none of this reduces the underlying violation — that of me invading your private space, surveilling your activities. In fact, these controls are antithetical to a core privacy principle: transparency. I'm hiding and obscuring my activity to prevent you from being harmed, but alas, I'm not doing anything to stop my actual violation of your privacy.
You may recall the incident with Target and their response to the issue of inferring a teenager's pregnancy. The situation allegedly arose when Target used big data analytics to deduce that a customer was likely pregnant based on her purchases. They then proceeded to send mailers to the customer offering various newborn related products. This inadvertently alerted a household member of the teenage customer's pregnancy.
Here we have two distinct violations: the secondary use of purchase information (to infer her pregnancy) and the disclosure to the household member. Target's response? To obfuscate the inference by designing their flyers with a mix of products, only slightly skewed towards those appealing to pregnant women.
While this reduced the probability of future household participants from learning of a Target customer's pregnancy via its targeted advertising, it did nothing to alleviate the violation of the secondary use of purchase information. Their mitigation actively defies the principle of transparency. Rather than sending a notice to the customer telling them "our analysis of your purchases suggests you are pregnant," or better yet alerting them prior to or at the time of purchase, the company actively obscured their inference in the flyers they sent (note others have suggested a less nefarious reason).
Other privacy-risk approaches also create incentives for organizations to hide privacy violations. Most "privacy" risk models are based on organizational risks (such as fines, lawsuits or reputational harm from being hauled before a legislative body). One recent example of obfuscation involves Facebook's hiding their Messenger App's collection of call data. As Damian Collins, member of the British Parliament, said, "To mitigate any bad PR, Facebook planned to make it as hard as possible for users to know that this was one of the underlying features"
For privacy-risk analysis to not create perverse anti-transparency incentives, one must incorporate a broader view of impact than harms.
In the risk framework I use in my book Strategic Privacy by Design, I do just this. The impact side of the equation incorporates both a notion of size of the affected population and secondary adverse consequences, which would be those financial and non-financial harms we're most familiar with.
Thus, while obscurity might reduce the chance an individual suffers the effects of embarrassment from finding a camera in their house, obscurity has a counter effect on the population affected (i.e. people would view a more covert camera in people's homes as more nefarious and offensive to social norms). If one knowingly invites my camera into their house, it's no longer a privacy violation. Transparency, control and consent all have the effect of transforming the activity (a camera in a home) from a violative act to one that is not.
There is a further factor, beyond awareness and consent, I use to determine whether an activity (a camera in a home) constitutes a violation (surveillance), namely benefit. For instance, a baby or a person with a severe disability may not be aware of or be able to consent, but if they are the primary beneficiary of their monitoring and that benefit outweighed the lack of awareness and consent, this activity does not constitute a privacy violation.
This is what makes privacy harder than security, because often there are broader social policy questions at play.
With various U.S. legislative proposals, NIST's new privacy risk framework, and the recently formed project committee for ISO to create a standard in consumer products all in the works, stakeholders should recognize that privacy-risk analysis must not be harm-focused but rather focused on the violation of people's privacy.
Privacy-risk analysis should induce violation reduction, not mask violations in the name of harm reduction.
library_books Is 2013 the Year of the Privacy Engineer?


Aquamarine belongs to the beryl gemstone family, and encompasses a range of blues and greens. Bead-grade aquamarine tends to have many interesting inclusions and numerous opaque areas. The more intense the color of this stone, the higher its value. Aquamarine is a hard gemstone, which makes it an excellent choice for jewelry that will be worn frequently. However, be sure to store aquamarine beads in a dark place, as they often become paler if left out in the sun. Aquamarine is one of the most popular birthstones for March.
6mm Aquamarine Gemstone Bead, Star Cut Round. 60-70 beads per 15-16" strand, with approx. 0.7-1.2mm hole. These pretty star-cut blue-green gemstones are a great look -- a little different than your standard faceted round gemstone beads!


Take a break from sightseeing and shopping at visit our hotel bar. Locals and visitors alike can be found socializing here with a glass of wine or the perfect pint. This eye-catching spot in the Malmö city center is ideal for sharing quality time over food and drinks.
This sophisticated space makes a great setting for leisurely gatherings with business associates or enjoying a simple drink with loved ones. The menu offers a great selection of foods and desserts as well.
Don't miss the opportunity to visit the hotel's Wine Lounge. Wines are served accompanied by a detailed description and available in a multitude of variations.


Bug decided that she wanted a softball themed birthday party this year. So of course I made it happen. I browsed pinterest and pinned several amazing ideas. Then I got to work.
For the invites I used a cd to trace my circle. Yellow paper and red sharpie equals a simple and quick softball invite.
For our game I had to enlist the hubs help. We made a corn hole type throwing game.
We used a piece of plywood and my hubs got to use his saw to cut out the circles. We spray painted the board white and then I used neon spray paint to make the ball. Red acrylic paint for the stripes and black paint pen for the words. We bought the corn bags off of amazon. We played 2 "innings" and broke the kids into 2 teams. Each kid had a turn to toss. Three "pitches or strikes" and if they left a "Man" on base the next batter(person throwing the bag) could advance the runner. We used a chalk board to help keep score.
The kids also made washer necklaces. We used acrylic paint and sponge brushes. I bought the washers (3/8 x 1-1/4) at Walmart. Once the paint was dry the kids could use paint pens. Once they were complete we sprayed them with sealant. I used silver ribbon to make the chain part of the necklace.
I went with simple decorations. Red and black streamers for her softball team colors. We blew up yellow balloons and I drew the lines on with a red sharpie. She wore her jersey and the hubs and I wore our coaches shirts. I really debated on having the other players that were attending to wear their jerseys, but I figured the people not from our team might have felt left out. With her being the only one it made her birthday more special I think.
Party favors were all food lol. They were made up of bubble gum, sunflower seeds and I made puppy chow with yellow sixlets sprinkled in. Plus the kids had their washer necklaces. No little cheap toys from this party!
Her cake was really simple to make. I used 2 round cake pans. I filled 1 pan a little fuller so the cake would round out a little. Iced it with yellow butter cream icing and added some red stripes. Boom cake was done. I used yellow sixlets around the base of the cake. It gave it a more finished look. Cupcakes were iced with chocolate butter cream icing. I was going to make them look like a bat, but that didn't look right as I had to written on the cupcakes.
I love that I can make my kids birthday parties special and individualized. I'm tired from the party, but so worth it!


Magnificent - with the colours and sensuous shapes reminded me of the Rolling Stones logo.
Beautiful and striking colors warning the cars following behind in the dark.
I love the almost graphic nature of this. And I'm digging how the right side fades to black and keeps the emphasis on the tail light.
Is this off of an Audi? I love how the photo fades to black. Was this taken at night? (probably not, now that I'm looking at the EXIF info). 70-200mm 2.8? Fantastic lens!
Wow, just went back one more photo and saw the Audi there... That answers my first question!


