# Why is $\operatorname{Gal}(E/F)\cong GL_2(F)/F^\ast$ when $E=F(t)$ for $t$ transcendental?

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. 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.