How to show that if $V$, or $W$ are finite dimensional vector spaces over $k$, then $V^{ \ast} \otimes_{k} W^{\ast} \rightarrow (V \otimes_{k} W)^{\ast}$ given by $(f\otimes g)(v\otimes w)=f(v)g(w)$ is an isomorphism?.Try: I checked that the map is well defined.

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    $\begingroup$ Write everything down in a basis. $\endgroup$ – Qiaochu Yuan Mar 23 '13 at 19:05

More generally, if $M,N,M',N'$ are $R$-modules, then we have a canonical $R$-linear map $\alpha : \hom(M,M') \otimes_R \hom(N,N') \to \hom(M \otimes_R N,M' \otimes_R N')$ given by $\alpha(f \otimes g)(m \otimes n)=f(m) \otimes g(n)$. It is easy to check that $\alpha$ is an isomorphism for $M=N=R$, since then both sides identify with $M' \otimes N'$ and $\alpha$ becomes the identity. For fixed $M',N,N'$, the set of all $M$s such that $\alpha$ is an isomorphism is closed under finite direct sums (use the usual interchange properties between $\hom,\otimes,\oplus$). The same for $N$. It follows that $\alpha$ is an isomorphism when $M$ and $N$ are finitely generated free (and also projective) $R$-modules. For $M'=N'=R$ it follows that $M^* \otimes_R N^* \to (M \otimes_R N)^*$ is an isomorphism.

For more details, see Bourbaki, Algebra I, Chapter II, §4.4.

Alternatively, if $M$ resp. $N$ have bases $\{e_i\}$ resp. $\{f_j\}$, then $\{e_i^* \otimes f_j^*\}$ is a basis of $M^* \otimes N^*$ (where $\{e_i^*\}$ denotes the dual basis of $\{e_i\}$), and $\{(e_i \otimes f_j)^*\}$ is a basis of $(M \otimes N)^*$. We have $\alpha(e_i^* \otimes f_j^*)=(e_i \otimes f_j)^*$, since $\alpha(e_i^* \otimes f_j^*)(e_u \otimes f_v)=e_i^*(e_u) \cdot f_j^*(f_v)=\delta_{iu} \delta_{jv}=\delta_{(i,j),(u,v)}$. Thus, $\alpha$ maps a basis to a basis, which implies that $\alpha$ is an isomorphism.

The case of vector spaces is where $R$ is a field, but I don't think that this special case offers any simplifications. Sometimes unnecessary assumptions make a problem seem to be more difficult than it really is.

  • $\begingroup$ My god! What a tough proof. $\endgroup$ – user Mar 23 '13 at 16:04
  • $\begingroup$ @Martin Brandenburg: Generality is nice, but maybe you should also write up an elementary proof for vector spaces over a field? I mean, it's much easier in that case... and if OP is doing e.g. linear algebra, then your argument should be quite confusing to him, I think! $\endgroup$ – Sam Mar 23 '13 at 16:08
  • $\begingroup$ @Martin: Funny, I have been downvoted today too. I never downvote by the way. What about you? $\endgroup$ – Georges Elencwajg Mar 23 '13 at 16:18
  • $\begingroup$ @Martin Brandenburg Ha!ha ha! $\endgroup$ – user Mar 23 '13 at 16:20
  • $\begingroup$ No, for vector spaces it is not easier. I have written down the most natural proof. Try to read it, it is not complicated at all. Every step is almost trivial. $\endgroup$ – Martin Brandenburg Mar 23 '13 at 16:34

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