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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.

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Do you know anything about Theory of Grobner Bases? –  Joseph Curwen Dec 6 '13 at 9:34

1 Answer 1

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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:

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

  2. 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:

  1. The number of points of $V(I)$.
  2. The dimension of the ring $k[X_1, \dots, X_n]/I$ as $k$- vector space.
  3. 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.

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