0
$\begingroup$

Suppose $E:F$ is a finite field extension of $F$. If $V$ is both an $F$-vector space and an $E$-vector space, then is there any relation between the dimension of $V$ over $F$ and the dimension of $V$ over $E$?

It seems like the dimension over $E$ should be smaller! But, in the case that $V=\{0\}$, the trivial vector space, the dimensions over both fields are the same. Is this the only case in which they are equal?

Any hints or references would be appreciated, Thank you!

$\endgroup$
  • $\begingroup$ Implicitly you have assumed that $V$ is a vector space over $E$. So you have to make it explicit in your 2nd sentence. The dimension over the smaller field will be product of appropriate quantities (guess it) and for zero this equality will hold. $\endgroup$ – P Vanchinathan Apr 29 '15 at 0:27
  • $\begingroup$ Do you mean to say $\dim_{F}(V)=\dim_{E}(V)\times [E:F]$? $\endgroup$ – User0112358 Apr 29 '15 at 0:44
1
$\begingroup$

We should assume that the structure of $V$ as an $E$-vector space is compatible with the structure of $V$ as an $F$-vector space, that is, $c\cdot_F v = c\cdot_E v$ for any $c\in F$. Otherwise, strange things can happen.

If $V$ has basis $\mathcal{B}$ as an $E$-vector space, then $V$ is the direct sum of $|\mathcal{B}|$ copies of $E$. But $E$ is the direct sum of $|E:F|$ copies of $F$, so it follows that $V$ is the direct sum of $|\mathcal{B}| \cdot |E:F|$ copies of $F$.

In other words, the dimension of $V$ over $F$ is just $|E:F|$ times the dimension of $V$ over $E$. So $\dim_F V > \dim_E V$, unless both dimensions are zero, or $\infty$.

$\endgroup$
  • $\begingroup$ Cheers Slade :) $\endgroup$ – User0112358 Apr 29 '15 at 1:48
1
$\begingroup$

The question is ill-posed, since an $F$ vector space isn't an $E$-vector space in any natural way. For example $F$ itself isn't an $E$-vector space.

$\endgroup$
  • $\begingroup$ Thanks Jeremy. That is absolutely right :| $\endgroup$ – User0112358 Apr 29 '15 at 0:27
  • $\begingroup$ I have edited the question so at least it should be well posed. $\endgroup$ – User0112358 Apr 29 '15 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.