Let $F,F'$ be finite fields of order $q,q'$. Show that if $\gcd(q,q')\neq1$, then both are isomorphic to sub-fields of some finite field $L$.

Since $\gcd(p,p^{'})\neq 1\implies q=p^r,\ q'=p^s$.

Should I take the field $L$ to be of order $p^{r+s}$? Even if I take it, I am not sure how to show that $L$ has a subfield of order $q,q'$. Please help me out here.

  • 2
    $\begingroup$ If the sizes of $F$ and $F'$ are coprime, then so are their characteristics, so no, they are not subfields of any larger field $L$. Perhaps you mean $|F|=p^r$ and $|F'|=p^s$ with a condition on $r$ and $s$? (In fact, then the conclusion would be true without any condition on $r$ and $s$.) $\endgroup$ – arctic tern Oct 8 '16 at 4:50
  • $\begingroup$ @arctictern;I have done the edits ;Please check the problem now $\endgroup$ – Learnmore Oct 8 '16 at 5:51
  • 1
    $\begingroup$ You will have better luck with $L$ of order $p^{rs}$. But actually all you need is to use the fact that both $F$ and $F'$ are simple extensions of $\Bbb{F}_p$. Do you see why? $\endgroup$ – Jyrki Lahtonen Oct 8 '16 at 6:23
  • $\begingroup$ Yes because $\Bbb F_p$ is their prime subfield@JyrkiLahtonen $\endgroup$ – Learnmore Oct 8 '16 at 6:38
  • 1
    $\begingroup$ Are you familiar with the result that ${\Bbb F}_{p^d}\subseteq{\Bbb F}_{p^n}$ if $d\mid n$? $\endgroup$ – arctic tern Oct 8 '16 at 14:37

If $F$ is a finite field, it must have positive characteristic $p$, which must be a prime number (else factors of a composite characteristic would be zero divisors), and thus the subfield generated by the multiplicative unit $1$ will be a prime subfield, a copy of $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$. Then since $F$ satisfies the definition of being a vector space over $\mathbb{F}_p$, it has a basis and thus is isomorphic to $\mathbb{F}_p^n$ as a vector space for some dimension $n$. This implies $|F|=p^n$, so all finite fields' sizes are powers of their prime characteristics.

Finite multiplicative groups within fields are cyclic, so if $F$ is any finite field of order $q$ then the group of units $F^\times$ is cyclic of order $q-1$. As a result, it is the splitting field of $x^{q-1}-1$ - or equivalently $x^q-x$ if you want to include $0$ as a root - over its prime subfield $\mathbb{F}_p$. Since splitting fields are unique up to isomorphism, we conclude there is a unique finite field (up to isomorphism) of every possible order $q=p^n$ (powers of primes). For simplicity we could simply speak of splitting fields as they exist within a fixed choice of algebraic closure.

If $d\mid n$, then solutions to $x^{p^d}=x$ are also solutions to $x^{p^n}=x$ (take both sides to the $p^d$th power $n/d$ times), hence there is a containment $\mathbb{F}_{p^{\large d}}\subseteq\mathbb{F}_{p^{\large n}}$. Conversely such a containment implies $\mathbb{F}_{p^{\large n}}$ is a vector space over $\mathbb{F}_{p^{\large d}}$, hence by the previous argument $p^n$ is a power of $p^d$, i.e. $d\mid n$. Thus,

Theorem. There is a unique finite field of every prime power order $p^n$ called $\mathbb{F}_{p^{\large n}}$, and moreover there is containment $\mathbb{F}_{p^{\large d}}\subseteq\mathbb{F}_{p^{\large n}}$ if and only if $d\mid n$.

This is a standard classification theorem to know. In any case, this means $\mathbb{F}_{p^{\large r}}$ and $\mathbb{F}_{p^{\large s}}$ will be both contained within $\mathbb{F}_{p^{\large n}}$ if and only if $r,s$ both divide $n$, or equivalently $n$ is a multiple of $\mathrm{lcm}(r,s)$; one particular choice is $n=rs$.


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.