# Dimension of an $F$-vector space, compared to dimension over $E:F$

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!

• 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. – P Vanchinathan Apr 29 '15 at 0:27
• Do you mean to say $\dim_{F}(V)=\dim_{E}(V)\times [E:F]$? – User0112358 Apr 29 '15 at 0:44

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 $\infty$.
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.