# Is the bilinear map on a tensor product nondegenerate?

In this question, all rings are unital commutative, and all modules are unital (i.e., $$1\cdot a=a$$ for all module element $$a$$).

## Definition of Inner Products

Let $$M$$ be an $$R$$-module. An inner product (This terminology may not be standard) on $$M$$ is a map $$\langle\cdot,\cdot\rangle:M\times M\to R$$ such that:

1. (Bilinearity) For all $$a,a',b\in M$$ and all $$r\in R$$, $$\langle a+a',b\rangle=\langle a,b\rangle+\langle a',b\rangle\\ \langle ra,b\rangle=r\langle a,b\rangle\text{;}$$
2. (Symmetry) For all $$a,b\in M$$, $$\langle a,b\rangle=\langle b,a\rangle\text{;}$$
3. (Nondegeneracy) The linear map $$M\to M^*,\quad (b\mapsto(a\mapsto\langle a,b\rangle))$$ is bijective ($$M^*$$ is the dual module of $$M$$).

An inner product module is a module equipped with an inner product.

Let $$M$$ and $$N$$ be inner product $$R$$-modules. Let $$M\otimes N$$ be the tensor product. Then one can show that there is a unique bilinear map $$\langle\cdot,\cdot\rangle:(M\otimes N)\times(M\otimes N)\to R$$ such that $$$$\langle a\otimes b,a'\otimes b'\rangle = \langle a,a'\rangle\langle b,b'\rangle$$$$ for all $$a,a'\in M$$ and all $$b,b'\in N$$.

## Question

Is the above bilinear map on $$M\otimes N$$ an inner product? If not, is there an easy sufficient condition?

## My Attempt

I proved symmetry, but I could not prove nondegeneracy. Nondegeneracy means that the map $$h:(M\otimes N)\to(M\otimes N)^*$$ defined by $$$$h(\alpha)(\beta)=\langle\alpha,\beta\rangle$$$$ is bijective.

My attempt to prove that $$h$$ is surjective: Let $$\varphi:M\otimes N\to R$$ be a linear map. It suffices to show that there are $$a\in M,\ b\in N$$ such that for all $$x\in M,\ y\in N$$, $$$$\langle a,x\rangle\langle b,y\rangle=\varphi(x\otimes y)\text{.}$$$$ I do not know what to do next.

My attempt to prove that $$h$$ is injective: Let $$\alpha\in\ker h$$. We should show that $$\alpha=0$$. Write $$\alpha=a_1\otimes b_1+\cdots a_m\otimes b_m$$. Then for all $$x\in M,\ y\in N$$, $$$$\langle a_1,x\rangle\langle b_1,y\rangle+\cdots+\langle a_m,x\rangle\langle b_m,y\rangle=0\text{.}$$$$ This was all I could do.

I do not think you will get anywhere trying to reason explicitly with elements. That is the inconvenient nature of tensor products. Let $$h_M\colon M\rightarrow M^{\ast}$$ and $$h_N\colon N\rightarrow N^{\ast}$$ be the isomorphisms that are adjoint to the inner products on $$M$$ and $$N$$ respectively. Let $$h\colon M\otimes_RN\rightarrow(M\otimes_RN)^{\ast}$$ be the adjoint to the induced bilinear form on $$M\otimes_RN$$. To understand whether $$h$$ is an isomorphism, which is what we are interested in, means to understand its codomain $$(M\otimes_RN)^{\ast}$$ and compare $$h$$ to the maps $$h_M,h_N$$, which we know to be isomorphisms.
There is a map doing just that, the canonical map $$f\colon M^{\ast}\otimes_RN^{\ast}\rightarrow(M\otimes_RN)^{\ast}$$, which is adjoint to the map $$(M^{\ast}\otimes_RN^{\ast})\otimes_R(M\otimes_RN)\cong(M^{\ast}\otimes_RM)\otimes_R(N^{\ast}\otimes_RN)\rightarrow R\otimes_RR\cong R$$, the first isomorphism interchanging the middle two factors, the second map being the tensor product of the canonical evaluation maps and the last isomorphism being scalar multiplication. Explicitly, if $$\varphi\in M^{\ast}$$ and $$\psi\in N^{\ast}$$ and $$x\in M$$ and $$y\in N$$, then $$f(\varphi\otimes\psi)(x\otimes y)=\varphi(x)\psi(y)$$. It's a good exercise to check that the following triangle commutes: $$\require{AMScd} \def\diaguparrow#1{\smash{\raise.6em\rlap{\scriptstyle #1} \lower.3em{\mathord{\diagup}}\raise.52em{\!\mathord{\nearrow}}}} \begin{CD} && (M\otimes_RN)^{\ast}\\ & \diaguparrow{f} @AAhA \\ M^{\ast}\otimes_RN^{\ast} @< The map $$h_M\otimes h_N$$ is an isomorphism, beacuse $$h_M$$ and $$h_N$$ are. The commutativity of the diagram then implies that $$h$$ is an isomorphism if and only if $$f$$ is an isomorphism. Thus, from this perspective, the question reduces a question about $$f$$, which has nothing to do with inner products anymore.
However, the question when $$f$$ is an isomorphism is quite hard and not well-understood in general. I can't do better than refer you to this MO question, which contains both sufficient conditions and counter-examples.