text:
KC Royals: Are lefties an endangered bullpen species?
By Mike Gillespie
(Photo by Steven Ryan/Getty Images) /
Two of baseball's new rules and the crowd of pitchers competing for relief jobs may leave little room for left-handers in the KC Royals' bullpen.
Among the many problems facing new manager Mike Matheny, and which he needs to solve before his KC Royals leave Spring Training, is the crowded mess that constitutes the club's bullpen. It's been too long since the Royals had an adequate, much less stellar, relief corps, a roster flaw figuring prominently in the Royal regression that began too soon after the club won its second World Series in 2015. The club hasn't recovered from the departures of Wade Davis, Greg Holland, Kelvin Herrera, Luke Hochevar, and others.
Holland, of course, is back, the beneficiary of an invitation to spring camp and a one-year deal, but isn't the sure roster bet he used to be; although he was serviceable for Arizona last season, he hasn't had a truly Hollandesque campaign since 2017 with Colorado, his first year back after Tommy John surgery and his departure from the Royals. Whether he recaptures enough of his former form to make the club remains to be seen.
Holland, though, is only one of many–perhaps too many–hurlers competing for the few bullpen jobs the Royals realistically have to offer. That crowd and two new rules may severely limit the relief spots available for left-handers.
The first of the new rules relevant here generally limits pitching staffs to 13 (although a player who meets certain criteria may be designated as a "two-way" player, capable of playing a position and pitching without violating the 13-pitcher limit; other position players can pitch in extra-inning games or if their club trails or leads by at least six runs). Assuming Matheny chooses a five-man rotation, the rule leaves room for eight relievers.
The second new rule requires pitchers to face at least three batters; excepted are situations in which the side is retired before three batters face the new pitcher, or the new pitcher is injured or becomes ill. Inevitably, the rule will further diminish (if not eliminate for all practical purposes) the role of the LOOGY–the so-called "lefty one-out guy" who exists almost exclusively to retire just one batter–a dangerous left-hander–in a tight situation.
Together, these two new rules complicate things for KC Royals' lefties. No one on the club qualifies for two-way player status because none have pitched at least 20 big league innings, and played at least 20 games in the field or as DH in which they batted three times, during a specific period. So the 13-pitcher limit applies. Their roster spots secured by virtue of their 2019 performances, and taking three of eight bullpen spots, will be closer Ian Kennedy (righty) and short relievers Tim Hill (lefty) and Scott Barlow (righty).
That leaves room for five in the pen. Jorge Lopez, a righty with stellar but inconsistent stuff, will more than likely make the club–it would be shocking if he doesn't–and Kevin McCarthy, another right-hander who worked regularly last season, is an excellent bet to do the same. By virtue of his status as a Rule 5 draftee, and barring a disastrous spring, righty Stephen Woods Jr. will probably at least break camp with the club, reducing the available relief jobs to two; if he doesn't make the team, the spot will likely go to Kyle Zimmer or Glenn Sparkman, both right-handers.
If, as the Royals clearly hope, Holland and fellow former right-handed lock-down closer and Spring Training invitee Trevor Rosenthal make the club (the fact both pitched for Matheny in St. Louis doesn't hurt), no room remains for any lefties other than Hill. But is that a bad thing considering the new three-batter rule? Maybe…maybe not.
Again, diminution of the LOOGY will be the rule's most immediate impact–unless there are already two outs, or there's one out with a double-play situation, or a lefty is on-deck, Matheny won't have any true LOOGY opportunities because any LOOGY he might have must pitch to at least three batters, thus exposing such a specialist to a right-handed batter, the usual nemesis of the typical LOOGY.
Additionally, the utility of carrying a LOOGY seems to have decreased as the number of innings starters typically pitch has decreased–clubs need relievers who can pitch more than they used to, thus making the LOOGY less important. This, and the new three-batter rule, mean the days of the LOOGY, once a unique baseball species, may be drastically numbered.
The prospect of a KC Royals bullpen full of righties, then, may not be as troublesome as it once was, especially considering Kennedy, Holland and Rosenthal all historically fare better against left-handers than right-handers, and also pitch well against righties, thus reducing the need for a LOOGY or near-LOOGY. (Lefty Hill, always effective against left-handers, is available to face a left-handed threat and is decent against righties).
But is Matheny, prone to follow conventional baseball wisdom and strategy during his St. Louis managerial tenure, ready for a bullpen with only one lefty? He seems ready to embrace analytics, a new personal wrinkle that might make it easier to accept such a pen, and the prevalence of right-handers in baseball is an inescapable fact. Nevertheless, he may justifiably want another left-hander, but may not have room unless Holland or Rosenthal–or both–falter. But there are so many right-handers that left-handers may stand little chance to make the club.
Assuming Matheny wants another left-hander or two and can find room for at least one, Richard Lovelady is probably the most likely choice. The KC Royals have always been high on him, and justifiably so–in four minor league seasons and 133 relief appearances, Lovelady is 10-8 with a 2.17 ERA, 1.01 WHIP, 9.75 SO9 and 2.54 BB9. Probably big league ready in 2018, the Royals waited until last season to call him up, and then seemed reluctant to pitch him regularly; in 25 games, he was 0-3 with a 7.65 ERA and 1.900 WHIP.
Those numbers may be misleading, however, because Lovelady underwent knee surgery shortly after the season ended, suggesting injury may have contributed to his subpar performance. His minor league record proves he can pitch and makes him a viable bullpen candidate.
Like Lovelady, lefty Gabe Speier made an unimpressive KC Royals and major league debut last season, but possesses minor league numbers that probably warrant another big league look. Although he had a 7.36 ERA in nine games, his was a small 7.1 inning sample size; his 3.41 ERA in seven minor league campaigns is probably more indicative of his potential than his short stint in KC. Some of his peripherals have improved in each of the last four seasons and his stuff is good.
Randy Rosario is a left-handed short reliever who's pitched for the Twins, Cubs and Royals in his three major league campaigns. He's 6-0, with one win coming with the Royals after they obtained him on waivers from Chicago last September, and has a 4.83 ERA in 63.1 innings, an otherwise troublesome number tainted by his 13-game 5.91 ERA with the Cubs before they waived him. If he can improve his control a bit, he may catch Matheny's eye; in any event, the club may want a longer look at him than they had last year.
And then there is lefty Eric Skoglund, a once-promising prospect who simply hasn't distinguished himself and whose Kansas City career may be approaching its conclusion. He has two impressive major league victories–a scoreless, two-hit 6.1 inning performance in his 2017 big league debut and a two-hit, nine strikeout, seven-inning triumph in 2018. Unfortunately, those are his only wins in a 2-11, 6.61 ERA three-year span. Without significant improvement this spring, his chances to return to KC are slim.
Next. 2020 player preview: Bubba Starling. dark
Tim Hill is the only left-handed reliever assured of a spot in the KC Royals' bullpen. Baseball's new 13-pitcher and three-batter rules, and the sheer number of pitchers competing for relief roles, will make it hard for any other lefty relievers to secure jobs with the club. In Kansas City, then, the left-handed relief pitcher may be an endangered species.


Augustus John (1878 – 1961) was born in Tenby and spent his early years living there and in Haverfordwest.
A brown plaque on the facade of what is now the Belgrave Hotel on the Esplanade in Tenby commemorates John's birth there on 4 January 1878, and another decorates 32 Victoria Street where he lived from 1884 to 1894. A blue plaque at 5 Victoria Place in Haverfordwest notes that he lived there from 1878 to 1884. John attended the Tenby School of Art but only for a short time. He left Tenby at the age of seventeen to attend the Slade School of Art in London, where his older sister Gwen was also a student. John's outstanding ability was soon noticed by his drawing teacher at the Slade and he came to be recognised as the most talented draughtsman of his generation.
Famed as much for his bohemian lifestyle as for his talent, John was fascinated by Romany culture and sought out the "gypsies" during his extensive travels through the UK and Europe. During the 1920s John was recognised as the leading portrait artist of his day, working on pictures of celebrities of the time including Dylan Thomas, T E Lawrence and George Bernard Shaw. John was, during his life, a Trustee of the Tate Gallery and President of the Royal Society of Portrait Painters. John died at his home in Fordingbridge, England, in October 1961.


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Home > USA Betting Sites > North Dakota Sports Betting
The Peace Garden state is not a friendly place if you like to bet on sports. North Dakota is not a traditional anti-gambling US state because it does allow a lot of ways to wager legally to its citizens. Thirty-two casinos and pari-mutuel facilities call North Dakota home, more of them commercial than tribal. That means North Dakota is home to one of the highest concentrations of legal gambling venues in this part of the United States.
But there's one thing you won't find at any of those casinos and race tracks – legal sports betting. Even the greyhound tracks don't allow sports wagering, though you'll find plenty of casino-style machine games and simulcast OTB-style racing events available for wagers. As I discuss a little later, even bets at US-facing sportsbook websites may be illegal in North Dakota.
Here's a look at the sports and betting scene in one of America's least-populated and most beautiful states.
The state of North Dakota isn't home to any pro sports teams. This has to do with demographics. The state's population is around 740,000, and since the state is pretty large, that comes out to just 11 people per square mile. Pro sports teams need population density to survive; even the state's largest city, Fargo, is tiny, with an estimated 100,000 citizens as of 2014. By comparison, the smallest city with a professional US team in the Big 4 is Salt Lake City, and that city is double the size of Fargo.
NCAA Teams in North Dakota
Two NCAA teams call North Dakota home. They both compete in all sports at the Division 1 level, which is a nice treat for the otherwise sports-starved people of the Rough Rider State. Both teams are famous for participating in NCAA football and NCAA basketball.
The North Dakota Fighting Sioux hail from the University of North Dakota in Grand Forks. The Sioux have made a lot of noise in NCAA hockey, which isn't a major NCAA sport outside of a few niche markets. Still, what the Sioux have done in hockey is impressive – they've been in league twelve championships, winning seven of them outright. Before North Dakota's move to Division 1, they were highly competitive in football and basketball, as well. The team won several D-2 national championships, including one in both sports the season before they transferred into the top division. Their participation in NCAA basketball is forgettable – they've had just one player recruited since the 1970s, and they've yet to win many games since switching divisions two years ago. But between the multiple championships and the fact that UND has produced a dozen super-famous current NHL players, The Fighting Sioux are primarily a hockey squad.
The North Dakota State Bison represent North Dakota State University in Fargo. They're often called "The Thundering Herd." NDSU has competed as a D-1 Subdivision school since 2004, and they've had quite a storied experience in the ten short years since they joined. NDSU is the first team to win back-to-back-to-back championships in NCAA Subdivision football history. Unfortunately, it can be difficult or even impossible to find Subdivision games to bet on. NDSU had a surprise run during the 2014 NCAA basketball tournament, making it to the third round and beating #5 Oklahoma along the way. Could The Thundering Herd be establishing a new tradition of basketball excellence?
North Dakota Law and Online Sports Betting
One thing that causes some frustration about North Dakota gaming law is its broad definition of what constitutes "gambling." In Section 12.1-28-01(1) of the state's criminal code, gambling is defined as "risking any money, credit, deposit, or other thing of value for gain, contingent, wholly or partially, upon lot, chance, the operation of gambling apparatus, or the happening or outcome of an event, including an election or sporting event, over which the person taking the risk has no control." It basically gives the state the right to define gambling as anything they want, which is partially responsible for the difficult environment of the state for sports bettors.
Some people interpret North Dakota law in a way that makes online gambling explicitly illegal. This quote from the office of the Attorney General is the best way to describe this argument: "Online gambling is prohibited under the state's Constitution, regardless of the amount wagered. In North Dakota, gambling may be conducted only by licensed non-profit organizations, Tribal casinos or through a state run lottery." I'm not so sure about the AG's interpretation, since there is no actual prohibition in the state code against placing a bet on a sporting event on the Internet.
Another exception to gambling laws in the state – so-called "social gambling." It's hard to say how far this allowance goes for sports bettors, but it probably covers things like office pools and March Madness bracket contests. An important restriction on legal social gambling in North Dakota – no one can wager more than $25 on any individual hand, game, or other outcome. Betting $26 or more makes you guilty of a misdemeanor, and you may be charged with a felony if you wager more than $500 in a social setting. A quick check reveals that no one has ever been arrested under this statute for placing a sports bet, so your office pool is probably safe, even if you bet $50 a square.
Please note that I am NOT a lawyer, and I do NOT intend anything in this article to be interpreted as actual legal advice. US state law is complicated enough as it is without throwing in gaming into the mix. I'm just giving my observations based on the facts I know about the state and its criminal code.
Citizens of North Dakota face some pretty strict gambling laws, but they have outlets for casino-style gambling and racebook bets at greyhound tracks. The state is fairly liberal about social, private, and charitable gaming as well. Sports betting is one area where North Dakota's laws seem stuck in the last century. Though you can play blackjack and slots in a dozen cities across this sparsely-populated state, you won't find a single legitimate bookmaker anywhere in the state.