text:
Sony ZV-1 Review | Trusted Reviews
The Sony ZV-1 is a huge leap forward in merging the accessibility of a smartphone and the proficiency of a dedicated camera into one device. It's not perfect, and we hope to see its overheating issues and convoluted menus resolved in its eventual successor, but as a first attempt, Sony has created the all-in-one device that vloggers have been waiting for.
Unbelievable mic quality
Top-tier autofocus
Instant bokeh mode
Can overheat indoors
Convoluted menus that are a pain to use
Micro-USB rather than USB-C
Review Price: £699
24-70mm f/1.8-2.8 lens
2.7x optical zoom
270 degree rotating screen
Footage up to 4K 25p
Looking to bridge the gap between smartphones and cameras, the Sony ZV-1 could be the Holy Grail that vloggers have been waiting for.
Of course, even with Sony's stellar track record, a first gen product is always guaranteed to have some quirks that'll no doubt be ironed out in the next iteration. Plus, at £699, the ZV-1 is no small purchase, especially for part-time vloggers.
Design – lovably light and compact for easy shooting
There's a case to be made that, if Sony genuinely wanted to attract the smartphone crowd, the ZV-1 would need to feel as portable and as lightweight as a smartphone – within reason of course.
The sight of bulky camera kits and elephant trunk lenses can be enough to send millennials running (myself included), so before a discussion about features could even take place, Sony would have to nail the design. Well, it's safe to say they've done it.
Despite its focus on video capture, the ZV-1 has more in common, design-wise, with the Sony RX series. While not quite as small as Sony's compact cameras however, the ZV-1 made my portable Canon M50 suddenly seem a lot larger by comparison.
At 10.5 x 4.4 x 6 cm, the ZV-1 can easily be placed in a pocket or a small camera bag, and you won't get tired holding it at arm's length. Despite its premium features, the ZV-1's grip is actually made of plastic and while it might not look like the most elegant option, it still did the job of keeping the camera firm in my hand.
Even though there's a solid built-in mic – and we'll talk more about that later – there is a hotshoe on the top if you want to go with your tried and tested set-up, alongside an expected tripod mount at the bottom. Sadly, there's no headphone input to be found here, although I've yet to see any vlogger constantly checking the audio levels whilst filming. What is baffling however is the inclusion of a micro-USB port over USB-C. Come on Sony, USB-C should be a given for any device of this price point.
Aiming to please its intended audience, there are fewer physical buttons to be found on the ZV-1 than your typical camera, with the rear-facing control panel featuring a scroll wheel that encompasses a larger button, surrounded by four smaller buttons for navigation and switching between different modes.
It's familiar territory for camera enthusiasts, but it doesn't scream 'accessibility' in the same way that its revamped top button layout can boast. Showing its intentions as a video-first device, the ZV-1 packs a huge red button for recording, alongside a smaller shutter button for photography, and a third shutter dedicated entirely to enacting the camera's bokeh mode.
Features and performance – a dream come true for vloggers
As an all-in-one device, the ZV-1 needed to cram a ton of features into a small amount of retail space. That's no easy feat, but leave it up to Sony to absolutely hit this aspect right out of the park.
Let's start with the aforementioned bokeh mode – certainly one of the more heavily publicised features of the ZV-1. On the top of the camera, nestled away from the shutter press and record button, there's a third shutter dedicated to enabling a live bokeh mode, where the background is blurred out in real time, keeping the focus solely on your face.
The feature works incredibly well – it's fast, easy to use and really sells the premise that the ZV-1 is an approachable camera for newcomers.
As impressed as I was with the bokeh mode, I was more interested in testing out the ZV-1's in-built mic. I'm a bit of a stickler when it comes to good audio quality, particularly during vlogging when there's more of a chance of too many competing sounds ruining a take.
The video below was filmed entirely on the ZV-1 and as you can no doubt hear – the audio quality is excellent. The camera's 3-capsule mic did a fantastic job of honing in on my voice despite the chaos of Central London occurring in the background.
The only time I had any noticeable issue was when a fleet of motorcyclists used the road encompassing Trafalgar Square, but even then their noisy engines still weren't enough to drown out my voice completely. There's also a wind screen attachment included in the box for an extra helping hand during some of those more blustery days.
As you can no doubt tell, I made the rookie mistake of forgetting to switch on the camera's stabilisation features, but they can be seen on full display in a video further down this review. There are two options here, standard and active stabilisation, although opting for the latter does crop the footage so you'll want a decent amount of distance between yourself and the camera if you decide to use it.
Appealing directly to beauty vloggers and tech journalists alike, Sony's also included a new 'Product Showcase' feature. When switched on, product showcase allows the lens to immediately focus on any item that you put in front of it. For example, if I wanted to go immediately from a shot of my face to the brand new Samsung Galaxy S20 Ultra, I could simply bring the device up to the camera and it would automatically shift the focus away from myself to the phone. Given that a fair amount of cameras struggle with this, or require some physical interaction to get the focus to shift, having this feature just makes the filming process so much easier.
As you'd expect, no vlogging camera is complete without a swivelling screen, and the ZV-1's no exception. You can flip the screen out 180 degrees so you can keep track of everything while you're filming, or swivel it within a 270 degree rotation instead if you're trying to go for a more artistic shot. While it's a minor addition here, the ability to switch off the ZV-1 by simply closing the screen is something that I became immediately addicted to and never looked back.
All of these features combine to make an unbelievably streamlined process – the only part of the camera that doesn't meet this standard is in its navigation. Diving into the ZV-1's menus, the camera falls into the same trap of having an abundance of pages and categories that, while fine for a professional, are bound to put off any amateurs.
This isn't a deal breaker by any means, but it does seem like a strange oversight when Sony's gone out of its way to make the rest of the ZV-1 as user-friendly as possible.
What could be a deal breaker however is the ZV-1's propensity for overheating. Taking the camera out and about isn't an issue, but I found things to be quite the contrary whilst filming indoors. Utilising a traditional filming set-up, with two softbox lights on either side of the camera, the ZV-1 had to call it quits after half an hour.
Granted, I do live in a top-floor apartment where the general temperature is going to be slightly warmer, but other reviews have also noted this issue, so if you are looking for a camera solely for indoor filming then the ZV-1 might not be your best bet.
Video and image quality – bright, vibrant and exactly what's needed
The Sony ZV-1 can film in up to 4K 25p 100m, which is perfectly fine for vlogging but not the full 60fps experience to satisfy any adventurers. In that regards, the GoPro Hero 8 Black still remains the top dog.
Throughout our testing, footage remained bright, vibrant and even more importantly, in focus. With other cameras, I've usually come across a threshold where the auto-focus reaches its limit and has a hard time adjusting, but I was unable to find such a threshold with the ZV-1.
The inclusion of a 24-70mm f/1.8-2.8 lens and a one-inch sensor is key to the ZV-1's success, as it's the ideal setup specifically for the vlogging format, even though it does sacrifice the ability to dive into wide, landscape shooting.
Being a camera, you can indeed take still shots with the Sony ZV-1, and for the most part, they do the trick just fine. There's not a huge amount of versatility but the aforementioned auto-focus allows users to snap between the points of a scene that they want to hone in on, just with a tap of the touchscreen.
Should you buy the Sony ZV-1?
It doesn't matter if you're a newcomer or an established online personality, the Sony ZV-1 offers a class-leading all-in-one experience that's near-perfect for vloggers. The unbelievable microphone quality, complemented by top-tier autofocus and a dedicated bokeh button makes it an absolute joy to use.
Until Sony can work out the overheating problem, the ZV-1 is best avoided for strictly indoor filming, and the convoluted menus and use of Micro-USB prevent it from being the future-proof camera we'd hoped for, but the ZV-1 has set a new bar for what a vlogging camera can be.
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text:
\section{Almost vanishing of local cohomology}

The concept of almost vanishing that we use here comes out of recent work on \emph{Almost Ring Theory} by Gabber and Ramero \cite{GR}. This theory was developed to give a firm foundation to the results of Faltings on \emph{Almost \'etale extensions} \cite{Faltings}, and these ideas have their origins in a classic work of Tate on \emph{$p$-divisible groups} \cite{Tate}. The use of the general theory, for our purposes, is comparatively straightforward, but it illustrates the main questions in looking at certain homological conjectures, as discussed later in the section. The approach is heavily influenced by Heitmann's proof of the Direct Summand Conjecture for rings of dimension three \cite{Heitmann-dim3}.

Let $A$ be an integral domain, and let $v$ be a valuation on $A$ with values in the abelian group of rational numbers; more precisely, $v$ is a function from $A$ to ${\mathbb Q}\cup\{\infty\}$ such that
\begin{enumerate}
\item $v(a)=\infty$ if and only if $a=0$,
\item $v(ab)=v(a)+v(b)$ for all $a,b\in A$, and
\item $v(a+b)\geqslant\min\{v(a),v(b)\}$ for all $a,b\in A$.
\end{enumerate}
We will also assume that $v(a)\geqslant 0$ for all elements $a\in A$.

\begin{definition}\label{def:almostzero}
An $A$-module $M$ is \emph{almost zero} if for every $m\in M$ and every real number $\varepsilon>0$, there exists an element $a$ in $A$ with $v(a)<\varepsilon$ and $am=0$. When it is necessary to specify the valuation, we say that \emph{$M$ is almost zero with respect to the valuation $v$}.
\end{definition}

We note some properties of almost zero modules:
\begin{enumerate}
\item For an exact sequence
\[
0\longrightarrow M'\longrightarrow M\longrightarrow M''\longrightarrow 0\,,
\]
the module $M$ is almost zero if and only if each of $M'$ and $M''$ is almost zero.

\item If $\{M_i\}$ is a directed system consisting of almost zero modules, then its direct limit $\varinjlim_iM_i$ is almost zero.
\end{enumerate}

In \cite{GR} Gabber and Ramero define a module to be almost zero if it is annihilated by a fixed ideal ${\mathfrak m}$ of $A$ with ${\mathfrak m}={\mathfrak m}^2$. This set of modules also satisfies conditions (1) and (2), though in many cases their condition is stronger than the one in Definition~\ref{def:almostzero}.

The \emph{absolute integral closure} $R^+$ of a domain $R$ is the integral closure of $R$ in an algebraic closure of its fraction field. An important situation for us will be where $(R,{\mathfrak m})$ is a complete local ring. In this case, fix a valuation $v\colon R\longrightarrow{\mathbb Z}\cup\{\infty\}$ which is positive on ${\mathfrak m}$. By Izumi's Theorem \cite{Izumi}, two such valuations are bounded by constant multiples of each other. Since $R^+$ is an integral extension, $v$ extends to a valuation $v\colon R^+\longrightarrow{\mathbb Q}\cup\{\infty\}$. Let $A$ be a subring of $R^+$ containing $R$; we often take $A$ to be $R^+$. Note that $v$ is positive on the maximal ideal of $A$. The ring $A$ need not be Noetherian, and by a \emph{system of parameters} for $A$, we shall mean a system of parameters for some Noetherian subring of $A$ that contains $R$.

The main question we consider is whether the local cohomology modules $H^i_{\mathfrak m}(A)$ are almost zero for $i<\dim A$. Let $x_1\dots,x_d$ be a system of parameters for $R$. Then the local cohomology module $H^i_{\mathfrak m}(A)$ is the $i$-th cohomology modules of the \v Cech complex
\[
0\longrightarrow A\longrightarrow\oplus A_{x_i}\longrightarrow\oplus A_{x_ix_j}\longrightarrow\cdots\longrightarrow A_{x_1\cdots x_d}\longrightarrow 0\,.
\]
The question whether $H^i_{\mathfrak m}(A)$ is almost zero for $i=0,\dots,d-1$ is closely related to the question whether the $x_i$ come close to forming a regular sequence in the following sense.

\begin{definition}
A sequence of elements $x_1,\dots,x_d\in A$ is an \emph{almost regular sequence} if for each $i=1,\dots,d$, the module
\[
((x_1,\dots,x_{i-1}):_Ax_i)/(x_1,\dots,x_{i-1})
\]
is almost zero. If every system of parameters for $A$ is an almost regular sequence, we say that $A$ is \emph{almost Cohen-Macaulay}.
\end{definition}

The usual inductive argument as in \cite[Theorem~IV.2.3]{Serre} shows that if $A$ is almost Cohen-Macaulay, then the modules $H^i_{\mathfrak m}(A)$ are almost zero for $i<\dim A$. However, we do not know whether the converse holds in general.

As motivation for the definitions introduced above, we discuss how these are related to the homological conjectures. Let $x_1,\dots,x_d$ be a system of parameters for a local ring $R$. Hochster's \emph{Monomial Conjecture} states that
\[
x_1^t\cdots x_d^t\notin\big(x_1^{t+1},\dots,x_d^{t+1}\big)R\qquad\text{ for all }
t\geqslant 0.
\]
This is known to be true for local rings containing a field, and Heitmann \cite{Heitmann-dim3} proved it for local rings of mixed characteristic of dimension up to three. It remains open for mixed characteristic rings of higher dimension, where it is equivalent to several other conjectures such as the Direct Summand Conjecture (which states that regular local rings are direct summands of their module-finite extension rings), the Canonical Element Conjecture, and the Improved New Intersection Conjecture; for some of the related work, we mention \cite{EG, Hochster-CBMS, Hochster-ds, PS} and \cite{Roberts-intersection}.

