Prove that the division ring is commutative if for every $x$, $x^7=x$

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).

• I'm not sure that it's necessarily $\Bbb F_7$. The same identity $a^7=a$ holds for $\Bbb F_2$ and $\Bbb F_3$. – Nurdin Takenov Aug 26 '14 at 4:30
• @NurdinTakenov Sorry you are of course correct. Still, it is clearly false for infinite fields, and finite fields of higher order than 7 (a degree 7 polynomial has at most 7 roots). It is also false for a field of order 5 (as there is an element of order 4 in the multiplicative group) . So then $\Bbb F_2,\Bbb F_3, \Bbb F_4, \Bbb F_7$ (all of which satisfy $x^7=x$) are the only possibilities and the rest of the answer is still valid. – PVAL-inactive Aug 26 '14 at 5:21
• @rschwieb I've included (edited in) the key observation that leads to the field being finite (it was already in my above comment). – PVAL-inactive Aug 26 '14 at 13:44
• What's $\Bbb F_n$? Is it just a ring with $n$ elements? – someonewithpc Aug 29 '16 at 21:54
• @someonewithpc It is the unique (up to isomorphism) field with $n$ elements. It only makes sense when $n$ is a prime power. – PVAL-inactive Aug 29 '16 at 22:25

I think the path you chose, complemented by PVAL's answer is a simple path:

1) Show the ring is semisimple

2) Point out the matrix rings must have dimension $1$

3) Observe the centers division rings involved must be finite, therefore $K$ is finite and the division rings are finite dimensional $K$ algebras.

4) Apply Wedderburn's Little theorem to conclude the division rings are commutative.

There is another way that bundles steps 2 and 4 into one (although it is not much simpler: we are just trading extra steps for use of a more powerful theorem)

1) Show the ring is semisimple

2') Apply Jacobson's generalization of WLT to conclude the ring is commutative, and note that a commutative semisimple ring is a finite product of fields.

3') Deduce that a field satisfying $x^7=x$ is finite, and determine what the possibilities for $K$ are, and then what finite extensions of $K$ are possible to appear in the factorization of the ring.

You should start by wondering about the characteristic of this ring. You will quickly see that the characteristic is $p$ such that $p-1|7-1$.

Assume now you that you are living in a division algebra $R$, such that $a^7-a=0$ for every $a$ in $R$. What can you say about $R$? Can you conclude that $R=k$ where $k$ is the ground field in the reduced list you established?

Hope this helps.

• I realised that shortly after writing my answer, and edited accordingly. – Theon Alexander Aug 26 '14 at 17:26
• Looks good now! – rschwieb Aug 26 '14 at 17:27