redpajama_set_name: RedPajamaCommonCrawl


I came across this magnificent piece of work at the Norton Simon museum in Los Angeles, shortly after the loss of a pregnancy.
I bought a print that now hangs in our bedroom. I used to stare at it, wanting to be that woman...working so hard for her family. Cleaning, cooking and doing laundry. All because she loved them so much.
My husband and I wanted to be parents more than anything. We would cry every time kids came to the door dressed up in cute costumes, for candy, on Halloween. Or clean behind the refrigerator on Christmas day, just to distract ourselves from the loud cry in our hearts of wanting a child.
Not knowing if it would ever be possible, I truly became ok if we were never able to have children. I could still appreciate this piece of art. This lady, who so joyfully serves her family.
God heard the desires of our heart. We now have a beautiful girl & boy. We have gone through the no sleep stage of infancy. The chasing toddler stage. We are now full force in the elementary school stage. Sleep may be more plentiful, but, there are new stages to grow through. Frustrating days. Days when counting to 10 is not effective. Times when we ask ourselves, "what did we do?"
These are the days I need to take some quiet time and gaze at the lady in the picture. She means something more to me now. She is still managing to find the joy in working for her family. Even in times she is not certain she is cut out for the hard emotional work it takes. She will continue to put one foot in front of the other. Doing what needs to be done. Trusting that the final outcome will bring her a deeper joy than she could have ever imagined.
Wonderful insight into your world of the past & present, Sandy! I love your thoughts!


text:
package network

import (
	"crypto/rand"
	"crypto/rsa"
	"fmt"
	"io/ioutil"
	"net"
	"os"
	"strings"

	"golang.org/x/crypto/ssh"
	"golang.org/x/crypto/ssh/agent"
)

const (
	defaultPort = 22
	defaultUser = "core"
	rsaKeySize  = 2048
)

// Dialer is an interface for anything compatible with net.Dialer
type Dialer interface {
	Dial(network, address string) (net.Conn, error)
}

// SSHAgent can manage keys, updates cloud config, and loves ponies.
// The embedded dialer is used for establishing new SSH connections.
type SSHAgent struct {
	agent.Agent
	Dialer
	User     string
	Socket   string
	sockDir  string
	listener *net.UnixListener
}

// NewSSHAgent constructs a new SSHAgent using dialer to create ssh
// connections.
func NewSSHAgent(dialer Dialer) (*SSHAgent, error) {
	key, err := rsa.GenerateKey(rand.Reader, rsaKeySize)
	if err != nil {
		return nil, err
	}

	addedkey := agent.AddedKey{
		PrivateKey: key,
		Comment:    "core@default",
	}

	keyring := agent.NewKeyring()
	err = keyring.Add(addedkey)
	if err != nil {
		return nil, err
	}

	sockDir, err := ioutil.TempDir("", "mantle-ssh-")
	if err != nil {
		return nil, err
	}

	// Use a similar naming scheme to ssh-agent
	sockPath := fmt.Sprintf("%s/agent.%d", sockDir, os.Getpid())
	sockAddr := &net.UnixAddr{Name: sockPath, Net: "unix"}
	listener, err := net.ListenUnix("unix", sockAddr)
	if err != nil {
		os.RemoveAll(sockDir)
		return nil, err
	}

	a := &SSHAgent{
		Agent:    keyring,
		Dialer:   dialer,
		User:     defaultUser,
		Socket:   sockPath,
		sockDir:  sockDir,
		listener: listener,
	}

	go func() {
		for {
			conn, err := listener.Accept()
			if err != nil {
				return
			}
			go agent.ServeAgent(a, conn)
		}
	}()

	return a, nil
}

// Close closes the unix socket of the agent.
func (a *SSHAgent) Close() error {
	a.listener.Close()
	return os.RemoveAll(a.sockDir)
}

// Add port to host if not already set.
func ensurePortSuffix(host string, port int) string {
	switch {
	case !strings.Contains(host, ":"):
		return fmt.Sprintf("%s:%d", host, port)
	case strings.HasPrefix(host, "[") && strings.HasSuffix(host, "]"):
		return fmt.Sprintf("%s:%d", host, port)
	case strings.HasPrefix(host, "[") && strings.Contains(host, "]:"):
		return host
	case strings.Count(host, ":") > 1:
		return fmt.Sprintf("[%s]:%d", host, port)
	default:
		return host
	}
}

func (a *SSHAgent) newClient(host string, user string, auth []ssh.AuthMethod) (*ssh.Client, error) {
	sshcfg := ssh.ClientConfig{
		User: user,
		Auth: auth,
	}
	addr := ensurePortSuffix(host, defaultPort)
	tcpconn, err := a.Dial("tcp", addr)
	if err != nil {
		return nil, err
	}

	sshconn, chans, reqs, err := ssh.NewClientConn(tcpconn, addr, &sshcfg)
	if err != nil {
		return nil, err
	}

	client := ssh.NewClient(sshconn, chans, reqs)
	err = agent.ForwardToAgent(client, a)
	if err != nil {
		client.Close()
		return nil, err
	}

	return client, nil
}

// NewClient connects to the given host via SSH, the client will support
// agent forwarding but it must also be enabled per-session.
func (a *SSHAgent) NewClient(host string) (*ssh.Client, error) {
	return a.NewUserClient(host, a.User)
}

// NewUserClient connects to the given host via SSH using the provided username.
// The client will support agent forwarding but it must also be enabled per-session.
func (a *SSHAgent) NewUserClient(host string, user string) (*ssh.Client, error) {
	client, err := a.newClient(host, user, []ssh.AuthMethod{ssh.PublicKeysCallback(a.Signers)})
	if err != nil {
		return nil, err
	}
	return client, nil
}

// NewPasswordClient connects to the given host via SSH using the
// provided username and password
func (a *SSHAgent) NewPasswordClient(host string, user string, password string) (*ssh.Client, error) {
	client, err := a.newClient(host, user, []ssh.AuthMethod{ssh.Password(password)})
	if err != nil {
		return nil, err
	}
	return client, nil
}

redpajama_set_name: RedPajamaGithub


"Gary Cohn and Steve Mnuchin RiskTheir Reputations" by pushing Trump tax plan, writes N.Y. Times columnist David Leonhardt: "Within the administration, there are real differences among how top officials have behaved and how they are perceived. Several — Tom Price, Reince Priebus, Sean Spicer and Rex Tillerson — ... turned themselves into punch lines."
"The clearest exception is Jim Mattis, the defense secretary. Mattis has done so partly by avoiding scandal and minimizing conflicts with Trump. But he has also been careful to set his own ethical boundaries... Cohn and Mnuchin have started to risk theirs. This column is a plea to them: Please stop, for everyone's sake, including yours."
Why it matters: "Neither one of them has yet turned 60 years old [Cohn is 57; Mnuchin, 54]. These won't be their last jobs."


text:
You are here: Home » Attractions » Royal Air Force Museum London
Royal Air Force Museum London
Opened in 1972 by Her Majesty the Queen and situated on the historic site of London Aerodrome at Hendon, our North London Museum is the only London attraction to house over 100 aircraft from around the world including some very early aircraft designs through to the latest modern day jets and military aircraft. With free admission plus free interactive and fun activities, including 3D Cinema and Our Finest Hour, our London museum offers entertaining and educational days out for all the family just 30 minutes from Central London. For further details check our What's On section.
Website http://www.rafmuseum.org
Email london rafmuseum.org
Address The National Cold War Exhibition, Hendon, Greater London, NW95LL
We don't currently have any trips departing for Royal Air Force Museum London. Check back soon!


You came in like a lamb and left like the dog days of July. I do not enjoy turning on my AC this early. But your 85 degrees with high humidity leave me little choice.
However, I do enjoy the lilacs you brought extra early. Our driveway smells like a dream. And now my dining room does, too.
Maybe we'll get some actual spring in April.
This entry was posted in Uncategorized on 04.02.12 by awastell.
Remember how we always wanted to build a secret little playhouse midst the lilac bushes in our backyard? Miss you and lilacs too.
Remember the hidden playhouse we always wanted to build midst the lilacs in the backyard? Miss you and lilacs too.


We were the only campers. Saw 3 other cars over 3 days.
Road in is a little rough. My Garmin said there was a back way out, but I couldn't find it.
I believe I saw a leach on my leg after wading in the lake, but I shook it off too quickly to positively ID it.
On the hill above there is a caldera with a pond and salt lick that elk and antelope frequent. Cows stay by the lake.
Pretty. Quiet. Not too buggy in mid-June. A little overgrown. Signs of hunters.


Inspired by Dear Data, the data drawing pen pal project, designers Josefina Bravo, Sol Kawage, and Tomoko Furukawa use the postcard medium to send each other weekly how-to instructions for a wide variety of everyday things. The only rule is that they can't use words.
As of writing this, they're on week 37, which covered how to roll maki, how to eat an apple like a boss, and how to make mayonnaise.


