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Ok, I'm reading some thesis of some former students, and come up with this proof, but it doesn't really look good to me. So I guess it should be wrong somewhere. So, here it goes:

Let $R$ be a unitary commutative ring, and $I$ be its ideal, and $N$ is an $R-$module. I'll now show that: $I \otimes_R N \cong IN$.

First, since $1 \in R$, and $I$ is an ideal of $R$, we must have that $I = RI$.

So: $I \otimes_R N = RI \otimes_R N = R \otimes_R IN \cong IN$.

Ok, this must be ssssoooo wrong, since if this proof is true, then I can prove that any $R-$module is flat...

How come? :((((((

Thank you guys so much,

And have a good day,

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Very nice! It reminds me of the paradoxes we read in high-school purporting to prove that all triangles are isoceles or that $0=1$ (by cleverly dividing by some number which was zero cleverly disguised). We need such brain teasers in more advanced mathematics too! – Georges Elencwajg Jun 3 '13 at 19:01
Despite this being an elementary question, I think it's really good. This can really confuse someone (I know it's gotten me at least once). The distinction between an ideal-as-subset and ideal-as-abstract-module is subtle. – Ryan Reich Jun 3 '13 at 19:43
up vote 20 down vote accepted

The equality $RI\otimes_R N=R\otimes_R IN$ is very subtly false: the point is that it does not hold in $I\otimes_RN$, which is the only place where it could hold.

But, since tensor product is $R-$bilinear, can't we write (for example) $1\cdot i\otimes n=1\otimes i\cdot n \:$?
No, we can't! Because $1\otimes i\cdot n$ does not make sense in $I\otimes_RN$, since $1\notin I$ and thus $1$ may not be put on the left-hand side of $\otimes_R$ in $I\otimes_RN$.

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Yup, I get it now. Thanks very much. :* But I still have one concern left, it's on page 2 of this text (i.e page 148)…, the third paragraph "where $\lambda \otimes 1$ is an epimorphism, with kernel $r(\lambda)A$", how can the author conclude that? I can manage to have $\mbox{ker} \lambda \otimes 1 = r(\lambda) \otimes A$, but I don't think that they are congruent. Is this false? – user49685 Jun 3 '13 at 19:05
Very nicely explained. +1 – DonAntonio Jun 3 '13 at 19:23
Dear user49685, the epimorphism claim in the linked article results from the right exactness of the tensor product. Translated in your notation (so that users don't have to go to the article) it says that for a fixed $r \in R$ and for the principal ideal $I=rR\subset R$, the morphism $N\to I\otimes_R N: n\mapsto r\otimes n$ is surjective. Nothing wrong with that! – Georges Elencwajg Jun 3 '13 at 19:24
As for the assertion on the kernel: If $J$ denotes the kernel of $r\cdot:R\to I$ we have $ 0\to J\to R\to I\to 0$ and we get (by right exactness of $\otimes_R$) the exact sequence $J\otimes _RN\to N=R\otimes_R N \to I\otimes_R N \to 0$ and the kernel of the second map, which is what you are asking about, is the image $J\cdot N$ of the first map, just as the author claims. – Georges Elencwajg Jun 3 '13 at 19:50
Nice! Given the second paragraph, it seems to me that what you are actually explaining is why the equality $RI \otimes_R N=R \otimes_R IN$ fails. (The other equality, suitably interpreted, seems fine to me). – Stephen Jun 3 '13 at 20:11

It is not true that $RI\otimes N = R\otimes IN$. You can't "factor out" elements of the ideal linearly because the ideal is being thought of as a module over $R$, and $1$ is (generally) not in the ideal. Try to write down what you think the isomorphism should be and you'll see it.

For concreteness, take $I=(a)$ where $a$ is some non-invertible element in some commutative ring $R$. What you're claiming is $(a)\otimes N\cong a(R\otimes N)$, which is clearly nonsense. For example $a\otimes n\neq a(1\otimes n)$, since $1\notin I$.