The connection between the Monomial Conjecture and the almost Cohen-Macaulay property is evident from the following proposition.

\begin{proposition}
Let $R$ be a local domain with an integral extension which is almost Cohen-Macaulay. Then the Monomial Conjecture holds for $R$, i.e., for each system of parameters $x_1,\dots,x_d$ of $R$, we have
\[
x_1^t\cdots x_d^t\notin\big(x_1^{t+1},\dots,x_d^{t+1}\big)R\qquad\text{ for all }t\geqslant 0\,.
\]
\end{proposition}

\begin{proof}
Let $A$ be an integral extension of $R$ which is almost Cohen-Macaulay with respect to a valuation $v$ which is positive on the maximal ideal of $R$. Then $v(x_i)>0$ for each $i=1,\dots,d$; let $\varepsilon$ be the minimum of these positive rational numbers. If $x_1^t\cdots x_d^t\in(x_1^{t+1},\dots, x_d^{t+1})R$ for some integer $t$, then
\[
x_1^t\cdots x_d^t = a_1x_1^{t+1}+\cdots+a_dx_d^{t+1}
\]
for elements $a_i$ of $A$. (The $a_i$ can be chosen in $R$, though we will only consider them as elements of $A$.) Rearranging terms in the above equation, we have
\[
x_1^t(x_2^t\cdots x_d^t-a_1x_1)\in\big(x_2^{t+1},\dots,x_d^{t+1}\big)A\,.
\]
Since $A$ is almost Cohen-Macaulay, the elements $x_1^t,x_2^{t+1},\dots,x_d^{t+1}$ form an almost regular sequence. Hence there exists $c_1\in A$ with $v(c_1)<\varepsilon/d$ and
\[
c_1(x_2^t\cdots x_d^t-a_1x_1)\in\big(x_2^{t+1},\dots,x_d^{t+1}\big)A\,.
\]
This implies that $c_1x_2^t\cdots x_d^t\in(x_1,x_2^{t+1},\dots,x_d^{t+1})A$. We now repeat the process for $x_2$, i.e., we have
\[
c_1x_2^t\cdots x_d^t=b_1x_1+b_2x_2^{t+1}+\cdots+b_dx_d^{t+1}
\]
with $b_i\in A$, so
\[
x_2^t(c_1x_3^t\cdots x_d^t-b_2x_2)\in\big(x_1,x_3^{t+1},\dots,x_d^{t+1}\big)A\,.
\]
By an argument similar to the one above, there is an element $c_2\in A$ with $v(c_2)<\varepsilon/d$ and
\[
c_1c_2x_3^t\cdots x_d^t\in\big(x_1,x_2,x_3^{t+1},\dots,x_d^{t+1}\big)A\,.
\]
Repeating this procedure $d-2$ more times, we obtain elements $c_1,c_2,\dots,c_d$ in $A$ with $v(c_i)<\varepsilon/d$ and
\[
c_1c_2\cdots c_d=u_1x_1+\cdots+u_dx_d
\]
for some $u_i\in A$. But then
\[
v(c_1\cdots c_d)=v(c_1)+\cdots+v(c_d)<d(\varepsilon/d)=\varepsilon
\]
whereas, since $v(u)\geqslant 0$ for all $u\in A$, we also have
\[
v(u_1x_1+\cdots+u_dx_d)\geqslant\min\{v(u_ix_i)\}\geqslant\min\{v(u_i)+v(x_i)\}
\geqslant\min\{v(x_i)\}=\varepsilon\,,
\]
which is a contradiction.
\end{proof}

To put the results of the remainder of the paper in context, we recall the situation in positive characteristic. Let $R$ be a complete local domain containing a field of characteristic $p>0$. We let $R_\infty$ denote the perfect closure of $R$, that is, $R_\infty$ is the ring obtained by adjoining to $R$ the $p^n$-th roots of all its elements.

\begin{proposition}\label{prop:Rinfinity}
Let $(R,{\mathfrak m})$ be a complete local domain containing a field of prime characteristic. Then $R_\infty$ is almost Cohen-Macaulay with respect to any valuation which is positive on ${\mathfrak m}$.
\end{proposition}

\begin{proof}
Let $v$ be such a valuation, and let $x_1,\dots,x_d$ be a system of parameters for $R_\infty$. Suppose that
\begin{equation}\label{eqn:Rinfinity}
ax_i=b_1x_1+\cdots+b_{i-1}x_{i-1}
\end{equation}
for $a,b_j\in R_\infty$. Let $R'$ be the Noetherian subring of $R_\infty$ generated over $R$ by $a$, $b_1,\dots,b_{i-1}$, and $x_1,\dots,x_d$. By Cohen's structure theorem, $R'$ is a finite extension of a power series ring $S=K[[x_1,\dots,x_d]]$, where $K$ is a coefficient field. Let $m$ be the largest integer such that $R'$ contains a free $S$-module of rank $m$. In this case, the cokernel of
\[
S^m\subseteq R'
\]
is a torsion $S$-module, so there exists a nonzero element $c\in S$ such that $cR'\subseteq S^m$. Taking $p^n$-th powers in equation~\eqref{eqn:Rinfinity} gives us
\[
a^{p^n}x_i^{p^n}\in\big(x_1^{p^n},\dots,x_{i-1}^{p^n}\big)R'\qquad\text{ for all }n\ge0\,.
\]
Multiplying the above by $c$ and using $cR'\subseteq S^m$, we get
\[
ca^{p^n}x_i^{p^n}\in\big(x_1^{p^n},\dots,x_{i-1}^{p^n}\big)S^m\,.
\]
Since $x_1^{p^n},\dots,x_i^{p^n}$ is a regular sequence on the free module $S^m$, it follows that
\[
ca^{p^n}\in\big(x_1^{p^n},\dots,x_{i-1}^{p^n}\big)S^m\subseteq\big(x_1^{p^n},\dots,x_{i-1}^{p^n}\big)R'\,.
\]
Taking $p^n$-th roots in an equation for $ca^{p^n}\in(x_1^{p^n},\dots,x_{i-1}^{p^n})R'$ gives us
\[
c^{1/p^n}a\in\big(x_1,\dots,x_{i-1}\big)R_\infty\qquad\text{ for all }n\ge0\,.
\]
Since the limit of $v(c^{1/p^n})$ is zero as $n\longrightarrow\infty$, it follows that $R_\infty$ is almost Cohen-Macaulay.
\end{proof}

In \cite{HHbig} Hochster and Huneke proved the much deeper fact that for an excellent local domain $R$ of positive characteristic, the ring $R^+$ is Cohen-Macaulay; see also \cite{HL}. We remark that the subring $R_\infty$ may not be Cohen-Macaulay in general: if $R$ is an $F$-pure ring which is not Cohen-Macaulay, then, since $R\hookrightarrow R_\infty$ is pure, $R_\infty$ is not Cohen-Macaulay as well.

If $R$ is a local domain containing a field of characteristic zero, then $R^+$ is typically not a big Cohen-Macaulay algebra. For example, let $R$ be a normal ring of characteristic zero which is not Cohen-Macaulay. Then the field trace map shows that $R$ splits from finite integral extensions. Consequently a nontrivial relation on a system of parameters for $R$ remains nontrivial in finite extensions, and hence in $R^+$. Specifically, for a ring $(R,{\mathfrak m})$ of characteristic zero, the map
\[
H^i_{\mathfrak m}(R)\longrightarrow H^i_{\mathfrak m}(R^+)
\]
is injective for all $i$. This leads to the following question.

\begin{question}\label{q1}
Let $(R,{\mathfrak m})$ be a complete local domain. For $i<\dim R$, is the image of natural map
\[
H^i_{\mathfrak m}(R)\longrightarrow H^i_{\mathfrak m}(R^+)
\]
almost zero?
\end{question}

The answer is affirmative if the ring $R$ contains a field of positive characteristic: this follows from Proposition~\ref{prop:Rinfinity}, or from either of the stronger statements \cite[Theorem~1.1]{HHbig} or \cite[Theorem~2.1]{HL}. If $R$ is a three-dimensional ring of mixed characteristic $p$, Heitmann \cite{Heitmann-dim3} proved that the image of $H^2_{\mathfrak m}(R)$ in $H^2_{\mathfrak m}(R^+)$ is killed by $p^{1/n}$ for all integers $n\ge0$; more recently, he proved the stronger statement \cite[Theorem~2.9]{Heitmann2005} that $H^2_{\mathfrak m}(R^+)$ is annihilated by $c^{1/n}$ for all $c\in{\mathfrak m}$ and $n\ge0$.
Hence the answer to Question~\ref{q1} is also affirmative for mixed characteristic rings of dimension less than or equal to three.

\section{Examples}

In this section, we present two nontrivial examples where local cohomology modules of characteristic zero rings are annihilated by elements of arbitrarily small positive order. The examples are ${\mathbb N}$-graded, and in such cases it is natural to use the valuation arising from the grading: $v(r)$ is the least integer $n$ such that the $n$-th degree component of $r$ is nonzero.

\begin{proposition}\label{prop:normalization}
Let $K$ be a field of characteristic zero, and consider the hypersurface $S=K[x,y,z,w]/(xy-zw)$. For distinct elements $\alpha_i$ of $K$, let $\eta$ be a square root of
\[
\prod_{i=1}^4(x-\alpha_iz)\,.
\]
Then the integral closure of $S[\eta]$ in its field of fractions is the ring
\[
R=S\left[\eta,\frac{w}{x}\eta,\frac{w^2}{x^2}\eta\right]\,.
\]
\end{proposition}

\begin{proof}
The element $(w^2/x^2)\eta$ is integral over $S[\eta]$ since
\[
\left(\frac{w^2}{x^2}\eta\right)^2=\prod_{i=1}^4(w-\alpha_iy)\,.
\]
A similar computation shows that $(w/x)\eta$ is integral over $S[\eta]$, and it remains to prove that the integral closure of $S[\eta]$ is generated by these elements. An element of the fraction field of $S[\eta]$ can be written as $a+b\eta$, with $a$ and $b$ from the fraction field of $S$. Now $a+b\eta$ is integral over $S$ if and only if its trace and norm of are elements of $S$. Since $2$ is a unit in $S$ this is equivalent to $a\in S$ and $b^2\eta^2\in S$. Thus the integral closure of $S[\eta]$ is $S\oplus I\eta$, where $I$ is the fractional ideal consisting of elements $b$ with $b^2\eta^2\in S$.

