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.
Consider the map $\mathbb{C}[x,y] \to \mathbb{C}$ given by $f(x,y) \mapsto f(i, -3)$, i.e. evaluating the polynomial at $x=i, y=-3$.
Show this is a surjective homomorphism. Show the kernel is exactly the ideal $I=(x-i, y+3)$. Then you can use the first isomorphism theorem to show that $\mathbb{C}[x,y]/I$ is a field, and thus $I$ is a maximal (and therefore prime) ideal.
In general, though, we cannot assume that starting with two prime ideals will give us a new prime ideal. For example, the ideals $(3)$ and $(5)$ are both prime in the integers, but their product is not.
Best Answer
I'm not sure what you mean when you ask whether $I$ is irreducible. It's probably helpful to use the third (and I guess also the second) isomorphism theorem, which will tell you: $$\mathbb{Z}[x]/(x^4+x^3+x^2+x+1,2) \cong \mathbb{F}_2[x]/(x^4+x^3+x^2+x+1).$$ Now if you show that the polynomial $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{F}_2$, then since $\mathbb{F}_2[x]$ is a UFD, this will give you a field.