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Let $\xi$=$\ \sqrt{\vphantom{\sum}2+{\sqrt2}} $

(1) Find the minimal polynomial $r(x)$ $\in$ $\mathbb{Q}$ of $\xi$

(2)Prove that $\ \sqrt{\vphantom{\sum}2-{\sqrt2}} $ is also a root of $r(x)$

(3)Prove that $\ \sqrt{\vphantom{\sum}2-{\sqrt2}} \in \mathbb{Q}(\xi)$

(4)Find a theorem that says ; $\xi \longrightarrow \sqrt{\vphantom{\sum}2-{\sqrt2}} $ that defines an automorphism $\zeta$: $\mathbb{Q}(\xi) \longrightarrow \mathbb{Q}(\xi)$

(5) Using the basis ${1,\xi,\xi^2,\xi^3}$ of $\mathbb{Q}(\xi)$ as a vector space over $\mathbb{Q}$, find the matrix representation of $\zeta$ as a linear transformation.

(6)Show that $Aut_{\mathbb{Q}}(\mathbb{Q}(\xi))$ is cyclic of order 4

SOLUTION

(1) By the defintion og Galois I must get all the different conjugates of the generator After much working I got $(x^2-2)^2-2=0 $ so the minimal polynomial is $x^4-4x^2+2$ and by Eisenstein's criterion it's not hard to see that it's irreducible.

(2) I reversed the process and I got the conjugates and after some working so this was easy.

(6) Using (2) the other conjugates are just additive inverses of the k=$\ \sqrt{\vphantom{\sum}2+{\sqrt2}} $ and so all the conjugates are in k. K is a Galois over $\mathbb{Q}$ and after some working we get that [k:$\mathbb{Q}$]=4 and I already showed (not here) that I get a cyclic group of order 4.

My burning question for (5) i'm a bit confused.Normally for linear algebra related problems I would be given something like a function for $v:\mathbb{R} \longrightarrow \mathbb{R}$ which is my $\zeta $ in this case but in order to find to that I always given a matrix to begin with and then and then I use the basis given to compute.

In this question I am not given any matrix but instead asked to find it which is the reverse process and I really don't even understand how to begin this part of the problem.

Can anyone help me understand what I need to do with part (5)?

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    $\begingroup$ You need to find $\zeta(1), \zeta(\xi), \zeta(\xi^2),$ and $\zeta(\xi^3)$ as linear combinations of $1, \xi, \xi^2,$ and $\xi^3$ $\endgroup$ – J. W. Tanner Mar 31 at 20:22
  • $\begingroup$ There are infinitely many rational numbers, so the finite-fields was not appropriate. $\endgroup$ – Jyrki Lahtonen Apr 2 at 21:29
  • $\begingroup$ Excuse me for being a font Nazi, but see Guidelines for good use of $\rm\LaTeX$ in question titles: "The primary rule for using $\LaTeX$ in titles is to be vertically terse. Try to make your title take up as little vertical space as possible (the height of $\sum\limits_{n=1}^\infty$ is an approximate upper bound for what is admissible)." Your \sqrt{\vphantom{\sum}}, while looking nice (it taught me something), takes up more vertical space than it ought to. $\endgroup$ – Calum Gilhooley Apr 2 at 22:19
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  1. You can compute powers of $\xi$ until you find some $\Bbb{Q}$-linear dependence between them, or more simply, note that $$\xi^2=2+\sqrt{2},$$ and hence that $(\xi^2-2)^2=2$, so the minimal polynomial divides $x^4-4x^2+2$.
  2. You can simply plug $\sqrt{2-\sqrt{2}}$ into the minimal polynomial to see that it is a root. Or even simpler, as above note that $$\sqrt{2-\sqrt{2}}^2=2-\sqrt{2},$$ and hence that $(\sqrt{2-\sqrt{2}}^2-2)^2=2$, so it is also a root of $x^4-4x^2+2$.
  3. Note that $\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}=\sqrt{4-2}=\sqrt{2}$, which means that $$\sqrt{2-\sqrt{2}}=\frac{\sqrt{2}}{\xi},$$ so it suffices to show that $\sqrt{2}\in\Bbb{Q}(\xi)$.
  4. I do not know what theorems you are familiar with or expected to choose from.
  5. From the above it is clear that $(\xi^0,\xi^1,\xi^2,\xi^3)$ is an (ordered) basis of $\Bbb{Q}(\xi)$, so there exist unique coefficients $c_{i,j}\in\Bbb{Q}$ such that \begin{eqnarray*} \zeta(\xi^0)&=&c_{1,1}\xi^0+c_{1,2}\xi^1+c_{1,3}\xi^2+c_{1,4}\xi^3,\\ \zeta(\xi^1)&=&c_{2,1}\xi^0+c_{2,2}\xi^1+c_{2,3}\xi^2+c_{2,4}\xi^3,\\ \zeta(\xi^2)&=&c_{3,1}\xi^0+c_{3,2}\xi^1+c_{3,3}\xi^2+c_{3,4}\xi^3,\\ \zeta(\xi^3)&=&c_{4,1}\xi^0+c_{4,2}\xi^1+c_{4,3}\xi^2+c_{4,4}\xi^3,\\ \end{eqnarray*} and then the matrix of $\zeta$ w.r.t. this ordered basis is $C:=(c_{j,i})_{1\leq i,j\leq 4}$. So this is a matter of determining the $c_{i,j}$. Of course $$\zeta(\xi^0)=\zeta(1)=1=1\cdot\xi^0+0\cdot\xi^1+0\cdot\xi^2+0\cdot\xi^3,$$ which already yields the first column of the matrix. The others are a bit more work; because $\xi$ is a root of $x^4-4x^2+2$ we find that $$\xi^{-1}=2\xi-\tfrac12\xi^3,$$ and we already saw in question 3 that $\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}=\sqrt{2}$, meaning that $$\xi^1\zeta(\xi^1)=\sqrt{2}=\xi^2-2\xi^0,$$ from which it follows that $$\zeta(\xi^1)=\xi^1-2\xi^{-1}=-3\xi+\xi^3,$$ yielding the second column. The third and fourth column I leave to you.
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  • $\begingroup$ I'm confused about something. Please forgive my ignorance but how did you arrive at the following (1) How did you compute $\zeta (\xi^0)$ ? Meaning what did you substitute the values into to get that. (2) How did you compute $\xi^{-1}$? $\endgroup$ – Jason Moore Apr 1 at 0:14
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    $\begingroup$ By definition of a field automorphism $\zeta(1)=1$. As for computing $\xi^{-1}$, because $\xi$ is a root of $x^4-4x^2+2$ it satisfies $$\xi^4-4\xi^2=-2,$$ and multiplying through by $-\tfrac12\xi^{-1}$ yields the expression I mention. $\endgroup$ – Servaes Apr 1 at 0:19

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