text:
Selective quotations
On 26 October 2018 24 December 2020 By DonaldIn Behaviour, Politics, Reformation
We have a very clear and recent example of selective quotation by the BBC in its commentary surrounding the use of parliamentary privilege to name Sir Philip Green as the anonymous person who secured a court injunction preventing The Daily Telegraph publishing allegations of sexual and racial harassment.
While other commentators quite rightly concentrate on the use and abuse of parliamentary privilege, I draw attention to another feature that will receive no attention from the main stream media.
There have been many complaints about bias in the main stream media (MSM), with American President Donald Trump leading the charge. It is common for people and political parties to complain about BBC bias. The BBC defends itself by claiming that as the complaints come from opposite sides of a debate, the BBC must be quite balanced. A better defence would be to point out that all programmes follow a line that is inherently biased, but as long as different views are allowed to be expressed in alternative programmes then the BBC performs its function as a public broadcasting corporation.
Selective quotation
However, selective quotation is another matter. This is very easy to do, is commonly done, and can be damaging. It manifests that the desire to win an argument is greater than the regard for truth. An old example that comes to mind because of its frequent recurrence is the selective curtailing of George Galloway's meeting with Saddam Hussein, and the most recent one is the clipping from BBC's This Week showing Alan Johnson last night questioning Peter Hain's use of Parliamentary privilege as 'undermining the law' or judiciary. News items have repeatedly shown it and Politics Live does the same today. Each time the video clip stops very abruptly. Why? Because it goes on immediately to show the BBC's Andrew Neil saying this view was very interesting and Michael Portillo agreed with this and said, "Well said, Alan." Why did the BBC in its reporting of this opinion confine this sentiment to Alan Johnson and not include Michael Portillo who was of a similar view? The answer is not very important, but it demonstrates the readiness of the BBC to cut clippings short and prevent viewers seeing the reaction to the point being made in context.
I remind people that there is a context to every story and if people were more experienced in exegeting the Bible and more familiar with the centuries of debate over biblical doctrine, as former generations were, they would be more alert to the use of quotations out of context.
The historian Dr David Starkey put parliamentary privilege in context, beginning in Henry VIII's reign, that it was to prevent the judiciary interfering with parliament. He also compared Henry VIII's breach with the pope in Rome to "a hard Brexit 16th-century style", each Brexit requiring the approval of parliament, and the BBC subtitled this section of the debate 'Court of Tudor Vs. Court of May'.
Alan Johnson said that parliamentary privilege should be used very carefully, recalling that Erskine May, the parliamentary manual of practice, explained that parliamentary privilege should be used as part of one's legislative duties. I find it difficult to see how Lord Haines' exposure of Philip Green is part of a legislative duty and it appears to me that he has taken the law into his own hands. This will lead to serious debate on the use and abuse of parliamentary privilege.
The Politics Live presenter Emma Barnett preceded the This Week clipping with: "someone who is very clear in their view that they didn't think it was the right thing to do is Alan Johnson", and after the clip she said: "he is very clear there." The problem is that he was not "very clear" and he was more nuanced in his opinion. He questioned it and correctly raised the topic for debate. Similarly, I am raising the topic of accurate, contextual quotation, but this will not be debated by those who are practised in the nefarious art of misrepresenting other people.
Meanwhile, this blogpost allows me to add examples of selective quotation by MSM as they come across my path.
BBC watch monitors the BBC's reporting of events in Israel.
4 Dec 2018: following the market turmoil on this politically significant day for Brexit, the BBC news at 6 p.m. reported the gloomy predictions of the Governor of the Bank of England, Mark Carney, and the BBC report finished with "the pound is at its lowest against the dollar for eighteen months". Was this meant to suggest that Brexit was the cause? Why did the BBC not compare the pound to the euro instead of the dollar? After all, this was the topic.
1. Only last month the euro was also at its lowest against the dollar for eighteen months. 2. the pound against the euro is not at its lowest in the past eighteen months, and only last month it was at its highest against the euro for eighteen months. Where does the BBC fact-checking fit into this?
So what was the BBC conveying to its audience?
Two hours later, the BBC reported that Wall Street fell 2% today, with shares in financial companies suffering the biggest declines. Presumably, even the BBC could not blame this on Brexit.
Sensibly, former Governor of the Bank of England Mervyn King has compared Theresa May's Brexit to pre-World War II appeasement policy of the Nazis. In his article in Bloomberg he has pointed out that Brexit preparations were inadequate and "incompetence on a monumental scale".
He was saddened that the Bank of England had been drawn into Project Fear.
11 Mar 2020: bold and deliberate truncating of Boris Johnston's statement on coronavirus, demonstrating the shameless activity of those who bear false witness.
16 Nov 2020: the BBC reported on the 'second' vaccine against COVID-19 on its 1 p.m., 6 p.m. and 10 pm. news. It was referring to the good news that Moderna has developed a vaccine that is nearly 95% effective and can be stored at a better temperature than the Pfizer vaccine announced on 9th November seven days ago.
The broadcast news went on to report about forthcoming vaccines, so it appears that the BBC cannot bring itself to refer to the Russian vaccine that was the 'second' vaccine to be developed. The online report does not repeat the mistake that Moderna's is the 'second' vaccine and it does mention the Russian vaccine, but why does the broadcast media repeatedly insist on ignoring the Russian vaccine? Is the BBC afraid of telling the truth? BBC Newsnight still repeated the 'second' claim tonight, possibly on the back of the Deputy Chief Medical Officer's refence to "the second penalty [shot] now, that's also gone into the back of the net." I am sure they can count, so why this sidelining of the Russian vaccine?
The Hadassah Medical Centre, Jerusalem, has ordered over 1.5 million doses of Sputnik V vaccine, whose Director suggested that it will be the first vaccine to market and expects 50% of the world to use it. However, the BBC cannot mention it, nor the two vaccines being tested in Cuba. Could politics possibly be behind the BBC 'news'?
11 Dec 2020: Europe's Astra-Zeneca lab is cooperating with Russian Sputnik V lab to trial a mixture of the vaccines to improve immunity. This should be rather embarrassing for the BBC who ignored the Russian vaccine in its reports, see above, but so few will call out the BBC that it can afford to carry on regardless.
BBCDonald Trumpmedia bias
Balancing individual rights with societal rights and the common good
Edward P. C. Greene 1937–2018


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text:
\section{Introduction}\label{Sect:Intro}

In characteristic $0$ algebraic geometry, Bertini theorems are often used to prove theorems using the idea of descending induction.  That is to say if a variety exhibits a given property $\mathcal{P}$, it is often desirable for the property to descend to a general choice of codimension 1 subvariety.  Many properties in characteristic $0$, such as being smooth or having relatively nice singularities, do descend to a general hyperplane section.  More generally, it was later shown that an important invariant, the multiplier ideal, also exhibits these Bertini type theorems for pairs (or even triples) $(X,\Delta)$.

The case of characteristic $p>0$ was less understood.  First, the notion of a multiplier ideal is not known to exist unless a log resolution of the pair $(X,\Delta)$ exists.  Furthermore, the positive characteristic analog of the multiplier ideal, the test ideal, was not known to satisfy such a property except in very specific circumstances such as the case where $\tau(X,\Delta) = \mathscr{O}_X$.  In this paper, it is shown in \autoref{prop:FailSB1} that the natural analog of the Bertini theorem for multiplier ideals is false for test ideals in dimension $n\geq 3$.

\begin{theorem*}
There exist $\mathbb{Q}$-divisors $\Delta$ on $\mathbb{P}^n$ with the property that for a general hyperplane $H$,
$$\tau(\mathbb{P}^n,\Delta)|_H \neq \tau(H,\Delta|_H).$$
\end{theorem*}

In the process of proving this proposition, it is shown in \autoref{thm:Cex} that the analog of Koll\'ar-Bertini theorem \cite[9.2.29]{Laz} is also false for a class of pairs.

\begin{theorem*}
	Given any $\epsilon>0$, and $n\geq 3$, there are pairs $(\mathbb{P}^n,\Delta)$ such that for a general hyperplane $H$,
	$$\tau(\mathbb{P}^n,\Delta) \neq \tau(\mathbb{P}^n,\Delta+\epsilon H).$$
\end{theorem*}

This in some sense eliminates any hope that a Bertini theorem can hold for test ideals. Thus for many types of F-singularities which are governed by the test ideal, taking a general hyperplane section will not preserve the type of singularity.  However, it is shown to hold in \cite{SZ} for pairs $(X,\Delta)$ which are sharply F-pure or strongly F-regular.  In \autoref{Sect:Questions}, some follow-up questions are posed regarding the open state of the question for non-pairs, the property of being F-rational, and whether there is something more geometric to consider regarding these types of questions.

\

\noindent{\it Acknowledgements}: I would like to thank my advisor Karl Schwede for suggesting this problem, as well as many valuable discussions that led to understanding the nature of the problem as well its the eventual solution.  His assistance with earlier drafts of this paper is also greatly appreciated.  The author is also very grateful for the comments of the referee which improved the paper through very detailed feedback.  I also would like to thank Stefano Filipazzi for discussions about many of the corresponding statements for multiplier ideals.  Finally, I am very appreciative for some helpful comments from Eric Canton on a earlier draft of this paper.

\section{Background and Motivation}\label{Sect:Motiv}

\subsection{Test Ideals}

Let $X=\Spec(R)$ be a normal variety over a perfect field $K$ of characteristic $p>0$ and let $\Delta$ be an effective $\mathbb{Q}$-divisor on $X$.  This is the data of a pair $(X,\Delta)$.  The (big) test ideal associated to $R$, denoted $\tau(X,\Delta)$, is defined as follows:

\begin{definition}
	$\tau(X,\Delta)$ is the smallest non-zero ideal $J$ of $R$ such that for every $e>0$ and every $R$-linear homomorphism $\phi: F_*^eR(\lceil (p^e-1)\Delta \rceil) \rightarrow R$, $J$ is compatible with $\phi$ in the following sense:
	$$\phi(F_*^e J)\subseteq J$$
\end{definition}

This ideal can be computed locally and can thus be extended to non-affine varieties.  It is a very important invariant which governs the singularities of a positive characteristic variety.  In general, $X$ is strongly F-regular (in particular normal and Cohen-Macaulay) if and only if $\tau(X) = \mathscr{O}_X$, and smaller test ideals imply more severe singularities.  In the case that $\Delta$ is a $\Q$-Cartier divisor, and $\Delta=t\cdot\Div(f)$, for simplicity one often writes $\tau(R,\Delta) = \tau(R,f^{t})$.

\subsection{Relating Multiplier Ideals to Test Ideals}

Let $X \subseteq \mathbb{P}^n$ be a smooth quasi-projective variety over $\mathbb{C}$.

\begin{definition}\label{DefnMult}
Let $\Delta\geq 0$ be a $\mathbb{Q}$-divisor such that $K_X+\Delta$ is $\Q$-Cartier, and let $\pi:X' \rightarrow X$ be a log resolution of $\Delta$.  The multiplier ideal sheaf is defined as
$$\mathcal{J}(X,\Delta) = \pi_* \mathscr{O}_{X'}(\lceil K_{X'} - \pi^*(K_X + \Delta)\rceil).$$
\end{definition}

This notion is independent of the log resolution chosen, and measures the singularities of the pair $(X,\Delta)$ in characteristic zero.  The primary motivating theorem is as follows (see \cite[9.2.29 and 9.5.9]{Laz}):

\begin{theorem}\label{KolBer}
Let $(X,\Delta)$ be a pair, with $X$ a smooth complex variety and $\Delta\geq 0$ a $\mathbb{Q}$-divisor.  Let $|V|$ be a base-point free linear series on $X$.  Then for a general choice of divisor $B\in |V|$, and any $0 \leq \epsilon < 1$,
$$\mathcal{J}(X,\Delta) = \mathcal{J}(X,\Delta+\epsilon B).$$

In addition, for general $B\in |V|$
$$\mathcal{J}(X,\Delta) \otimes_{\mathscr{O}_X}\mathscr{O}_B \cong \mathcal{J}(B,\Delta|_B).$$
\end{theorem}

So multiplier ideals behave as well as one could expect when taking general hyperplane sections.  Moreover, multiplier ideals are intimately related to test ideals.  Let $X = \Spec(R)$ be a complex variety, with $R=\mathbb{C}[x_0,\ldots,x_n]/I$ with $D\geq 0$ an effective divisor.  If $I=\langle f_1,\ldots,f_m\rangle$, then let $A$ be the $\mathbb{Z}$-algebra generated by all of the coefficients of the $f_i$ and the equations of a log resolution $\pi: \tilde{X} \rightarrow X$.  Then $X$ and $\pi$ are defined over $A$.  For each maximal ideal $\mathfrak{m}\subseteq A$, $A/\mathfrak{m}$ is a finite field, and we can consider $X_\mathfrak{m}$ obtained by base change.  The following theorem motivates the idea that many of the well studied properties of multiplier ideals have a chance to descend to test ideals:

\begin{theorem}\label{thm:Red}
\cite{Tak}, cf. \cite{Smi},\cite{HY},\cite{Hara} Let $X$ be a complex projective variety, and $\Delta$ a $\mathbb{Q}$-divisor on $X$ with $K_X+\Delta$ a $\Q$-Cartier divisor.  Then for a given model $A$ of $X$ and an open dense collection of closed points $\mathfrak{m} \in \mSpec(A)$

$$\mathcal{J}(X,\Delta)_\mathfrak{m} = \tau(X_\mathfrak{m},\Delta_\mathfrak{m}).$$
\end{theorem}

