7
$\begingroup$

Perhaps a silly question. I'm trying to understand trascendental field extensions, but I can't find a lot of instructive examples.

Consider the extension $\mathbb{Q}(\pi)/\mathbb{Q}$. What is its group of automorphism, $\mathrm{Aut}(\mathbb{Q}(\pi)/\mathbb{Q})$?

$\endgroup$
2
  • 1
    $\begingroup$ Note that any automorphism is determined by its action on pi. $\endgroup$
    – Vik78
    Apr 7, 2016 at 3:38
  • 1
    $\begingroup$ The point of $\pi$ being transcendental over $\mathbb{Q}$ is that $\mathbb{Q}(\pi)$ is isomorphic as a field to univariate rational functions $\mathbb{Q}(X)$ over $\mathbb{Q}$. $\endgroup$
    – hardmath
    Apr 7, 2016 at 3:43

1 Answer 1

4
$\begingroup$

If $F$ is a field and $K = F(\alpha)$ is a transcendental extension, then every $F$-automorphism of $K$ is of the form $$\alpha \mapsto \frac{a\alpha + b}{c\alpha + d},$$ where $a,b,c,d \in F$ with $ad - bc \neq 0$.

$\endgroup$
7
  • $\begingroup$ Nice. So $\mathrm{Aut}(\mathbb{Q}(\pi)/\mathbb{Q})=GL_2(\mathbb{Q})$. $\endgroup$ Apr 7, 2016 at 3:51
  • 1
    $\begingroup$ Is there a reference for you result? Thank for your answer! $\endgroup$ Apr 7, 2016 at 3:51
  • 1
    $\begingroup$ It can be deduced from the results of Section 13.2 #18 in Dummit & Foote. $\endgroup$ Apr 7, 2016 at 3:54
  • $\begingroup$ Thanks, I have a copy. I'll check it out. $\endgroup$ Apr 7, 2016 at 3:55
  • 3
    $\begingroup$ @SamuelPlath: It's not quite $GL_2(\mathbb{Q})$, because if $b=c=0$ and $a=d$ then the automorphism you get is just the identity. So you actually get the quotient of $GL_2(\mathbb{Q})$ by the subgroup of matrices that are scalar multiples of the identity (this quotient is called $PGL_2(\mathbb{Q})$). $\endgroup$ Apr 7, 2016 at 3:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .