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Let $F$ be a field, and consider the polynomial ring $F[X_1,...,X_n]$. I am trying to prove that every power of the ideal $(X_1,...,X_k)$ is primary for $k\leq n$. For $k = n$, we have that $(X_1,...,X_n)$ is maximal, and $(X_1,...,X_n)^j$ has radical $(X_1,...,X_n)$, so is primary. I also know that if $\mathfrak p$ is a prime ideal of a ring $A$, then $\mathfrak p[X]$ is a prime ideal of $A[X]$. However, I do not see a way to use this result to say anything about powers of a prime ideal. Does anyone have any suggestions?

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It may be worth noting here that, in general, powers of prime ideals are not primary; see Example 3 on page 51 of Atiyah-Macdonald. – mbrown Mar 24 '13 at 20:24
up vote 4 down vote accepted

We have $F[X_1,\dotsc,X_n]/(X_1,\dotsc,X_k)^p = F[X_1,\dotsc,X_k]/(X_1,\dotsc,X_k)^p [X_{k+1},\dotsc,X_{n}]$. Therefore it suffices to prove: If $A$ is a commutative ring with the property that every zero divisor is nilpotent, then this property also holds for $A[T]$. But this is easy using the well-known classification of zero divisors of $A[T]$: they are killed by some $0 \neq a \in A$, so that in particular the coefficients are zero divisors. But then the coefficients are nilpotent, and therefore also the polynomial is nilpotent.

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