# Why the morphisms of vector spaces, over different fields is not interesting?

Suppose $V_\mathbb{F_V}$ and $W_\mathbb{F_W}$ are two vector spaces over fields $\mathbb{F}_V$ and $\mathbb{F}_W$. Then a homomorphism of these vector spaces consists of maps $f:V\rightarrow W$ and $f_\mathbb{F}:\mathbb{F}_V\rightarrow \mathbb{F}_W$ satisfying: $$f\left(a.v+b.u\right)= f_{\mathbb{F}}\left(a\right)f\left(v\right)+f_{\mathbb{F}}\left(b\right)f\left(u\right)$$ for all $a,b \in \mathbb{F}_V$ and $v,u \in V$. With such morphisms we can talk about the category of all vector spaces over arbitrary fields. But, I have never seen such examples. Why is that? Is it because the category of fields is not very welcoming of a place.

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There is indeed a category of all vector spaces with morphisms as you describe. It has many interesting properties – first of all, notice that it comes equipped with a projection to the category of all fields, $p : \textbf{Vect} \to \textbf{Fld}$. Let $\textbf{Vect}(K)$ be the non-full subcategory of $\textbf{Vect}$ of objects $V$ such that $p V = K$ and morphisms $f$ such that $p f = \textrm{id}_K$. This is easily seen to be isomorphic to the usual category of $K$-vector spaces. Given any field homomorphism $\phi : K \to L$, we get a functor $\phi^\sharp : \textbf{Vect}(L) \to \textbf{Vect}(K)$, and it is not hard to check that the operation $(-)^\sharp$ is strictly functorial. The category $\textbf{Vect}$ is then seen to be the Grothendieck construction applied to $(-)^\sharp$, and therefore $p : \textbf{Vect} \to \textbf{Fld}$ is a Grothendieck fibration.

Why is this interesting? Well, it gives us a way to consider vector spaces over all fields on equal grounds, and the universal property of some familiar constructions is best expressed in terms of this Grothendieck fibration. For example, if $W$ is a $L$-vector space, then $\phi^\sharp W$ is a $K$-vector space $V$ and a morphism $f : V \to W$ lying over $\phi : K \to L$ such that for all morphisms $h : U \to W$ in $\textbf{Vect}$ lying over $\chi : F \to L$ in $\textbf{Fld}$ and all factorisations $\chi = \phi \circ \psi$, there is a unique morphism $g : U \to W$ lying over $\psi : F \to K$ such that $h = g \circ f$. If you draw the diagram you will see this is basically the universal property of a pullback, but two different categories are involved here.

On the other hand, $\phi^\sharp : \textbf{Vect}(L) \to \textbf{Vect}(K)$ has a well-known left adjoint $\phi_\sharp : \textbf{Vect}(K) \to \textbf{Vect}(L)$, namely the tensor product $\phi_\sharp V = L \otimes_K V$. This makes $p : \textbf{Vect} \to \textbf{Fld}$ into a Grothendieck bifibration, and again this means $\phi_\sharp V$ can be described in terms of a universal property.

You are quite right that $\textbf{Fld}$ isn't a category with particularly good properties – and unfortunately that means $\textbf{Vect}$ also lacks the same properties. For example, there is no terminal object in either $\textbf{Fld}$ or $\textbf{Vect}$. In this respect, the category $\textbf{Mod}$ of all modules over all commutative rings is more well-behaved. $\textbf{Mod}$ has some remarkable properties – in addition to being a Grothendieck bifibration, it (or rather $\textbf{Mod}^\textrm{op}$) is what is known as a stack for the faithfully flat topology on $\textbf{CRing}^\textrm{op}$. This is studied at length in SGA1 and is the motivating example behind the whole theory of fibred categories in general.

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Your conditions imply that $F_V$ can be viewed as a subfield of $F_W$ and via that $W$ is a $F_V$-vector space and $f$ is $F_V$-linear. For example nothing prevents you from considering the linear maps from the real vector space $C([0,1])$ of continuous function on the interval $[0,1]$ to the complex vector space $\mathbb C^{42}$. But what you get is the same as if you considered $\mathbb C^{42}$ just as $\mathbb R^{84}$.

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But, $f_F$ is not necessarily injective, why $F_V$ can be viewed as a subfield of $F_W$? –  Hooman Oct 4 '12 at 16:34
While you didn't say it, to have any chance of sensible properties we should take $f_\mathbb{F}$ to be a homomorphism of fields, and all of these are injective. –  Matt Pressland Oct 4 '12 at 16:54
It's not exactly the same – you have first choose an embedding of $\mathbb{R}$ into $\mathbb{C}$, and there might be more than one. –  Zhen Lin Oct 5 '12 at 6:33
I think the question is: For finite dimensional vector spaces $V_F$ and $W_{F'}$, if $End(V_F)\cong End(W_{F'})$ as rings, what can you conclude about the relationship of $V_F$ to $W_{F'}$?
The answer is that $V$ and $W$ are semilinearly isomorphic, which is a generalization of a linear isomorphism. You might be interested in semilinear transformations :)