Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
Is it not generated by the generators of $I$ that are in $A$ already? –  Joe Tait Feb 21 '13 at 13:20
    
No, for example, if $a+bX$ is in $I$ and $b^2=0$, then $(a+bX)(a-bX) = a^2-b^2X^2=a^2\in I\cap A$. –  Lior B-S Feb 21 '13 at 13:22
    
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 $\{\phi(a_1),\ldots,\phi(a_n)\}$. –  J. Gaddis Feb 21 '13 at 13:38
    
@linearfish: It is surely contained in the ideal you mention, but only seldom is it actually the same. –  tomasz Feb 21 '13 at 13:40
    
@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 $(1 + x)$. –  Jim Feb 21 '13 at 17:01
add comment

1 Answer

up vote 7 down vote accepted

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.

share|improve this answer
2  
The first paragraph repeats the well-known fact that the kernel of the localization $A \to A_t$ is $\cup_n \mathrm{Ann}(t^n)$. –  Martin Brandenburg Feb 21 '13 at 16:35
    
+1 for this answer. –  user26857 Feb 22 '13 at 1:42
    
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 $A[X]/(1-tX) \to A_t$. –  Lior B-S Feb 22 '13 at 6:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.