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I am trying to prove that the group $Gal( \mathbb{Q}(\pi), \mathbb{Q})$ (some denote it more correctly as $Aut ( \mathbb{Q}(\pi), \mathbb{Q})$) is infinite, knowing only that $\pi$ is trancedental. I tried to construct automorphisms but I didn't really manage to do so. Any answers-hints?

user128787
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2 Answers2

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This is a pure transcendental extension. $\mathbb{Q}(\pi)$ is the field of rational functions of one variable; any additional 'arithmetic' meaning of $\pi$ is completely irrelevant.

It's well-known that $\operatorname{Aut}(F(t) / F)$ is the group of linear fractional transformations $t \mapsto \frac{a+bt}{c+dt}$ where $a,b,c,d \in F$ and $ad-bc \neq 0$, which in turn is isomorphic to $\operatorname{PGL}(2, F)$.

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For instance, for any $p, q\in \Bbb Q$ with $p\neq 0$, $\pi\mapsto p\pi+q$ is an automorphism.

Arthur
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  • If we have an automorphism of $\Bbb Q(\pi)$ over $\Bbb Q$, then by definition of "over" it must send any rational number to itself. In fact, by the definition of ring homomorphism, it must send any rational to itself. – Arthur May 27 '17 at 11:12
  • yes, it was stupid, I deleted my comment but you were quick enough. I have another question; from the definition you gave you seem to know that every element of $\mathbb{Q}(\pi)$ is generated by $1$ and all the powers of $\pi$. Is that right? – user128787 May 27 '17 at 11:17
  • Not quite. For instance, we have $\frac{1}{1+\pi}\in \Bbb Q(\pi)$, but that can't be written as a finite sum of the powers of $\pi$ with rational coefficients. But by the above automorphism, it's mapped to $\frac1{1+q+p\pi}$. – Arthur May 27 '17 at 11:18
  • nevertheless, my question is whether the value of $\sigma$ at $\pi$ is enough to know that it is an automorphism well defined for any other element – user128787 May 27 '17 at 11:22
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    Yes, that it is. You simply take an element of $\Bbb Q(\pi)$, write it out, substitute every single instance of $\pi$ with $p\pi + q$ (remember parentheses if the old $\pi$ was involved in a multiplication of any kind), and there you have it. – Arthur May 27 '17 at 11:23
  • you define it for $\pi$ and extent "rationally", that is what I understand from that. Right? – user128787 May 27 '17 at 11:25
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    You could say so. The other answer had a more thorough reason for this before it was edited. Basically, any homomorphism from $\Bbb Q(\pi)$ to any ring may be built from a homomorphism from $\Bbb Q[\pi]$ (which is generated by $1$ and all positive powers of $\pi$) into that same ring with the specific property that any non-zero element is sent to something invertible. – Arthur May 27 '17 at 11:30