# Finite field extensions of $\mathbb C(t)$ are quotient fields of $\mathbb C[X,Y]$ modulo an irreducible polynomial?

Given a field of the form $\mathbb C(t)[g]$, where $t$ is transcendental over $\mathbb C$ and $g$ is algebraic over $\mathbb C(t)$ do I always find an irreducible polynomial $F\in \mathbb C[X,Y]$ such that $Quot(\mathbb C[X,Y]/(F))\cong \mathbb C(t)[g]$?

I thought about taking the minimal polynomial of $g$ over $\mathbb C(t)$ and killing the denominators, but I couldn't manage writing a serious proof.

• I know the primitive element theorem I don't see, why it should give me a solution. Beeing a little bit more precise would be helpful. Thanks. – max Dec 27 '12 at 23:52
• I misread the problem statement. You do not need the primitive element theorem for the body question (but the fact that the body and title question are equivalent uses the primitive element theorem). You are correct that you should use the minimal polynomial of $g$ over $\mathbb{C}(f)$ and then clear denominators. – Qiaochu Yuan Dec 28 '12 at 0:07
• @beginner I've changed $f$ by $t$ suggesting "transcendental". – user26857 Dec 29 '12 at 0:15

Let $R$ be an UFD, $K$ its field of fractions, $p\in K[T]$ an irreducible monic polynomial and $a\in R$ such that $ap\in R[T]$ is primitive (there is always an $a\in R$ with this property!) and therefore irreducible. Then the field of fractions of $R[T]/(ap)$ is isomorphic to the field $K[T]/(p)$.
In order to prove this note that there exists an injective homomorphism $R[T]/(ap)\to K[T]/(p)$. Then it's easy to show that this is also surjective.
Remark. For another proof of the desired isomorphism one can uses the usual properties of localization (with respect to the multiplicative system $S=R-\{0\}$).