Suppose $F$ is a finite field of characteristic $p$ ($p$ a prime). Prove $\exists u \in F$ such that $F = \mathbb{F}_{p}(u)$. Here, $\mathbb{F}_{p}$ denotes the field with $p$ elements.

Here is what I know:

$F^{\times}$, the group under multiplication consisting of $F - \{ 0 \}$, is cyclic since $F$ is a finite field. I also know since $F$ is of characteristic $p$ that $\mathbb{F}_{p}$ is a subgroup of $F$.

Now, if we let $u$ be the generator of $F^{\times}$, then every nonzero element can be written as $u^{m}$ for some $m \in \mathbb{N}$. This shows that $F \subseteq \mathbb{F}_{p}(u)$. Also, $F$ is field that contains $u$ and $\mathbb{F}_{p}$, so it must contain the smallest field containing both of these, which is $\mathbb{F}_{p}(u)$. Thus, we have $\mathbb{F}_{p}(u) = F$.

So, if I can prove the claim, then what is my question? Well, we proved the sets are equal. But we didn't prove they are isomorphic. Are they? What is the homomorphism? Do you just assume the operations on $\mathbb{F}_{p}(u)$ and $\mathbb{F}$ behave the same?

Proving two sets are equal is not enough to prove the underlying structure is the same, and this is what I am doubting. Is the field $\mathbb{F}_{p}(u)$ the same as the field $F$?

  • $\begingroup$ All you did happens within $\;\Bbb F_p\;$ and thus the sets are equal, which is more than isomorphic. $\endgroup$ – Timbuc Oct 23 '14 at 23:06
  • $\begingroup$ @Timbuc Equal is more than isomorphic? We know every element in each field is in the opposite field (the definition of equality). But do we know that the elements behave the same way in both fields? I guess you're right -- in my proof I had already assumed they do behave the same way. $\endgroup$ – layman Oct 23 '14 at 23:07
  • $\begingroup$ Yes we know, @Math : operations are all the time modulo $\;p\;$ . $\endgroup$ – Timbuc Oct 23 '14 at 23:09
  • $\begingroup$ You should probably change the title to match the question ("Suppose $F$ is a finite field of characteristic $p$...") I clicked through expecting to comment that the result you are asked for is false. $\endgroup$ – Stephen Oct 23 '14 at 23:13
  • 1
    $\begingroup$ Equality is an isomorphism, but isomorphisms aren't always equalities. So equality is stronger. $\endgroup$ – Arthur Oct 23 '14 at 23:13

Think about $\mathbb F_p$ as being the field generated by $1$ in $F$. That means that $$\mathbb F_p = \{ 1_F, 1_F+1_F,..., 1_F+1_F+...+1_F \}$$

Then $\mathbb F_p \subset F$ and $u \in F$. Now when you construct $\mathbb F_p(u)$ you construct it as a subfield of $F$. If the underlying set is the same, it follows that the subfield contains all elements.

  • $\begingroup$ Yes, you are right. $\mathbb{F}_{p}$ as a subfield of $F$ is precisely the elements $\{1_{F}, 1_{F} + 1_{F}, \dots \}$. Thanks for the insight. $\endgroup$ – layman Oct 23 '14 at 23:11

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.