Since $S$ is a normal domain, $b^2\eta^2$ belongs to $S$ if and only if $v_{\mathfrak p}(b^2\eta^2)\ge0$ for all valuations $v_{\mathfrak p}$ corresponding to height one prime ideals ${\mathfrak p}$ of $S$. Note that $v_{\mathfrak p}(\eta^2)>0$ precisely for the primes ${\mathfrak p}_0=(x,z)$ and ${\mathfrak p}_i=(x-\alpha_iz,w-\alpha_iy)$ for $1\leqslant i\le4$. Since $v_{{\mathfrak p}_0}(\eta^2)=4$ and $v_{{\mathfrak p}_i}(\eta^2)=1$ for $1\leqslant i\le4$,
the condition for $b$ to be an element of $I$ is that
\[
v_{{\mathfrak p}_0}(b)\geqslant-2\qquad\text{ and }\qquad v_{\mathfrak p}(b)\ge0\text{ for all }{\mathfrak p}\neq{\mathfrak p}_0\,.
\]
This implies that $v_{\mathfrak p}(bx^2)\ge0$ for all height one primes ${\mathfrak p}$, i.e., that $bx^2\in S$. Let $b=s/x^2$. Then $v_{(x,w)}(b)\ge0$ implies that $s$ must be in the ideal $(x,w)^2$. Hence $I$ is generated over $S$ by $1$, $w/x$, and $w^2/x^2$.
\end{proof}

\begin{example}\label{ex1}
We continue in the notation of Proposition~\ref{prop:normalization}, i.e., $R$ is the normalization of $S[\eta]$. The ring $R$ is normal by construction, and has dimension three. It follows that $H^0_{\mathfrak m}(R)=0=H^1_{\mathfrak m}(R)$, where ${\mathfrak m}$ is the homogeneous maximal ideal of $R$. We show that there are elements of $R^+$ of arbitrarily small positive order annihilating the image of $H^2_{\mathfrak m}(R)$ in $H^2_{\mathfrak m}(R^+)$.

Note that $x$, $y$, $z+w$ form a homogeneous system of parameters for the hypersurface $S$, and hence also for $R$. In the ring $R$, we have a relation on these elements given by the equation
\[
\frac{w}{x}\eta\cdot(z+w)=\eta\cdot y+\frac{w^2}{x^2}\eta\cdot x\,.
\]
This is a nontrivial relation in the sense that $(w/x)\eta$ does not belong to the ideal generated by $x$ and $y$, so the ring $R$ is not Cohen-Macaulay. Consider the element of $H^2_{\mathfrak m}(R)$ given by this relation; it turns out that $H^2_{\mathfrak m}(R)$ is a one-dimensional $K$-vector space generated by this element, see Remark~\ref{rem:Segre}.

Let $v$ be the valuation defined by the grading on $R$, i.e., $v$ takes value $1$ on $x$, $y$, $z$, and $w$, and $v(\eta)=2$. We construct elements $x_n$ in finite extensions $R_n$ of $R$ with $v(x_n)=1/2^n$ and $x_n(w/x)\eta\in(x,y)R_n$; it then follows that each $x_n$ annihilates the image of the map $H^2_{\mathfrak m}(R)\longrightarrow H^2_{\mathfrak m}(R^+)$.

Let $R_1$ be the extension ring of $R$ obtained by adjoining $\sqrt{x-\alpha_iz}$ for $1\leqslant i\leqslant 4$ and normalizing. We claim that the element $x_1=\sqrt{x-\alpha_1z}$ multiplies $(w/x)\eta$ into the ideal $(x,y)R_1$. To see this, note that
\begin{multline*}
x_1\frac{w}{x}\eta=x_1\frac{w}{x}\prod_{i=1}^4\sqrt{x-\alpha_iz}
=(x-\alpha_1z)\frac{w}{x}\prod_{i=2}^4\sqrt{x-\alpha_iz}\\
=x\left(\frac{w}{x}\prod_{i=2}^4\sqrt{x-\alpha_iz}\right)
-y\left(\alpha_1\prod_{i=2}^4\sqrt{x-\alpha_iz}\right)\,.
\end{multline*}
The element $x_1$ has $v(x_1)=1/2$. To find an annihilator $x_2$ with $v(x_2)=1/4$, we first write
\[
x-\alpha_3z=\beta(x-\alpha_1z)-\gamma(x-\alpha_2z)
\]
for suitable $\beta,\gamma\in K$, and then factor as a difference of squares to obtain
\begin{multline*}
x-\alpha_3z\\
=\left(\sqrt{\beta(x-\alpha_1z)}+\sqrt{\gamma(x-\alpha_2z)}\right)
\left(\sqrt{\beta(x-\alpha_1z)}-\sqrt{\gamma(x-\alpha_2z)}\right)\,.
\end{multline*}
We let
\[
x_2=\sqrt{\sqrt{\beta(x-\alpha_1z)}+\sqrt{\gamma(x-\alpha_2z)}}\,,
\]
which is an element with $v(x_2)=1/4$. Now
\[
x_2\sqrt{x-\alpha_3z}=\lambda\left(\sqrt{\beta(x-\alpha_1z)}+\sqrt{\gamma(x-\alpha_2z)}\right)
\]
where
\[
\lambda=\sqrt{\sqrt{\beta(x-\alpha_1z)}-\sqrt{\gamma(x-\alpha_2z)}}\,,
\]
and so
\begin{multline*}
x_2\eta=\lambda(x-\alpha_1z)\sqrt{\beta(x-\alpha_2z)(x-\alpha_4z)}\\
+\lambda(x-\alpha_2z)\sqrt{\gamma(x-\alpha_1z)(x-\alpha_4z)}\,.
\end{multline*}
Using this, we get
\begin{align*}
x_2\frac{w}{x}\eta
&=x\left(\lambda\frac{w}{x}\sqrt{\beta(x-\alpha_2z)(x-\alpha_4z)}\right)
-y\left(\lambda\alpha_1\sqrt{\beta(x-\alpha_2z)(x-\alpha_4z)}\right)\\
&\quad+x\left(\lambda\frac{w}{x}\sqrt{\gamma(x-\alpha_1z)(x-\alpha_4z)}\right)
-y\left(\lambda\alpha_2\sqrt{\gamma(x-\alpha_1z)(x-\alpha_4z)}\right)
\end{align*}
and consequently $x_2(w/x)\eta\in(x,y)R_2$, where $R_2$ is the finite extension of $R$ obtained by adjoining the various roots occurring in the previous equation and normalizing.

We describe briefly the process of constructing $x_n$ for $n\geqslant 3$. The first step is to write $\sqrt{x-\alpha_4z}$ in terms of $\sqrt{x-\alpha_1z}$ and $\sqrt{x-\alpha_2z}$ as we did for $\sqrt{x-\alpha_3z}$ above. This enables us to write $\sqrt{x-\alpha_3z}\sqrt{x-\alpha_4z}$ as a product of four square roots, each of which is a linear combination of $\sqrt{x-\alpha_1z}$ and $\sqrt{x-\alpha_2z}$. We can now repeat the process used to construct $x_2$, essentially replacing $x$ by $\sqrt{x-\alpha_3z}$ and $z$ by $\sqrt{x-\alpha_4z}$. Finally, we can repeat this process indefinitely, obtaining elements $x_n$ with $v(x_n)=1/2^n$ which annihilate the given element of local cohomology.
\end{example}

\begin{remark}\label{rem:Segre}
The ring $R$ in the previous example can be obtained as a Segre products of rings of lower dimension, and we briefly discuss this point of view.

Let $A$ and $B$ be ${\mathbb N}$-graded normal rings which are finitely generated over a field $A_0=B_0=K$. Their \emph{Segre product} is the ring
\[
R=A\#B=\bigoplus_{n\ge0}A_n\otimes_KB_n\,,
\]
which inherits a natural grading where $R_n=A_n\otimes_KB_n$. If $K$ is algebraically closed then the tensor product $A\otimes_KB$ is a normal ring, and hence so is its direct summand $A\#B$. If $M$ and $N$ are ${\mathbb Z}$-graded modules over $A$ and $B$ respectively, then their Segre product is the $R$-module
\[
M\#N=\bigoplus_{n\in{\mathbb Z}}M_n\otimes_KN_n\,.
\]
Using ${\mathfrak m}$ to denote the homogeneous maximal ideal of $R$, the local cohomology modules $H_{\mathfrak m}^k(R)$ can be computed using the K\"unneth formula due to Goto and Watanabe, \cite[Theorem 4.1.5]{GW}:
\begin{multline*}
H^k_{\mathfrak m}(R)=\left(A\#H^k_{{\mathfrak m}_B}(B)\right)\ \oplus\
\left(H^k_{{\mathfrak m}_A}(A)\#B\right)\\
\oplus\bigoplus_{i+j=k+1}\left(H^i_{{\mathfrak m}_A}(A)\#H^j_{{\mathfrak m}_B}(B)\right)\,.
\end{multline*}
It follows that if $A$ and $B$ have positive dimension, then
\[
\dim(A\#B)=\dim A+\dim B-1\,.
\]

