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.
It is equivalent to show that every non-minimal ideal is maximal. If $P\subsetneq Q\subset R$ are prime ideals, we can view $Q$ as an ideal of the integral domain $R/P$. The additive group of $R/P$ is also finitely generated, so it suffices to show that if $R$ is an integral domain with finitely generated additive group, then every non-zero prime ideal of $R$ is maximal.
Suppose $R$ is an integral domain whose additive group is finitely generated, and let $n$ be the rank of the additive group. Suppose $P\subset R$ is a non-zero prime ideal, and take $x\in P\backslash\{0\}$. Since multiplication by $x$ is a bijection of $R$ onto $xR$, the rank of $xR$ is $n$, so the rank of $P$ is also $n$. The quotient of a finitely generated abelian group by a subgroup with the same rank is finite, so $R/P$ is a finite integral domain, hence a field. It follows that $P$ is maximal.
Best Answer
The easiest way to remember which is which is probably to think that a field is also a domain and a maximal ideal is also prime. So when you quotient by a maximal ideal you get a field, but also a domain, because the ideal was also prime.
You should know that the, calling $\pi:R\to R/I$ projection map, then an ideal $J\subset R/I$ is prime (resp. maximal) if and only if $\pi^{-1}(J)$ is prime (resp. maximal) in $R$. In particular, $\{0\}$ is prime (resp. maximal) in $R/I$ if and only if $I$ is prime (resp. maximal) in $R$. So it is enough to consider the case $I=\{0\}$.
It can be useful to learn and understand many equivalent definitions of domain/field and maximal/prime ideal. For $R\neq \{0\}$ a commutative ring with $1$:
$R$ is a field $\iff$ every $0\neq a\in R$ is invertible $\iff$ the only ideals of $R$ are $\{0\}$ and $R$ $\iff \{0\}$ is maximal
and
$R$ is a domain $\iff$ for $a,b\in R$ if $ab=0$ then either $a=0$ or $b=0$ $\iff$ for $a,b\in R$ if $ab\in\{0\}$ then either $a\in\{0\}$ or $b\in\{0\}$ $\iff$ $\{0\}$ is prime
Remember that by definition the improper ideal $R$ is not maximal nor prime.
A little aside you didn't ask for, but it fits nice in this answer. A ring is called reduced if there are no nonzero nilpotents ($a\in R$ is called nilpotent if $a^n=0$ for some $n\in\mathbb{N}$). A proper ideal $I\subsetneq R$ is called radical if $\sqrt I = I$, that is if every $a\in R$ such that $a^n\in I$ for some $n\in \mathbb N$ is already in $I$. It follows that every prime ideal is radical and that a ring is reduced iff $\{0\}$ is radical. So:
$\{$maximal ideals$\}\subset\{$prime ideals$\}\subset\{$radical ideals$\}$
$\{0\}$ is maximal $\implies$ $\{0\}$ is prime $\implies$ $\{0\}$ is radical
$R$ is a field $\implies$ $R$ is a domain $\implies$ $R$ is reduced