Let $A$ be a commutative ring with $1$ and $A[X]$ the ring of polynomials in one variable over $A$. Assume $I$ is a finitely generated ideal of $A[X]$. My question is

Is $I\cap A$ necessarily finitely generated?

(If $A$ has zero divisors, I couldn't even prove that if $I=(f)$ is principal, then $I\cap A$ is finitely generated.)

  • $\begingroup$ Is it not generated by the generators of $I$ that are in $A$ already? $\endgroup$
    – Joe Tait
    Feb 21, 2013 at 13:20
  • 1
    $\begingroup$ No, for example, if $a+bX$ is in $I$ and $b^2=0$, then $(a+bX)(a-bX) = a^2-b^2X^2=a^2\in I\cap A$. $\endgroup$
    – Lior B-S
    Feb 21, 2013 at 13:22
  • $\begingroup$ Let $\phi:A[X] \rightarrow A$ be the map sending $a+bX \mapsto a$. If $\{a_1,\ldots,a_n\}$ are the generators of $I$, then $I \cap A$ should be finitely generated by the elements $\{\phi(a_1),\ldots,\phi(a_n)\}$. $\endgroup$
    – J. Gaddis
    Feb 21, 2013 at 13:38
  • $\begingroup$ @linearfish: It is surely contained in the ideal you mention, but only seldom is it actually the same. $\endgroup$
    – tomasz
    Feb 21, 2013 at 13:40
  • $\begingroup$ @linearfish: $f \in I$ doesn't imply $\phi(f) \in I$. Take for example the principal ideal $(1 + x) \in \mathbb C[x]$. $\phi(1 + x) = 1$ is not in $(1 + x)$. $\endgroup$
    – Jim
    Feb 21, 2013 at 17:01

1 Answer 1


A counterexample: Let $f = tX-1$, where $t \in A$. Given $g = \sum_{i=0}^n a_i X^i \in A[X]$, then $fg = -a_0 +(ta_0-a_1)X + \ldots +(ta_{n-1}-a_n)X^n +(ta_n) X^{n+1}$. Thus $a \in A$ is $fg$ for some $g$ if and only if there are $a_0,\ldots, a_n \in A$ such that $a = -a_0$, $a_1 = ta_0$, ..., $a_n = ta_{n-1}$, $0 = ta_n$. This is equivalent to $t^{n+1}a = 0$. Thus $A \cap (f) = \bigcup_{n=1}^\infty\mathrm{Ann}_A(t^n)$. It now suffices to find an example of $A$ and $t \in A$ such that $\bigcup_{n=1}^\infty\mathrm{Ann}_A(t^n)$ is not finitely generated.

Let $K$ be a field; for every $i \in \mathbb{N}$ let $A_i = K[T]/(T^i)$ and let $t_i \in A_i$ be the image of $T$. Then $\mathrm{Ann}_{A_i}(t_i^n)$ is $A_i$, if $n \ge i$, and a proper ideal of $A_i$, if $n < i$. Let $$A = \prod_{i=1}^\infty A_i = \{(a_1,a_2, \ldots) \| a_i \in A_i, \ i = 1, 2, \ldots \}$$ (with addition and multiplication coordinate-wise) and let $t = (t_1, t_2, \ldots) \in A$. Then $\mathrm{Ann}_A(t^n) = \prod_{i=1}^\infty \mathrm{Ann}_{A_i}(t_i^n)$. Clearly $\mathrm{Ann}_A(t^{n-1}) \subset \mathrm{Ann}_A(t^{n})$ for every $n$, and the inclusion is strict, because both sides have distinct projections on the $n$-th coordinate. If $\bigcup_{n=1}^\infty\mathrm{Ann}_A(t^n)$ were finitely generated, we would have $\bigcup_{n=1}^\infty\mathrm{Ann}_A(t^n) = \mathrm{Ann}_A(t^k)$ for some $k$ (such that the right hand side contains the generators). A contradiction.

  • 2
    $\begingroup$ The first paragraph repeats the well-known fact that the kernel of the localization $A \to A_t$ is $\cup_n \mathrm{Ann}(t^n)$. $\endgroup$ Feb 21, 2013 at 16:35
  • $\begingroup$ On behalf of Haran: @Martin Brandenburg: It would be more appropriate to say that the first paragraph proves that the kernel of $A \to A[X]/(1-tX)$ is the kernel of $A \to A_t$; but this immediately follows from the isomorphism $A[X]/(1-tX) \to A_t$. $\endgroup$
    – Lior B-S
    Feb 22, 2013 at 6:05

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