In particular, it seems reasonable to ask if the same Bertini-type theorem holds for test ideals as well.  Note that in characteristic $p>0$, one typically needs to replace the consideration of a base point free linear series with the assumption that the condition holds on fibers of a map.  A particular case of this more general conjecture is known in the case of a strongly F-regular pair.  This corresponds to the case of $\tau(X,\Delta) = \mathscr{O}_X$, and the theorem is restated here:

\begin{theorem}\label{thm:SZ}
	\cite[Theorem 6.7]{SZ} Suppose that $X$ is a variety over an algebraically closed field $k$, let $\Delta \geq 0$ be a $\mathbb{Q}$-divisor on $X$, and let $\phi:X \rightarrow \mathbb{P}^n_k$ be a $k$-morphism with separably generated residue field extensions\footnote{Note that one cannot expect the same statement to hold for a general member of an arbitrary basepoint free linear system.}.  Suppose $(X,\Delta)$ is a strongly F-regular pair.  Then for an open dense subset $U\subseteq (\mathbb{P}^n_k)^\vee$ and all $H\in U$, $\left(\phi^{-1}(H),\Delta|_{\phi^{-1}(H)}\right)$ is strongly F-regular.
\end{theorem}

Furthermore, one can easily test many lower dimensional hypersurface examples of a fixed degree in a finite field, using Macaulay2 \cite{M2}.  There is a script created by the author available as \cite{M2Code}, which produces a list of hyperplanes for which the restriction fails.  Tests examples included hypersurface degree smaller than $6$ in characteristics $p\leq 11$ (as well as small field extensions) in $\P^n$ for $n\leq 4$.  This yielded only examples for which a few special problematic hyperplanes for restriction theorems (can be excluded as a component of a closed set upon taking the perfection of the underlying field).

\section{Computation of Test Ideals for $(\mathbb{A}^n,f^{\frac{1}{p}})$}\label{Sect:Comp}

Consider the pair $(\mathbb{A}_K^n,f^{\frac{1}{p}})$.  As noted above $f^{\frac{1}{p}}$ represents the divisor defined by $f$ with coefficient $\frac{1}{p}$, and $f$ is an element of the ring $S=K[x_1,\ldots,x_n]$.  Assume $K$ is an $F$-finite field with a basis for $F_*K$ over $K$ given as $\langle F_*u_1,\ldots,F_*u_m \rangle_K$.  Let
$$F_* f=\sum_{\boldsymbol{\alpha},i} s_{\boldsymbol{\alpha},i} F_* u_i \boldsymbol{x^\alpha}.$$
In general, $\boldsymbol{\alpha}$ will denote a multi-index in $\Lambda = \Lambda_{p,n} := \left\{0,1,\ldots,p-1\right\}^{\oplus n}$, coming from the set of exponents occurring in the standard basis of monomials $F_*S$ over $S$.  Similarly, $\boldsymbol{x^\alpha}$ is shorthand for $x_1^{\alpha_1}\cdots x_n^{\alpha_n}$.  The index $i$ will range from $1$ to $m$, to indicate the elements of the basis of $F_*K$ as above.  In general, a boldface symbol will denote a multi-index. Also note that $s_{\boldsymbol{\alpha},i}\in S$ is viewed naturally as a subset of $F_*S$ under the inclusion $S\hookrightarrow F_*S: x \mapsto xF_*1 = F_*x^p$.  Let $\Phi$ denote the generator of $\Hom_S(F_*S,S)$ as an $F_*S$-module under pre-multiplication
$$\Phi: F_*S \rightarrow S: F_* u_1\boldsymbol{x^{p-1}} \mapsto 1$$
\noindent and $\Phi$ sends all of the other elements of the standard basis $\langle F_*u_i\boldsymbol{x^\alpha} \ | \ \boldsymbol{\alpha} \in \Lambda,i=1,\ldots,m\rangle_S$ to $0$.

  The test ideal $\tau(\mathbb{A}^n,f^{\frac{1}{p}})$ can then be realized as the image of $\Phi(F_*f\cdot -)$ which can be computed explicitly as follows.  This was written down as early as \cite{HT}, but the proof following the notation outlined above is provided here for convenience:

\begin{proposition}\label{prop:Comp}
	\cite{BMS} $\tau(\mathbb{A}^n,f^{\frac{1}{p}}) = \Phi(\langle F_*  f \rangle_{F_*S}) = \langle s_{\alpha,i} \ | \ \boldsymbol{\alpha} \in \Lambda,i=1,\ldots,m\rangle_S$.
\end{proposition}
  \begin{proof}
  The first equality follows by definition and the fact that $\Phi^{\circ e}$ generates $\Hom_S(F_*^e S,S)$ for each $e>0$, so it only remains to show that $\Phi(\langle F_*  f \rangle) = \langle s_{\alpha,i}  \ | \ \boldsymbol{\alpha},i\rangle_S$.  Consider an element $g$ of $\Phi(\langle F_*  f \rangle_{F_*S})$.  Then $S$-linearity of $\Phi$ implies
	$$g=\Phi(F_*h\cdot f) = \sum_{\boldsymbol{\alpha},m} s_{\boldsymbol{\alpha},i} \Phi(F_* (u_i \cdot h\cdot \boldsymbol{x^\alpha}))\subseteq \langle s_{\alpha,i}  \ | \ \boldsymbol{\alpha},i\rangle_S.$$

	\noindent Similarly, considering the element $\Phi\left(F_* (u_i^{-1}\cdot\boldsymbol{x^{p-1-\alpha}} \cdot f)\right) = s_{\alpha,i} \in \Phi(\langle F_* f \rangle_{F_*S})$ yields the reverse inclusion.\end{proof}

Let $H$ be a hyperplane section in $\mathbb{A}^n$ defined by $l=c_0 + c_1x_1 + \ldots + c_nx_n$. Considering the pair $(\mathbb{A}^n_K,(f\cdot l)^{\frac{1}{p}})$ where we can conclude identically to the previous case that the corresponding test ideal $\tau(\mathbb{A}^n_K,(f\cdot l)^{\frac{1}{p}})$ is given as the image of $\Phi(F_* (f\cdot l \cdot S))$, or equivalently
$$\left\langle c_{i,0} s_{\boldsymbol{\alpha},i} + \sum_{j=1,\ldots,n} c_{i,j} \cdot s_{\boldsymbol{\alpha}-\boldsymbol{1}_j,i} \cdot x_j^{\left\lfloor\left(\frac{\boldsymbol{(\boldsymbol{\alpha}-\boldsymbol{1}_j)+\boldsymbol{1}_j}}{p}\right)_j\right\rfloor} \ | \ \boldsymbol{\alpha}\in\Lambda,i=1,\ldots,m \right\rangle_S$$
Note that $\boldsymbol{1}_j$ is representative of a multi-index with $1$ in position $j$ and $0$ elsewhere. Moreover when $\alpha_j=0$, $\boldsymbol{\alpha}-\boldsymbol{1}_j$ is treated as $p-1$ in the $j^{\text{th}}$ entry.  Therefore, if $\alpha_j=0$, then the exponent is treated as
$$\left\lfloor\left(\frac{(\boldsymbol{\alpha}-\boldsymbol{1}_j)+\boldsymbol{1}_j}{p}\right)_j\right\rfloor = \left\lfloor\frac{(p-1) + 1}{p}\right\rfloor =1.$$
Finally, $c_{i,j}\in K$ are elements s.t. $F_*(c_j) = \sum_{i} c_{i,j}F_*(u_i)$.

So rephrasing the statement from above, to disprove the direct analog of \autoref{KolBer}, it is enough to show that for a non-closed (or even open) set of $H=V(l)$, the two ideals $\Phi(F_*f\cdot S)$ and $\Phi(F_*f\cdot l\cdot S)$ do not agree.  This will be shown below as \autoref{thm:Cex}.

One can further consider the case of the test ideal sheaf $\tau(\mathbb{P}^n,f^{\frac{1}{p}})$, which on each element of the standard affine cover is $\tau(\mathbb{A}^n,f^{\frac{1}{p}}|_{x_i=1})$ since the big test ideal is defined locally.  This is recorded in \autoref{cor:proj} and \autoref{thm:HHfalse} for the sake of completeness.

\section{Examples and Counterexamples}\label{Sect:CountEx}

Let $K=K^{p}$ be a perfect field, and $S = K[x_1,\ldots,x_n]$, and consider $\mathbb{A}^n = \Spec(S)$ in which case the trace map can be view as
$$\Phi: F_*S \rightarrow S.$$

Continuing the use of the multi-index notation $\boldsymbol{x^\alpha} = x_1^{\alpha_1}\cdots x_n^{\alpha_n}$, $\Phi$ is defined by

\[ \Phi(\boldsymbol{x^\alpha}) = \begin{cases}
      \boldsymbol{x^{\frac{\alpha-p+1}{p}}} & p \ | \ \alpha_i+1 \ \ \forall i \\
      0 & \text{otherwise}
   \end{cases}
\]

The best case scenario for extending Bertini theorems to characteristic $p>0$ would be a direct generalization of the Bertini Theorem for Multiplier Ideals.  That is to say that for $I$ homogeneous, $R = S/I$, and $\Delta$ a $\mathbb{Q}$-Cartier Divisor on $X=\Spec(R)$, is it true that for $0\leq\epsilon<1$ and a general hyperplane $H$, the analog of \cite[9.2.29]{Laz} holds true?
$$\tau(R,\Delta) \stackrel{?}{=} \tau(R,\Delta+\epsilon H)$$
$$\tau(X,\Delta)|_{H} \otimes_{\mathscr{O}_X} \mathscr{O}_H \stackrel{?}{=} \tau(H,\Delta|_H)$$

The following examples shows that this is not the case.  All explicit counter-examples are worked out in the affine case $S=K[x_1,\ldots, x_n]$.

\begin{theorem}\label{thm:Cex}
	Suppose that $S = K[x_1,\ldots,x_n]$ is a polynomial ring over an infinite perfect field of characteristic $p>0$, with $n\geq 3$.  Consider an element $f\in S$ of the form
	$$F_*f = f_{p-1}\cdot F_*(x_1\cdots x_{n-1})^{p-1}x_n^{p-1} + f_{p-2}\cdot F_*(x_1\cdots x_{n-1})^{p-1}x_n^{p-2} + \ldots + f_0\cdot F_*(x_1\cdots x_{n-1})^{p-1}$$

	\noindent where the $f_i$ are polynomials in $K[x_1,\ldots,x_{n-1}]$ with the additional independence property that for each $i$,
	\begin{equation}f_i \notin \langle f_0,\ldots,f_{i-1}\rangle_S+ \mathfrak{p}\cdot \tau(S,f^{\frac{1}{p}}) = \langle f_0,\ldots,f_{i-1},x_j f_i,\ldots, x_jf_{p-1}\rangle_S \tag{$\star$}\end{equation}