We claim that the ring $R$ in Example~\ref{ex1} is isomorphic to the Segre product $A\#B$, where
\[
A=K[a,b,c]/\big(c^2-\prod_{i=1}^4(a-\alpha_ib)\big)
\]
is a hypersurface with $\deg a=\deg b=1$ and $\deg c=2$, and $B=K[s,t]$ is a standard graded polynomial ring. The map
\begin{align*}
x&\longmapsto as\,,&y&\longmapsto bt\,,&z&\longmapsto bs\,,&w&\longmapsto at\,,\\
\eta&\longmapsto cs^2\,,&(w/x)\eta&\longmapsto cst\,,&(w/x)^2\eta&\longmapsto ct^2
\end{align*}
extends to a $K$-algebra homomorphism $\varphi\colon R\longrightarrow A\#B$. This is a surjective homomorphism of integral domains of equal dimension, so it must be an isomorphism. Since $A$ and $B$ are Cohen-Macaulay rings of dimension $2$, the K\"unneth formula for $H^2_{\mathfrak m}(R)$ reduces to
\[
H^2_{\mathfrak m}(R)=\left(A\#H^2_{{\mathfrak m}_B}(B)\right)\ \oplus\
\left(H^2_{{\mathfrak m}_A}(A)\#B\right)\,.
\]
The module $H^2_{{\mathfrak m}_B}(B)$ vanishes in nonnegative degrees, which implies that $A\#H^2_{{\mathfrak m}_B}(B)=0$. The component of $H^2_{{\mathfrak m}_A}(A)$ in nonnegative degree is the one-dimensional vector space spanned by the degree $0$ element
\[
\left[\frac{c}{ab}\right]\in H^2_{{\mathfrak m}_A}(A)\,.
\]
Hence $H^2_{\mathfrak m}(R)$ is the one-dimensional vector space spanned by $[c/ab]\otimes 1$. The search for elements $x_n\in R^+$ of small degree annihilating the image of $H^2_{\mathfrak m}(R)$ in $H^2_{\mathfrak m}(R^+)$ is essentially the search for homogeneous elements of $A^+$, of small degree, multiplying $c$ into the ideal $(a,b)A^+$.
\end{remark}

\begin{example}\label{ex2}
Let $K$ be an algebraically closed field of characteristic zero, $\theta\in K$ a primitive cube root of unity, and set
\[
A=K[x,y,z]/\big(\theta x^3+\theta^2 y^3+z^3\big)\,.
\]
Let $R$ be the Segre product of $A$ and the polynomial ring $K[s,t]$. Then $R$ is a normal ring of dimension $3$, and the elements $sx$, $ty$, $sy+tx$ form a homogeneous system of parameters for $R$. Using the K\"unneth formula as in Remark~\ref{rem:Segre}, the local cohomology module $H^2_{\mathfrak m}(R)$ is a one-dimensional vector space spanned by an element corresponding to the relation
\[
sztz(sy+tx)=(sz)^2ty+(tz)^2sx\,.
\]
To annihilate this relation by an element of $R^+$ of positive degree $\varepsilon\in{\mathbb Q}$, it suffices to find an element $u\in A^+$ of degree $\varepsilon$ such that
\[
uz^2\in(x,y)A^+\,;
\]
indeed if $uz^2=vx+wy$ for homogeneous $v,w\in A^+$ of degree $1+\varepsilon$, then
\[
(s^{\varepsilon}u)(sztz)=(s^{\varepsilon}tv)(sx)+(s^{1+\varepsilon}w)(ty)\,,
\]
and $s^{\varepsilon}tv$ and $s^{1+\varepsilon}w$ are easily seen to be integral over $S$.

We have now reduced our problem to working over the hypersurface $A$, where we are looking for elements $u\in A^+$ of small degree which annihilate
\[
\left[\frac{z^2}{xy}\right]\in H^2_{{\mathfrak m}_A}(A^+)\,.
\]
Let $A_1$ be the extension of $A$ obtained by adjoining $x_1$, $y_1$, $z_1$, where
\[
x_1^3=\theta^{1/3}x+\theta^{2/3}y\,,\qquad y_1^3=\theta^{1/3}x+\theta^{5/3}y\,,\qquad z_1^3
=\theta^{1/3}x+\theta^{8/3}y\,.
\]
Note that $x$ and $y$ can be written as $K$-linear combinations of $x_1^3$ and $y_1^3$. Moreover,
\begin{multline*}
(x_1y_1z_1)^3=\big(\theta^{1/3}x+\theta^{2/3}y\big)\big(\theta^{1/3}x+\theta^{5/3}y\big)\big(\theta^{1/3}x+\theta^{8/3}y\big)\\
=\theta x^3+\theta^2y^3=-z^3\,,
\end{multline*}
so $z$ belongs to the $K$-algebra generated by $x_1$, $y_1$, and $z_1$. Now
\begin{multline*}
\theta x_1^3+\theta^2 y_1^3+z_1^3=\theta\big(\theta^{1/3}x+\theta^{2/3}y\big)+\theta^2\big(\theta^{1/3}x+\theta^{5/3}y\big)+\big(\theta^{1/3}x+\theta^{8/3}y\big)\\
=\big(\theta^{4/3}+\theta^{7/3}+\theta^{1/3}\big)x+\big(\theta^{5/3}+\theta^{11/3}+\theta^{8/3}\big)y=0\,,
\end{multline*}
which implies that
\[
A_1=K[x_1,y_1,z_1]/\big(\theta x_1^3+\theta^2 y_1^3+z_1^3\big)
\]
is a ring isomorphic to $A$. Thus $A\subset A_1$ gives a finite embedding of $A$ into itself under which the generators of degree $1$ go to elements of degree $3$; or, in terms of the original degree, the new generators of the homogeneous maximal ideal have degree $1/3$. Since $[H^2_{{\mathfrak m}_A}(A)]_0$ is annihilated by all elements of positive degree, the image of $[H^2_{{\mathfrak m}_A}(A)]_0$ in $H^2_{{\mathfrak m}_A}(A_1)$ is annihilated by elements of degree $1/3$. Iterating this construction, we conclude that there are elements of arbitrarily small positive degree annihilating the image of $[H^2_{{\mathfrak m}_A}(A)]_0$ in $H^2_{{\mathfrak m}_A}(A^+)$. Quite explicitly, we have a tower of extensions
\[
A=A_0\subset A_1\subset A_2\subset\dots\quad\text{ where }\quad
A_n=K[x_n,y_n,z_n]/\big(\theta x_n^3+\theta^2 y_n^3+z_n^3\big)\,.
\]
The maps $H^2_{\mathfrak m}(A_n)\longrightarrow H^2_{\mathfrak m}(A_{n+1})$ preserve degrees, so $[H^2_{{\mathfrak m}_A}(A)]_0$ maps to the socle of $H^2_{\mathfrak m}(A_n)$ which is killed by all elements of $A_n$ of positive degree, e.g., by the elements $x_n,y_n,z_n$ which have degree $1/3^n$.
\end{example}

\begin{remark}
In \cite[Theorem~2.9]{Heitmann2005} Heitmann proves that if $(R,{\mathfrak m})$ is a mixed characteristic excellent local domain of dimension three, then the image of $H^2_{\mathfrak m}(R)$ in $H^2_{\mathfrak m}(R^+)$ is annihilated by arbitrarily small powers of every non-unit. The corresponding statement is false for three-dimensional domains of characteristic zero: for the ring $R$ of Example~\ref{ex2}, we claim that $\sqrt{sz}$ does not annihilate the image of $H^2_{\mathfrak m}(R)\longrightarrow H^2_{\mathfrak m}(R^+)$. Because of the splitting provided by field trace, it suffices to verify that
\[
\sqrt{sz}\left(sztz\right)\notin(sx,ty)T\,,
\]
where $T$ is any normal subring of $R^+$ containing $R[\sqrt{sz}]$. Take $T$ to be the Segre product of $\widetilde{A}=A[\sqrt{x},\sqrt{y},\sqrt{z}]$ and $\widetilde{B}=B[\sqrt{s},\sqrt{t}]$. Note that $\widetilde{B}$ is a polynomial ring in $\sqrt{s}$ and $\sqrt{t}$, and that $\widetilde{A}$ is the hypersurface
\[
K[\sqrt{x},\sqrt{y},\sqrt{z}]/\left(\theta(\sqrt{x})^6+\theta^2(\sqrt{y})^6+(\sqrt{z})^6\right)\,.
\]
It is enough to check that $\sqrt{sz}\left(sztz\right)\notin(sx,ty)(\widetilde{A}\otimes_K\widetilde{B})$, and after specializing $\sqrt{s}\longmapsto 1$ and $\sqrt{t}\longmapsto 1$ to check that
\[
(\sqrt{z})^5\notin\left((\sqrt{x})^2,(\sqrt{y})^2\right)\widetilde{A}\,,
\]
which is immediately seen to be true. The same argument shows that $\sqrt{sx}$, $\sqrt{sy}$, etc. do not annihilate the image of $H^2_{\mathfrak m}(R)\longrightarrow H^2_{\mathfrak m}(R^+)$. The situation is quite similar with Example~\ref{ex1}.
\end{remark}

\section{Annihilators using the Albanese map}

For an ${\mathbb N}$-graded domain $R$ which is finitely generated over a field $R_0$, let $R^{+\operatorname{GR}}$ be the ${\mathbb Q}_{\ge0}$-graded ring generated by those elements of $R^+$ which can be assigned a degree such that they satisfy a homogeneous equation of integral dependence over $R$. If $R_0$ is a field of prime characteristic, Hochster and Huneke \cite[Theorem~5.15]{HHbig} proved that the induced map
\[
H^i_{\mathfrak m}(R)\longrightarrow H^i_{\mathfrak m}(R^{+\operatorname{GR}})
\]
is zero for all $i<\dim R$. Translating to projective varieties, one immediately has the vanishing theorem:

\begin{theorem}\cite[Theorem~1.2]{HHbig}
Let $X$ be an irreducible closed subvariety of ${\mathbb P}^n_K$, where $K$ is a field of positive characteristic. Then for all integers $i$ with $0<i<\dim X$, and all integers $t$, there exists a projective variety $Y$ over a finite extension of $K$ with a finite surjective morphism $f\colon Y\longrightarrow X$, such that the induced map
\[
H^i(X,{\mathcal O}_X(t))\longrightarrow H^i(Y,f^*{\mathcal O}_X(t))
\]
is zero.
\end{theorem}

