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I was having linear algebra class and we have been discussing about a possible group homomorphism that might allow mapping between two vector spaces over two different fields

This is also an extension of this question

Suppose we have vector spaces $V$ and $W$ over some general field $\mathbb{F}_1$ and $\mathbb{F}_2$ and $T$ is a (linear) map from $V$ to $W$

In order to get around the issue of this vector space axiom becoming undefined because of c being in different fields

$$T(c\mathbf{x})=cT(\mathbf{x})$$

What's the issue in doing this (adapting the definition of group homomorphism, where there are two groups $(G,@)$ and $(H,*)$)?

$$\phi (a @b)=\phi(a)*\phi(b)$$

to the context of vector space (where the fields are defined as $(\mathbb{F}_1,+,*)$ and $(\mathbb{F}_2,",@)$)

$$T(c_\mathbb{F_1}*\mathbf{x})=T(c_\mathbb{F_1})@T(\mathbf{x})=c_\mathbb{F_2}@T(\mathbf{x})$$

(The two cs are different because they are elements of different fields)

It seems valid as long every element in $\mathbb{F}_2$ can be mapped from at least one in $\mathbb{F}_1$. What subtleties have we overlooked?

If this is valid is this still a linear algebra?

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Secret
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  • Fields have a group structure given by addition and, if you take away $0$, one given by multiplication. Looks like you thought only of multiplication. What becomes out of $(a+b)v = av+bv$? (If you ask instead for a ring homomorphism, there will be only injective ones.) – j.p. Mar 16 '15 at 13:47
  • The notations $\mathbb F_1$ and $\mathbb F_2$ are rather misleading.... – darij grinberg Mar 16 '15 at 13:47
  • $$T((a_\mathbb{F_1}+b_\mathbb{F_1})*\mathbf{x})=T(a_\mathbb{F_1}+b_\mathbb{F_1})@T(\mathbf{x})=(T(a_\mathbb{F_1})"T(b_\mathbb{F_1}))@T(\mathbf{x})=(a_\mathbb{F_2}"b_\mathbb{F_2})@T(\mathbf{x})=a_\mathbb{F_2}@T(\mathbf{x})"b_\mathbb{F_2}@T( \mathbf {x})$$will this work??? – Secret Mar 16 '15 at 13:53
  • @Secret Are $;a_{F_1}, b_{F_1};$ scalars? Because if they are then you first have to tell us what can possibly be $;T(a_{F_1}+b_{F_1});$ ... – Timbuc Mar 16 '15 at 14:25
  • Yes they are, and (for illustration) T can be a map from the reals (where $a_{\mathbb{F}1}$ and $b{\mathbb{F}1}$ were in) to the complex numbers (where $a{\mathbb{F}2}$ and $b{\mathbb{F}_2}$ were in) and at the same time maps elements in $V$ to elements in $W$.

    One example of T can be $$T: (a, \mathbf{v} ) \rightarrow (a+e^a i,\mathbf{w})$$

    However the scalars can be from more abstract fields, they are not necessary have to be from the reals or complex respectively

     – Secret Mar 16 '15 at 22:02
  • So for your case, using this example of T, it will become

    $$T(a_{ \mathbb{F}1}+b{ \mathbb{F}2})=(a{\mathbb{F}1}+e^{a{\mathbb{F}1}}i)+(b{\mathbb{F}1}+e^{b{\mathbb{F}1}}i)=a{ \mathbb{F}2}+b{ \mathbb{F}_2}$$

     – Secret Mar 16 '15 at 22:10

1 Answers1

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A map between two vector spaces over different fields cannot be linear (see this question), but can be semilinear . In this case there exists an homomorphism between the two fields $ \phi:\mathbb{F}_1 \to \mathbb{F}_2$ that is also an homomorphism between the multiplicative groups of the fields.

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Emilio Novati
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