	\noindent where $\mathfrak{p}=\langle x_1,\ldots,x_{n-1}\rangle$ and $j=1,\ldots,n-1$.  Then for a general hyperplane $H=V(l)$,
	$$\tau(S,f^{\frac{1}{p}}) \neq \tau(S,(l\cdot f)^{\frac{1}{p}}).$$
\end{theorem}
\begin{proof}
	Following \autoref{prop:Comp}, one can conclude directly that $\tau(S,f^{\frac{1}{p}}) = \langle f_0,\ldots,f_{p-1}\rangle_S$ and giving the linear form $l$ a presentation of $l=c_0+c_1x_1 + \ldots + c_nx_n$, we can conclude $\tau(S,(f\cdot l)^{\frac{1}{p}})$ is generated by elements of the form:

\

	\begin{itemize}
		\item $c_0f_{i} + c_nf_{i-1}$ for $i=1,\ldots,p-1$, the coefficient of $F_*(x_1\cdots x_{n-1})^{p-1}x_n^i$.
		\item $c_0f_0 + c_nx_nf_{p-1}$, the coefficient of $F_*(x_1\cdots x_{n-1})^{p-1}$.
		\item $c_jx_jf_i$ for $i=0,\ldots,p-1$ and $j=1,\ldots,n-1$, the coefficient of $F_*(x_1\cdots \hat{x}_j\cdots x_{n-1})^{p-1}x_n^i$.
	\end{itemize}

\

	Restrict to considering only those $l$ for which all of the $c_j$ are non-zero, which will stand as the open condition in the theorem.  To prove the claim, note that it is enough to show that $f_{p-1} \notin \tau(S,(f\cdot l)^{\frac{1}{p}})$, which by considering the generators of the first type above is equivalent to showing every $f_i\notin \tau(S,(f\cdot l)^{\frac{1}{p}})$. Suppose (aiming for a contradiction) $f_{p-1}$ has a presentation as an element of $\tau(S,(f\cdot l)^{\frac{1}{p}})$:
	$$f_{p-1} = g_0\cdot(c_0f_0 + c_nx_nf_{p-1}) + \sum_{i\leq p-1} g_i\cdot(c_0f_{i} + c_nf_{i-1}) + \sum_{i,j} h_{i,j} \cdot x_jf_i.$$

	\noindent where $g_i,h_{i,j}\in S$ are some coefficients, and the indices $i,j$ come from the list of generators above. Without loss of generality, we may assume that $g_i\in K[x_n]$ by taking any monomial terms of the original presentation of $g_i$ in $\langle x_1,\ldots,x_{n-1}\rangle_S$, and instead include them in some combination of the $h_{i,j}$. Explicitly, if each $g_i = g_{i,0} + x_1g_{i,1} + \ldots + x_{n-1}g_{i,n-1}$ with $g_{i,0}\in K[x_n]$, then one can replace $g_i$ with $g_{i,0}$, replace $h_{p-1,j}$ with $h_{p-1,j} + c_0g_{p-1,j}+c_nx_ng_{0,j}$, and replace each $h_{i,j}$ with $h_{i,j}+c_0g_{i,j}+c_ng_{i+1,j}$ for each $i<p-1$.  Rearranging the original representation of $f_{p-1}$ yields
	\begin{equation}(1-c_nx_ng_0-c_0g_{p-1})f_{p-1} = \sum_{i\leq p-2} (c_0g_i+c_ng_{i+1})f_i + \sum_{i,j} h_{i,j} \cdot x_jf_i.\tag{$\dagger$}\end{equation}

	Since each $f_i$ are polynomials only in the variable $x_1,\ldots,x_{n-1}$, the condition $(\star)$ implies that $x_n^af_i \notin x_n^a\cdot\langle f_0,\ldots,f_{i-1}\rangle_S+ x_n^a\cdot\mathfrak{p}\cdot \tau(S,f^{\frac{1}{p}})$ for any $a>0$.  This allows one to consider $(\dagger)$ and the equations that follow filtered by their $x_n$-degree.  Indeed, suppose \linebreak $1-c_nx_ng_0-c_0g_{p-1} \neq 0$ and consider the smallest $x_n$ degree of $1-c_nx_ng_0-c_0g_{p-1}$, and write it as $Cx_n^a$ with $C\in K^\times$.  Then considering the whole of equation $(\dagger)$ at $x_n$ degree $a$,
	$$Cf_{p-1} = C_0f_0 + \ldots + C_{p-2}f_{p-2} + \sum_{i,j} h_{i,j}'x_jf_i.$$

	\noindent with $C_i\in K$, and $h_{i,j}'\in K[x_1,\ldots,x_{n-1}]$ is the $x_n$-degree $a$ component of $h_{i,j}$.  However, ($\star$) implies directly that this is not possible, providing a contradiction to the fact that $C\neq 0$, which is to say that there is no lowest $x_n$-degree component of $1-c_nx_ng_0-c_0g_{p-1}$.  Thus, $g_{p-1} = c_0^{-1}(1-c_nx_n g_0)$.

	\begin{claim}\label{thm:claim}
	The condition $(\star)$ implies that $c_ng_{i} = -c_0g_{i-1}$ for each $i=1,\ldots,p-1$.
	\end{claim}

	\begin{proof}
	The claim is proved by descending induction on $i$, starting with the case $i=p-1$.  Rearranging the presentation above again using the fact that $1-c_nx_ng_0-c_0g_{p-1}=0$ yields
	$$-(c_ng_{p-1}+c_0g_{p-2})f_{p-2} = \sum_{i\leq p-3} (c_0g_i+c_ng_{i+1})f_i + \sum_{i,j} h_{i,j} \cdot x_jf_i$$

\noindent The same technique used on ($\dagger$) above, by considering the lowest $x_n$-degree piece, implies that $c_ng_{p-1}+c_0g_{p-2}=0$.	Now, assume $k\geq 1$ and the claim is true for $k+1,\ldots,p-1$.  Then

	$$-(c_ng_{k+1}+c_0g_{k})f_{k} = \sum_{i\leq k-1} (c_0g_i+c_ng_{i+1})f_i + \sum_{i,j} h_{i,j} \cdot x_jf_i.$$

\noindent The same argument used in ($\dagger$) again shows that $c_ng_{k+1}+c_0g_k = 0$, or equivalently $g_{k+1} = -c_n^{-1}c_0g_{k}$.\end{proof}

	Combining all of the data provided by \autoref{thm:claim}, one sees that
	\begin{equation}c_0^{-1}(1-c_nx_n g_0) = g_{p-1} = (-c_0c_n^{-1})g_{p-2} = (-c_0c_n^{-1})^2g_{p-3} = \ldots = (-c_0c_n^{-1})^{p-1}g_{0}.\tag{$\dagger\dagger$}\end{equation}

\noindent On one hand, this implies in particular that no $g_i$ is $0$ as that would imply that all $g_i$ were $0$ and the left most side of ($\dagger\dagger$) would read $c_0^{-1} = 0$.  On the other hand, ($\dagger\dagger$) is impossible since the $x_n$-degree of the left hand side of ($\dagger\dagger$) is exactly one larger than that of the right hand side.  Thus no such presentation can exist, and
	$$f_{p-1} \in \tau(S,f^{\frac{1}{p}}) \setminus \tau(S,(l\cdot f)^{\frac{1}{p}}).$$\end{proof}

\begin{remark}
This proof can easily be extended to the case where $K$ is any F-finite infinite field and $F_*^ef\in F_*^eS$, and the conclusion being that for a general choice of hyperplane $H=V(l)$, $\tau(\mathbb{A}^n,f^{\frac{1}{p^e}})\neq  \tau(\mathbb{A}^n,(l\cdot f)^{\frac{1}{p^e}})$, with nearly identical  assumptions and proof.  Namely, choose
$$f = \sum_{i,j} f_{i,j}\cdot F_*^e(u_j x_1\cdots x_{n-1})^{p^e-1}x_n^i$$

\noindent meeting the assumption
$$f_{i,j} \notin \langle f_{0,k},\ldots,f_{i-1,k},f_{i,1},\ldots,\hat{f}_{i,j},\ldots,f_{i,m}\rangle_S+ \mathfrak{p}\cdot \tau(S,f^{\frac{1}{p^e}})$$

\noindent here, $\mathfrak{p} = \langle x_1,\ldots,x_{n-1}\rangle_S$, $i=0,\ldots,p^e-1$, and $j,k=1,\ldots,m$ represent the basis of $F_*K$ over $K$.  Thus in dimension greater than $2$, this provides examples of pairs with the property that for a given $t>0$, $\tau(S,f^{\frac{1}{p^e}}) \neq \tau(S,l^t\cdot f^{\frac{1}{p^e}})$, simply by taking $e\gg 0$ such that $\frac{1}{p^e} \leq t$.
\end{remark}

This gives a somewhat algorithmic way to produce counterexamples in dimension greater than 2.  Indeed, once the dimension is 3 or larger one can always find $f_i$ meeting the condition $(\star)$ in the theorem.

Additionally, one can homogenize the equation for $f$ with respect to an additional variable $x_0$ and produce a more geometric counterexample:

\begin{corollary}\label{cor:proj}
Given any $\epsilon>0$ and $n>2$, there exist pairs $(\mathbb{P}_K^n,\Delta)$ such that for an open set $U\in (\mathbb{P}_K^n)^\vee$ and all $H\in U$,
	$$\tau(\mathbb{P}_K^n,\Delta) \neq \tau(\mathbb{P}_K^n,\Delta + \epsilon H)$$
\end{corollary}

Utilizing \autoref{thm:Cex}, one can easily produce explicit counterexamples.

\begin{corollary}[Dimension 4]\label{thm:Ex1}
Let $K=K^p$ be an infinite perfect field of characteristic $p>0$, and let $S = K[x,y,z,w]$. Then there exists $f\in S$ for which a general hyperplane $H=V(l)$ has the property
$$\tau(\mathbb{A}^4,f^\frac{1}{p}) \neq \tau(\mathbb{A}^4,(l\cdot f)^\frac{1}{p}) $$
\end{corollary}
\begin{proof}
	Let $F_*f = F_*(xyz)^{p-1}\left[xF_*w^{p-1} + y^{p-1}zF_*w^{p-2}+y^{p-2}z^2F_*w^{p-3} + \ldots + yz^{p-1}\right]$.

