13
$\begingroup$

Let $R$ be a commutative ring. If I am not mistaken, there is the following fact:

For a finitely generated abelian group $A$, the $R$-module $A\otimes R$ is free if and only if we can write the torsion part of $A$ as a direct sum of cyclic groups of the form $\mathbb{Z}/k$, where $k$ is invertible or zero in $R$

While elementary, I found this surprisingly tricky to prove and my proof takes about half a page. But I assume, this fact should be known. As I need it for a paper of mine, I want to ask whether someone knows a reference for it. In absence of a reference, I would also be happy with a slick 5-line proof!

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Half a page seems reasonable. There are little subtleties: e.g., for $R=\mathbf{Z}/6\mathbf{Z}$ (which is somewhat a degenerate case, since all its modules are projective), and denoting by $C_k$ the cyclic group of order $k$, $A\otimes R$ is free for $A=C_2\times C_3$ but not for $C_2\times C_3^2$. $\endgroup$
    – YCor
    Nov 21, 2017 at 16:45
  • 2
    $\begingroup$ If you had asked when the module is projective the answer would be much cleaner, because projectivity is preserved and reflected by direct sums ($M = \bigoplus M_i$ is projective iff each $M_i$ is projective), but freeness isn't reflected. After tinkering for a bit I agree that half a page is reasonable. $\endgroup$
    – Qiaochu Yuan
    Nov 21, 2017 at 17:03
  • $\begingroup$ @Qiaochu: if you replace "free" by "projective", the question boils down to: for which primes $p$, integer $m\ge 1$, and commutative ring $R$, is $R/p^kR$ a projective module? What do you call a "much cleaner answer" in this case? I don't see how to do without referring to decomposition of $R$ as a product. $\endgroup$
    – YCor
    Nov 21, 2017 at 17:52
  • 1
    $\begingroup$ $R/nR$ is a projective $R$-module iff the map $R \to R/nR$ splits iff $(n) = (n^2)$ in $R$. $\endgroup$
    – Qiaochu Yuan
    Nov 21, 2017 at 17:54
  • 1
    $\begingroup$ ... iff $nR$ is generated by an idempotent. $\endgroup$
    – YCor
    Nov 21, 2017 at 18:12

1 Answer 1

Reset to default
19
$\begingroup$

Here is a 7-line proof of your statement.

Claim. Let $R$ be a commutative ring with identity $1_R$. Let $A \simeq \mathbb{Z}/d_1\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/d_k\mathbb{Z}$ be a finitely generated Abelian group given with its invariant factor decomposition, i.e., $d_i \ge 0, d_i \neq 1$ and $d_i$ divides $d_{i + 1}$ for every $1 \le i \le k - 1$ (note that $d_i = 0$ is allowed).

Then $A \otimes_{\mathbb{Z}} R$ is free over $R$ if and only if either $d_i 1_R$ is zero or a unit of $R$ for every $i$.

Proof (7 lines). Let $M \Doteq A \otimes_{\mathbb{Z}} R \simeq R/d_1 R \oplus \cdots \oplus R/d_k R$. If $d_i 1_R$ is zero or a unit of $R$ for every $i$, then $M$ is clearly free over $R$. Assume now that $M$ is free over $R$. We reason by contradiction and consider the smallest $j \ge 1$ such that $d_j 1_R$ is neither zero nor a unit. Then we have $M \simeq R/d_j R \oplus \cdots \oplus R/d_k R$. Considering $M/\mathfrak{m}M$ for a maximal ideal $\mathfrak{m}$ which contains $d_jR$, we see that the minimal number of generators of $M$ over $R$ is $s \Doteq k -j + 1$. Therefore $M \simeq R^s$ and consequently $d_jM$ cannot be generated by less than $s$ elements. As $d_jM \simeq (d_jR/d_{j + 1} R) \oplus \cdots \oplus (d_jR/d_k R)$, we get a contradiction.

You may turn the above proof into a one-liner if you refer to this more general result of Irving Kaplansky [1, Theorem 9.3].

Kaplansky's Theorem. Let $R$ be a ring in which every one-sided ideal is two-sided. In particular, $R$ may be any commutative ring. Suppose the $R$-module $M$ is isomorphic to the direct sum of cyclic modules $R/S_1, \dots, R/S_m$, and also to the direct sum of $R/T_1, \dots, R/T_n$, where $S_i,\, T_¡$ are ideals each containing its successor, $S_1, T_1\neq R$. Then $m = n$ and $S_i = T_i$ for all $i$.

