Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is an example on page $235$ of Jacobson's Algebra I that I'm reading. I quote

Let $F$ be a field and $E=F(t)$ where $t$ is transcendental over $F$. $u\in E$ is a generator of $E/F$ if and only if it has form $$ u=\frac{at+b}{ct+d},\quad ad-bc\neq 0$$ Since an automorphism of $E/F$ sends generators into generators, it follows that $\operatorname{Gal}(E/F)$ is the set of maps $$ f(t)/g(t)\to f(u)/g(u)$$ We can see that $\operatorname{Gal}(E/F)$ is isomorphic to the factor group $GL_2(F)/F^\ast$ where $F^\ast$ is the set of matrices $\operatorname{diag}\{a,a\}, a\neq 0$.

What explicitly is the isomorphism? I see that generators $u$ can be identified with matrices in $GL_2(F)$, but how does that give an isomorphism of $\operatorname{Gal}(E/F)$ with $GL_2(F)/F^\ast$, and why quotient by $F^\ast$? It seems like this means we consider two automorphisms to be the same if they differ by a scalar multiple, but why do that? Thanks.

share|cite|improve this question
The keyword here is "projective linear group." – Qiaochu Yuan Mar 28 '13 at 19:45
More generally, $\mathrm{Aut}_K(K(t_1,\dotsc,t_n))=\mathrm{PGL}_{n-1}(K) := \mathrm{GL}_n(k) / k^* = \mathrm{Aut}_k(\mathbb{P}^{n-1}_k)$. – Martin Brandenburg Mar 28 '13 at 19:55

There are two things that you need to check. Neither is hard, just do them! Good luck!

  1. That the composition of the automorphism matches the product of matrices. So if $$ A=\pmatrix{ a&b\cr c&d\cr}, \qquad A'=\pmatrix{a'&b'\cr c'&d'\cr} $$ are two invertible matrices, then the composition $\phi_A\circ \phi_{A'}$of the automorphisms $$ \phi_A:t\mapsto \frac{at+b}{ct+d},\qquad\text{and}\qquad \phi_{A'}:t\mapsto \frac{a't+b'}{c't+d'} $$ is equal to the automorphism $\phi_{AA'}$.
  2. The automorphism $\phi_A$ is the identity mapping, i.e. satisfies $\phi_A(t)=t$ if and only if $A$ a (non-zero) multiple of the identity matrix.
share|cite|improve this answer
Thanks Jyrki, that makes sense. I was confused on what the desired identification would be. This looks like it will be routine. – Chelsea Dirks Mar 28 '13 at 19:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.