Are $I\otimes_{R}J$ and $IJ$ isomorphic as $R$-modules?

Question

Let $I$ and $J$ be ideals of a commutative ring $R,$ considered as $R$-modules. The product ideal $$IJ=\left\{\sum_{i=1}^{n}a_{i}b_{j}:n\in\mathbb{N},a_{i}\in I,b_{j}\in J\right\}$$ is also an $R$-module.

Question: Is it true that $I\otimes_{R}J\cong IJ?$

This seems intuitive to me, since I like to think of the tensor product $M\otimes_{R}N$ as "sums of things of the form $m\otimes n$ where $m\in M$ and $n\in N,$" and this sounds a lot like a description of $IJ.$

I came up with the following (now seen to be incorrect) argument, which originally I had posted as an answer:

Let $f\colon I\times J\to IJ$ be given by $(a,b)\mapsto ab.$ This is an $R$-bilinear map, so there exists a (unique) corresponding $R$-linear map $h\colon I\otimes_{R} J\to IJ$ with $h(a\otimes b)=ab.$ We would like to prove that this map $h$ is bijective.

To prove that $h$ is surjective: the element $\sum_{i=1}^{n}a_{i}b_{j}\in IJ$ is mapped to by $\sum_{i=1}^{n}a_{i}\otimes b_{j}.$

To prove that $h$ is injective, we consider its kernel. Suppose $x=\sum_{i=1}^{n}a_{i}\otimes b_{j}\in I\otimes_{R}J$ is such that $h(x)=0.$ Letting $y=\sum_{i=1}^{n}a_{i}b_{j}\in IJ,$ we see that $$x = \sum_{i=1}^{n}(a_{i}b_{j}\otimes1) = \left(\sum_{i=1}^{n}a_{i}b_{j}\right)\otimes1 = y\otimes 1.$$ Therefore $h(y\otimes1) = 0.$ But by the defining property of $h$ we have $h(y\otimes 1)=y\cdot1=y,$ whence $y = h(x) =0.$ Therefore we must have had $x=0\otimes 1=0$ all along, so the kernel of $h$ is trivial and this completes the proof.

However, the above argument does not work, as pointed out to my by @xyzzyz, since we don't know that $1\in J$ (and indeed this is true if and only if $J=R,$ which changes the nature of the question).

Answer 1

This is not true. For instance, let $k$ be a field, $R=k[x]/(x^2)$, and $I=J=(x)\subset R$. Then $IJ=(x^2)=0$. But we also have $I\cong R/I$ (since $I$ is the annihilator of $x$ and $x$ generates $I$), so $I\otimes_R J\cong I\otimes_R R/I\cong I/I^2\cong I.$

In general, the map $I\times J\to IJ$ given by $(i,j)\mapsto ij$ induces a homomorphism $f:I\otimes_R J\to IJ$ which is surjective but need not be injective. Indeed, tensoring the short exact sequence $0\to I\to R\to R/I\to 0$ with $J$ gives a long exact sequence $$\operatorname{Tor}_1^R(R,J)\to\operatorname{Tor}_1^R(R/I,J)\to I\otimes_R J\stackrel{g}\to R\otimes_R J\to R/I\otimes_R J\to 0$$ where the map $g$ is just the composition of $f$ with the inclusion map $IJ\to J\cong R\otimes_R J$, so $f$ has the same kernel as $g$. Since $R$ is flat as an $R$-module, $\operatorname{Tor}_1^R(R,J)=0$, so the kernel of $g$ is $\operatorname{Tor}_1^R(R/I,J)$. Thus the map $f$ is an isomorphism iff $\operatorname{Tor}_1^R(R/I,J)=0$.