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Let $R$ be a polynomial ring over a field $k$, $R = k[x_1, \dots, x_n]$. Suppose $R'$ is an associative $R$-algebra and it has the property that there exists a degree $m<n$ monomial in the $x_i$'s which vanishes. Consider the (two-sided) ideal $I$ in $R'$ given by $\bigcap_{i = 1}^n (x_i)$.

On the one hand, this ideal $I$ is necessarily nilpotent: in fact, $I^m = 0$.

On the other hand, if the $x_i$ form a regular sequence on $R'$, then the intersection is also the product, so that $I = 0$ itself.

Are there intermediate conditions (if none of the $x_i$'s are nonzerodivisors) that will enable one to say something better than the naive bound $I^m = 0$? In fact, I am happy with being able to conclude that all $m'$th powers of elements in $I$ are zero for some $m' \ll m$.

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  • $\begingroup$ "A degree $m<n$ monomial vanishes" means that there is a monomial of degree $m<n$ which vanishes or that all monomials with this property are $0$ in $R'$? $\endgroup$
    – user26857
    Jul 7, 2015 at 12:20
  • $\begingroup$ @user26857: The former. I've edited to clarify this point. $\endgroup$ Jul 7, 2015 at 12:26

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