3

I understand that an homomorphism between vector spaces must preserve the sum between vectors and the scalar multiplication.

By example, let vector spaces $(E,\Bbb Q)$ and $(F,\Bbb Q[\sqrt 2])$, then we can define something like

$$f:E\to Q$$

such that $f(\lambda v+\mu w)=\lambda\sqrt2 f(v)+\mu\sqrt2 f(w)$ for $\lambda,\mu\in\Bbb Q$ and $\lambda\sqrt 2,\mu\sqrt2\in\Bbb Q[\sqrt 2]$. This would be a homomorphism between vector spaces but not a linear map, right?

Or if we take some vector spaces $(A,\Bbb R)$ and $(B,\Bbb C)$ and we set a function

$$g:A\to B$$

such that $g(r a+s b)=r g(a)+sf(b)$ for $a,b\in A$ and $r,s\in \Bbb R$ this preserve the operations of vector spaces too because $\Bbb R\subset\Bbb C$.

These examples would be correctly called as homomorphisms between vector spaces (with no necessarily the same field)?

Masacroso
  • 30,417

2 Answers2

6

By definition, linear maps of vector spaces can only exist between vector spaces over the same field. You could define a new class of maps between pairs of the form $(V,K)$ where $V$ is a vector space over $K$ with the property you mention (or more precisely, you would probably want pairs of maps $(f,g) : (V,K)\to(W,L)$, where $f : V\to W$ is a homomorphism of abelian groups, $g : K\to L$ is a homomorphism of rings and $f(av + bw) = g(a)f(v) + g(b)f(w)$ for $a,b\in K$, $v,w\in V$), but they would no longer be linear maps of vector spaces (although the data of such a map $(f,g) : (V,K)\to (W,L)$ would be equivalent to the data of a linear map $V\to W$ when $K = L$ and the map $g : K\to L$ is the identity).

There are other sorts of ways you might try to make this work as well: if $M$ is an $R$-module and $N$ is an $S$-module, and you have a morphism of rings $R\to S$, you can give $N$ the structure of an $R$ module via the homomorphism, and then you could talk about a morphism $M\to N$ of $R$-modules. In the world of vector spaces, this would be the same as starting with a vector space $V/K$ and a vector space $W/L$, where $L$ is an extension of $K$, and then via restriction, considering $W$ as a vector space over $K$, and looking at maps $V\to W$ where $W$ is considered as a $K$-vector space.

Stahl
  • 23,212
2

Homomorphisms between vector spaces $V$, $W$ (over the same field $k$) are by definition the linear maps between those vector spaces. Linearity itself, is also frequently mentioned with respect to the underlying field i.e. we say "$k$-linear map". In the level of the Category theory, these are the morphisms in the category of $k$-vector spaces.

The maps you are speaking about in your post, are neither homomorphisms nor linear. In fact, they may face various problems in the following sense: Let such a map $g:A\to B$, between $(A,\Bbb R)$ and $(B,\Bbb C)$ and let it be bijective. Then it possesses an inverse map $g^{-1}$ (in the set-theoretic sense). Would this inverse, be "linear" in the sense defined in your post? It would actually not even be well-defined. Let alone the case in which the underlying fields might be completely different (as sets).

KonKan
  • 7,344
  • Homomorphisms dont need to be bijective. The $g$ that you describes is a try for an isomorphism, what it is impossible because $\Bbb R$ is not isomorphic to $\Bbb C$. – Masacroso Oct 17 '16 at 06:30
  • Yes, this is what I am talking about. I only wanted to mention possible pitfalls in the definitions of such maps. And I mentioned the case of bijective homomorphisms as an example. – KonKan Oct 17 '16 at 06:33
  • Well, a bijective homomorphism not having an inverse isn't really a problem. Although categories like $\mathsf{Grp}$ and $\mathsf{Ring}$ satisfy the property that if a homomorphism in the category is bijective on the underlying sets it is an isomorphism, many other categories do not have this property. Even other concrete categories (e.g. $\mathsf{Top}$) need not satisfy this property. – Stahl Oct 17 '16 at 07:02
  • Yes but such maps, map between objects of different categories. – KonKan Oct 17 '16 at 07:04
  • I'm not sure what you mean. A continuous bijection (take the identity map from an infinite set equipped with the discrete topology to the same infinite set but with the indiscrete topology) need not be an isomorphism, although it is a continuous bijection. This does not involve any category other than $\mathsf{Top}$. – Stahl Oct 17 '16 at 07:06
  • Oh, I think I see what you mean, I'm sorry. As in, they're maps in categories other than $\mathsf{Vect}$. – Stahl Oct 17 '16 at 07:08
  • No, I was refering to the maps mentioned in the OP. I meant that, vector spaces over different fields "live" in different categories. – KonKan Oct 17 '16 at 07:08