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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}$2012-09-21
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    Or even more boring, $\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots$ over $\mathbb{Z}$.2012-09-21
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    $A[X]$ over $A$ is enough, you don't need infitintely many variables :)2012-09-21
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    @Michalis Please consider converting the three comments to an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-16

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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.2013-06-17
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    ...except the zero ring, that is :)2013-06-17
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    @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?2013-06-17
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    @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 $m_i$.2013-06-17
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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.2013-06-17
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    @Manny: thanks! That makes sense.2013-06-17