Blue was correct, I need to fix my understanding of this:

Finite fields have cardinality of a prime order because they have a prime subfield that has finite characteristic.

I do not completely understand why finite field of each characteristic is unique. Perhaps we look at a polynomial and express the finite field as a splitting field? And we use uniqueness of splitting fields?

Then, why is it that $\mathbb F_{p^m} \subseteq \mathbb F_{p^n}$ if and only if $m$ divides $n$?

  • $\begingroup$ You were stumbling over the solution? Does that mean you have access to a solution but you don't understand it? Then you need to post what you have in front of you and say what you don't understand about it. Otherwise, you still need to provide your thoughts and your work so far. What do you know about finite fields that you might suspect are helpful here? $\endgroup$ – blue Aug 12 '14 at 3:20
  • $\begingroup$ Do you know how the polynomial $x^{p^k} - x$ behaves in $\mathbb{F}_{p^k}$? $\endgroup$ – Henry Swanson Aug 12 '14 at 3:20
  • $\begingroup$ This fact was used in the solution to the problem, stated as a minor detail. I don't know how to verify it while reading the solutions. $\endgroup$ – anonymous489 Aug 12 '14 at 3:20
  • 3
    $\begingroup$ He pointed out that polynomial because that's how you answer the question. Indeed, you hinted that you've already seen a solution that uses that fact, although you are currently withholding it from us. Why are you trying to answer an exam question about finite fields when you don't know the basics about them? Also, I did not ask you why splitting fields are unique, I asked you why there is a unique finite field of a given order. The answer does use splitting fields, but you've already indicated you don't know how the appropriate polynomials are involved. $\endgroup$ – blue Aug 12 '14 at 3:49
  • 4
    $\begingroup$ You asked the question without informing us you already had a solution, without telling us what about the solution you didn't get, without telling us any of your own thoughts or ideas about it, and when asked questions to ascertain what you do know about finite fields versus what you don't know, so that we can identify what to say, you've put up a fight. Not in all of this time have you corrected any of these issues, or even acknowledged them. I downvoted and I deleted my answer to your question, but I wasn't the one who voted to close because I was patient to let you fix them. No more. $\endgroup$ – blue Aug 12 '14 at 4:01

The notation $\Bbb F_q$, where $q$ is a prime power, refers to the finite field of order $q$. If there were multiple finite fields of order $q$, then the notation $\Bbb F_q$ would be ambiguous! The existence and uniqueness of finite fields of given prime power order are one of the first things proven about them, and both parts can done by showing that $\Bbb F_q$ is the splitting field of $x^q-x$ over $\Bbb F_p$. This is because in general field theory we know that splitting fields exist and are unique up to isomorphism.

Another general fact from field theory we know is that finite groups of units within $F^\times$ for any field $F$ are always cyclic. Let $F$ be a field of order $q$. Then $F^\times$ is cyclic of order $q-1$, hence all of its elements satisfy $x^{q-1}=1$, hence satisfy $x^q=x$. But even $x=0$ satisfies the latter, so $x^q-x$ has every element of $F$ as a root, and since $|F|=q=\deg(x^q-x)$, these are precisely all of its roots, hence $F$ is the splitting field of $x^q-x$ over $\Bbb F_p$. Conversely if $F$ is the splitting field of that polynomial, then $F$ is generated over $\Bbb F_p$ by some primitive $(q-1)$th root of unity (since all of the nonzero roots of $x^q-x$ are powers of it), so $F^\times$ is generated by $\Bbb F_q^\times$ and this primitive root, which means that $F^\times$ must be cyclic of order $q-1$, hence $|F|=q$.

So we know for every prime power $p^m$ there exists a unique field $\Bbb F_{p^m}$ of order $p^m$, which is the splitting field of $x^{p^m}-x$. If $m\mid n$ then $(x^{q^m}-x)\mid(x^{q^n}-x)$ (can you prove this? what does the quotient look like?). Another fact from general field theory: if $f(x)\mid g(x)$ then $f$'s splitting field is contained in $g$'s splitting field (technical note: the splitting fields have to exist in a fixed algebraic closure). So $\Bbb F_{q^m}\subset \Bbb F_{q^n}$. Conversely, if $\Bbb F_{q^m}\subset\Bbb F_{q^n}$, then $q^n$ is a power of $q^m$, because the larger field is a vector space over the smaller field (general fact about vector spaces).

  • $\begingroup$ @anonymous489, you're thanking blue for "bullying" you? That doesn't make any sense. $\endgroup$ – Joel Reyes Noche Mar 4 '15 at 4:18
  • $\begingroup$ "so $F^\times$ is generated by $\Bbb F_q^\times$ and this primitive root"... why? $\endgroup$ – darij grinberg Sep 27 '15 at 23:25

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.