Addendum. I found the discussion of Qiaochu Yuan and YCor about projective modules (see comments above) particularly enlightening. This is why I would like to assemble all its fragments here.

Qiaochu Yuan's Claim on projectives. Let $R$ be a commutative ring with identity and let $A$ be a direct sum of cyclic groups $\mathbb{Z}/k_i\mathbb{Z}$ where $k_i \in \mathbb{Z}$ and $i$ ranges in some arbitrary set. Then $A \otimes_{\mathbb{Z}} R$ is a projective module over $R$ if and only if $k_iR = k_i^2R$ for every $i$.

Proof. Let $M \Doteq A \otimes_{\mathbb{Z}} R \simeq \bigoplus_i R/k_i R$. Then $M$ is projective over $R$ if and only if $R/k_iR$ is projective for every $k$. This holds if and only if the natural map $R \twoheadrightarrow R/k_iR$ splits for every $i$, which is in turn equivalent to the fact that $k_iR$ is an idempotent ideal. As observed by YCor, $k_iR$ is generated by an idempotent in this case, see e.g, [2, Exercise 2.1], a classical application of Nakayama's lemma.

If $\mathbb{Z}1_R \simeq \mathbb{Z}/n\mathbb{Z}$ for some $n > 1$, then the condition $\text{gdc}(k, n) = \text{gdc}(k^2, n)$, or equivalently $\text{gcd}(k, \frac{n}{\text{gcd}(n, k)}) = 1$, implies $kR = k^2R$. This is certainly a necessary condition when $R = \mathbb{Z}/n\mathbb{Z}$.

YCor outlined the fact that every module over $\mathbb{Z}/6\mathbb{Z}$ is projective. He further extends this remark by mentioning below that every module over a commutative ring $R$ with identity is projective if and only if $R$ is the direct product of finitely many fields, see this MO post for references and a more general statement.

Remark on projective modules over $\mathbb{Z}/n\mathbb{Z}$. Let $n$ be a positive integer. Then the following are equivalent:

  • The integer $n$ has no square factor.

  • Every module over $\mathbb{Z}/n\mathbb{Z}$ is projective.

Proof. By the first Prüfer Theorem, a module over $R = \mathbb{Z}/n\mathbb{Z}$ is a direct sum of cyclic factors $\mathbb{Z}/k\mathbb{Z}$ where $k$ divides $n$. If $n$ is square-free, any such factor is projective over $R$ by Qiaochu Yuan's Claim. Hence any module over $R$ is projective. If $n$ has a square factor $k^2$ with $k > 1$, then $R/kR$ is not projective over $R$ by Qiaochu Yuan's Claim.

[1] I. Kaplansky, "Elementary divisors and modules", 1949.
[2] H. Matsumura, "Commutative Ring Theory", 1986.

Improve this answer
$\endgroup$
9
  • $\begingroup$ Very nice, thank you! May I include this in my paper, of course with proper attribution to you? $\endgroup$
    – Lennart Meier
    Nov 22, 2017 at 9:47
  • 2
    $\begingroup$ It's an additional half-line, but the contradiction comes with $d_jM\simeq (d_jR/d_{j+1}R)\oplus\cdots\oplus (d_jR/d_kR)$. $\endgroup$
    – YCor
    Nov 22, 2017 at 9:53
  • $\begingroup$ @YCor Thanks, I followed your advice in my last edit. Another thing which might not be crystal clear at first sight: $d_j R\simeq R/I$ for some proper ideal $I$. Thus $d_jR^s$ surjects onto $(R/\mathfrak{m})^s$ for some maximal ideal $\mathfrak{m}$ containing $I$. As a result, $d_jR^s$ cannot be generated by less than $s$ elements. It has been omitted because it is too similar to a previous observation. $\endgroup$
    – Luc Guyot
    Nov 22, 2017 at 17:22
  • $\begingroup$ @LennartMeier Sure, I'll be honored! $\endgroup$
    – Luc Guyot
    Nov 22, 2017 at 17:29
  • 1
    $\begingroup$ I didn't claim what you call "YCor's claim" (which is true and an easy particular case of the fact that for a commutative ring $A$, every $A$-module is projective iff $A$ is a finite product of fields, see mathoverflow.net/questions/62464) $\endgroup$
    – YCor
    Nov 22, 2017 at 23:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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