I am trying to prove the following:
Let $R$ be a field and $M,N,L$ be vector spaces over $R$. We have a pairing: \begin{align*} \langle .,. \rangle: M \times N \rightarrow L \end{align*} We can also consider a pairing as a $R$-linear map, i.e. homomorphism: \begin{align*} &\Phi_1: M \rightarrow \hom_R(N,L)\\ & m \mapsto (n \mapsto \langle m,n \rangle) \end{align*} or as the $R$-linear map: \begin{align*} &\Phi_2: N \rightarrow \hom_R(M,L)\\ & n \mapsto (m \mapsto \langle m,n \rangle) \end{align*} We call the pairing perfect if $\Phi_1$ is an isomorphism. Now I am trying to prove that if $M,N,L$ are finite dimensional: $\Phi_1$ is an isomorphism iff $\Phi_2$ is an isomorphism.
This is how far I came: Suppose $\Phi_1$ is an isomorphism. To show that $\Phi_2$ is injective, we need to show that the kernel of $\Phi_2$ is zero. So we want to show that if $\langle m,n \rangle=0$ for all $m$, $n$ must be zero (non-degeneracy). Suppose $\langle m,n \rangle=0$ for all $m$, then $\Phi_1(m)(n)=0$ for all $m$. Now if $n\neq 0$, and $L$ is a non-zero vector space there must be a non-zero constant function $\psi \in \hom_R(N,L)$ such that $\psi(n) \neq 0$. But because $\Phi_1$ is surjective, $\psi=\Phi_1(m)$ for some $m$, so $\psi(n)=\Phi_1(m)(n)=0$. So $n$ must be zero. Therefore $\Phi_2$ is injective.
To show that $\Phi_2$ is surjective, I want to use a dimension argument, if $\dim(M)=\dim(N)$ we are done. But we only know that $\dim(M)=\dim(\hom(N,L)) \geq\dim(N)$ and $\dim(N) \leq \dim(\hom(M,L))$ since $\Phi_2$ is injective. And also: What if $L$ is a zero vectorspace? Then assuming that $\Phi_1$ is an isomorphism, $M$ must be zero, but $N$ can still be non-zero.
It feels like the thing I am trying to prove only holds for the case that $L=R$ (this would solve the above problems since if $L=0$,then $M,N=0$ and $\dim(M)=\dim(N)$.) If this is the case, can anyone give a counterexample to the case where $L \neq R$?
This can only work if $L=R$ or $L=\{0\}$. Suppose it is not. You are assuming that $\Phi_1$ is an isomorphism and you want to prove that $\Phi_2$ is an isomorphism too. Suppose that this is so. Let $m=\dim M$, $n=\dim N$, and $l=\dim L$. From the fact that $\Phi_1$ is an isomorphism, you get that $m=nl$ and from the fact that $\Phi_2$ is an isomorphism you get that $n=ml$. But $l>1$ and so this is impossible, unless $m=n=0$.