MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $R$ be an local Artinian ring, with maximal ideal $\mathfrak{m}$.

Let $e$ be the smallest positive integer for which $\mathfrak{m}^e=(0)$.

Let $t$ be the smallest positive integer for which $x^t=0$ for all $x \in \mathfrak{m}$.

We know $t \leq e$, with equality holding whenever $\mathfrak{m}$ is a principal ideal (i.e., $R$ is a principal ideal ring). Moreover, equality holds whenever $e \leq 2$.

What (else) is known about the relationship between these two integers?

What about the case when $R$ is the Artinian ring associated to a point of an algebraic curve that is contained in two distinct irreducible components?

share|cite|improve this question
In the general case, the minimal number $g$ of generators of $\mathfrak m$ should also play a role. For instance, $e \leq 1+(t-1)g$. – Neil Epstein May 24 '14 at 4:12

If $R$ contains a field of characteristic zero, then $e=t$. This follows from the fact that if $V$ is a finite dimensional vector space over a field of characteristic zero, the image of the map $V\to S^dV$, $v\mapsto v^d$ generates $S^dV$ as a vector space for any $d$. In your case, suffices to prove that $\mathfrak{m}^t=0$. If not, consider $V=\mathfrak{m}/\mathfrak{m}^2$ and $d=t$ composed with the surjective map $S^tV\to\mathfrak{m}^t/\mathfrak{m}^{t+1}$ to get the desired contradiction.

share|cite|improve this answer
It may be too early in the morning for this, but isn't $k[[x,y]]/(x^2,y^2)$ a counterexample? Here $\mathfrak{m}=(x,y)$, so $\mathfrak{m}^2=(xy)\neq 0$ (i.e. $e>2$), but $x^2=y^2=0$ (i.e. $t=2$). – Ketil Tveiten Jun 5 '14 at 7:46
@Ketil Assuming the field isn't of characteristic 2, the above is not a counterexample, since $(x+y)^2 = 2xy \neq 0$, so $t \neq 2$. – Neil Epstein Jun 5 '14 at 17:50

To complement Mohan's answer, it is worth noting that there are counterexamples when $R$ contains a field $k$ of prime characteristic $p$. Indeed, when $p\geq 3$, let $R=k[\![X,Y]\!]/(X^p, Y^p)$, and denote the images of $X$, $Y$ in $R$ by $x$, $y$ respectively. Then I claim that $t=p$ but $e\geq 2p-2>p$. To see this, note that any element of $f\in\mathfrak m$ is of the form $f=xg+yh$, and then by Freshman's Dream, $f^p = x^p g^p + y^p h^p = 0$, whereas clearly $x^{p-1} \neq 0$, showing that $t=p$. On the other hand, $0 \neq x^{p-1} y^{p-1} \in {\mathfrak m}^{2p-2}$.

A characteristic 2 counterexample is given by $k[\![X,Y]\!]/(X^4, Y^4)$ ($k$ any field of char $2$), in which case $t=4$ but $e\geq 6$.

To summarize, your question of equality has a 'yes' answer if you are willing to assume the ring contains $\mathbb Q$, but can be 'no' if $R$ contains a field of any other characteristic. I don't know what happens in mixed characteristic.

EDIT: Equality fails in any mixed characteristic $(p^c, p)$. To see this, let $A := {\mathbb Z}/(p^c)$ and $R := A[X,Y]/(X^p, Y^p)$. First note that $0\neq p^{c-1} (xy)^{p-1} \in {\mathfrak m}^{c+2p-3}$, whence $e>c+2p-3$. However, I claim that $t \leq c+2p-3$. To see this, note that any element of $\mathfrak m$ has the form $pf+xg+yh$. We have $(xg+yh)^{2p-1}=0$ since every term in the expansion is divisible by $x^p$ or $y^p$, and by a similar computation we have $$ (xg+yh)^{2p-2} = {2p-2 \choose p-1} (xygh)^{p-1}. $$ We have $$ (pf+xg+yh)^{c+2p-3} = \sum_{i=0}^{c+2p-3} {c+2p-3 \choose i} (pf)^i (xg+yh)^{c+2p-3-i}, $$ and by the above considerations, the only term that potentially survives is the term where $i=c-1$. That is, $$ (pf+xg+yh)^{c+2p-3} = {c+2p-3 \choose c-1} (pf)^{c-1} (xg+yh)^{2p-2} = {c+2p-3 \choose c-1} (pf)^{c-1} {2p-2 \choose p-1} (xygh)^{p-1}. $$ But it is elementary to check that $p \mid {2p-2 \choose p-1}$, whence $p^c$ divides the displayed term, which is then $0$ in $R$.

share|cite|improve this answer
Thank you. I am definitely interested in the mixed characteristic ($p^c,p)$ case. We also have $c \leq t$ to use/understand. – user51197 Jun 4 '14 at 18:17
@user51197 See my edit above for an example where $e\neq t$ in the mixed characteristic case. – Neil Epstein Jun 5 '14 at 19:54
@user51197 As you will see in the above edit, equality fails for any mixed characteristic pair $(p^c,p)$, which then completely resolves the question of equality. I hope this answer is now sufficient. – Neil Epstein Jun 5 '14 at 20:27
Doesn't your counterexample in the equicharacteristic $p$ case also work when $p=2$? In that example one has (for any prime $p$, including $p=2$) $e≥2p−1$, it seems. Thus, $e>p=t$. – user51197 Jun 7 '14 at 16:01
@user51197 Huh. I guess you're right; I was making things a bit too complicated. We have $e>2p-2$ since $0 \neq (xy)^{p-1} \in {\mathfrak m}^{2p-2}$, just as you say. – Neil Epstein Jun 8 '14 at 21:56

Your Answer


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.