Let $\mathtt{F}$ and $\mathtt{F'}$ be two finite fields of order $q$ and $q'$ respectively. Then:

  1. $\mathtt{F'}$ contains a subfield isomorphic to $\mathtt{F}$ if and only if $q\le q'$

If part is true, only if part fails for $q=p_1^a,q'=p_2^b$ such that $q\le q',p_1\ne p_2$, are two prime numbers.

  1. $\mathtt{F'}$ contains a subfield isomorphic to $\mathtt{F}$ if and only if $q$ divides $q'$

For $q=2,q'=p_1^b$,where $p_1$ is an odd prime, $q\not|q'$, but converse is true.

  1. If $\gcd(q,q')\ne1$, then both are isomorphic to subfields of finite field $\mathtt{L}$

True! Revolves around the fact that order of a finite field is power of a single prime.

  1. Both $\mathtt{F}$ and $\mathtt{F'}$ are quotient rings of the ring $\mathbb{Z}[X]$

Isn't it how finite fields are explicitly generated?

Is it correct? Specially on saying that for (1) if part is true and for (2) converse is true? Else, are there any counter examples?


1 Answer 1


For 1, you could just take a more accurate counter-example $q=2$ and $q'=3$. It seems to me that this contradicts the "if" part.

For 2, this is false, because of Jyrki's comment. The good property is the following if $q=p^m$ and $q'=p^n$ then $F$ is contained in $F'$ if and only if $m$ divides $n$. A counter-example to the proposition 2 is then $F:=\mathbb{F}_4$ and $F':=\mathbb{F}_8$ (respectively the field with $4$ and $8$ elements). A quick argument to show this is the multiplicativity of degrees, assume that $F'$ contains $F$ then, because we have $K:=\mathbb{F}_2$ (the primitive field) in both fields we have :

$$[F':K]=[F':F][F:K] $$

Now $[F':K]=dim_K(F')=3$ and $[F:K]=dim_K(F)=2$ hence $2$ divides $3$ which is clearly false.

For 3, you are right, they are contained in a common finite extension of their common primitive field.

For 4, you are right. It is however a good and easy exercise to show this properly.

  • 2
    $\begingroup$ Recheck 2. The field of eight elements does not have a subfield with 4 elements. $\Bbb{F}_{p^m}$ is a subfield of $\Bbb{F}_{p^n}$ if and only if $m\mid n$. Mind you, this has been covered many times on our site. For example here (This was the top item on the Related list in the right margin). $\endgroup$ Jun 8, 2015 at 18:53
  • $\begingroup$ Totally right @JyrkiLahtonen. $\endgroup$ Jun 8, 2015 at 19:51
  • $\begingroup$ @ClementGuerin, few comments are oddly disappearing, can you add Jyrki's comment to the answer? $\endgroup$ Jun 9, 2015 at 7:13
  • 1
    $\begingroup$ @JessePFrancis, I edited my answer, Jyrki's comment is still here. It is a common mistake (I made myself) to consider 2 true, but the multiplicativity of degree clearly shows that the good condition is not the divisibility of the cardinals but the divisibility of their respective $p$-logarithms. $\endgroup$ Jun 9, 2015 at 7:46

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