Powers of elements in an Artinian Ring

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?

• In the general case, the minimal number $g$ of generators of $\mathfrak m$ should also play a role. For instance, $e \leq 1+(t-1)g$. 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.

• 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. $t=2$). 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 $t \neq 2$. 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 $R$.

• Thank you. I am definitely interested in the mixed characteristic ($p^c,p)$ case. We also have $c \leq t$ to use/understand. Jun 4 '14 at 18:17
• @user51197 See my edit above for an example where $e\neq t$ in the mixed characteristic case. 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. 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, $e>p=t$. 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. Jun 8 '14 at 21:56