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?

  • 6
    $\begingroup$ In the general case, the minimal number $g$ of generators of $\mathfrak m$ should also play a role. For instance, $e \leq 1+(t-1)g$. $\endgroup$ May 24, 2014 at 4:12

2 Answers 2


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.

  • $\begingroup$ 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. $t=2$). $\endgroup$ Jun 5, 2014 at 7:46
  • $\begingroup$ @Ketil Assuming the field isn't of characteristic 2, the above is not a counterexample, since $(x+y)^2 = 2xy \neq 0$, so $t \neq 2$. $\endgroup$ Jun 5, 2014 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 $R$.

  • $\begingroup$ Thank you. I am definitely interested in the mixed characteristic ($p^c,p)$ case. We also have $c \leq t$ to use/understand. $\endgroup$
    – user51197
    Jun 4, 2014 at 18:17
  • $\begingroup$ @user51197 See my edit above for an example where $e\neq t$ in the mixed characteristic case. $\endgroup$ Jun 5, 2014 at 19:54
  • $\begingroup$ @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. $\endgroup$ Jun 5, 2014 at 20:27
  • 1
    $\begingroup$ 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, $e>p=t$. $\endgroup$
    – user51197
    Jun 7, 2014 at 16:01
  • $\begingroup$ @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. $\endgroup$ Jun 8, 2014 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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