text: # Field extensions, proving an element is in a field
  
  1. May 18, 2010
  
  ### Firepanda
  
  I can do all the question apart from the 'Conclude rt(2) in an element of Q(rt(2) + rt(3))'
  
  That's all I need help on.
  
  I tried
  
  rt(2) = a + b(rt(2) + rt(3))
  
  squaring it and equating you get
  
  2 = a2 + 5b2 + (some other terms of rt(2) rt(3) rt(5))
  
  but this cannot be true for a, b in Q, how else are you supposed to do this?
  
  Thanks
  
  2. May 18, 2010
  
  ### Tedjn
  
  Solve for the square root of 2.
  
  3. May 18, 2010
  
  ### Firepanda
  
  Solve what exactly?
  
  4. May 18, 2010
  
  ### Tedjn
  
  You know that $(\sqrt{2}+\sqrt{3})^2 - 2\sqrt{2}(\sqrt{2}+\sqrt{3}) - 1 = 0$. Here, we're fortunate because there is a $\sqrt{2}$ available that we can isolate.
  
  5. May 18, 2010
  
  ### Firepanda
  
  I tried that before but it left me with something pretty messy when all i wanted to show was it could be written as a + b(rt(2) + rt(3))
  
  i ended up with rt(2) = -(2 + rt(5)) / (rt(2) + rt(3))
  
  6. May 18, 2010
  
  ### Tedjn
  
  Remember that this is $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, not $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$. Also, I don't see from where the $\sqrt{5}$ comes.
  
  7. May 18, 2010
  
  ### Firepanda
  
  hmm whats the difference?
  
  guess I was slacking when jotting down notes and didn't realise when they were curvy brackets and not :(
  
  8. May 18, 2010
  
  ### Tedjn
  
  The former is the smallest field containing (root 2 + root 3), which means there are inverses. I've slacked before in math classes, so I know the feeling.
  
  9. May 18, 2010
  
  ### Firepanda
  
  Then that sort of ruins my answer for part i) of
  
  isn't k = {0, 1, T, 1+T} since k = {a + bT | a, b in {0,1}}?
  
  and so g is irreducible since no element of k is a root
  
  but you're saying k isn't {a + bT | a, b in {0,1}}
  
  10. May 18, 2010
  
  ### Tedjn
  
  It turns out that F2(T) contains all quotients of polynomials in F2[T]. Interpret it as the field of fractions of F2[T]. What have you learned that you can apply relating irreducible polynomials over the two?
  
  11. May 18, 2010
  
  ### Firepanda
  
  Probably missed your point but ill take a stab
  
  if it's irreducible in Fp[T], p prime, then it's irreducible in Z[X]
  
  12. May 18, 2010
  
  ### Tedjn
  
  It can be tough to wrap your head around, but the indeterminate variable is X. So from what you proved we know that our polynomial g(X) is irreducible in (F2[T])[X]. It is not even in Z[X]. However, we would like to say something about its irreducibility in (F2(T))[X]. This might remind you of the connection between irreducibility in Z[X] and Q[X].
  
  13. May 18, 2010
  
  ### Firepanda
  
  If it's irreducible in Q[X] then it's irreducible in Z[X] is gauss's lemma, and them im stumped
  
  Is this point crucial to the solution?
  
  14. May 18, 2010
  
  ### Tedjn
  
  I think it's the opposite, that irreducibility in Z[X] implies irreducibility in Q[X]. I've also just finished studying these kinds of things, so I can't be sure that this is crucial, but it seems so to me. We can use Gauss's lemma to conclude what you need passing from the ring to the field of fractions.
  
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explanation: The document is a discussion thread from a math forum where a user named Firepanda is seeking help with a problem related to field extensions, specifically proving that \(\sqrt{2}\) is an element of \(\mathbb{Q}(\sqrt{2} + \sqrt{3})\). The thread includes a series of responses from another user, Tedjn, providing hints and guidance. The main question is how to show that \(\sqrt{2}\) can be expressed in terms of \(\sqrt{2} + \sqrt{3}\) within the field \(\mathbb{Q}(\sqrt{2} + \sqrt{3})\).
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # Field extensions, proving an element is in a field
  
  1. May 18, 2010
  
  ### Firepanda
  
  I can do all the question apart from the 'Conclude rt(2) in an element of Q(rt(2) + rt(3))'
  
  That's all I need help on.
  
