What is the number of elements in $Aut(Q(\pi)/Q)$?

Question

I tried to prove that $|Aut(E/F)|$ is finite, then $E/F$ is a finite extension, but then now I think $Q(\pi)/Q$ would be a counterexample for this.

I can see that there are two automorphisms $\pi\mapsto \pi$ and $\pi\mapsto -\pi$, but how do I prove that these are the only elements of $Aut(Q(\pi)/Q)$?

Answer 1

There are many more than that. As $\pi$ is transcendental, $\mathbf Q(\pi)\simeq \mathbf Q(x)$, and it is known that $$\operatorname{Aut}(\mathbf Q(x))=\mathbf{PGL}_2(\mathbf Q), $$ the projective linear group of order $2$ over $\mathbf Q$ which is the set of homographic transformations: $$x\mapsto \frac{ax+b}{cx+d}, \quad ad-bc\neq 0.$$

Counter-example:

Take any finite field $\mathbf F_q$ $\,(q=p^n)$. Then $\mathbf Q(x)$ is not a finite extension, but its groups of automorphisms is finite, since $\lvert \mathbf{GL}_2(\mathbf Q)\rvert=q(q-1)(q^2-1)$, hence $\lvert \mathbf{PGL}_2(\mathbf Q)\rvert=q(q^2-1)$.

Answer 2

This is actually not true; $\mathrm {Aut}(\mathbb{Q}(\pi)/\mathbb Q)$ is infinite. In general, if we have a field of rational functions $\mathbb{Q}(x_1,x_2,\ldots,x_n)$ in indeterminates $x_1,x_2,\ldots,x_n$ (for which $\mathbb{Q}(\pi)$ is with $n=1$) the automorphism group of this field over $\mathbb{Q}$ contains the general linear group over $\mathbb{Q}$.

This is a bit easier to see for $\mathbb{Q}[x_1,x_2,\ldots,x_n]$. Since this ring is the free commutative $\mathbb{Q}$-algebra on $\{x_1,x_2,\ldots,x_n\}$, for any linear transformation $A$ on the vector space spanned by $x_1,x_2,\ldots,x_n$ there is a homomorphism of rings $\mathbb{Q}[x_1,x_2,\ldots,x_n]\to \mathbb{Q}[x_1,x_2,\ldots,x_n]$ given by $x_i\mapsto Ax_i$ for all $i$. Since this is surjective on the degree $1$ polynomials and these generate the ring, this is a surjective homomorphism. If the kernel were nontrivial we'd have a nonzero polynomial $p(x_1,x_2,\ldots,x_n)$ such that $p(Ax_1,Ax_2,\ldots,Ax_n)=0$. We can use induction on $n$ to show that $p$ is the zero polynomial, which is a contradiction. Thus the automorphism group at least contains the general linear group.

To move this to the field of rational functions, we simply define $A\left(\frac{p}{q}\right)=\frac{Ap}{Aq}$. Once you show that this indeed defines a homomorphism, it follows immediately that it is injective by the fact that the ring is a field, and surjectivity follows from the fact that the image includes all polynomials. Thus in particular your automorphism group is infinite. This is a bit of overkill, but for the case you are considering it means we can send $\pi\mapsto r\pi$ for any nonzero rational $r$ to obtain an automorphism.

Answer 3

To the implied question about finite automorphism groups and infinite extensions: $\mathbb{Aut}(\mathbb{R}/\mathbb{Q})$ is trivial, but $\mathbb{R}$ is clearly not a finite extension of $\mathbb{Q}$. (To see this note that in $\mathbb{R}$, $x > y$ iff there is a $z$ with $x - y = z^2$. Hence an automorphism of $\mathbb{R}$ must be order-preserving, from which it is easy to see that the only automorphism is the identity.)