How is the field extension finite

Question

I was reading the book Arithmetic of Elliptic Curves by Silverman and I came across the below proposition

Proposition.Let C/K be a curve(Projective variety of dimension 1), and let t ∈ K(C) be a uniformizer at some nonsingular point P ∈ C(K). Then K(C) is a finite separable extension of K(t).

The proof says that K(C) is a finite extension of K(t) since the transcendence degree of K(C) is 1, but I don't see how it follows directly.Can some explain this?Here,K(C) is the function field of the curve C.

Answer 1

$K(X)$ is a finitely generated extension of $K$ for any (irreducible) variety, because it's the quotient field of a finitely generated $K$-algebra, namely $\mathcal O_X(U)$ for some non-empty affine open.

Secondly, $t$ is not algebraic over $K$, because $\mathcal O_{C,P}$ is a discrete valuation ring containing $K$: If $t$ was algebraic, then one could conclude that it is invertible in $\mathcal O_{C,P}$, which a uniformizer in a DVR never is.

So in your situation $K(C)/K(t)$ is also finitely generated. Because $K(C)$ and $K(t)$ both have transcendence degree $1$ over $K$ the extension $K(C)/K(t)$ must be algebraic (transcendence degrees add up in tower of extensions). Finally, finitely generated + algebraic implies finite.