Over fields of characteristic zero, the corresponding statements are false because of the splitting provided by field trace. However, the following graded analogue of Question~\ref{q1} remains open.

\begin{question}\label{q:gr1}
Let $R$ be an ${\mathbb N}$-graded domain, finitely generated over a field $R_0$ of characteristic zero. For $i<\dim R$, is every element of the image of
\[
H^i_{\mathfrak m}(R)\longrightarrow H^i_{\mathfrak m}(R^{+\operatorname{GR}})
\]
killed by elements of $R^{+\operatorname{GR}}$ of arbitrarily small positive degree?
\end{question}

This question, when considered for Segre products, leads to the following:

\begin{question}\label{q:gr2}
Let $R$ be an ${\mathbb N}$-graded domain of dimension $d$, finitely generated over a field $R_0$ of characteristic zero. Is the image of
\[
\left[H^d_{\mathfrak m}(R)\right]_{\ge0}\longrightarrow H^d_{\mathfrak m}(R^{+\operatorname{GR}})
\]
killed by elements of $R^{+\operatorname{GR}}$ of arbitrarily small positive degree?
\end{question}

In Examples~\ref{ex1} and~\ref{ex2}, we obtained affirmative answers to Question~\ref{q:gr1} by explicitly constructing the annihilators. In this section, we obtain an affirmative answer for the image of $[H^2_{\mathfrak m}(R)]_0$ and also settle Question~\ref{q:gr2} for rings of dimension two. We first recall some basic facts about the relationship between graded rings and very ample divisors.

If $R$ is a standard graded ring, the associated projective scheme $X=\operatorname{Proj} R$ has a very ample line bundle ${\mathcal O}_X(1)$ with sections defined by elements of degree one, which generate the line bundle. Conversely, a very ample line bundle defines a standard graded ring and an embedding of $X$ into projective space.

The strategy is to find, for an arbitrarily large positive integer $n$, a finite surjective map from an integral projective scheme $Y$ to $X$, together with an ample line bundle ${\mathcal L}$ on $Y$, such that ${\mathcal L}^{\otimes n}$ is the pullback of ${\mathcal O}_X(1)$ and such that a section of ${\mathcal L}$ annihilates the pullback of the given element of cohomology. This will essentially be accomplished by mapping $X$ to its Albanese variety, and pulling back by the multiplication by $N$ map for large integers $N$. The precise result that we prove is as follows.

\begin{theorem}\label{thm:main}
Let $R$ be an ${\mathbb N}$-graded domain which is finitely generated over a field $R_0$ of characteristic $0$. Let $X=\operatorname{Proj} R$ and let $\eta$ be an element of $H^1(X,{\mathcal O}_X)$. Then, for every $\varepsilon>0$, there exists a finite extension $R\subseteq S$ of ${\mathbb Q}$-graded domains such that the image of $\eta$ under the induced map
\[
H^1(X,{\mathcal O}_X)\longrightarrow H^1(Y,{\mathcal O}_Y)\qquad\text{ where }Y=\operatorname{Proj} S
\]
is annihilated by an element of $S$ of degree less that $\varepsilon$.
\end{theorem}

We remark that since $H^1(X,{\mathcal O}_X)$ corresponds to the component of $H_{\mathfrak m}^2(R)$ of degree zero, this theorem only implies that we can annihilate elements of $H^2_{\mathfrak m}(R)$ of degree zero by elements of small degree. If $H^2_{\mathfrak m}(R)$ is generated by its degree zero elements---and this happens in several interesting examples---we can deduce the result for all elements of $H^2_{\mathfrak m}(R)$.

\begin{proof} Replacing $R$ by its normalization, it suffices to work throughout with normal rings. We also reduce to the case where $R$ is a standard ${\mathbb N}$-graded ring as follows. Using \cite[Lemme~2.1.6]{EGA2}, $R$ has a Veronese subring $R^{(t)}$ which is generated by elements of equal degree. The local cohomology of $R^{(t)}$ supported at its homogeneous maximal ideal ${\mathfrak m}$ can be obtained by \cite[Theorem~3.1.1]{GW} which states that
\[
H^i_{\mathfrak m}(R^{(t)})=\bigoplus_{n\in{\mathbb Z}}[H^i_{\mathfrak m}(R)]_{nt}\,.
\]
In particular, we have
\[
[H^2_{\mathfrak m}(R^{(t)})]_0=[H^2_{\mathfrak m}(R)]_0=H^1(X,{\mathcal O}_X)\,.
\]

If elements of this cohomology group can be annihilated in graded finite extensions of $R^{(t)}$, then the same can be achieved in extensions of $R$.

We next treat the special case where $\operatorname{Proj} R$ is itself an abelian variety, which we denote $A$. For each integer $N$, let $[N_A]\colon A\longrightarrow A$ be the morphism corresponding to multiplication by $N$. Assume further that the very ample sheaf ${\mathcal O}_A(1)$ defining the graded ring $R$ satisfies the condition that

\begin{equation}\label{eq:neg1}
[(-1)_A]^*({\mathcal O}_A(1))={\mathcal O}_A(1)\,.
\end{equation}
Note that, if ${\mathcal L}$ is any very ample line bundle on $A$, the new very ample line bundle \[{\mathcal O}_A(1)={\mathcal L}\otimes [(-1)_A]^*{\mathcal L}\] satisfies this further assumption.

We recall two facts about abelian varieties from Mumford \cite{Mumford}:
\begin{enumerate}

\item $H^1(A,{\mathcal O}_A(1))=0$. By ``The Vanishing Theorem" \cite[page 150]{Mumford}, given a line bundle ${\mathcal L}$, there is a unique integer $i$ such that $H^i(A,{\mathcal L})$ is nonzero. Since ${\mathcal O}_A(1)$ is very ample, this integer must be $0$.

\item $[N_A]^*({\mathcal O}_A(1))={\mathcal O}_A(N^2)$. This follows from \cite[Corollary~II.6.3]{Mumford} since we are assuming \eqref{eq:neg1}.
\end{enumerate}

The theorem in this case follows from these two properties: the morphism $[N_A]$ induces a map
\[
R=\bigoplus_n\Gamma(A,{\mathcal O}_A(n))\longrightarrow\bigoplus_n\Gamma(A,[N_A]^*{\mathcal O}_A(n))\,,
\]
and, by the second property above,
\[
\Gamma(A,[N_A]^*{\mathcal O}_A(n))=\Gamma(A,{\mathcal O}_A(N^2n))\,.
\]

Thus we have a map of graded rings from $R$ to itself, that takes an element of degree $1$ to an element of degree $N^2$. Denote the new copy of $R$ by $S$, and regrade $S$ with a ${\mathbb Q}$-grading such that the map $R\longrightarrow S$ preserves degrees. This implies that $S$ has elements $s$ of degree $1/N^2$ under the new grading. Such an element $s$ must annihilate the image of $\eta\in H^1(A,{\mathcal O}_A)$, since the product $s\cdot\eta$ lies in $H^1(A,{\mathcal O}_A(1))=0$. Hence for each positive integer $N$, we have found a finite extension of $R$ with an element of degree $1/N^2$ that annihilates the image of $\eta$.

The remainder of the proof is devoted to reducing to the previous case. Let $R$ be a graded domain such that $X=\operatorname{Proj} R$ is normal. Let $A$ be the \emph{strict Albanese variety} of $X$ as defined in \cite{Chevalley}. It is the dual abelian variety to the Picard variety of $X$ (in the sense of Chevalley-Grothendieck) which parametrizes line bundles algebraically equivalent to 0. Let $\varphi\colon X\longrightarrow A$ be the corresponding Albanese morphism. Then (since the ground field has characteristic 0) $\varphi$ induces an isomorphism
\[
H^1(A,{\mathcal O}_A)\cong H^1(X,{\mathcal O}_X)\,,
\]
see Chevalley \cite{Chevalley}.

Since $A$ is an abelian variety, it has a very ample invertible sheaf ${\mathcal O}_A(1)$, see for example \cite[pp.~60--62]{Mumford}. After replacing ${\mathcal O}_A(1)$ by ${\mathcal O}_A(1)\otimes [(-1)_A]^*{\mathcal O}_A(1)$ if necessary, we may assume that
\[
[(-1)_A]^*({\mathcal O}_A(1))\cong {\mathcal O}_A(1)\,.
\]
We let ${\mathcal O}_X(1)$ denote the very ample invertible sheaf defined by the grading on $R$. Let $\pi_1\colon Y_1\longrightarrow X$ be the pullback of multiplication by $N$ on $A$, and let $\varphi_1\colon Y_1\longrightarrow A$ be the map induced by $\varphi$, so that we have the fiber product diagram below.
\[
\xymatrix{Y_1\ar[r]^{\varphi_1}\ar[d]_{\pi_1}& A\ar[d]^{[N_A]}\\
X\ar[r]^{\varphi} & A}
\]

Let ${\mathcal M}_1=\varphi_1^*({\mathcal O}_A(1))$. Then
\[
\pi_1^*(\varphi^*({\mathcal O}_A(1)))=\varphi_1^*([N_A]^*({\mathcal O}_A(1)))=\varphi_1^*({\mathcal O}_A(N^2))={\mathcal M}_1^{\otimes N^2}.
\]

Now let $m$ be an integer such that $\varphi^*({\mathcal O}_A(-1))\otimes {\mathcal O}_X(m)$ is globally generated; such an $m$ exists since ${\mathcal O}_X(1)$ is ample, and we fix one such $m$.

Since the sheaf $\varphi^*({\mathcal O}_A(-1))\otimes {\mathcal O}_X(m)$ is globally generated, there exists a map $\psi\colon X\longrightarrow{\mathbb P}^n$ such that
\[
\psi^*({\mathcal O}_{{\mathbb P}^n}(1))=\varphi^*({\mathcal O}_A(-1))\otimes {\mathcal O}_X(m)\,.
\]

Let $\alpha\colon{\mathbb P}^n\longrightarrow{\mathbb P}^n$ be a finite map such that $\alpha^*({\mathcal O}_{{\mathbb P}^n}(1))={\mathcal O}_{{\mathbb P}^n}(N)$; for example, we can take $\alpha$ to be the map defined by the ring homomorphism on a polynomial
ring that sends the variables to their $N$-th powers.

