Assume $F$ is a field and $f$ is an irreducible polynomial in $F[X,Y]$ which involves the variable $Y$.

Then, by Gauss's lemma, $f$ is irreducible also in $F(X)[Y]$ so that $F(X)[Y]/(f)$ is a field (where $(f)$ is the ideal generated by $f$).

I'm looking for a simple way to see that the fraction field of the integral domain $F[X,Y]/(f)$ is isomorphic to $F(X)[Y]/(f)$.


Let $A$ be a UFD, and $f\in A[Y]$ irreducible with $\deg f\ge 1$. Then the field of fractions of $A[Y]/(f)$ is $K[Y]/(f)$, where $K$ is the field of fractions of $A$.

Set $S=A-\{0\}$. Then $K[Y]/(f)=S^{-1}A[Y]/S^{-1}(f)\simeq S^{-1}(A[Y]/(f))$, so $K[Y]/(f)$ is a ring of fractions of the integral domain $A[Y]/(f)$ and moreover it is a field, so necessarily $K[Y]/(f)$ is the field of fractions of $A[Y]/(f)$.

  • $\begingroup$ Thank you. This is just the kind of a clear and concise proof I was looking for. $\endgroup$ – feebly Aug 14 '15 at 0:15

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