4
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ I'm not sure that it's necessarily $\Bbb F_7$. The same identity $a^7=a$ holds for $\Bbb F_2$ and $\Bbb F_3$. $\endgroup$ – Nurdin Takenov Aug 26 '14 at 4:30
  • 1
    $\begingroup$ @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. $\endgroup$ – PVAL-inactive Aug 26 '14 at 5:21
  • $\begingroup$ @rschwieb I've included (edited in) the key observation that leads to the field being finite (it was already in my above comment). $\endgroup$ – PVAL-inactive Aug 26 '14 at 13:44
  • $\begingroup$ What's $\Bbb F_n$? Is it just a ring with $n$ elements? $\endgroup$ – someonewithpc Aug 29 '16 at 21:54
  • $\begingroup$ @someonewithpc It is the unique (up to isomorphism) field with $n$ elements. It only makes sense when $n$ is a prime power. $\endgroup$ – PVAL-inactive Aug 29 '16 at 22:25
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.