1

I was working through the following https://math.stackexchange.com/a/426300/299525 and I could not justify one of the steps.

It seems to be a consequence of a general result, as discussed in the comments.

Proposition : If $K/L$ is an extension of fields, $V$ a vector space over $L$, then $$ \textrm{Hom}_L (V,K) \cong \textrm{Hom}_L (V,L) \otimes _L K$$

So it seems most natural to define a map going the other way out of the tensor product, perhaps $(\varphi , k) \mapsto k (\psi \circ \varphi)$, where $\psi : L \to K$. This is well-defined and bilinear so it induces a map from the tensor. Injectivity is also clear since $\psi$ is an injection.

However, I am not sure why every $L$-linear homomorphism $V \to K$ must factor through $\psi$, up to a multiple of $K$ though, i.e. why it should be surjective.

Community
  • 1
Future
  • 4,502
  • You need to construct the inverse map. Define it by first choosing a basis for $V$ and doing what comes naturally. – Alex G. Dec 27 '15 at 03:45

1 Answers1

3

This is only true (at least in a natural way) if either $K$ or $V$ is finite-dimensional over $L$. If $K$ is finite-dimensional, pick a basis $\{e_1,\dots,e_n\}$ of $K$. Given a linear map $\alpha:V\to K$, write $\alpha(v)=\alpha_1(v)e_1+\dots+\alpha_n(v)e_n$ with $\alpha_i(v)\in L$. Then each $\alpha_i$ is a linear map $V\to L$, and $\alpha$ is the image of $\alpha_1\otimes e_1+\dots+\alpha_n\otimes e_n$ under the map you have described.

If $V$ is finite-dimensional, pick a basis $\{v_1,\dots,v_n\}$ for $V$, and let $\{\alpha_1,\dots,\alpha_n\}$ be the dual basis of $\operatorname{Hom}_L(V,L)$, so $\alpha_i(v_j)=\delta_{ij}$. Then any $\alpha:V\to K$ is the image of $\alpha_1\otimes \alpha(v_1)+\dots+\alpha_n\otimes\alpha(v_n)$ under the map you have described.

(Note that in particular, it is not true that every map $V\to K$ factors through $\psi$, up to a multiple of $K$. Rather, every map is a linear combination of maps that factor through $\psi$. This is not true when $K$ and $V$ are both infinite-dimensional over $L$, essentially because you can't take "infinite linear combinations".)

Eric Wofsey
  • 330,363
  • This is a pretty elegant proof, I like it. However, I am concerned about the additional finiteness hypothesis added. Were you able to get a chance to read the link I posted about the origins of this problem? I wonder if the finiteness condition might affect that result. – Future Dec 27 '15 at 03:54
  • 1
    Well, I'm not saying anything that wasn't said already by Georges Elencwajg in the comments there. If you want to know whether there actually is a map of schemes such that the map $T_{X,x}\to\mbox{Hom}{k(y)}({\frak{m}}_y/{\frak{m}}_y^2,k(x))$ does not factor through the image of $T{Y,y}\otimes k(x)$, I suggest you ask that as a new question. – Eric Wofsey Dec 27 '15 at 04:02
  • 1
    @T.S.L The discussion you linked also assumed the field extension is finite. – Alex G. Dec 27 '15 at 04:04
  • I didn't even notice that there were more comments to be expanded. One last question then: the comments in the previous post all seemed to concur that this isomorphism is in fact canonical. However, if we are to choose a basis, then isn't that a non-canonical thing to do? – Future Dec 27 '15 at 04:06
  • 1
    In the question, you described the map itself in a completely canonical way. You only have to choose a basis to prove that the map happens to be surjective. – Eric Wofsey Dec 27 '15 at 04:09
  • Ah yes, that's a good point. Thanks again! – Future Dec 27 '15 at 04:39