Structure of maximal ideals of the quotient $\mathbb{C}[x,y,z]/ I$

I am trying to understand the general approach to the problems of the following type:

Problem.

a) Let $I\subset\mathbb{C}[x,y,z]$ be an ideal generated by $$\langle \ (x^2+y^2)^3+zx+3y^2z^3\ ,\ y^4-1,\ z^4x,\ 3x^2+2y^2-1\rangle.$$

Find the number of maximal ideals of the quotient $\mathbb{C}[x,y,z]\ /\ I$ and their generators.

b) What are the generators of the radical $\sqrt{I}?$

My general question is: where do the maximal ideals of $\mathbb{C}[x,y,z]\ /\ I$ come from and what is a good (not too advanced!) reference for problems of such type?

My guess is that I need to use Hilbert's Nullstellensatz. So I do have to find the set of common zeros.

This question is most easily approached geometrically. First recall the correspondence theorem for rings, which says that there is a bijection between the ideals of $R/I$ and the ideals of $R$ that contain $I.$ In fact, more is true: There is a bijection between the maximal ideals of $R/I$ and the maximal ideals of $R$ that contain $I.$ See this question for more details.

We want to interpret the problem geometrically, so we want to work with a ring that is the coordinate ring of some affine algebraic set. The coordinate ring of an affine algebraic set $V(J)\subseteq \mathbb{C}^n$ is $\dfrac{\mathbb{C}[x_1,\cdots, x_n]}{\sqrt{J}}.$ The ideal $I$ in your question is not a radical ideal, so the quotient ring is not the coordinate ring of an affine algebraic set. That is fine, because we only want to count the maximal ideals and find generators for the radical. A key observation is that a maximal ideal $M$ contains an ideal $I$ if and only if it contains $\sqrt{I}.$ Therefore our problem can be seen as counting the maximal ideals in $\dfrac{\mathbb{C}[x,y,z]}{\sqrt{I}}.$

There is a bijection between the points of an affine algebraic set and the maximal ideals of its coordinate ring. The ring $\dfrac{\mathbb{C}[x,y,z]}{\sqrt{I}}$ is the coordinate ring of the affine algebraic set $V(I),$ so now we just need to count the number of points on $V(I).$ That is, we need to count the points $(x,y,z)$ that make the generators of $I$ vanish.

Here is a guide: To make the 2nd generator vanish you need $y^4=1,$ so $y^2=\pm 1$ and $y=\pm 1, \pm i.$

To make the 3rd generator vanish, you need $3x^2+2y^2-1=0.$ From our previous working, we see there are two cases to consider: If $y^2=1$ or $y^2=-1.$ If $y^2=1$ then $3x^2+1=0$ so $x=\pm i/\sqrt{3}$ and if $y^2=-1$ then $x=\pm 1.$

Now finally put in all these cases into the first generator. Then count up all the solutions you found and that is the number of maximal ideals in $\dfrac{\mathbb{C}[x,y,z]}{\sqrt{I}},$ which is the number of maximal ideals in $\dfrac{\mathbb{C}[x,y,z]}{I}.$

I'm not quite certain how to tackle b) yet. The problem is that even once we've found all the points $(x',y',z'),$ we know that $\sqrt{I} = \sqrt{\cup (x-x',y-y',z-z')}$ but finding generators for this ideal is not clear to me.

An elementary source for this material is the first 17 pages of Fulton's Algebraic Curves.

• Thanks! There is a minor problem in your arguments, namely, we have 4 generators of I not three (the third one being $z^4x$). So it does make sense to start analyzing this one first. I got that $V(I)=(+-1,+-i,0)$ so we have 4 maximal ideals in $C[x,y,z]/I$
– Bob
Commented Oct 10, 2013 at 16:34
• @Sandra: that's right. Now note that computing the radical is quite easy in this case since there are so few maximal ideals to contend with. Geometrically, you know $z=0,y^2=-1,x^2=1$...
– user641
Commented Oct 11, 2013 at 14:48
• I think to get generators of the ideal of a finite set of points, you could start off with $(x - x_1) (x - x_2) (x - x_r)$ with $x_1, \ldots, x_r$ distinct and similarly for $y, z$. Then, select some linear functional $\phi$ which distinguishes all points of the cartesian product from each other, and add $(\phi(x,y,z) - a_1) \cdots (\phi(x,y,z) - a_r)$ which selects just the desired points. Commented Feb 21, 2020 at 18:53

This is not a complete answer but I guess it contains good hint to attack problem of this sort.

Let's $R$ be a commutative unital ring and $I \subseteq R$ an ideal.

• There's a one on one correspondence between ideals of $R/I$ and the ideals of $R$ containing $I$: namely the one that send every ideal in $R$ containing $I$ in its image through the projection $\pi \colon R \to R/I$. This correspondence restrict to a correspondence between maximal ideal of $R$ containing $I$ and maximal ideals of $R/I$.

• Hilbert's Nullstellensatz is equivalent to the following statement:

Every maximal ideal in an algebraically closed field is of the form $$I(\{p\}) = \{f \in K[x_1,\dots,x_n] \mid f(p)=0\}$$ for some $p \in \mathbb A^n_K$ the $n$-th dimensional affine space over the (algebraically closed) field $K$.

• The Nullstellensatz tells that

For an algebraically closed $\mathbb C$ and for every $n \in \mathbb N$ an ideal $I \subseteq \mathbb C[x_1,\dots,x_n]$ then $$I(V(I)) = \sqrt{I}$$

So basically to address the point a) is sufficient to find the points $p \in \mathbb A^3_\mathbb{C}$ that are contained in $V(I)$, then the maximal ideals in $\mathbb C[x,y,z]/I$ are the $I(\{p\})/I$ for $p \in V(I)$.

To solve point b) you can find out $\sqrt(I)$ as the ideal associated to the affine variety $V(I)$, i.e. as the ideal $I(V(I))$.

Hope this helps.

• Thanks! So to find the generators of the maximal ideals of $C[x,y,z] / I$ I need to find the image of the generators of $I({p}) / I$ under projection $\pi?$ Is there any explicit way to determine them directly?
– Bob
Commented Oct 10, 2013 at 16:39
• @Sandra Sure: for every point $p \in \mathbb A_K^n$ the ideal $I(p)=\langle x_1 - p_1,\dots,x_n - p_n\rangle$, where by $p_i$ I mean the $i$-th coordinate of $P$. This determinate the generator in $K[x_1,\dots,x_n]$ and so you can apply this result in the case $K= \mathbb C$. Then passes the generators to quotient. Commented Oct 10, 2013 at 16:44