  I tried
  
  rt(2) = a + b(rt(2) + rt(3))
  
  squaring it and equating you get
  
  2 = a2 + 5b2 + (some other terms of rt(2) rt(3) rt(5))
  
  but this cannot be true for a, b in Q, how else are you supposed to do this?
  
  Thanks
  
  2. May 18, 2010
  
  ### Tedjn
  
  Solve for the square root of 2.
  
  3. May 18, 2010
  
  ### Firepanda
  
  Solve what exactly?
  
  4. May 18, 2010
  
  ### Tedjn
  
  You know that $(\sqrt{2}+\sqrt{3})^2 - 2\sqrt{2}(\sqrt{2}+\sqrt{3}) - 1 = 0$. Here, we're fortunate because there is a $\sqrt{2}$ available that we can isolate.
  
  5. May 18, 2010
  
  ### Firepanda
  
  I tried that before but it left me with something pretty messy when all i wanted to show was it could be written as a + b(rt(2) + rt(3))
  
  i ended up with rt(2) = -(2 + rt(5)) / (rt(2) + rt(3))
  
  6. May 18, 2010
  
  ### Tedjn
  
  Remember that this is $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, not $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$. Also, I don't see from where the $\sqrt{5}$ comes.
  
  7. May 18, 2010
  
  ### Firepanda
  
  hmm whats the difference?
  
  guess I was slacking when jotting down notes and didn't realise when they were curvy brackets and not :(
  
  8. May 18, 2010
  
  ### Tedjn
  
  The former is the smallest field containing (root 2 + root 3), which means there are inverses. I've slacked before in math classes, so I know the feeling.
  
  9. May 18, 2010
  
  ### Firepanda
  
  Then that sort of ruins my answer for part i) of
  
  isn't k = {0, 1, T, 1+T} since k = {a + bT | a, b in {0,1}}?
  
  and so g is irreducible since no element of k is a root
  
  but you're saying k isn't {a + bT | a, b in {0,1}}
  
  10. May 18, 2010
  
  ### Tedjn
  
  It turns out that F2(T) contains all quotients of polynomials in F2[T]. Interpret it as the field of fractions of F2[T]. What have you learned that you can apply relating irreducible polynomials over the two?
  
  11. May 18, 2010
  
  ### Firepanda
  
  Probably missed your point but ill take a stab
  
  if it's irreducible in Fp[T], p prime, then it's irreducible in Z[X]
  
  12. May 18, 2010
  
  ### Tedjn
  
  It can be tough to wrap your head around, but the indeterminate variable is X. So from what you proved we know that our polynomial g(X) is irreducible in (F2[T])[X]. It is not even in Z[X]. However, we would like to say something about its irreducibility in (F2(T))[X]. This might remind you of the connection between irreducibility in Z[X] and Q[X].
  
  13. May 18, 2010
  
  ### Firepanda
  
  If it's irreducible in Q[X] then it's irreducible in Z[X] is gauss's lemma, and them im stumped
  
  Is this point crucial to the solution?
  
  14. May 18, 2010
  
  ### Tedjn
  
  I think it's the opposite, that irreducibility in Z[X] implies irreducibility in Q[X]. I've also just finished studying these kinds of things, so I can't be sure that this is crucial, but it seems so to me. We can use Gauss's lemma to conclude what you need passing from the ring to the field of fractions.
  
  Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread from a math forum where a user named Firepanda is seeking help with a problem related to field extensions, specifically proving that \(\sqrt{2}\) is an element of \(\mathbb{Q}(\sqrt{2} + \sqrt{3})\). The thread includes a series of responses from another user, Tedjn, providing hints and guidance. The main question is how to show that \(\sqrt{2}\) can be expressed in terms of \(\sqrt{2} + \sqrt{3}\) within the field \(\mathbb{Q}(\sqrt{2} + \sqrt{3})\).
  
  Label: question-and-answer