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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.

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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

1 Answer 1

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.
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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

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