Is $k[x,y]/{(f,g)}$ an integral domain, where $f$ and $g$ are irreducibles? Let $k$ be an (algebraically closed) field of characteristic zero.
Let $f,g \in k[x,y]$ be two irreducible polynomials.

Is $\frac{k[x,y]}{(f,g)}$ an integral domain? (probably no?).

In the one variable case the answer is positive, namely,
for an irreducible $h \in k[x]$, $\frac{k[x]}{(h)}$ is an integral domain (since $(h)$ is a prime ideal).
Thank you very much!
 A: Let $k=\mathbb C$, $f=x^2+y$, $g=x^2-y$. These are irreducible, and $x\not\equiv 0$, $x\cdot x\equiv0\pmod{f,g}$.
A: From a geometric perspective, $f$ and $g$ each define irreducible curves in the plane,  and $k[x,y]/(f,g)$ will be the coordinate ring of the intersection of these curves.  So, to get an example where $k[x,y]/(f,g)$ is not a domain, you just need an example where the intersection of $f$ and $g$ has more than one point.  An intersection of two (distinct) irreducible curves will just be a finite set of points, and this intersection will be reducible if there is more than one point.  So, for instance, if you take $f=y-x^2$ and $g=y-1$, then they intersect at both $(1,1)$ and $(-1,1)$, so $k[x,y]/(f,g)$ will not be a domain.
(There are other ways to get counterexamples as well.  In the example in Mike Earnest's answer, the two curves intersect at just one point, but the intersection is not transverse and so $k[x,y]/(f,g)$ ends up not being a reduced ring, with the nilpotent elements somehow capturing the "multiplicity" of the intersection.  Or, if you take two curves that don't intersect at all like $f=x$ and $g=x+1$, then $k[x,y]/(f,g)$ will be the zero ring which is not a domain.)
A: Recall that if $R$ is a commutative ring and $\mathfrak{I} \triangleleft R$ and ideal, then $\mathfrak{I}$ is prime if and only if $R/\mathfrak{I}$ is a domain. In your case, then, $\mathbb{k}[x,y]/(f,g)$ will be a domain if and only if $(f,g)$ is prime. Certainly this does not happen always.
