How can I decide if $\Bbb{C}[X,Y]/(X^2,Y^2)$ is a local ring or not?

Question

I want to check if $\Bbb{C}[X,Y]/(X^2,Y^2)$ is a local ring or not.

My claim is that it is a local ring.

I wanted to do it as follows:

If I remember correctly we had the fact that maximal ideals of $R/I$ where $R$ is a ring and $I$ an ideal are in one to one correspondence with maximal ideals of $R$ containing $I$.

So using this fact in my opinion it is enough to find all the maximal ideals in $\Bbb{C}[X,Y]$ containing $(X^2,Y^2)$. I know that maximal ideals of $\Bbb{C}[X,Y]$ are of the form $(X-a,Y-b)$ for some $a,b\in \Bbb{C}$. Now since $(X^2,Y^2)\subset (X-a,Y-b)$ needs to hold it is equivalent to say that $$X^2\in (X-a,Y-b)~~~\text{and}~~~~Y^2\in (X-a,Y-b)$$But this means that $X^2=P(X-a)+Q(Y-b)$ and $Y^2=U(X-a)+V(Y-b)$ for some $P,Q,U,V\in \Bbb{C}[X,Y]$. But this only holds iff $a=b=0$ and $P=X, Q=U=0, V=Y$.

So in my opinion this shows that our ring is local.

Is this correct so or did I wrote completely nonsense?

Answer 1

What you have written is indeed correct! But there's an even easier argument that you can use, which doesn't rely on the fact that $\mathbb{C}$ is algebraically closed (and in particular doesn't rely on the characterization of the maximal ideals of $\mathbb{C}[x,y]$, which is a nontrivial theorem). Let $F$ be any field; then I claim that $F[x,y]/(x^2,y^2)$ is local. As in your argument, it suffices to show that there is only one maximal ideal of $F[x,y]$ containing $(x^2,y^2)$. In fact, we can say even more – there is only one prime ideal of $F[x,y]$ containing $(x^2,y^2)$! For suppose $P<F[x,y]$ is such a prime ideal. Since $x^2\in P$, $x$ must lie in $P$, and since $y^2\in P$ also $y\in P$. Thus $P\geqslant(x,y)$. But $(x,y)$ is a maximal ideal, since $F[x,y]/(x,y)\cong F$. Thus in fact $P=(x,y)$ and we are done.