# Show that if $\gcd(q,q')\neq1$, then fields are isomorphic to sub-fields of some finite field

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.

• 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 $s$.) – arctic tern Oct 8 '16 at 4:50
• @arctictern;I have done the edits ;Please check the problem now – Learnmore Oct 8 '16 at 5:51
• 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? – Jyrki Lahtonen Oct 8 '16 at 6:23
• Yes because $\Bbb F_p$ is their prime subfield@JyrkiLahtonen – Learnmore Oct 8 '16 at 6:38
• Are you familiar with the result that ${\Bbb F}_{p^d}\subseteq{\Bbb F}_{p^n}$ if $d\mid n$? – 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 $n=rs$.