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.
Your responses for the first three are exactly correct. To see why $\langle 0 \rangle$ is not maximal in $\mathbb{Z}$, notice that $\langle 0 \rangle \subsetneq \langle x \rangle$ for any nonzero $x \in \mathbb{Z}$.
For the fourth, the definition of a maximal ideal $I$ is that there are no other ideals $J$ such that $I \subsetneq J \subsetneq R$. For example, $\langle 4 \rangle$ is not maximal since $\langle 4 \rangle \subsetneq \langle 2 \rangle \subsetneq R$. If you can convince yourself that this is true, then you should be able to proceed with your problem.
As an aside, there are some very useful facts you should be aware of (and prove, if possible):
- An ideal $I$ of a commutative ring $R$ is maximal $\iff$ $R/I$ is a field.
- An ideal $I$ of a commutative ring $R$ is prime $\iff$ $R/I$ is an integral domain.
- All maximal ideals are prime.
- If $R$ is a principal ideal domain ($\mathbb{Z}$, e.g.), then all nonzero prime ideals are maximal.
- If $R$ is a field, then $\langle 0 \rangle$ is the only maximal ideal.
Best Answer
Let $P$ be a prime ideal of $\mathbb C[x,y]$ containing $I=\langle x^2 + 1, y + 3\rangle$. Then $x^2+1\in P$ and $y+3\in P$. Since $x^2+1=(x+i)(x-i)$ we have $x+i\in P$ or $x-i\in P$. In conclusion, $\langle x + i, y + 3\rangle\subseteq P$ or $\langle x - i, y + 3\rangle\subseteq P$. Since $\langle x \pm i, y + 3\rangle$ are maximal ideals (why?) then $P=\langle x \pm i, y + 3\rangle$.