I have from a book
(b) Let $E = \Bbb{C}(X)$. Then $\operatorname{Aut}(E / \Bbb{C})$ consists of the maps $X \mapsto \dfrac{aX + b}{cX + d}, ad-bc \neq 0\ldots$
Not sure what $\Bbb{C}(X)$ is. Thanks.
I have from a book
(b) Let $E = \Bbb{C}(X)$. Then $\operatorname{Aut}(E / \Bbb{C})$ consists of the maps $X \mapsto \dfrac{aX + b}{cX + d}, ad-bc \neq 0\ldots$
Not sure what $\Bbb{C}(X)$ is. Thanks.
On
$E = \mathbf{C}(X)$ is the fractions field of the domain $\mathbf{C}[X]$, ring of polynomials in the formal variable $X$ with coefficients in $\mathbf{C}$. Every element of $E$ can be written $\frac{P}{Q}$ with $P,Q\in\mathbf{C}[X]$ coprime polynomials and $Q\not=0$, which in this case means that $P$ and $Q$ have no common roots. The definition is valid for any field $k$ instead of $\mathbf{C}$.
Well, first, what is $\mathbb{C}[X]$? It is, by definition, the set $\{ a_{0} + a_{1}X + a_{2} X^{2} + a_{3}X^{3} + \dots + a_{n}X^{n} | a_{i} \in \mathbb{C}, n \in \Bbb N \}$. In other words, it is the ring of polynomials with coefficients in $\Bbb C$ (and it's actually an integral domain).
Now, $\Bbb C(X)$ is the fractional field of the integral domain $\mathbb{C}[X]$. That is, $\Bbb C(X)$ is the set of elements that look like $\frac{a}{b}$ where $a, b \in \mathbb{C}[X]$ and $b \neq 0$. Of course, we are really talking about equivalence classes (just like in $\mathbb{Q}$, $\frac{2}{3} = \frac{4}{6}$, for example). It's easy to prove that $\mathbb{C}(X)$ satisfies the axioms of a field. We call it the fractional field of the ring $\Bbb C[X]$ since it's constructed by taking all possible "fractions" made from the elements of $\Bbb C[X]$.