Let $Y_2$ be the fiber product of $\psi$ and $\alpha$, which gives us a diagram
\[
\xymatrix{Y_2 \ar[r]^{\varphi_2}\ar[d]_{\pi_2}& {\mathbb P}^n\ar[d]^{\alpha} \\
X \ar[r]^{\psi} & {\mathbb P}^n\,.}
\]
Let ${\mathcal M}_2=\varphi_2^*({\mathcal O}_{{\mathbb P}^n}(1))$. We then have
\begin{multline*}
\pi_2^*(\varphi^*({\mathcal O}_A(-1))\otimes{\mathcal O}_X(m))\cong\pi_2^*(\psi^*({\mathcal O}_{{\mathbb P}^n}(1)))=\varphi_2^*(\alpha^*({\mathcal O}_{{\mathbb P}^n}(1)))\\
\cong \varphi_2^*({\mathcal O}_{{\mathbb P}^n}(N))\cong {\mathcal M}_2^{\otimes N}\,.
\end{multline*}
Note that the above morphisms $\pi_i\colon Y_i\longrightarrow X$, $i=1,2$ are \emph{finite and surjective}.

Let $Y$ be a component of the normalization of the reduced fibre product of $\pi_1\colon Y_1\longrightarrow X$ and $\pi_2\colon Y_2\longrightarrow X$, with the property that $Y\longrightarrow X$ is surjective. Since $Y_1\times_XY_2\longrightarrow X$ is finite and surjective, some irreducible component of this fiber product maps onto $X$ via a finite morphism, and we may take $Y$ to be the normalization of any such component.

We then have an induced finite surjective map $\pi\colon Y\longrightarrow X$, and induced maps $\mu_1\colon Y\longrightarrow Y_1$ and $\mu_2\colon Y\longrightarrow Y_2$, giving a commutative diagram
\[
\xymatrix{Y \ar[r]^{\mu_1}\ar[d]_{\mu_2}\ar[dr]^{\pi}& Y_1\ar[d]^{\pi_1} \\
Y_2 \ar[r]^{\pi_2} & X\,.}
\]

By construction, we have
\begin{multline*}
\pi^*{\mathcal O}_X(m)=\pi^*(\varphi^*{\mathcal O}_A(1)\otimes\varphi^*{\mathcal O}_A(-1)\otimes{\mathcal O}_X(m))\\
=\mu_1^*\pi_1^*(\varphi^*({\mathcal O}_A(1)))\otimes \mu_2^*\pi_2^*(\varphi^*({\mathcal O}_A(-1))\otimes{\mathcal O}_X(m))\\
=\mu_1^*({\mathcal M}_1^{\otimes N^2})\otimes \mu_2^*({\mathcal M}_2^{\otimes N}) = {\mathcal M}^{\otimes N}\,,
\end{multline*}
where ${\mathcal M}=\mu_1^*{\mathcal M}_1^{\otimes N}\otimes \mu_2^*{\mathcal M}_2.$

Now ${\mathcal M}_1= \varphi_1^*({\mathcal O}_A(1))$ is generated by global sections of the form $\varphi_1^*(u)$ with $u\in H^0(A,{\mathcal O}_A(1))$. Choose any such element $u$ such that its image in $H^0(Y,\mu_1^*{\mathcal M}_1)$ is nonzero. Let $v$ be a nonzero element of $H^0(Y,\mu_2^*{\mathcal M}_2)$, and let $s$ be the image of $\mu_1^*\varphi_1^*(u^N)\otimes v$ in $H^0(Y,{\mathcal M})$. Then $s$ is a nonzero section of ${\mathcal M}$, and since $\pi^*({\mathcal O}_X(m))={\mathcal M}^{\otimes N}$, the degree of $s$ in the grading induced from that on $R$ is $m/N$. We claim that the composition
\[
\CD
H^1(X,{\mathcal O}_X)@>\pi^*>>H^1(Y,{\mathcal O}_Y)@>\cdot s>>H^1(Y,{\mathcal M})
\endCD
\]
vanishes. For this, it suffices to show that the composition
\[
\CD
H^1(X,{\mathcal O}_X)@>\pi^*>>H^1(Y,{\mathcal O}_Y)@>\mu_1^*\varphi_1^*(u)>>H^1(Y,\mu_1^*{\mathcal M}_1)
\endCD
\]
vanishes, which in turn reduces to showing that
\[
\CD
H^1(X,{\mathcal O}_X)@>\pi_1^*>>H^1(Y_1,{\mathcal O}_{Y_1})@>\varphi_1^*(u)>>H^1(Y_1,{\mathcal M}_1)
\endCD
\]
vanishes. Since $\varphi^*\colon H^1(A,{\mathcal O}_A)\longrightarrow H^1(X,{\mathcal O}_X)$ is an isomorphism, we further reduce to showing that
\[
\CD
H^1(A,{\mathcal O}_A)@>[N_A]^*>>H^1(A,{\mathcal O}_A)@>\cdot u>>H^1(A,{\mathcal O}_A(1))
\endCD
\]
vanishes. But this is true since $H^1(A,{\mathcal O}_A(1))=0$.

Since we can make $m/N$ arbitrarily small by choosing $N$ large, this completes the proof.
\end{proof}

As a corollary, we see that the answer to Question~\ref{q:gr2} is affirmative for rings of dimension two:

\begin{corollary}\label{cor:dim2}
Let $R$ be an ${\mathbb N}$-graded domain of dimension $2$, which is finitely generated over a field $R_0$ of characteristic zero. Then the image of
\[
\left[H^2_{\mathfrak m}(R)\right]_{\ge0}\longrightarrow H^2_{\mathfrak m}(R^{+\operatorname{GR}})
\]
is killed by elements of $R^{+\operatorname{GR}}$ of arbitrarily small positive degree.
\end{corollary}

\begin{proof}
By adjoining roots of elements if necessary, we may assume that $R$ has a system of parameters $x,y$ consisting of linear forms. Theorem~\ref{thm:main} implies that the image of $[H^2_{\mathfrak m}(R)]_0$ is killed by elements of $R^{+\operatorname{GR}}$ of arbitrarily small positive degree, so it suffices to prove that $[H^2_{\mathfrak m}(R)]_{\ge0}$ is the $R$-module generated by $[H^2_{\mathfrak m}(R)]_0$.

Since $x$ and $y$ are linear forms, we have $[R_{xy}]_{n+1}=[R_{xy}]_n\cdot R_1$ for all integers $n$. Computing $H^2_{\mathfrak m}(R)$ using the \v Cech complex on $x,y$, it follows that
\[
\left[H^2_{\mathfrak m}(R)\right]_{n+1}=\left[H^2_{\mathfrak m}(R)\right]_n\cdot R_1
\qquad\text{ for all }n\in{\mathbb Z}\,.\qedhere
\]
\end{proof}

\section{Closure operations}

The issues discussed here are closely related to closure operations considered by Hochster and Huneke, and by Heitmann. The \emph{plus closure} of an ideal ${\mathfrak a}$ of a domain $R$ is defined as ${\mathfrak a}^+={\mathfrak a} R^+\cap R$. It has desirable properties for rings of prime characteristic, e.g., it bounds colon ideals on systems of parameters: if $x_1,\dots,x_d$ is a system of parameters for an excellent local domain $R$ containing a field of prime characteristic, then
\[
(x_1,\dots,x_{i-1}):_Rx_i\subseteq (x_1,\dots,x_{i-1})^+\qquad\text{ for all }i\,.
\]
In general, plus closure does not have this colon-capturing property for rings of characteristic zero or of mixed characteristic. Several alternative closure operations are defined by Heitmann in \cite{Heitmann2001} including the \emph{extended plus closure}. Building on these ideas, he settled the Direct Summand Conjecture for mixed characteristic rings of dimension three \cite{Heitmann-dim3}. In \cite[Theorem~1.3]{Heitmann2005} Heitmann proved that extended plus closure has the colon-capturing property for arbitrary sets of three parameters in excellent domains of mixed characteristic.

Let $(R,{\mathfrak m})$ be a complete local domain and fix, as usual, a valuation $v\colon R\longrightarrow{\mathbb Z}\cup\{\infty\}$ which is positive on ${\mathfrak m}$ and extend to $v\colon R^+\longrightarrow{\mathbb Q}\cup\{\infty\}$. In \cite{HHJPAA} Hochster and Huneke define the \emph{dagger closure} ${\mathfrak a}^\dagger$ of an ideal ${\mathfrak a}$ as the ideal consisting of all elements $x\in R$ for which there exist elements $u\in R^+$, of arbitrarily small positive order, with $ux\in{\mathfrak a} R^+$. In \cite[Theorem~3.1]{HHJPAA} it is proved that the dagger closure ${\mathfrak a}^\dagger$ agrees with the tight closure ${\mathfrak a}^*$ for ideals of complete local domains of prime characteristic, see also \cite[\S\,6]{HHJAMS}. While tight closure is defined in characteristic zero by reduction to prime characteristic, the definition of dagger closure is characteristic-free.

\begin{question}[Hochster-Huneke]\label{q:dagger}
Does the dagger closure operation have the colon-capturing property, i.e., if $x_1,\dots,x_d$ is a system of parameters for a complete local domain $R$, is it true that
\[
(x_1,\dots,x_{i-1}):_Rx_i\subseteq(x_1,\dots,x_{i-1})^\dagger\,?
\]
\end{question}

According to Hochster and Huneke \cite[page~244]{HHJPAA} ``it is important to raise (and answer) this question.'' If Question~\ref{q:dagger} has an affirmative answer, then so does Question~\ref{q1}.