	It immediately meets the conditions of the theorem:
	$$x \notin \langle y^{p-1}z,y^{p-2}z^2,\ldots,yz^{p-1},x^2,xy,xz\rangle_S$$
	$$y^{p-1}z \notin \langle y^{p-2}z^2,\ldots,yz^{p-1}\rangle + \langle y^pz,x^2,xy,xz\rangle_S$$
	$$\vdots$$
	$$yz^{p-1} \notin \langle x,y,z\rangle\cdot \langle x,y^{p-1}z,y^{p-2}z^2,\ldots,yz^{p-1}\rangle_S.$$

	\noindent Thus for any $l=c_0+c_xx+c_yy+c_zz+c_ww$ with no $c_\cdot = 0$, one concludes directly that $\tau(\mathbb{A}^4,f^{\frac{1}{p}})\neq \tau(\mathbb{A}^4,(f\cdot l)^{\frac{1}{p}})$. Just for comparison's sake, in this case we are comparing $\langle x, y^{p-1}z,y^{p-2}z^2,\ldots,yz^{p-1}\rangle_S$ with
	$$\langle c_0x+c_wy^{p-1}z, c_0y^{p-1}z + c_wy^{p-2}z^2,\ldots, c_0y^2z^{p-2} + c_wyz^{p-1},c_0yz^{p-1}+c_wwx, $$
	$$ x^2,xy^{p-1}z, xy^{p-2}z^2,\ldots,xyz^{p-1},y^pz, y^{p-1}z^2,\ldots,y^2z^{p-1},yz^p \rangle_S$$

	From the monomials it is clear that these generators can not yield a $xw^j$ nor $y^iz^{p-i}w^j$ for any $j>0$ or $i=1,\ldots,p-1$.  Thus, the hyperplanes that satisfy restriction are a subset of the closed subset of hyperplanes with at least one coefficient $0$.  To be more precise, $c_w$ must be $0$ and $c_0\neq 0$ for the equations to work out in such a way that the two ideals agree.\end{proof}

\begin{corollary}[Dimension $n\geq 3$]\label{thm:Ex2}
	Consider $S= K[x_1,\ldots,x_n]$ with $K$ infinite perfect of characteristic $p>0$, and $n\geq 3$.  Let $H_0,\ldots,H_{p-1}\subseteq\mathbb{A}^{n-1}$ be general hyperplanes through the origin, viewed as $V(x_n)\subset \mathbb{A}^n$ with $H_i=V(l_i)$ (thus $l_i = c_{i,1}x_1+ \ldots c_{i,n-1}x_{n-1}$ for some $c_{i,j}\in K$). Consider $f_i$ the product of all but the $i^{\text{th}}$ of these hyperplanes:
	$$f_i = \prod_{j=0,\ldots,\hat{i},\ldots, p-1} l_j$$

Then these $f_i$ satisfy the conditions of the theorem.  Thus $F_*f = F_*(x_1\cdots x_{n-1})^{p-1}\sum f_iF_*x_n^i$ yields a $n$-dimensional counterexample in any positive characteristic to Bertini for test ideals.
\end{corollary}
\begin{proof}
As each $f_i$ is homogeneous of degree $p-1$, the condition $(\star)$ is equivalent to
	$$f_i\notin \langle f_0,\ldots, f_{i-1}\rangle$$

\noindent This is arranged by construction, since
$$V(\langle f_0,\ldots, f_{i-1}\rangle) = \bigcap_{j=0,\ldots,i-1}V(f_j) = \bigcap_{j=0,\ldots,i-1}\left(\bigcup_{k\neq j} H_k\right) = H_i\cup\ldots\cup H_{p-1}$$

\noindent In particular, it contains $H_i$ which $V(f_i)$ does not, so $V(f_i)\not\supseteq V(\langle f_0,\ldots, f_{i-1}\rangle)$.  Therefore, ($\star$) is met (all ideals involved are radical), and one can conclude that for a general hyperplane $H=V(l)$,
	$$\tau(\mathbb{A}^n,f^\frac{1}{p}) \neq \tau(\mathbb{A}^n,(l\cdot f)^\frac{1}{p}).$$\end{proof}

In addition, I show that the direct analog of Bertini's Second Theorem for test ideals fails in general.  Namely, there exists $(X,\Delta)$ with the property that for a general hyperplane $H$, one has
$$\tau(X,\Delta)\cdot\mathscr{O}_H \neq \tau(H,\Delta|_H).$$

\begin{proposition}\label{prop:FailSB1}
	(Dimension 4) Consider \autoref{thm:Ex1}.  Let $D$ be the divisor associated to $f$.  Then for a general choice of $H=V(l)$,
	$$\tau(\mathbb{P}^4,\frac{1}{p}D)|_H \neq \tau(H,\frac{1}{p}D|_H)$$
\end{proposition}
\begin{proof}
	Let $l=c_0+c_xx+c_yy+c_zz+c_ww$ with each $c_i\neq 0$.	By \autoref{prop:Comp}, It is enough to show that $x\notin\tau(H,\bar{f}^{\frac{1}{p}})$.  Begin by eliminating $w$ from all equations by virtue of the relation $w = -c_w^{-1}(c_0 + c_xx + c_yy + c_zz)$.  Then
	$$F_*\bar{f} = \left[(c_w^{-1}c_0)^{\frac{p-1}{p}}x + (c_w^{-1}c_0)^{\frac{p-2}{p}}y^{p-1}z + \ldots + yz^{p-1} \right]F_*(xyz)^{p-1} + \sum_{\boldsymbol{\alpha}<\boldsymbol{p-1}}s_{\boldsymbol{\alpha}}F_*x^{\boldsymbol{\alpha}}$$
Now, utilizing the fact that the test ideal can be computed as in \autoref{prop:Comp}, but with respect to any basis of $F_* (S/\langle l) \rangle$ over $S/\langle l \rangle$, such as the $(xyz)^{\boldsymbol{\alpha}}$ with $\boldsymbol{0}\leq \boldsymbol{\alpha} < \boldsymbol{p}$ basis, it is easily seen that
	$$\tau(H,\bar{f}^{\frac{1}{p}}) = \langle (c_w^{-1}c_0)^{\frac{p-1}{p}}x + (c_w^{-1}c_0)^{\frac{p-2}{p}}y^{p-1}z + \ldots + yz^{p-1},s_{\boldsymbol{\alpha}} \rangle$$
	It is also easy to see that each $s_{\boldsymbol{\alpha}} \in \langle x,y,z\rangle \cdot \left(\tau(S,f^{\frac{1}{p}}) + \langle l \rangle\right)$, since the replacement of $w$ with $-c_w^{-1}(c_0 + c_xx + c_yy + c_zz)$ yields either a constant coefficient (accounted for in the explicitly written down generator) or some higher multiple of $x$, $y$, or $z$, and further every term of the original $f$ had $F_*(xyz)^{p-1}$ in it already.  Thus, it is clear that $0\neq x\notin \tau(H,\bar{f}^{\frac{1}{p}})$, whereas $x\in \tau(\P^4,f^{\frac{1}{p}})$.\end{proof}

Similar considerations apply to \autoref{thm:Ex2}.  However, in the case where the $f_i$ are chosen homogeneous of fixed degree, there is an easy method to detect failure of Bertini:

\begin{theorem}\label{thm:BertiniType}
	Let $S = K[x_1,\ldots,x_n]$, for $K$ a perfect infinite field and let $$F_*f = \sum_{i=0}^{p-1}f_iF_* x_1^{p-1}\cdots x_{n-1}^{p-1} x_n^i$$ with each $f_i\in K[x_1,\ldots,x_{n-1}]$, homogeneous of the same degree $d$, and such that the $f_i$ span a $K$-vectorspace of $S_d$ of dimension at least 2.  Then for a general choice of hyperplane $H$,
$$\tau(S,f^\frac{1}{p})\cdot S/\langle l \rangle \neq \tau(S/\langle l \rangle,\bar{f}^\frac{1}{p})$$
\end{theorem}

\begin{proof}
	Let
$$l=c_0^p+c_1^px_1+\ldots+c_n^px_n.$$
$\tau(S,f^{\frac{1}{p}}) = \langle f_i \ | \ i=0,1,\ldots,p-1\rangle$, as usual.  Assume that $c_n\neq 0$ in the presentation of $l$.  Then $S/\langle l \rangle \cong K[x_1,\ldots,x_{n-1}]$ via
$$x_i\mapsto x_i \ \ \ \ i=1,\ldots,n-1$$
$$x_n \mapsto -c_n^{-p}(c_0^p+c_1^px_1\ldots+c_{n-1}^px_{n-1}).$$
Note that $\bar{f_i}$ are also homogeneous of degree $d$ using the standard grading of $K[x_1,\ldots,x_{n-1}]$. Moreover, one can compute $\tau(S/\langle l \rangle,\bar{f}^\frac{1}{p})$ with respect to any basis of $F_*R$, such as $F_*x_1^{\alpha_1}\cdots x_{n-1}^{\alpha_{n-1}}$, where $0\leq \alpha_j\leq p-1$. So \begin{align*}
F_*\overline{f} = & \sum_{i=0}^{p-1} f_iF_* x_1^{p-1}\cdots x_{n-1}^{p-1} x_n^i\\
= & \sum_{i=0}^{p-1} f_iF_* x_1^{p-1}\cdots x_{n-1}^{p-1}\cdot(-c_n)^{-ip}(c_0^p+c_1^px_1+\ldots+c_{n-1}^px_{n-1})^i.\end{align*}
Therefore, $\bar{f}$ expressed in this basis has non-zero coefficients associated to $F_*\boldsymbol{x^{p-1+\theta}}$, where exponents are taken $\mod p$ and $0\leq |\boldsymbol{\theta}|\leq p-1$.  Applying trace of $S/\langle l\rangle$, the generators come in two types:
\begin{itemize}
	\item $\sum_{i=0}^{p-1} (-c_0c_n^{-1})^i\cdot f_i$, the coefficient of $F_*\boldsymbol{x^{p-1}} = F_*x_1^{p-1}\cdots x_{n-1}^{p-1}$.
	\item $\boldsymbol{x^{\lceil \frac{\theta}{p}\rceil}}\sum_{i=|\boldsymbol{\theta}|}^{p-1} d_{\boldsymbol{\theta},i}\cdot f_i$ the coefficient of $F_*\boldsymbol{x^{p-1+\theta}}$ ($\mod p$ exponents)
\end{itemize}
where $d_{\boldsymbol{\theta},i}=c_0^{i-|\boldsymbol{\theta}|}(-c_n)^{-i}\boldsymbol{c^\theta}\in K$.  It is clear that $\tau(S/\langle l \rangle,\bar{f}^{\frac{1}{p}})$ is homogeneous and
$$\tau(S/\langle l \rangle,\bar{f}^{\frac{1}{p}}) \subseteq \langle \sum_i (c_0c_n^{-1})^if_i\rangle + \mathfrak{p}\cdot \tau(S, f^{\frac{1}{p}})\cdot S/\langle l \rangle$$
where $\mathfrak{p} = \langle x_1,\ldots,x_{n-1} \rangle$.  Intersecting $\tau(S/\langle l \rangle, \bar{f}^\frac{1}{p})$ with $(S/\langle l\rangle)_d$ yields a 1-dimensional vector space $\langle \sum_i (c_0c_n^{-1})^if_i  \rangle_K$.  Therefore, since both ideals are homogeneous, one can compare them degree-wise, and note that
$$\left(\tau(S,f^\frac{1}{p})\cdot S/\langle l \rangle\right)_d \neq \tau(S/\langle l \rangle,\bar{f}^{\frac{1}{p}})_d.$$
and therefore, the two ideals are distinct.  This completes the proof.
\end{proof}

\begin{corollary}\label{prop:FailSB2}
(Dimension $n\geq 3$) Consider the situation of \autoref{thm:Ex2}.  If $D$ is the divisor associated to $f$, then for a general choice of $H=V(l)$,
$$\tau(\mathbb{P}^n,\frac{1}{p}D)|_H \neq \tau(H,\frac{1}{p}D|_H)$$
\end{corollary}

\begin{proof}
Choosing each of the $l_i$ as in \autoref{thm:Ex2} implies directly that the $f_i$ are homogeneous of degree $p-1$.  Therefore, since the $l_i$ are chosen generally, any two of the $f_i$ are linearly independent, and thus the condition of \autoref{thm:BertiniType} is satisfied.
\end{proof}

This answers \cite[Question 8.3]{SZ} in the negative in any dimension larger than 2.  Moreover, a similar argument provides a counterexample to a natural extension of a question of Hochster and Huneke.

\begin{theorem}
\cite[Theorem 7.3(c)]{HH} Let $\varphi:R\rightarrow S$ be a flat homomorphism of Noetherian rings in characteristic $p>0$.  Suppose that $R$ is F-regular, S is excellent, and that $\varphi$ has regular fibers.  Then $S$ is also F-regular.
\end{theorem}

This brings about the following question:

\begin{Question}If $\pi: X\rightarrow Y$ is a flat morphism of characteristic $p>0$ \ Noetherian schemes with regular fibers, with $\Delta$ a $\Q$-Cartier divisor on $Y$, is it true that $\tau(Y,\Delta_Y)\cdot\mathscr{O}_X = \tau(X,\pi^*\Delta_Y)$?
\end{Question}

