My proof of $I \otimes N \cong IN$ is clearly wrong, but where have I gone wrong?

Question

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 an 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 then $$I \otimes_R N = RI \otimes_R N = R \otimes_R IN \cong IN\,.$$

But this must be so wrong, since if this proof is true, I could prove that any $R$-module is flat. So how come this proof is wrong?

Answer 1

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 $I\otimes_RN$.

Answer 2

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 $1\notin I$.

Answer 3

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.