0
$\begingroup$

Let $t^2-2$ be the minimal polynomial of $\alpha$ over $\mathbb Q$ and $t^2-4t+2$ be the minimal polynomial of $\beta$ over $\mathbb Q$. I want to define an isomorphism $i:\mathbb Q\to\mathbb Q$ such that $i(t^2-2)=i(1)t^2+i(-2)=t^2-4t+2$. How can I do this?

According to theorem 5.16 of Stewart's Galois theory, this shows that the extensions $\mathbb Q(\alpha)$ and $\mathbb Q(\beta)$ are isomorphic. Another way that I've tried is as $\mathbb Q(\alpha)\cong\frac{\mathbb Q[t]}{\langle t^2-2\rangle}$ and $\mathbb Q(\beta)\cong\frac{\mathbb Q[t]}{\langle t^2-4t+2\rangle}$, I need to show that $\langle t^2-2\rangle=\langle t^2-4t+2\rangle$. let $t=s-2$, then

$$\langle t^2-2\rangle=\langle(s-2)^2-2\rangle=\langle s^2-4s+2\rangle.$$ But I'm not sure if this justifies that $\langle t^2-2\rangle=\langle t^2-4t+2\rangle$? Could you please say me if this works or not? What other methods can I use to show that $\mathbb Q(\alpha)$ and $\mathbb Q(\beta)$ are isomorphic? Thanks!

$\endgroup$
  • $\begingroup$ Surely you meant $i:\Bbb Q[t]\to\Bbb Q[t]$? $\endgroup$ – David C. Ullrich Apr 20 '18 at 18:42
  • $\begingroup$ it's easier to find the roots of both polynomials. Then see how you can take a root of the first to a root of the second by a simple map (translation works). $\endgroup$ – Orest Bucicovschi Apr 21 '18 at 5:39
2
$\begingroup$

$\newcommand{\Q}{\mathbb{Q}}$You are trying to do something impossible. The only automorphism of the field $\Q$ is the identity.

Rather, note that if $\alpha$ is a root of $t^{2} - 2$, then $\alpha + 2 \in \Q(\alpha)$ is a root of $t^{2} - 4 t + 2$, as $$ (\alpha + 2)^{2} - 4 (\alpha + 2) + 2 = \alpha^{2} + 4 \alpha + 4 - 4 \alpha - 8 + 2 = 0. $$

$\endgroup$
  • $\begingroup$ This shows that $\langle t^2-2\rangle=\langle t^2-4t+2\rangle$? $\endgroup$ – user547800 Apr 20 '18 at 18:15
  • 1
    $\begingroup$ $\newcommand{\Q}{\mathbb{Q}}$No, this isn't true. It rather shows that $\Q(\alpha) $ contains a root $\beta$ of $t^{2} - 4 t + 2$, so you may say that $\Q(\alpha) = \Q(\beta) $. $\endgroup$ – Andreas Caranti Apr 20 '18 at 19:00
  • $\begingroup$ So in fact, $\mathbb Q(\alpha)=\mathbb Q(\alpha+2)\cong\mathbb Q(\beta)$ and so $\mathbb Q(\alpha)\cong\mathbb Q(\beta)$. $\endgroup$ – user547800 Apr 21 '18 at 7:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy