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In my class of algebraic topology, a friend of mine stated the following:

If $R\ne 0$ is a commutative ring with unit and $I\subset R\oplus R$ is a submodule such that $(R\oplus R)/I\cong R$, then $I\cong R$ as $R$-modules.

I believe that it is true for PIDs, but (if it is true) how to prove it for any commutative ring?

Thank you!

user26857
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Renan Mezabarba
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4 Answers4

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If it is an isomorphism of $R$-modules, we have the exact sequence $$ 0 \to I \to R \oplus R \to R \to 0 $$ where $R \oplus R \to R$ is the map $R \oplus R \to (R \oplus R)/I \simeq R$. Since $R$ is a free $R$-module, it is also projective, hence this sequence splits ; this means we have an isomorphism $R \oplus R \simeq I \oplus R$. Since this means $I$ is a direct summand of a free module, this proves that $I$ has to be projective.

In the case of PIDs, since any projective module is free, we are done. In general, the implication "projective $\Rightarrow$ free" is false. It is also true in the case of a local ring (not easy to prove that projective modules over a local ring are free), so at least you can say $I_P \simeq R_P$ for all prime ideals $P \in \mathrm{Spec} \, R$ (since the localization of a projective module is projective over the localization ; this is an easy statement). Therefore, in full generality, I think the best statement you can get is that $I$ is locally free in the sense defined above ($I_P \simeq R_P$ for all $P$).

Edit : In the comments lies a sketch of a proof that if $R$ is a noetherian integral domain, then $I \simeq R$.

Second edit : Lemma 19.18 in Eisenbud's "Commutative Algebra with a view towards Algebraic Geometry" is what you want. The proof is easy if you assume that projective modules over a local ring are free. Thanks to user26857 for that, I learned something today!

Hope that helps,

Patrick Da Silva
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  • Do you have an example when the cancellation property for projectives (in this concrete situation) fails by any chance? – Pavel Čoupek May 25 '15 at 20:17
  • You have shown that $I$ is stably free, which in general is much stronger than projective (= locally free). – Julian Rosen May 25 '15 at 20:17
  • What "cancellation property" are you talking about? – Patrick Da Silva May 25 '15 at 20:17
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    @Julian Rosen : projective and locally free are not the same thing. The wikipedia page gives you the example of a boolean ring which is locally free and non-projective. But yeah, stably free implies locally free, although not projective. – Patrick Da Silva May 25 '15 at 20:18
  • Oh, right. I guess this is only true for finitely generated modules. – Julian Rosen May 25 '15 at 20:21
  • Well, it seems that you claim (at least if I understand correctly) that one can get only some projective in general, and not $R$. That means at least that there should be a ring $R$ and a projective $Q$ such that $R\oplus R \simeq R \oplus Q \not \Rightarrow R \simeq Q$. – Pavel Čoupek May 25 '15 at 20:26
  • @PavelČoupek : I honestly don't have anything in mind. The only scenario in which I explicitly worked out projective modules were Dedekind domains, and in this case the cancellation law you described actually holds. This is because if $A \oplus Q \simeq A^2$, then $$ A = \Lambda^2(A^2) \simeq \Lambda^2(A \oplus Q) \simeq \Lambda^1(A) \otimes_A \Lambda^1(Q) \simeq Q. $$ (I used the second exterior power, in case you don't recognize the notation. Note that projective modules $M$ of rank $n$ satisfy $\Lambda^m(M) = 0$ for $m > n$.) – Patrick Da Silva May 25 '15 at 20:44
  • @PavelČoupek : I checked out my Dedekind domain notes, the only thing I need in this case is that $Q$ is projective of rank $1$ (which follows from $A \oplus A \simeq A \oplus Q$) and that $A$ is a noetherian integral domain, so that $\Lambda^2 A = \Lambda^2 Q = 0$. – Patrick Da Silva May 25 '15 at 21:02
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    @PavelČoupek $R\oplus R\simeq R\oplus Q\Rightarrow R\simeq Q$ for any commutative ring $R$ and any $R$-module $Q$. – user26857 May 25 '15 at 21:04
  • @user26857 : Any proof? I saw you delete your last answer! Note that $\Lambda^2 Q = 0$ holds only for a noetherian integral domain $R$ and a projective rank $1$ $Q$ in general. If you know it holds more generally I'd be interested! – Patrick Da Silva May 25 '15 at 21:05
  • http://math.stackexchange.com/a/1285560/121097 – user26857 May 25 '15 at 21:06
  • @user26857 : I have a problem with your link, I'm not sure it's correct! – Patrick Da Silva May 25 '15 at 21:07
  • Try a little harder (you spent less than one minute to think about it!) or check Eisenbud, CA, Lemma 19.18. (Btw, meanwhile @JulianRosen posted a full proof of the OP's result.) – user26857 May 25 '15 at 21:09
  • @user26857 : Right. I confused my memory with the statements of my notes, so I was doubting your statement. But you're right! (P.S. : I deleted my comment about Dedekind domains and the class group, that was me confusing statements again!) – Patrick Da Silva May 25 '15 at 21:22
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    You can also check Matsumura, CRT, for an elementary proof when $R$ is an integral domain. (This is one of the main ingredients for a proof that regular local rings are UFDs.) – user26857 May 25 '15 at 21:27
  • @PatrickDaSilva At first I chose your answer, because it seemed simpler to me (at the beginning the result appeared to be false in general). However, Rosen's answer is more basic, which is preferable for me, since I'm not a algebraist. Anyway, thank you. – Renan Mezabarba May 25 '15 at 22:03
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    @RenanManeliMezabarba : I agree with you, so no offense taken ;) – Patrick Da Silva May 25 '15 at 22:14
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Let $\varphi$ denote the surjection $R\oplus R\to (R\oplus R)/I\cong R$. Define $a=\varphi(1,0)$, $b=\varphi(0,1)\in R$. Then $\varphi(x,y)=ax+by$. Since $\varphi$ is surjective, we can find $x_0$, $y_0\in R$ with $ax_0+by_0=1$.

