# Showing existence of a field extension of degree $n$ for a finite field $F$

EDIT: Just mentioning that this is a homework question.

This is my first time posting a question on math.stackexchange, so I hope you find it in your hearts to forgive any stylistic or rule transgressions I make. I have searched through quite a few of the similar threads that popped up but nothing answered my question.

The problem is as the title suggests; given a finite field $F$ and some $n > 0$, show that it has a finite field extension of degree $n$.

My attempt at a solution is as follows:

Let $|F| = p^{m}$.

Consider the splitting field of $x^{p^{mn}}-x$ over the integers modulo $p$ for some prime $p$; call it $G$. This is a finite field of order $p^{mn}$. Then it contains a subfield of size $p^{m}$, say $G'$. This is isomorphic to $F$. However, I am pretty sure $G$ does not constitute an extension of $F$.

I have tried constructing a field extension of $F$ isomorphic to $G$ by considering the image of a map $\varphi: G \rightarrow Im(\varphi)$ such that $\varphi$ restricted to G' is the isomorphism from $G$ to $F$, but I hit a wall there in showing that it was an isomorphism (briefly, I consider a n-basis for $G$ and tried defining it accordingly but wasn't able to complete it because I could not prove bijectivity).

If it is not too much trouble, I would simply prefer a tiny hint that pushes me in a promising direction.

Thanks!

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Why not look at a splitting field of $x^{p^{nm}} - x$ over $F$? – Dylan Moreland Jul 17 '12 at 1:36
I am currently looking for a blunt object to strike myself with, but in the meantime, thank you, I see it clearly now - existence, extension, Tower Law. – Tim Zhou Jul 17 '12 at 1:59

Why do you think $\,G\,$ doesn't constitute an extension of your original $\,F\,$? Because it actually is, or perhaps more formally: $\,G\,$ is a vector space of dimension $\,mn\,$ over the prime field of characteristic $\,p\,\,,\,\Bbb F_p\,$ , and $\,F\,$ is a linear space of dimension $\,m\,$ over the same prime subfield.

Let $K\leq G\,$ be a subfield of dimension $\,m\,$ over $\,\Bbb F_p\,$ . But it's not hard to show both $\,H\,\,,\,F$ are splitting fields over $\,\Bbb F_p\,$ of the same polynomial, namely $\,x^{p^m}-x\in\Bbb F_p[x]\,$ and, thus, they're isomorphic as fields, not only as vector spaces of the same dimension over the same field.

We, in fact, have just passed above over one of the proofs (or the proof) that there's only one field of a given finite cardinality u[p to isomorphism.

Well, there you have your extension of $\,F\,$...which, in fact, you did find.

Added: I couldn't see how to give "just a hint" as the OP already solved the problem. What was lacking is to get her/him convinced that she/he actually did solve the problem and, hopefully, the above will help.

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Thank you for your answer, and I am not disputing your correctness - my only issue is that I seem to have a gap in accepting this notion of "up to isomorphism". I understand $F$ and $G$ are isomorphic splitting fields. My instructor has been pretty over the map with how they use "is" and "unique up to isomorphism" in class - might be the source of my confusion. However, even if isomorphic, why is an extension of one necessarily an extension of the other?I might be pedantic with details, but I am not seeing how the extension of an isomorphic copy of $F$ is itself an actual extension of $F$. – Tim Zhou Jul 17 '12 at 2:16
Hmmm....do you accept that if $\,F\,$ is a field and $\,f(x)\in F[x]\,$ is an irreducible polynomial of degree $\,\geq 1\,$, then the field $\,K:=F[x]/(f(x))\,$ is an extension of $\,F\,$? Because if you do, and the above is one of the most basic and elementary theorems in fields extensions, then it is basically the same argument: one has that we can embed $\,F\,$ into $\,K\,$ via $\,a\to a+(f(x))\in K\,$, which is an injective ring homomorphism and etc. – DonAntonio Jul 17 '12 at 2:19
Yes the first theorem guarantees the existence of a field extension given an irreducible polynomial. Okay, I see what you mean... the theorem proves the existence of a field extension by viewing the image of F under that embedding as being F contained in the extension. So extension in this context means what you have defined; is this necessarily equivalent to the definition of an extension using the intersection? – Tim Zhou Jul 17 '12 at 2:53
What intersection are you talking about? Unless you talk of the splitting field (of a family of polynomials over a field) as the intersection of all the extension fields that contain all the roots of all the polynomials, I'm not sure I follow you...and if you meant this then yes: in your particular case it is pretty easy to show that the above mentioned intersection has to be the field $\,\Bbb F_{p^{nm}}\,$ as it has the minimal ammount of elements required to be a splitting field for $\,x^{p^{nm}}-x\in\Bbb F_p[x]\,$... – DonAntonio Jul 17 '12 at 3:37
I meant something else but I realize that it was irrelevant. I think I have a clearer picture now. Thanks for your patience with my dumb questions, it really helped. One last item: in all contexts from here on out, when we say a field $F'$ is an extension of another field $F$, we are actually saying that $F$ embeds into $F'$; $F'$ does not necessarily actually contain the actual elements of $F$, correct? For example, if $K \cong G$ and $F$ is a field containing $G$, then am I valid in saying that $F$ is an extension of $K$ as well? – Tim Zhou Jul 17 '12 at 4:08