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Suppose that $V_1$ is a vector space over the field $K_1$ and $V_2$ is a vector space over the field $K_2$. What is the definition of an isomorphism between these two vector spaces?

My best guess would be that $K_1$ and $K_2$ have to be isomorphic as fields. Say $\xi:K_1\to K_2$ is an isomorphism. And furthermore, we must have a bijective function $\sigma:V_1 \to V_2$ with

$$\sigma(\lambda v_1 + \mu v_2) = \xi(\lambda)\sigma(v_1) + \xi(\mu)\sigma(v_2)$$ for all $v_1,v_2 \in V_1$ and all $\lambda, \mu \in K_1$.

Is this correct, or have I missed or added extra conditions?

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It seems correct but silly to consider different but isomorphic scalar fields. – lhf Apr 25 '12 at 16:14
I have never encountered a notion of isomorphism of vector spaces over different fields. However, I agree that this is a good definition (and essentially the case you are considering does not use two different fields). Why would you want to use two different base fields? – Daan Michiels Apr 25 '12 at 16:14
Is it possible to have two vector spaces be isomorphic if their underlying fields are NOT isomorphic? – nullUser Apr 25 '12 at 16:16
No, unless you want to introduce this notion yourself. – Daan Michiels Apr 25 '12 at 16:17
@nullUser It wouldn't make any sense. One of the best properties of vector spaces is that they are classified (up to isomorphism) by their dimension, making a given vector space isomorphic to a direct sum of copies of your base field. Different base fields would give you different decompositions of your vector space...? – M Turgeon Apr 25 '12 at 16:30
up vote 3 down vote accepted

I don't really know what it means for a definition to be correct. Certainly this is a sensible definition (e.g. it defines an "equivalence relation" on the "set" of vector spaces). It is the notion of isomorphism that one gets in the following category: objects are pairs $(k, V)$ of a field and a vector space over that field, and morphisms are pairs $(\phi, T)$ of field morphisms $\phi : k_1 \to k_2$ and maps $T : V_1 \to V_2$ such that $$T(av + bw) = \phi(a) T(v) + \phi(b) T(w).$$

However, when most people talk about isomorphism of vector spaces, they almost always work implicitly with a fixed base field $k$. This means more than that $k_1, k_2$ are isomorphic; it means that we have fixed an isomorphism between them.

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One reason to restrict to the case of a fixed field is that it makes $k$-vector spaces into algebras (in the sense of universal algebra), with one unary operation for every scalar in the field, and the category of $k$-vector spaces into a variety (with all that this entails). – Arturo Magidin Apr 25 '12 at 16:55
@Qiaochu: The category you describe is a reasonably interesting category. It can be obtained by applying the Grothendieck construction to the strict (!) 2-functor $\textbf{Vect}(-) : \textbf{Fld}^\textrm{op} \to \mathfrak{Cat}$, and is thus a fibred category over $\textbf{Fld}$. It's a sub-fibred-category of the stack of quasicoherent sheaves over the category of schemes... – Zhen Lin Apr 25 '12 at 17:52

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