The coordinate cross V(xy) is reducible and its coordinate ring has zero divisors

Let $k$ be an algebraically closed field and let $X=V(xy)\subseteq {\mathbf{A}}_k^2\mathrm{.}$ By Zero loci in affine space are the closed sets of the classical Zariski topology, $V(xy)=V(x)\cup V(y),$ so $X$ is the union of the two coordinate axes. Each axis is a proper closed subset of $X$, hence $X$ is reducible.

If a polynomial $f\in k[x,y]$ vanishes on $X$, then $f(x,0)$ vanishes at every element of the infinite field $k$, so it is the zero polynomial and $y$ divides $f$. Likewise, $f(0,y)=0$ identically, so $x$ divides $f$. Since $x$ and $y$ are coprime, $xy$ divides $f$. Thus $I(X)=(xy)$.

Its coordinate ring is $k[X]=k[x,y]\mathrm{/}(xy)\mathrm{.}$ The classes $\bar{x}$ and $\bar{y}$ are both nonzero: for instance, $x$ does not vanish at $(1,0)$ and $y$ does not vanish at $(0,1)$, so neither lies in the ideal $(xy)$. But $\bar{x}\bar{y}=\overline{xy}=0.$ Thus $k[X]$ has zero divisors, exactly as the irreducible-domain criterion predicts.