  The result of Hochster and Huneke holds for Strongly F-regular pairs, where both test ideals are their respective sheaves of rings.  However, \autoref{prop:FailSB1} (or similarly \autoref{prop:FailSB2}) can be used to show that this question is false more generally.

\begin{theorem}\label{thm:HHfalse}
Let $K$ be a perfect field, and let $Z$ be the reduced closed subscheme of $\P^n\times_K(\P^n)^\vee$ defined to be the closure of the set $\{(P,H) \ | \ P\in H\}$.  Then $Z\rightarrow \P^n$ is a flat morphism, with regular fibers, such that there exists $\Delta$ a $\Q$-divisor on $\P^n$ with the property that
$$\tau(\P^n,\Delta) \cdot \O_Z \neq \tau(Z,\pi^*\Delta).$$
\end{theorem}

\begin{proof}
The statement that  $Z\rightarrow \mathbb{P}^n$ is a flat morphism is proved in the course of \cite[Theorem 3.7]{SZ}.  Let $\Delta = \frac{1}{p}\text{div}(f)$, where $f$ is chosen to satisfy the assumptions of \autoref{thm:BertiniType}.  Note that to show that the two ideals of $\mathcal{O}_Z$ differ, it suffices to show that their localizations differ.  Let $P\in \P^n$, and consider the maximal ideal corresponding to a point $\mathfrak{m} = (P,H)\in Z$.  Then the localization map has the form
$$K[x_1,\ldots,x_n]_{\langle \underline{x}\rangle} = \mathcal{O}_{\P^n,P} \rightarrow \mathcal{O}_{Z,(P,H)} \cong K[x_1,\ldots,x_n,y_1,\ldots,y_n]/\langle y_1x_1+\ldots+y_nx_n\rangle_{\langle \underline{x},\underline{y}\rangle}.$$
Here the underline is representing $\underline{x} = x_1,\ldots,x_n$, and similarly for $\underline{y}$.  Note this is the case since every hyperplane passing through the origin necessarily has this form. Since test ideals localize, we can perform a further localization at the prime ideal ${\langle \underline{x}\rangle \subseteq \mathcal{O}_{Z,(P,H)}}$.  It suffices to prove that
$$\tau(\P^n,\Delta)\cdot \mathcal{O}_{Z,\langle \underline{x}\rangle}=\tau(\P^n,\Delta)_{\langle \underline{x}\rangle}\cdot \mathcal{O}_{Z,\langle \underline{x}\rangle} \neq \tau(Z,\pi^*\Delta)_{\langle \underline{x}\rangle}$$

The left hand side is simply $\tau(S,f^{\frac{1}{p}})_{\langle \underline{x}\rangle} = \langle f_i \ | \ i=0,1,\ldots,p-1\rangle\cdot S_{\langle \underline{x}\rangle}$.  The right hand side is $$\tau\left(k(y_1,\ldots,y_n)[x_1,\ldots,x_n]/\langle y_1x_1+\ldots+y_nx_n\rangle_{\langle \underline{x}\rangle},f^{\frac{1}{p}}\right).$$
Call this ring $R$ for simplicity.  This ideal can be computed by applying the trace operator $\Phi_R$ to the ideal $\langle f\rangle_R$.  Similar to the proof of \autoref{thm:BertiniType}, we can compute the test ideal with respect to any basis of $F_*R$ over $R$.  Noting that
$$x_n = y_n^{-1}\left(y_1x_1+\ldots+y_{n-1}x_{n-1} \right)$$
the basis of interest is
$$\left\lbrace F_*\left(y_1^{\beta_1}\cdots y_n^{\beta_n}\cdot x_1^{\alpha_1}\cdots x_{n-1}^{\alpha_{n-1}}\right) \ : \ 0\leq \alpha_i,\beta_i < p\right\rbrace$$
Therefore, ones concludes $\tau(Z,\pi^*\Delta)_{\langle \underline{x}\rangle}$ is homogeneous in $x_1,\ldots,x_{n-1}$ and has at most one generator of $x$-degree $d$, whereas $\tau(\P^n,\Delta)$ has at least $2$ by the assumptions imposed on $f$.  This completes the proof.
\end{proof}
As a final consideration, we ask about the Bertini statement in dimension 2.  It is immediately clear that \autoref{thm:Cex} cannot be directly extended to dimension 2, as finding a collection of $f_i$ meeting $(\star)$ is clearly impossible using only a single variable.  In fact, some simple geometry allows us to conclude that it is in fact true.

\begin{theorem}\label{thm:dim2}
Let $X\subseteq \mathbb{P}^n$ be a normal projective surface over an infinite perfect field $K$ of characteristic $p>0$, and let $\Delta\geq 0$ be an effective $\Q$-divisor, with $K_X+\Delta$ $\Q$-Cartier.  Then for a general hyperplane $H\in (\P_K^n)^\vee$,
$$ \tau(X,\Delta)\cdot \mathscr{O}_H = \tau(X\cap H,\Delta|_H)$$

Furthermore, for every $0\leq \epsilon < 1$,
$$ \tau(X,\Delta) = \tau(X,\Delta + \epsilon H)$$
\end{theorem}

\begin{proof}
	Let $\Sigma$ be the locus containing singularities of $X$ and singularities of $\Delta$.  Then by normality, $\Sigma$ is a finite set of closed points.  Take $H$ any hyperplane not intersecting $\Sigma$, with $X\cap H$ regular.  Then by completing the local ring at a point of $x\in \Delta\cap H$, by regularity Cohen structure theorem implies $\mathscr{O}_{X,x} = K\llbracket x,y\rrbracket$, and by a linear change of coordinates, we can assume $H=V(y)$ and $\Delta = t\cdot V(x)$ for some positive rational number $t$.  In this complete local setting, $\mathscr{O}_H = K\llbracket x \rrbracket$, and $\Delta|_H$ is simply $t\cdot V(x)$.  Finally, the following computations follow directly from \autoref{Sect:Comp}:
	$$\tau(K\llbracket x,y\rrbracket,x^t) \otimes_{K\llbracket x,y\rrbracket} K\llbracket x\rrbracket = \langle x^{\lfloor t \rfloor}\rangle = \tau(K\llbracket x\rrbracket,x^t)$$

	Finally, since every point of $X\cap H$ and $\Delta \cap H$ is smooth in $X,\Delta$ respectively, the test ideal sheaves agree as well.

	The second statement follows by the same logic, as $\tau(X,\Delta + \epsilon H) = \langle x^{\lfloor t \rfloor} \rangle$ in $K\llbracket x,y\rrbracket$.\end{proof}

\section{Further Questions}\label{Sect:Questions}

The previous section implies that an analog of Bertini's Theorem for test ideals is not possible for pairs $(X,\Delta)$ in general.  However, there are also classes for which such a theorem is possible.  For example, if $f$ is of the form $f_\alpha F_*x^\alpha$, or if the $F_*x^\alpha$ are in sufficiently low degrees, or even in many cases where the $f_i$ are not quite as independent as condition $(\star)$ enforces.  A more general class comes from \autoref{thm:SZ}, which states that for Sharply F-pure or Strongly F-regular pairs $(X,\Delta)$, an analog of the second theorem of Bertini in fact holds.  So a general question one could pose would be which pairs $(X,\Delta)$ have such a property?  Or more specifically, is there some geometric or arithmetic property that is governing whether or not the theorem holds?

  One particular thing to note is that \autoref{thm:Ex2} has $\Delta$ of a fairly large degree; $(p+n)\cdot(p-1)$ where $n$ is the dimension.  It is possible that the theorem holds up to a particular degree for a given dimension of variety.  One could also ask if there is some type of locus determined by the test ideal for which a hyperplane not intersecting the said locus is sufficient to guarantee that the restriction statement holds, and what properties this locus has.  In particular, if it was a finite collection of points, a general hyperplane would always miss this loci and the Bertini theorem would hold.

Two geometric properties of interest in positive characteristic Algebraic Geometry and Commutative Algebra, which are related to the test ideal, are F-injectivity and F-rationality.  The direct analog of the second theorem of Bertini for F-injectivity is known to be false \cite[Proposition 7.4]{SZ} (still open for normal F-injective, as the counter-example used the Weakly normal construction of \cite{CGM}), however it is currently unknown whether the same is true for F-rationality.

Finally, a very important question that remains open is whether or not one can use this type of technique to find counterexamples to the original Hochster-Huneke question.  Namely, could it still be true when considering a variety instead of pairs?

\begin{Question}
	For an irreducible projective variety $X$, and a very ample linear series $|D|$, is it true that for a general choice of $\Delta\in |D|$, one has that $\tau(X) = \tau(\Delta\cap X)$?
\end{Question}

The answer is very likely no based on the findings of this paper, but an explicit counterexample is still unknown to the author.  In particular, the class of $f$ provided from \autoref{thm:Cex}, $X=V(f)$, for which the trace map has a particularly nice form $\Phi_X(g) = \Phi(g F_*f^{p-1})$, cannot be used because none of these $X$ are irreducible varieties.

\bibliographystyle{skalpha}

redpajama_set_name: RedPajamaArXiv


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The Swedish detective Beck returns to UK screens for season 9
It's been 25 years since Beck first aired on Swedish television, making this one of the most enduring crime shows in European television and Scandinavia's longest-running crime drama export. The ninth season is four episodes in extent, and began airing on BBC Four in the 9pm subtitled crime slot on Saturday 7 May. It's also available on iPlayer.
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However, in the rather elastic timeline of the show, he's had three months off and is now raring to go, brushing aside the paperwork and heading out to solve crimes. Still, Beck's tentative health has become an ongoing issue in the series.
If you've watched Beck before you'll know that comings and goings are fairly frequent in the murder squad. Beck's number two, Alex Beijer (Jennie Silfverhjelm) arrived a few seasons ago and competition with her is one of the reasons Steinar Hovland (Kristofer Hivju) returned to Norway. That, and the fiery Norseman came into conflict with former undercover cop Josef Eriksson (Martin Wallstöm). In many ways, Steinar and Josef are similar – risk-takers, perhaps filling the same role as Beck's original number two, Gunvald Larsson (Mikael Persbrandt).
It's reassuring to have Jenny Bóden (Anna Asp) and Oskar Bergman (Måns Nathanaelson) both as investigators and for their side stories. Oscar is still trying to master Tinder…
Left to right: Josef, Oskar, Alex, Jenny and Steinar.
But finding dates isn't the focus. The first episode in the series, Haunted by the Past, begins when a man linked to a powerful Danish drugs syndicate is found floating in the canal with a shotgun wound to the neck. The chief suspects are members of a gang of dealers that the man was selling to, or perhaps his outfit took him out? The clues aren't adding up and when Beck's team ask a Stockholm kingpin who is in prison to explain what's going on, he says he'll speak but only to Steinar. This cues the Norwegian detective's return. However, the case has several layers to it as the dead man had a volatile male lover and a Danish syndicate member is in Sweden looking for revenge.
Unlike most crime dramas these days, Beck continues to cover an entire case in a single 90-minute episode. You feel like you get your money's worth each time, so to speak. In Sweden the series was broadcast once a month between December 2021 and March 2022 – we get to watch them all this May.
The second episode of season nine has the interesting title Rage Room. A man has been found dead in the woods, and another man has died of a heart attack after being woken up in the night by an intruder… apparently. Meanwhile, within the team we have three high-voltage characters – Alex, Steinar and Josef – offsetting Beck's gentler approach. Now we'll find out just how well they settle down together.
The introduction of Alex Beijer to the series has arguably brought with it more depth than either Steinar or Josef. Formerly, she has worked covertly overseas in Iraq and her experiences there drove the harrowing Death in Samarra story during season eight – one of the most powerful episodes of the last few years. In episode three of the new series she is reminded of the horror of past events. When she's about to take part in a daytime TV show, a hostage situation arises in the studio. How can she use her experience and guide Beck's team to resolving it?
The final episode of the season, The Crying Cop, to some extent revives the social commentary aspect of Beck, which was so prominent in earlier seasons. The police are under scrutiny after the shooting of a 14-year-old, and protests are taking place. Then, there's a shocking massacre at the police academy. This enables us to see more of Beck's grandson, Vilhelm, who attends the academy and who may have information valuable to the investigation.
Could this be the genesis of a new character for the series. Another Beck to chart a fresh course for the series when Martin Beck eventually retires? It's an interesting proposition.
Beck and many of the themes of the series are based loosely on the original Martin Beck novels by Maj Sjöwall and Per Wahlöö written in the 1970s and 80s. You can find out more about the TV series with our guide here, which currently runs up to season eight.
Good old Grannen (Ingvar Hirdwall). Get ready for a story…
Beck crime show Martin Beck Scandinavian crime fiction Stockholm Sweden
Max bosworth
19 May, 2022 at 11:12 am
I like watching beck
Jenny Mayfield
28 July, 2022 at 4:03 am
It gave an insight in the life of people working in that industry. Non aggressive talking unlike the Irish or British police.
Tammara
14 December, 2022 at 2:03 am
In the USA, your synopsis of season 9 is what we watched as season 8. Will we be getting another season here?
The Cook by Ajay Chowdhury
The Last to Disappear by Jo Spain
Biker gangs, drugs, guns, cops, infiltrators, murder – get ready for organised crime, Scandinavian style, as the Danish crime show Grow arrives on Channel 4 and on Walter Presents. The first episode will air at 11:30pm on Sunday, 22 January, and you'll be able to…
Happy Valley returns to the BBC for season three
Happy Valley might not be the best title for this hit crime drama. It's not about a Chinese takeaway and nor does the Calder Valley, where it's set, seem a particularly happy place. Not in a show dealing with violence against women, rape, suicide, addiction,…
Best crime show of 2022 shortlist
As crime fiction lovers, we all know there's nothing better than settling down with a new book. These days, however, there is a wealth of well scripted crime shows to attract our attention too, many of them based on books that we know and love….


