# A problem of field in abstract algebra

If $V$ is a finite-dimensional vector space over the field $K$, and if $F$ is a subfield of $K$ such that $[K:F]$ is finite, show that $V$ is a finite-dimensional vector space over $F$ and that moreover $\dim_F V=(\dim_K V) [K:F]$.

I know that $\dim_F V=(\dim_K V) [K:F]$ can be changed into $[V:F]=[V:K][K:F]$, and it is similar to the theorem

Let $F\subseteq K\subseteq V$, then $[V:F]=[V:K][K:F]$.

But is it able to drive the relation $K\subseteq V$?

• @JessePFrancis, in what way is $1\in V$ for an abstract vector space $V$? – WSL Apr 15 '15 at 10:30
• @WSL, my bad. And why is the question and answer voted down? Not just this, many new questions are voted down without any reason! – Jesse P Francis Apr 15 '15 at 10:41
• no idea... I agree that it is without reason. – WSL Apr 15 '15 at 10:43
• For quite some time there's a serial downvoter in the site, both of questions and of answers, and moderators either won't do anything or can't do anything. Either way, this situation sucks, and as many times has been brought up: until it will be compulsory to leave a little note explaining the reason of a downvote and the downvoter's nick/name, this will continue...and "this" is constantly distracting lots of members of the site's main issue: mathematics! I found this a worth question, thus I am upvoting it. Some get confused and downvote when the work shown in question is wrong: big mistake. – Timbuc Apr 15 '15 at 10:48
• @Timbuc, I agree to that! If you see the questions asked around this question, all questions and answers were down voted, for no reason. Probably making comment explaining why compulsory for down vote will help! – Jesse P Francis Apr 15 '15 at 11:02

Well, as far as the final relation is concerned, it is not true that you can embed $K\subset V$ in a canonical way.

I would suggest looking at a basis $\{ v_1, v_2, \ldots, v_k\}$ of $V$ as a $K$-vector space. Here $\dim_K(V) = k$.

Since $K$ is an $F$-vector space as well, it has a basis, say $\{a_1, a_2, \ldots , a_n\}$.

Now show that $\{a_1v_1, \ldots, a_nv_1, a_1v_2, \ldots, a_nv_k\}$ gives a basis for $V$ as an $F$-vector space. Since the $a_i$ form a basis over $F$, they are linearly independent over $F$. I leave the details that this forces this set to be $F$-linearly independent to you.

Then, this basis has $\dim_F(V) = k\cdot n = \dim_K(V)[K:F]$ elements.

• $\;\{a_1v_1\,,\,a_2v_1\}\;$ may already be linearly dependent, as $$\;(-a_2)\cdot a_1v_1+a_1\cdot a_2v_1=0$$ . Some care is required here. – Timbuc Apr 15 '15 at 10:51
• Not as an $F$-basis. As the $a_i$'s give a basis of $K$, they are $F$-linear independent, so this issue does not arise with $F$ coefficients. – WSL Apr 15 '15 at 10:53
• That is right, but then it may be hinted or explictly said that $\;a_i\;$ cannot be in $\;F\;$ . Again, somebody downvoted this answer. – Timbuc Apr 15 '15 at 10:56
• I see. I was curious as to why the downvote. My intent was for that to be worked out, but I'll edit. I am new here and assumed full details might not be appropriate. – WSL Apr 15 '15 at 10:58
• Don't mind the downvote:we have several losers around the place who serial downvote or, without any explanation, just downvote. Either way, you're right: you don't need to add any details, as your answer is fine as it is. +1 – Timbuc Apr 15 '15 at 11:00