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


2 Answers 2


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 $\mathbb C[t]/(t^\ell)$.

  • $\begingroup$ 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. $\endgroup$
    – ren l
    Mar 14, 2011 at 9:24
  • $\begingroup$ "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. $\endgroup$
    – rschwieb
    Dec 16, 2011 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 $\mathbb{C}[[t]]$.


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