I'm trying to solve a problem and I'm stuck.
Here is the original problem:
Let $A$ be a finite-dimensional algebra over a field $K$, such that for every $a\in A$, $a^7=a$. Show that $A$ is a direct product (sum?) of fields. What fields can arise?
We see that $A$ is Artinian and therefore its Jacobson radical is nilpotent. However from the fact that $a^7=a$ we see that there are no nilpotents, so Jacobson radical is zero. Therefore $A$ is semisimple and is a direct product of a matrix rings over division algebras. Since there are no nilpotents all matrix rings are 1-dimensional, so $A$ is a direct product of division rings.
Now we have to prove that all these division rings are fields. And that's where I am stuck. Can you give a hint what to do next? If I can prove that these division rings are finite I'm done, but I don't know how.
Any field $K$ where $x^7=x$ for all $x\in K$ is isomorphic to $\Bbb F_2$ ,$\Bbb F_3$, $\Bbb F_4$ or $\Bbb F_7$ by basic field theory ($|K|\leq7$, as a degree 7 polynomial has at most 7 roots). So your base field must be one of those 4. So your division rings are finite dimensional over a finite field, therefore they are finite. Now just apply Wedderburn's little theorem (a proof of this is outlined in some exercise in Dummit and Foote).