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Yes, thank you very much for pointing that out, but I have one question left, on page 2 of this text (i.e page 148)…, the third paragraph "where $\lambda \otimes 1$ is an epimorphism, with kernel $r(\lambda)A$", how can the author conclude that? I can manage to have $\mbox{ker} \lambda \otimes 1 = r(\lambda) \otimes A$, but I don't think that they are congruent. – user49685 Jun 3 '13 at 19:02
@user49685 You should ask that as a new question. – Potato Jun 3 '13 at 19:11
+1 Very nice indeed. – DonAntonio Jun 3 '13 at 19:29
@MarcvanLeeuwen Yes, sorry. – Potato Jun 4 '13 at 17:15

While it's true that in $M\otimes_RN$ you can do $$ xr\otimes y=x\otimes ry $$ you can't "exchange ideals" across the tensor product. A simple example should make this clear: set $R=\mathbb{Z}$, $I=2\mathbb{Z}$ and $N=\mathbb{Z}/2\mathbb{Z}$. Since, as $\mathbb{Z}$-modules we have $\mathbb{Z}\cong 2\mathbb{Z}$, we have $$ \mathbb{Z}\otimes N\cong (2\mathbb{Z})\otimes N $$ which is isomorphic to $N$. On the other hand, $IN=0$, because $I=2\mathbb{Z}$ is precisely the annihilator of $N$. So your reasoning is faulty to begin with.

When you have doubts about tensoring, drawing a commutative diagram can help.

Consider the exact sequence $0\to I\to R\to R/I\to 0$ and tensor it with $N$: you get the diagram with exact rows $$\require{AMScd} \begin{CD} 0@>>>\mathrm{Tor}^R_1(R/I,N)@>>> I\otimes_R N@>>> R\otimes_R N@>>> (R/I)\otimes_R N@>>> 0 \\ @. @. @VVV @VVV @VVV \\ {} @. 0 @>>> IN @>>> N @>>> N/IN @>>> 0 \end{CD} $$ where $R\otimes_R N\to N$ and $(R/I)\otimes_R N\to N/IN$ are isomorphisms and $I\otimes_R N\to IN$ is surjective. This means that the kernel of $I\otimes_RN\to IN$ is isomorphic to $\mathrm{Tor}^R_1(R/I,N)$ which is not necessarily zero.

It doesn't matter whether you know about Tor; just remember that tensoring doesn't (necessarily) preserve monomorphisms, so you can simply put in the kernel $K$ of the induced morphism $I\otimes N\to R\otimes N$. You now see clearly that proving $I\otimes N\to IN$ being an isomorphism is equivalent to proving that $K=0$.

In the example, we have $IN=0$, so clearly $I\otimes N$ is not isomorphic to $IN$. Actually, the induced morphism $I\otimes N\to R\otimes N$ is the zero map.

Saying that $\mathrm{Tor}^R_1(R/I,N)=0$ for all ideals $I$ is just saying that every $R$-module is flat.

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I think OP already realised that the given "proof" would imply that every $R$-module is flat, and that for this reason it could not possibly be correct. The question however was where the proof went wrong, and I don't see how this answer addresses that question. – Marc van Leeuwen Jun 5 '13 at 7:14
@MarcvanLeeuwen The "proof" has nothing of a proof; saying $RI\otimes_R N=R\otimes_R IN$ is simply utterly wrong. – egreg Jun 5 '13 at 8:31
I agree, but the question is clearly to explain that (as others have done); OP was obviously confused. Saying "utterly" does not take away the confusion or explain anything. – Marc van Leeuwen Jun 5 '13 at 8:36
@MarcvanLeeuwen I added an example. – egreg Jun 5 '13 at 9:07
A very simple and useful example, +1. Thanks. – Marc van Leeuwen Jun 5 '13 at 9:16

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