Dimension of division ring over center cannot be prime

Question

Let $D$ be a division ring. Let $F$ be its center. Note that $F$ is a field, so $D$ is a vector space over $F$.

How do we show that $\dim_F D$ cannot be a prime?

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What I tried is proving by contradiction. Suppose $\dim_F D=p$, then there exists a basis $\{x_1,\dots,x_p\}$ of $D$.

Then $1=a_1x_1+\dots+a_px_p$ for some $a_i\in F$.

Perhaps use the fact that characteristic of $F$ is either 1 or prime here? I am not sure how to continue.

Thanks for any help.

Answer 1

I think this will help. Hint:

Think along the lines of a field extension. $F=D$ there's nothing to prove. If $\alpha\in D\setminus F$, then $F[\alpha]\subseteq D$ is a field extension of $F$ with index $[F[\alpha]: F]>1$.

Now, $D$ is a vector space over this field as well. How do $[F[\alpha]: F]$, $[D:F[\alpha]]$ and $[D:F]$ relate?