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Lex Promotions Awarded the 2014 "Label Rep of the Year" at Duval Diamond Awards
Lex Promotions and Marketing Group Receives S.I.N. Nomination for 2013
LPMG Takes Over Ultra Festival & Winter Music Conference 2014
MIAMI, FL—March 24, 2014—Many of Florida and Georgia's urban indie artists, tastemakers, and DJs gathered in Jacksonville, Florida last night for the 2014 Duval Diamond Awards where Lex Promotions and Marketing Group (LPMG) was honored with the 2014 "Label Rep of the Year" award.
Company CEO Alex "Lex" Pierre-Louis gave a brief acceptance speech and some insight on his beginnings in the music industry. "I would borrow my Mom's car to go pass out flyers…make $20 and have to put $15 back in my Mom's [gas] tank," Pierre-Louis said candidly. LPMG has since gone on to become the label representative for Def Jam, Atlantic, Interscope, and Maybach Music Group (MMG). Pierre-Louis accepted the award along with LPMG team members Hugo Richardson and Terri Sherman. Richardson heads up the NewMusic4DJs department, as well as project management for LPMG clients. Sherman is the company liaison, who handles new client inquiries and business relations.
LPMG's promotional support of MMG CEO/artist Rick Ross's latest album Mastermind has resulted in another number one album for the beginning of 2014, as LPMG also marks its 10-year anniversary.
The LPMG team will begin taking new client submissions from March 31, 2014-April 11, 2014. For more information on Lex Promotions and Marketing Group, contact 305.974.4568 or email inquiries@lexpromotions.com.


Serenade issue 8 "Serenade" Discounts Apply !
Serenade, continues to piece her life back together. Going back to her singing career. She finds the club called the Flaming Wing. Only to find dark secrets lurking within its walls, and the ghost of Jonathan haunts Serena.


text:
Pseudochazara pelopea är en fjärilsart som beskrevs av Herrich-Schäffer 1852. Pseudochazara pelopea ingår i släktet Pseudochazara och familjen praktfjärilar. Inga underarter finns listade i Catalogue of Life.

Bildgalleri

Källor

Externa länkar

Praktfjärilar
pelopea

redpajama_set_name: RedPajamaWikipedia


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The 8 Most Awkward Coffee Dates Baristas Have Witnessed
PHOTOGRAPHED BY NICOLAS BLOISE.
Once upon a time, a friend told me that no one really takes people out on coffee dates. But I (ever the least fun person at a party) proved her wrong. Because I almost exclusively ask first dates to meet me at a coffee shop. Something about the atmosphere just makes me feel relaxed. Somehow, chatting over a latte or a cup of tea on Saturday afternoon feels way less loaded than sipping on a glass of wine at 9 p.m. on Friday night.
I can't pretend that most of my coffee dates weren't awkward, as first dates tend to be. But, they did and do still happen. I've actually witnessed several other people on awkward coffee dates just by popping into Starbucks for a black tea. So, in the spirit of continuing to prove my friend wrong, I asked baristas to spill the beans (pun intended) about the most awkward coffee dates they've witnessed. Ahead, eight baristas spin some of the saddest and most cringe-worthy tales you'll ever read.
"This isn't awkward! Just sad. I saw a girl sit down and wait for at least two hours for someone to show up. She sat in my section, and ordered coffee after coffee, just waiting. Eventually, I stopped by to see if she needed anything else. And she just started crying. She said, 'I guess he's not coming after all.' I gave her a cookie on the house, but she didn't eat it."
Hanna, 17
"This guy waited three hours for his date to show up, and once she finally did show up she was very rude to him and kept getting mad that he wasn't talking much. She proceeds to leave after about five minutes there. He comes up and says to us, 'Well that was the most awkward first date ever.'"
Molly, 23
"I was working front register when a girl walked in, clearly dressed up for a date, and sat in the cafe immediately. A few moments later, a man I'd gone to high school with who was a complete jerk walked in and sat with her. They'd obviously never met before.
"I could hear him talking about how much of a player he was back then, and how he'd turned his life around to respect women. The girl's face went from horrified to bored while he said, 'Oh I love women so much! Women are how this society gets going. Men wouldn't be anywhere without their wives to have children!' I've never seen someone book it from a date as fast as she did. He didn't even look concerned, just stood up, ordered a black coffee from me, and went on his way."
Keely, 20
"Oh god, so this is what happened: He went up and ordered this giant-ass drink, which came to like $7 and change. She ordered a small latte. When I read the total, she looked at him and he wouldn't look up from his phone. She pulled out her debit card, looked at me, and started laughing. We made their drinks, she took his giant one and left."
Jordan, 23
"One time, a younger regular (guy, mid-20s, athletic and conventionally attractive) came into my café all dressed up, which I complimented him on when he ordered (he usually stopped in for iced tea and a chat after hitting the gym, still dripping sweat), and he thanked me and told me he was going on a Tinder date with the 'girl of his dreams' that he had been chatting with for the past two months.
"After an hour, I checked in on him and he told me his date was running late and asked me for a phone charger. Being in a 'Tinder success story' relationship myself, I happily let him borrow my charger and got back to work, knowing all too well how he was feeling. Two more hours go by, and it's time to close up. It's been three hours since his date was supposed to arrive, and I had developed a sinking feeling he got ghosted since our prior conversation. I think he knew, too, because his entire demeanour had changed from nervous to utter despair in the time since he sat down.
"I finished cleaning out our pastry display and offered him a croissant and a large cup of coffee for free before informing him he had to go since it was time to close up. He looked at me with the biggest, saddest eyes (I could tell he was barely holding back tears at this point) and whispered, 'Oh… okay… thanks,' and walked out. It was the worst I'd ever felt telling someone they had to leave, and we never spoke about that night again. I heard he's in a very happy relationship now, though. So it looks like it all worked out in the end."
Kat, 26
"I once saw a girl and guy make out outside for a solid 20 minutes, then come into the store. They paid for their drinks separately, sat down, and talked for what was like a half hour more. Apparently, the guy was breaking up with the girl, judging from the fact that by the end of it she was crying and they left separately."
Alison, 22
"A former male barista that we considered a cool surfer dude and a blonde girl were sitting in the café, and you could tell from the moment they walked in that date was a nope. The girl looked uncomfortable, as he was staring at her like a lion planning to chase a baby gazelle, barely blinking. And she looked at us as if communicating with her eyes that she couldn't wait to leave.
"The date lasted 30 minutes, and my coworkers and I kept an eye on the girl to ensure she was okay. Then he said to her, 'Are we going over to your place or…' She replied as fast as she possibly could, 'NO. I have to go.' He proceeded to attempt several times to guilt-trip her as if she owed him time in a sexual manner. The more he talked and talked about how much he wanted to have sex with her, the more the girl, my coworkers, and every woman in the cafe felt disgusting and made signals for the girl to realise that it was definitely time to go. As she stood to leave, he decided to follow her out, and both got into their own cars. Neither one of them left the parking lot for 15 minutes, until he decided to accept that he failed and drove away. Minutes later, she went home.
"After that, my shift on duty came in and said, 'Oh I missed my crushtumer, isn't he cute?' And all of us looked at each other and answered, 'Hell no.'"
"A teen came in to meet up with a girl from school. Girl never showed up and was crying in the bathroom for a good 30 minutes. She got a Frappuccino and left. About 20 minutes later, a girl came in asking for the teen who had just left."
Originally published at https://www.refinery29.com.
Refinery29 is the #1 new-media brand for smart, creative…
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Also, Brooklyn maps are live on Mr. Beller's Neighborhood.
If you move your cursor over the big landmass at the bottom of the main map, Brooklyn lights up. If you click it, the word "Brooklyn" appears, like a sign at the entrance of a haunted house, and then you are looking at whole new topography. There are lot of green locator dots on the map, and not many red story dots. Hopefully that will begin to change. If you or anyone you know has been contemplating writing a Brooklyn based story, now is the time. There will also be a reading next month at Barbe's in Park Slope to commemorate this new development. Information to come.


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Singapore to extend major insurance initiatives to 2025
by Gabriel Olano 20 Feb 2020
Singapore to extend major insurance initiatives to 2025 | Insurance Business
During Singapore's announcement of the 2020 budget on February 18, the government announced that the Insurance Business Development (IBD) umbrella scheme and the IBD-Captive Insurance scheme would be extended up to 2025. Both schemes were set to end next month.
The Insurance Business Development scheme gives qualified insurers a concessionary tax rate of 10% for 10 years on qualifying income derived from onshore and offshore life reinsurance business, a report by the Straits Times said. The scheme also includes onshore and offshore general insurance and reinsurance, excluding the following business lines: motor, fire, personal accident, health, and work injury compensation.
Meanwhile, the Singaporean government also extended the Maritime Sector Incentive up to December 2026. The incentive provided several tax benefits to shipping operators, maritime lessors and other providers of shipping-related support services.
The moves are part of Singapore's efforts to elevate its status as a re/insurance and marine industry hub in the region. In previous years, Singapore developed several mechanisms to boost its insurance industry, such as advancement of alternative risk transfer mechanisms including insurance-linked securities (ILS) and government pools.
MAS grant boosts Singapore's bid for insurance superiority
Steering committee for Singapore insurance industry forms
Insurance industry facing strong headwinds – Minister Rajah