I want to solve this problem:

Let $K$ be a field. Let $X_1,\ldots,X_n$ be indeterminates. Prove that the ideal generated by $(X_1-t_1)^{s_1},\ldots,(X_r-t_r)^{s_r}$ is a primary ideal for all $0\leq r \leq n$, for all $t_1,\ldots,t_r \in K$ and all $s_1,\ldots,s_r\in \mathbb{N}$.

My attempt: as the problem seems hard to approach in my opinion, I want to prove a problems which looks more simple but is definitely a stronger result:

Let $R$ be an integral domain. Let $I$ be a primary ideal of $R$. Then $I[X]$ is a primary ideal of $R[X]$.

which is exactly the name of of this post.

If this result is true, then by induction and the fact that an ideal whose radical is maximal is primary, we solve the original problem.

I tried to prove the second problem by definition of primary ideal, but failed.

Can you help me solve the second problem? There is a chance this result is wrong, so can you show me the right way to solve Prob 1? Thank you.


2 Answers 2


If $f(x)h(x)\in I[x]$ and $h(x)\notin I[x]$, we have to show that $f^n(x)\in I[x]$ for some $n$.

Suppose $f(x)=a_nx^n+\dots+a_0$, $h(x)=b_mx^m+\dots+b_0$.

Consider the following two cases.

i) $b_m\in I$, then $h(x)-b_mx^m$ doesn't belong to $I[x]$ and have a lower degree than $h(x)$, and $f(x)h(x)-f(x)b_mx^m\in I[x]$.

ii) $b_m\notin I$. Suppose $a_ih(x)\notin I[x], a_{i+1}h(x), \ldots, a_mh(x)\in I[x]$. Then we have $(a_0+\ldots+a_ix^i)h(x)\in I[x]$. Thus $a_ib_m\in I$. Now consider $a_ih(x)-a_ib_mx^m$. It doesn't belong to $I[x]$ and has lower degree than $h(x)$, and $f(x)(a_ih(x)-a_ib_mx^m)\in I[x]$.

By induction, we can find some $b\notin I$ such that $bf(x)\in I[x]$. Then every coefficient $a_i$ of $f(x)$ belongs to $r(I)$, since $I$ is primary and $b\notin I$. Thus $f^n(x)\in I[x]$ for some $n$.


First, we do not need $R$ to be an integral domain for the second problem. In general, $I$ being a primary ideal is equivalent to every zero divisor in $R/I$ being nilpotent. Recall also that a polynomial is nilpotent if and only if it has nilpotent coefficients, and is a zero divisor if and only if there is a single nonzero element of the coefficient ring that annihilates all the coefficients of the polynomial. Finally, it is clear that $R[x]/I[x]$ is isomorphic to $(R/I)[x]$.

Hence, it suffices to show that if every zero divisor in a commutative ring $R$ is nilpotent, then the same is true for the polynomial ring $R[x]$. But this clear because a zero divisor in $R[x]$ must have coefficients that are zero divisors in $R$, hence nilpotent by the assumption on $R$, making the polynomial nilpotent.

  • $\begingroup$ why a zero divisor of R[x] must have coefficients that are zero divisors? $\endgroup$
    – wyhorgyh
    Commented Jan 29, 2022 at 14:05
  • 1
    $\begingroup$ See math.stackexchange.com/questions/83121 $\endgroup$
    – user26857
    Commented Jan 29, 2022 at 20:33
  • $\begingroup$ Why do zero divisors have to be nilpotent? It is the other way around, right? $\endgroup$
    – kubo
    Commented Nov 2, 2023 at 17:01
  • $\begingroup$ @kubo It is assumed that every zero divisor in $R$ is nilpotent. $\endgroup$ Commented Nov 3, 2023 at 1:00
  • $\begingroup$ Why is that? That is not necessarily true. If $R=k[x,y]/(xy)$ then $x$ is a zero divisor but not nilpotent $\endgroup$
    – kubo
    Commented Nov 3, 2023 at 10:18

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