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:
- (Bilinearity) For all $a,a',b\in M$ and all $r\in R$, $$\langle a+a',b\rangle=\langle a,b\rangle+\langle a',b\rangle\\ \langle ra,b\rangle=r\langle a,b\rangle\text{;}$$
- (Symmetry) For all $a,b\in M$, $$\langle a,b\rangle=\langle b,a\rangle\text{;}$$
- (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
\begin{equation}
\langle a\otimes b,a'\otimes b'\rangle = \langle a,a'\rangle\langle b,b'\rangle
\end{equation}
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
\begin{equation}
h(\alpha)(\beta)=\langle\alpha,\beta\rangle
\end{equation}
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 $x\in M,\ y\in N$,
\begin{equation}
\langle a,x\rangle\langle b,y\rangle=\varphi(x\otimes y)\text{.}
\end{equation}
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 $x\in M,\ y\in N$,
\begin{equation}
\langle a_1,x\rangle\langle b_1,y\rangle+\cdots+\langle a_m,x\rangle\langle b_m,y\rangle=0\text{.}
\end{equation}
This was all I could do.
Best Answer
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: $$\require{AMScd} \def\diaguparrow#1{\smash{\raise.6em\rlap{\scriptstyle #1} \lower.3em{\mathord{\diagup}}\raise.52em{\!\mathord{\nearrow}}}} \begin{CD} && (M\otimes_RN)^{\ast}\\ & \diaguparrow{f} @AAhA \\ M^{\ast}\otimes_RN^{\ast} @<<h_M\otimes h_N< M\otimes_RN \end{CD}.$$ 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.