# Determine all maximal ideals in the following ring

Determine all maximal ideals in $$R=\frac{\mathbb{C}[x,y]}{(x^3-x^2y,xy^2+xy+x+1)}$$

Attempt: By the correspondence theorem, it suffices to determine all maximal ideals in $$\mathbb{C}[x,y]$$ containing $$(x^3-x^2y,xy^2+xy+x+1)$$. In this ideal we have $$x^3-x^2y=0\implies x(x^2-xy)=0$$

If $$x=0$$, the ideal reduces to $$(1)=R$$ which is not proper and therefore not maximal.

If $$x^2=xy$$, then $$x=0$$ or $$x=y$$. In the second case, the ideal reduces to $$(x^3+x^2+x+1)$$. It therefore suffices to determine all linear factors of $$x^3+x^2+x+1$$ since $$\mathbb{C}[x]$$ is a PID. Since $$x^3+x^2+x+1=(x+1)(x-i)(x+i)$$ the maximal ideals in $$\mathbb{C}[x,y]$$ containing $$(x^3-x^2y,xy^2+xy+x+1)$$ are given by $$(x+1),(x-i),(x+i)$$

Is this approach correct? Now I realize I took out a $$y$$, and I probably need to put it back in the ideal somewhere. I am confused on this process of "eliminating the $$y$$" and then having to put it back in the ideal in the end.

For example, I used the relation $$x^3-x^2y=0$$ in order to eliminate one of the variables, in order to simplify the process of finding maximal ideals. However, how do I "keep track" of the eliminated variable correctly?

$$(x+1,x^3-x^2y),(x+i,x^3-x^2y),(x-i,x^3-x^2y)$$

instead? Do you sort of, have to put back in whatever generator you "zeroed out" or eliminated in the ideal, in the end? I am confused on this process.

Your approach is good. You want to describe maximal ideals in $$\mathbb{C}[x,y]$$ that contains the given ideal. Any maximal ideal in $$\mathbb{C}[x,y]$$ is of the form $$(x-a,y-b)$$ so it suffices to describe possible values for $$a$$ and $$b$$. Equivalently, from a geometric point of view, pairs $$(a,b)$$ of complex numbers satisfying the two equations in the given ideal. Now as you point out, $$a^2(a-b)=0$$ implies that $$a=b$$ or $$a=0$$ and the only possible case is $$a=b$$. We are left to determine what values of $$a$$ satisfies $$a^3+a^2+a+1=0$$. What you conclude is that the maximal ideals containing the given ideal are exactly
$$(y+1,x+1),(y-i,x-i),(y+i,x+i).$$ The image of these under the quotient is the maximal ideals you are looking for.
Note: These are the same ideals as you have written down, just substitute your value for $$x$$ in the second equation.
• Because the equation $xy^2+xy+x+1=0$ has no solution where $x=0$