Ideals in a polynomial ring

Question

(Sorry for the title being so vague, as I didn't know how to summarize it)

I have to solve the following problem:

Let $F=P_{1}^{e_{1}}\cdots P_{r}^{e_{r}}$ be the prime decomposition of a polynomial $F\in A=\mathbb{C}[X_{1},\dots,X_{n}]$. Compute the following:

1. The single generated prime ideals of the quotient ring $A/(F)$.
2. The same as 1. for the localization of $A$ by the powers of $F$.
3. The radical of the ideal $(F)$.
4. The nilradical of the above sets.

And I have got this:

1. I know that the prime ideals of $A/(F)$ are the prime ideals of $A$ that contain $(F)$. But I am having a hard time finding a general form of the prime ideals of $A=\mathbb{C}[X_{1},\dots,X_{n}]$.

2. In this case, I think the prime ideals of the localization are the prime ideals of $A$ such that $(F)$ is not contained in them. But I find it strange that this is the complementary of the case before (it can be, but I thought they were unrelated things).

3. The radical of $(F)$ is $\sqrt F = (P_1 P_2 \cdots P_r)$.

4. The nilradical is the intersection of the prime ideals, so I have to find the prime ideals first.

But now I am stucked in parts 1. and 2.

Answer 1

Algebraic geometry may help. Prime ideals in $A=\mathbb{C}[X_{1},\ldots,X_{n}]$ correspond to algebraic varieties (irreducible) in $\mathbb{C}^{n}$. The polynomial $F$ corresponds geometrically to its zero locus (same as the zero locus of its radical), which is an hypersurface $H$ that can be expressed as the union of its irreducible components $H_{1},\ldots,H_{r}$, which are the varieties corresponding to the prime factors $P_{1},\ldots, P_{r}$ (again, we don't care about the exponents). For example, for $n=2$ you get a bunch of plane curves as the zero locus of $F$. I think this point of view may clarify a bit your problem, as I will try to demonstrate now elaborating a bit on points 1. and 2.

1. Prime ideals containing $(F)$ correspond to subvarieties of $H$. If by single generated you mean principal prime ideals, then again you will get a subvariety with codimension $1$ in $H$. So in our example, if $n=2$, then $H$ is a curve and the prime ideals in $A/(F)$ correspond to points in the curve, i.e. prime ideals $(X_{1}-a,X_{2}-b)$ such that $F(a,b)=0$. If $n=3$, then $H\subset \mathbb{C}^{3}$ is a surface and prime ideals in $A/(F)$ correspond to (irreducible) algebraic curves in $\mathbb{C}^{3}$ contained in your surface $H$.

2. Prime ideals of the localisation $A_{F}$ correspond to prime ideals in $A$ which do not contain $F$, hence prime ideals in $A$ which do not contain any of the prime factors $P_{1},...,P_{r}$. If you are still interested in principal prime ideals, then again you will get (irreducible) hypersurfaces which are not contained in any irreducible component of $H$.

As a remark on what you said, if both in 1. and 2. you insist on principal ideals, then the two cases are not complementary. As the example for $n=2$ shows, principal prime ideals in $A/(F)$ are points (i.e. ideals in 1. correspond to points) whereas principal prime ideals in $A_{F}$ are still plane curves.

Answer 2

1. There's no general form of prime ideals in $\mathbf C[X_1,\dots, X_n]$, except for maximal ideals (Hilbert's Nullstellensatz). All you can say is prime ideals which contain $F$ is those prime ideals which contain one of $P_1,\dots, P_r$.

2. On the contrary, the prime ideals in $A_F$ correspond to the prime ideals in $A$ which contain none of $P_1, \dots, P_r$.