Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $E$ and $F$ are subfields of a finite field $K$ and $E\cong F$, prove that $E = F$.

A finite field is a simple extension of each of its subfields and $\mathbb{Z}_p$ is a subfield of every finite field. Hence $E\cong \mathbb{Z}_p(u)$ and $F\cong \mathbb{Z}_p(v)$ for some $u,v\in K$. Proving that $u = v$ given $\mathbb{Z}_p(u)\cong \mathbb{Z}_p(v)$ may be stronger than I need though, since $u$ and $v$ could be different generators for the same set.

Can anyone help me with this? Many thanks.

Edit: This is a Galois Theory free zone.

share|cite|improve this question
Fyi: using $\mathbb{Z}_p$ to denote the field with $p$ elements is sort of like pushing mongo. Use $\mathbb{F}_p$. – Kevin Jul 7 '12 at 8:49
up vote 7 down vote accepted

The isomorphism in particular implies that the orders of $E$ and $F$ are equal. That's all we need. Let us assume that you know that a finite field with $q=p^k$ elements is a splitting field for the polynomial $x^q-x=0$. Thus each of $E$ and $F$ consists of the roots (in $K$) of $x^q-x=0$. That implies that $E=F$.

share|cite|improve this answer
It seems intuitively obvious, but how do we know all the roots of $x^q - x$ are distinct? – Thoth Jul 7 '12 at 4:58
@Nollie: The derivative of $x^q-x$ is $$qx^{q-1}-1=0\cdot x^{q-1}-1=-1$$ which is relatively prime to $x^q-x$. Therefore the polynomial $x^q-x$ is separable, i.e. its roots are distinct. – Zev Chonoles Jul 7 '12 at 5:02
By looking at the derivative. However, the question makes me think you have not seen the result about $x^q-x$. – André Nicolas Jul 7 '12 at 5:02
If we have a field extension $L/K$ and the polynomial $f\in K[x]$ splits completely in $L$ (i.e. $L$ contains every root of $f$), then there is a unique subfield $F\subseteq L$ that is the splitting field of $f$ over $K$ inside $L$, namely, $$F=K(\alpha_1,\ldots,\alpha_t)$$ where the $\alpha_i$ are the roots of $f$. Because both $E$ and $F$ are splitting fields for $x^q-x\in \mathbb{F}_q[x]$ inside the finite field $K$, they must be equal. – Zev Chonoles Jul 7 '12 at 5:07
The theorem that a field of $q$ elements is a splitting field of $x^q-x$ over the ground field makes the fact that there are no multiple roots irrelevant. Which is why it is not mentioned in my answer. – André Nicolas Jul 7 '12 at 5:08

Do you know Galois Theory? If so, it's easy: The Galois group of any finite field over any other is cyclic, so it has only one subgroup of any given order, so only one field of any given index.

share|cite|improve this answer
Sorry I should have mentioned, I have no Galois Theory at my disposal. Group Theory up through Sylow Theorems, and Field Theory through splitting fields and general results on the classification of finite fields. Also undergrad ring theory, but that's probably not important. – Thoth Jul 7 '12 at 4:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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