Consider the hypersurface $K[[x,y,z]]/(x^3+y^3+z^3)$. If $K$ has prime characteristic, a straightforward calculation---performed in many an introductory lecture on tight closure theory---shows that $z^2\in(x,y)^*$. If $K$ has characteristic zero, the ``reduction modulo $p$'' nature of the definition of tight closure \cite[\S\,3]{HHJAMS} immediately yields $z^2\in (x,y)^*$ once again. In contrast, the computation that $z^2\in (x,y)^\dagger$ is quite delicate and, aside from the linear change of variables, is the computation we performed in Example~\ref{ex2}. While concrete descriptions of the multipliers of small order are available in this and some other examples, dagger closure remains quite mysterious even in simple examples such as diagonal hypersurfaces:

\begin{question}
Let $K$ be a field of characteristic zero, and let
\[
R=K[[x_0,\dots,x_d]]/(x_0^n+\cdots+x_d^n), \qquad\text{where $n>d$}.
\]
Does $x_0^d$ belong to the dagger closure of the ideal $(x_1,\dots,x_d)$?
\end{question}

A routine computation of tight closure shows that $x_0^d\in(x_1,\dots,x_d)^*$. In the case $d=2$, we have $x_0^2\in(x_1,x_2)^\dagger$ by Corollary~\ref{cor:dim2}.

redpajama_set_name: RedPajamaArXiv


before we end our mermaid week here at Happy Soirée, of course I had to share mermaid themed party tablescape!
I love putting together themed parties. however what I really love is finding ways to create a theme without using the obvious themed party supplies.
there really is so many different ways to create a mermaid themed party.
just be sure to add cake!!!
and I really really hope you & your mermaids will pop some bubbly with some pretty party decor & have some fun soon- like ASAP!


These luxury design brands are the masterminds behind some of the most incredible collections in the UK. As you know,British furniture industry combines an unparalleled history of craftsmanship, high- quality materials and eccentricity, with a more modern design style. They have been conquered new status in the interior design industry because the furniture designers are taking risks bringing clean-cut, inspired or edgy pieces that are more than objects of aesthetic value. Discover with the Luxury Bathrooms team which are the top luxury design brands that have the most inspirational design ideas in the country!
Harrods finally has its own original bespoke furniture collection that draws its aesthetic from Art Deco jewelry and tailoring. Designed in-house by Harrods Interiors features an incredible selection of upholstery, occasional pieces and case goods with timeless style and unique detailing.
Broom manufactures and retails his own collections, making him a supporter of British manufacturing techniques. Since 2007, Lee Broom is one of the UK's leading product & interior designers that have created over 75 furniture and lighting products, which he designs and manufactures under his own label.
Their handmade furniture designs are inspired by the techniques, the materials, and the finishes that characterize Bentley's renowned car interiors. Bentley Home is a partnership between Bentley and Club House Italia and features an exclusive collection of furniture and homeware designs. Their bespoke designs combine the brand's craftsmanship and material selection with the exacting design of one of Europe's leading luxury furniture makers.
Besides creating stunning interiors for major art collectors, Francis Sultana also produces annual collections of bespoke and limited edition furniture and textiles under the Francis Sultana brand.
The history of Tom Dixon as a brand goes beyond its conception in 2002. Created by the renowned interior designer, the award-winning British design brand continues to represent the heritage of British industry.
Established as a luxury brand since 2004, Jimmie Martin is based in London, United Kingdom. The brand's main focus is on hand painted pieces of furniture. They are known for transforming pieces of furniture into unique one-off pieces that become personal to their clients.
For over two centuries, the luxury brand celebrates the art of hand tools and green woodworking techniques. From the traditional Windsor chair to the iconic Arts and Crafts furniture, their Modern British collection takes these shapes and details, evolving them for a fresh take on a classic look.
The luxury brand showcases the finest British design and craftsmanship through their bespoke furniture, as well as interiors, collectible pieces, and luxury accessories. In Founded by David Snowdon, in 1985, the luxury design brand creates and manufactures their collections in the UK, relying on a network of 260 master craftsmen who all share David's love of handcrafted design and attention to detail.
Founded by John Crowell, their creative furniture designs reflects a dramatic theatrical style. Known for their stunning handmade furniture created by the traditional methods and materials, the luxury brand has the "made in England" stamp!
Max Rollitt classical, elegant aesthetic is teamed with an informal edge, a style that has become his trademark. His incredible experience in the field has developed an inherent sensitivity towards the materials and classical architectural elements that are condensed in the decorative arts.
The veteran of luxury British interiors seeks to evolve with clients, adopting new looks and modern twists on traditional classics, such as contemporary Ralph chair in Duccio Indigo. The furniture designs created by the famous interior designer constitutes all that was missing when sourcing furniture for her client projects.
Created in Preston, Lancashire (1968) the luxury design brand is one of the UK's finest upholstery businesses, whose reputation is known for producing fine quality sofas, chairs, and stools with hardwood frames built to last, offering a variety of suspension options and high-quality fabrics and hides.


text:
Netflix Inc (NFLX), The Boeing Company (BA), and Seven Stocks With Surging Earnings Estimates
Published on April 16, 2013 at 12:17 pm by SA in Stock Analysis
There is nothing more important to a stock than earnings. That makes earnings season the most important time of the year for the S&P 500.
The companies that miss expectations will be punished while those that deliver big surprises will be rewarded. But with the economy showing signs of weakness in the past month, analysts aren't expecting a blowout earnings season.
Earnings in the private sector have been trending higher for years, aggressively rebounding out of the financial crisis in 2009. But four years down the line, earnings growth is at the top of a multi-year cycle. Companies have cut all the fat they can cut. Earnings growth needs to come from sales growth, and that has become increasingly difficult with weakness in GDP (gross domestic product).
That's already played out with two bellwethers, with FedEx Corporation (NYSE:FDX) and Oracle Corporation (NASDAQ:ORCL) both falling short of estimates.
But there is an upside to these tempered expectations. With early results mixed and analysts' expectations lowered, it's going to be easier to impress the Street with a big surprise. Companies that deliver big earnings surprises will be rewarded.
According to research by financial economists, the post-earnings drift is the tendency for a stock's cumulative abnormal return to drift in the direction of an earnings surprise for several weeks, and sometimes months, following an earnings announcement.
That means an earnings surprise is twice as valuable: a sharp jump in the short run and a tendency to drift higher in the weeks and months to come. Companies that deliver earnings surprises are good for investors.
The six stocks listed below have seen the largest upward revisions in estimates in the S&P 500 going into first-quarter earnings season. That makes them great candidates for big earnings surprises. These are the kind of big gains my colleague Elliott Gue features in his Top 10 Stocks newsletter.
From the list, I have chosen to highlight Netflix, Inc. (NASDAQ:NFLX) because of its incredible upward momentum and The Boeing Company (NYSE:BA) because of its valuation and dividend.
Netflix, Inc. (NASDAQ:NFLX)
This may seem like a crazy pick considering Netflix is up a market-crushing 74% on the year. But Netflix, Inc. (NASDAQ:NFLX) has been making big adjustments to its business model, adjustments that many investors are confident will produce big results. That sentiment is being driven by Netflix's entry into the more lucrative, higher-margin content business, inking an exclusive deal with Disney and its subsidiaries and producing original content.
This reversal has already had a stunning effect on earnings projections. Ninety days ago, analysts were calling for 2013 earnings of 44 cents per share. Today, they are projecting earnings of $1.19. Their estimate for next year is even more bullish, pegged at $2.71, another 127% increase in earnings. Netflix, Inc. (NASDAQ:NFLX) is still expensive at this level, trading with a forward price-to-earnings (P/E) ratio of 145. But this stock has clear upward momentum.
Rackspace Hosting, Inc. (RAX), Oracle Corporation (ORCL): Should We Take Advantage... Oracle Corporation (ORCL), Hewlett-Packard Company (HPQ): Who Gets Hit in the... Netflix, Inc. (NFLX), LinkedIn Corp (LNKD): Protective Puts to Allow Your Gains... The Blackstone Group L.P. (BX), Apollo Global Management LLC (APO)…Revealed... The Underappreciated Metric That Spotlights Bargain Stocks: Wal-Mart Stores... Atlas Resource Partners, L.P. (ARP), BreitBurn Energy Partners L.P. (BBEP),...
Dreamliner Elliott Gue FedEx Corporation (FDX) Inc. (NASDAQ:NFLX) NASDAQ:NFLX NASDAQ:ORCL NYSE:BA NYSE:FDX Oracle Corporation (ORCL) P/E Ratio S&P 500 The Boeing Company (BA)


text:
Dr. Ralph R. Chase West Texas Collection
About Dr. Ralph R. Chase
Resources Sought
University Archive
Records Retention Program
Records Retention FAQs
Williwood Meador Collection
Museum of Telephony
Friends of the Library and WTC
Virtual Display: Runnels County
E.H. Danner Museum of Telephony
The E.H. Danner Museum of Telephony is located in Officers' Quarters #4 at Fort Concho National Historic Landmark in San Angelo, TX. The museum opened 13 July 1990. Inside the museum, visitors can see over 100 telephones and other equipment related to the history of the telephone.
This site is dedicated to telephone history. The site features the telephones from the Danner Museum of Telephony in San Angelo, Texas. By viewing this website you will see early telephones and how technology has changed through the years. You will learn how phones work and the impact they have made on people's lives. You can investigate the telephone's role in world events. Throughout the site, you will meet people that were involved in the industry. After visiting the site we hope you will imagine what a phone of the future might look like or how people will communicate in the future and what changes are still to come.
Parent Teacher Guide
Creation of this site was a collaborative effort between Verizon, the Dr. Ralph R. Chase West Texas Collection and the Information Technology Department at Angelo State University in San Angelo, Texas. Special thanks go to the members of the GTE Pioneers Club who were instrumental in the creation of the site. Two books were used extensively in the research of the site:
The Heritage of Time: The People and Times of GTE Southwest, 1876-1988 by Larry Johnson
The History of GTE: The Evolution of One of America's Great Corporations by Thomas McCarthy
Before Telephones
1990-20??