Define $\psi:R\to R\oplus R$ by $\psi(r)=(-b r,a r)$. Now:

  • $\varphi\circ\psi=0$, so the image of $\psi$ is contained in $I$,

  • The map $R\oplus R\to R$, $(r,s)\mapsto x_0 s-y_0 r$ is left-inverse to $\psi$, so $\psi$ is injective, and

  • For every $(r,s)\in I$, we have $ar+bs =0$, so $$ \psi(-y_0 r+x_0 s)= (by_0r-bx_0 s,-ay_0r+ax_0 s) = (by_0r+ax_0r, by_0s+ax_0 s) = (r,s), $$ and we conclude $\psi$ maps onto $I$.

We have shown $\psi$ is an isomorphism of $R$ onto $I$.

Julian Rosen
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The conclusion holds for any commutative ring.

We have a short exact sequence of $R$-modules $$0\to I\to R\oplus R\to (R\oplus R)/I\to 0.$$ Since $(R\oplus R)/I\simeq R$ the above sequence is split, so $R\oplus R\simeq R\oplus I$. This gives $$\bigwedge^2(R\oplus R)\simeq\bigwedge^2(R\oplus I)\Rightarrow R\simeq I.$$ (For the last isomorphism see this answer.)

Edit. The arguments above can be easily generalized in order to show (for $n\ge2$) $$R^n/I\simeq R^{n-1}\Rightarrow I\simeq R,$$ that is, stably-free ideals are free.

Community
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user26857
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It's certainly not true without further assumptions: take $R=\prod_{i\in\mathbb{N}} \mathbb{Z}$, and $I=\mathbb{Z}$.

EDIT: See Pavel's comment; this answer is not correct unless the isomorphism $(R\oplus R)/I\cong R$ is just an isomorphism of rings.

If we demand an isomorphism of modules, then the statement is true: see http://en.wikipedia.org/wiki/Invariant_basis_number.

FURTHER EDIT: Now I'm not sure that my previous edit is really true; oh well. I'm leaving this answer, with edits, up for posterity, but this should not be accepted.

Noah Schweber
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  • Yes, I meant "$\prod$" - I'm tired. And $I$ is supposed to be one of the factors of $R$, so $R\cong I\times R$. – Noah Schweber May 25 '15 at 19:56
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    Then it still does not work, basically since $R\oplus R \not\simeq R$ as $R$-modules. In fact, this is a property of all commutative rings (so-called invariant basis number property). – Pavel Čoupek May 25 '15 at 19:58
  • Ah, I interpreted "$R\oplus R/I\cong R$" to mean that the isomorphism was an isomorphism of rings. I'm leaving this answer up, just so - with your comment - it is clear that it matters where the isomorphism lives. – Noah Schweber May 25 '15 at 19:59