Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $(A,m)$ be a local commutative associative algebra over the field of complex numbers, $m^n\ne 0$, $m^{n+1}=0$ for some $n>0$, and

(1) $A$ is finite dimensional as vector space

(2) for any nonzero ideal $I$ of $A$, we have $m^n\subset I$

What can we say about such an $A$? For example, whether it is always a quotient algebra of some $\mathbb C[t^{m_1},\ldots,t^{m_k}]$?

I wonder whether there are some results charactered this kind of algebra more explicitly

share|improve this question
    
I added $ to your math stuff just so I can easily read your question. :) –  Sándor Kovács Mar 14 '11 at 8:16
1  
To Torsten Ekedahl, we have assumed that $A$ is a commutative algebra. –  ren l Mar 14 '11 at 8:25
    
To Sándor Kovács, thanks! –  ren l Mar 14 '11 at 8:26

2 Answers 2

up vote 4 down vote accepted

Such a ring has simple socle, namely $\mathfrak m^n$. It follows easily that $A$ is self-injective, Gorenstein and, of course, of dimension $0$. You can construct them using Macaulay's method of inverse system; this is explained in Eisenbud's book on commutative algebra, if I recall correctly.

If $A_4=\mathbb C[x,y]/(x^2,y^2)$, a four-dimensional example, then you cannot embed it in an algebra of the form $\mathbb C[t]/(t^\ell)$.

share|improve this answer
    
You are right on the second question. Thank you. Possiblly i need to restate the question. It is easy to see that $A=\C[t^2,t^3]/(t^4)$ satisfies all assumptions. Is all $A$ like this? or can it be written as a subquotient of $\C[t]$? I wonder whether there are some results character this kind of algebra more explicitly. –  ren l Mar 14 '11 at 9:24
    
"Commutative zero dimensional Gorenstein rings" also have another popular name: Frobenius algebras(/rings). If memory is serving me correctly, then among commutative Artinian local rings, the rings with a unique minimal ideal are exactly the Frobenius rings. –  rschwieb Dec 16 '11 at 14:44

If $A$ is an Artin local ring of the form you have, we may write it as the quotient of a power series ring. Now, by Flenner's Bertini theorem (I believe it applies here), you have a one-dimensional domain quotient $B$ of the power series ring and $A$ is a quotient of $B$. The integral closure of $B$ is the power series ring in one variable. So, $A$ is the subquotient of $\mathbb{C}[[t]]$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.