Prime/maximal ideals of $\mathbb{C}[x, y]$ containing a given ideal Remember that (i) every maximal ideal is a prime ideal, (ii) for proper ideals $I$ of rings $R$, the factor ring $R/I$ is a field iff $I$ is a maximal ideal of $R$, and that (iii) whenever $F$ (for example $F=\mathbb{C}$) is a field, $F[x]/(p(x))$ is a field iff $p(x)$ is irreducible over $F$ (by definition, $p(x)$ is nonconstant). 
Now, consider $I \subset \mathbb{C}[x, y]$ with $I=(x^2-1, y^3-1)$; computations show that (for $\alpha_1=(-1+i\sqrt{3})/2$ and $\alpha_2=(-1-i\sqrt{3})/2$)
$$I=((x+1)(x-1), (y-1)(y-\alpha_1)(y-\alpha_2)).$$
Since $\mathbb{C}$ is a field, (iii) implies (?) that 
$\mathbb{C}[x, y]/(p(x), q(y))$ is a field iff $p(x), q(y)$ are irreducible over $\mathbb{C}[x, y]$. It follows that the maximal ideals that contain $I$ are by (ii)
$$(x-1, y-1) \qquad (x-1, y-\alpha_1) \qquad (x-1, y-\alpha_2)$$
$$(x+1, y-1) \qquad (x+1, y-\alpha_1) \qquad  (x+1, y-\alpha_2).$$
From (i) they are all prime ideals.
In the problem, we are given that there are six maximal ideals and prime ideals.
My question is: if "$\mathbb{C}[x, y]/(p(x), q(y))$ is a field iff $p(x), q(y)$ are irreducible over $\mathbb{C}[x, y]$" follows from (iii), and how can we prove that there are no more than six prime and maximal ideals that contain $I$?
I am grateful for all your help. 
 A: The answer to your first question is "Yes", in fact if you consider $(p(x), g(y)) \cap \mathbb{C}[x]$, this must be a prime ideal (is an easy exercise) and then $p(x)$ must be irreducible.
The same argument holds with $(p(x), g(y)) \cap \mathbb{C}[y]$.
Looking at the problem of count how many maximal ideals contains $I$, I think you can follow this way:


*

*Call $M_i$ the maximal ideals you have found, then prove that $I=\cap M_i$

*Suppose there's another maximal ideal $N$ such that $I \subset N$, then $\cap M_i=I=N \cap I  = N \cap \bigcap M_i $. You can use now the following Lemma to conclude:
Lemma Let $P$ be a prime ideal. If $I_1 , \dots , I_n$ are ideals such that $\cap I_i \subseteq P$, then there's an index $j$ such that $I_j \subseteq P $. (Try to prove it!)
If I well remember there's another way to compute such number of maximal ideals, using a bit of Algebraic Geometry and Theory of Grobner Basis.
We starts from the observation that, in an algebrically closed field, a point of a variety $V(I)$ corresponds to a maximal ideal containig $I$. Follows that, if $V(I)$ is finite, it is contained in a finite number of maximai ideals, many as its points.
Now, we link tha finiteness of $V(I)$ to the Grobner Bases by the following result.
Theorem A variety $V(I)$ han a finite number of points iff there's only a finite number of monomials not contained in the Leading Terms  Ideal of $I$.
Then the following lemma (It's a vague image in my memory : I hope there's no mistakes in its assert ) can easily solve you problem:
Lemma Let $k$ me an algebricaly closed field and $I$ an ideal of the ring $k[X_1, \dots, X_n]$ such that $V(I)$ is finite. The following integers are equal:


*

*The number of points of $V(I)$.

*The dimension of the ring $k[X_1, \dots, X_n]/I$ as $k$- vector space.

*The number of monomials not contained in the ideal of the leading terms of $I$


It seems to be more complicated, but with this theorem is immediate, just calculating a Grobner Bases, to obtain a lot of information about the ideal $I$ and its variety.
In your case, for example, is very easy to check that $x^2-1$ and $y^3-1$ are a Grobner Bases of $I$ and that (for example) there's only 6 monomials not containden in the Leading Terms Ideal of I.
I'm conscious of the fatc that, if you're a student and don't know anything of this argumens, this is only an unintelligible speech, but I think is, in every case, very interesting.
