If $f(x,y)\in R[x,y]$ is an irreducible polynomial, is $R[x,y]/f(x,y)$ a field?

Question

If $f(x,y)\in R[x,y]$ is an irreducible polynomial, is $R[x,y]/f(x,y)$ a field?

I know that this is true for $R[x]$, because of the Euclidean algorithm. However, is it also true for polynomial rings with multiple variables?

Answer 1

I'll assume you mean that $R$ is a field since $R[X]$ is not euclidean otherwise. Then by Gauss' lemma, $R[X,Y]=R[X][Y]$ is a UFD since $R[X]$ is, and in a UFD irreducible elements are prime. Hence your factor ring is a domain. It may not be a field though. For instance, if $R=\mathbb{C}$ and $f=XY-1$, then your factor ring is isomorphic to the ring of rational functions with poles at zero (i.e. The ring of Laurent polynomials), which is not a field - what is $(X-1)^{-1}$?

Answer 2

For any commutative ring $R$ with unity, and any polynomial $f\in R[X,Y]$, $K=R[X,Y]/f$ cannot be a field.

Suppose that $R$ is a field (which is probably implicit in the question anyway). If $f$ is a unit, there is nothing to prove, so assume that $f$ is not a unit. Then there is a root $(\alpha,\beta)$ of $f$ over the algebraic closure of $R$. Let $s(X), t(Y)$ be the minimum polynomials of $\alpha$ and $\beta$.

The images of $s(X)$ and $t(Y)$ in $K$ cannot be units, because if $f(X,Y)$ and $s(X)$ generate the unit ideal, then plugging in $(\alpha,\beta)$ yields $0=1$, and similarly for $t(Y)$. Since $K$ is a field, we have $s(X),t(Y)\in (f(X,Y))$. But it is impossible for $(f(X,Y))$ to include both a nonzero polynomial in $X$, and a nonzero polynomial in $Y$, unless $f(X,Y)$ is a unit.

If $R$ is not assumed to be a field, I believe that we can reduce to that case by localizing at a maximal ideal containing the kernel of $R\to K$, then modding out by that ideal.

Answer 3

For any commutative ring $R$ with unity, and any polynomial $f\in R[X,Y]$, $R[X,Y]/(f)$ cannot be a field.

Suppose the contrary, and let $\mathfrak p=(f)\cap R$, a prime ideal. Then $\mathfrak pR[X,Y]\subset (f)$ and since $\mathfrak pR[X,Y]$ is not maximal the inclusion is strict. Now we mod out by $\mathfrak pR[X,Y]$ and thus we may assume that $R$ is an integral domain and $(0)=(f)\cap R$. Since $R[X,Y]/(f)$ is a field all its rings of fractions are also fields. In particular, $K[X,Y]/(f)$ is a field, where $K$ is the field of fractions of $R$. But this is not possible as Slade's answer shows. (Here I have an alternative elementary argument. Write $f$ as a polynomial in $y$ with coefficients in $k[x]$, and choose an irreducible polynomial $p∈k[x]$ which does not divide the leading coefficient of $f$. There exists such a polynomial since $k[x]$ has infinitely many irreducible polynomials. Then $(f)⊊(p,f)⊊k[x,y]$.)