Annihilator of a maximal ideal in a polynomial ring[updated]

Question

Let $R = k[x, y]$ be the polynomial ring in two variables and $I$ an ideal of $R$ such that $R/I$ is finite dimensional. I wonder if the following statement is correct?

If $\mathfrak{m}$ is a maximal ideal in $R/I$, then its annihilator in $R/I$ is nonzero.

Here, I assume $k$ is algebraically closed. Thanks!

Annihilator with respect to which ring? The ring $R$ or the ring $R/I$? — Simon Marynissen, Mar 17 '17 at 14:56
In an Artinian ring every maximal ideal has a non-zero annihilator, unless it's a field. — user26857, Mar 17 '17 at 22:02

score 1 · Answer 1 · answered Mar 17 '17 at 14:54

1

Suppose $I=0$ and take $M=(x-a,x-b)$ its annihilator is zero.

answered Mar 17 '17 at 14:54

Tsemo Aristide

87,475

Oops. That's correct. What if I assume $I \neq 0$? – Steve Mar 17 '17 at 15:33
You can take $I$ to be the ideal which defines an integral curve $C$. In this case $k[x,y]/I$ is integral. If you take the ideal which defines a point of $C$, its annihilator is zero. – Tsemo Aristide Mar 17 '17 at 15:42
Thanks, Tsemo. If I further assume $R/I$ is finite dimensional, would my statement be true? – Steve Mar 17 '17 at 15:47
If $R/I$ is finite dimensional, then it is a product $k^n$ since $I$ defines a variety which is a finite set of points. – Tsemo Aristide Mar 17 '17 at 15:49
1

I guess you meant $R/I$ modulo its radical? $R/(x - a, y- b)^2$ is finite dimensional, but it is not isomorphic to $k$. Am I right? – Steve Mar 17 '17 at 16:02
Yes, I have assumed that $I$ is a radical ideal. – Tsemo Aristide Mar 17 '17 at 16:03

score 1 · Accepted Answer · answered Mar 17 '17 at 20:44

1

If $R/I$ is finite-dimensional, the product of all (finitely many) maximal ideals is the nilradical, in particular the annihilator of any maximal ideal is non-zero.

answered Mar 17 '17 at 20:44

MooS

31,390

This is much simpler. Thanks! – Steve Mar 18 '17 at 19:59

Annihilator of a maximal ideal in a polynomial ring[updated]

2 Answers2