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Can you give me some examples of infinitely generated modules over commutative rings, other than $A[x_1,\ldots,x_n,\ldots]$?

Thanks a lot!

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$\mathbb{Q}$ over $\mathbb{Z}$ –  user8268 Sep 21 '12 at 7:02
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Or even more boring, $\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots$ over $\mathbb{Z}$. –  Zhen Lin Sep 21 '12 at 7:27
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$A[X]$ over $A$ is enough, you don't need infitintely many variables :) –  Michalis Sep 21 '12 at 10:39
    
@Michalis Please consider converting the three comments to an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 16 '13 at 8:09

1 Answer 1

One variable is enough! $A[X]$ over $A$ is infinitely generated as an $A$-module.

Other suggestions in the comments: $\mathbb{Q}$ over $\mathbb{Z}$ or an infinite direct sum of copies of $\mathbb{Z}$ over $\mathbb{Z}$.

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@QiaochuYuan, there is no unital ring (commutative or not) that is isomorphic to a direct sum of infinitely many copies of itself. You're correct that this requires a proof, but it has nothing to do with invariant basis number. The failure of the latter property only involves isomorphisms between direct sums of finitely many copies of the ring. –  Manny Reyes Jun 17 '13 at 1:40
    
...except the zero ring, that is :) –  Manny Reyes Jun 17 '13 at 1:44
    
@Manny: hmm. The example I had in mind was non-unital. Anyway, the argument I know showing that commutative rings have IBN also shows the stronger statement that free modules of arbitrary rank are determined by their rank (namely tensor with $R/m$ where $m$ is a maximal ideal, reducing to the case of fields). I don't know how to show that there doesn't exist a unital noncommutative ring $R$ such that $R$ is isomorphic to a countable direct sum of copies of $R$ as a left $R$-module; can you sketch a proof? –  Qiaochu Yuan Jun 17 '13 at 4:51
    
@Qiaochu, sure. First notice that $R$ is finitely generated as a left $R$-module, assuming $R$ is unital. Let's show that $M = \bigoplus_{\mathbb{N}} R$ cannot be finitely generated. Assume for contradiction that it has a finite generating set. Each $m_i$ can be expressed using finitely many coordinates. Thus there exists some coordinate $k \in \mathbb{N}$ such that each $m_i$ has zero $k$th coordinate. If we let $e_k$ be the $k$th "standard basis vector" in $M$. Then it's easy to see that $e_k \notin \sum R m_i$, so $M$ is not generated by the $m_i$. –  Manny Reyes Jun 17 '13 at 11:17
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This method can be generalized to show that the rank of an infinitely generated modules over any (unital) ring is well-defined, so that IBN is only an issue for finitely generated free modules. –  Manny Reyes Jun 17 '13 at 11:24

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