7
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ 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.) $\endgroup$
    – j.p.
    Mar 16, 2015 at 13:47
  • $\begingroup$ The notations $\mathbb F_1$ and $\mathbb F_2$ are rather misleading.... $\endgroup$ Mar 16, 2015 at 13:47
  • $\begingroup$ $$T((a_\mathbb{F_1}+b_\mathbb{F_1})*\mathbf{x})=T(a_\mathbb{F_1}+b_\mathbb{F_1})@T(\mathbf{x})=(T(a_\mathbb{F_1})"T(b_\mathbb{F_1}))@T(\mathbf{x})=(a_\mathbb{F_2}"b_\mathbb{F_2})@T(\mathbf{x})=a_\mathbb{F_2}@T(\mathbf{x})"b_\mathbb{F_2}@T( \mathbf {x})$$will this work??? $\endgroup$
    – Secret
    Mar 16, 2015 at 13:53
  • $\begingroup$ @Secret Are $\;a_{F_1}, b_{F_1}\;$ scalars? Because if they are then you first have to tell us what can possibly be $\;T(a_{F_1}+b_{F_1})\;$ ... $\endgroup$
    – Timbuc
    Mar 16, 2015 at 14:25
  • $\begingroup$ 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 $\endgroup$
    – Secret
    Mar 16, 2015 at 22:02

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .