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It's well known there exists only one finite field with $p^n$ elements, up to isomorphism, where $p$ is a prime and $n \geq 1$.

Let $n = m t$, where $m, t > 1$.

Then $F = Z_p[X]/(f(X))$ is a field with $p^n$ elements, provided $f$ is some irreducible polynomial of degree $n$ in $Z_p$.

By the same token, $G = Z_p[X]/(g(X))$ is a field with $m$ elements ($g$ irreducible of degree $m$). Then $G$ has $p^m$ elements.

Finally let: $H = G[X]/(h(X))$ where $h$ is an irreducible polynomial of degree $t$ with coefficients in $G$.

Now, in my understanding $F$ and $H$ have both $p^n$ elements. So my question is:

Are $F$ and $H$ isomorphic?

kimchi lover
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user3084125
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    Yes. Like you said, it's unique up to isomorphism. You answered your own question! – anon Dec 24 '20 at 19:51
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    Unless you are focused on programming finite fields into a computer, it's not always a good idea to imagine finite fields so explicitly as a quotient of a polynomial ring. But if you do, consider not always writing the indeterminate of the polynomial ring as $X$: the $X$ in $\mathbf F_p[X]/(f(X))$ would not be playing the same role as the $X$ in the other two finite fields: the isomorphism need not map the congruence class of $X$ to the congruence class of $X$. – KCd Dec 24 '20 at 19:55
  • This is a FAQ in a way, but it may be difficult to realize that if you have not internalized the basic existence/uniqueness result of finite fields. It is possible to prove this without invoking the result of uniqueness of splitting fields (see egreg's textbook answer), but those are a bit kludgy (which is the reason those textbooks resort to that argument). An alternative way is hinted at in this recent question. If I find the time, I will add an answer there, but the question already has a link to lecture notes, where it is discussed. – Jyrki Lahtonen Dec 25 '20 at 08:19

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This relies on a general fact about splitting fields.

Let $F$ be a field and $f(X)\in F[X]$ be a monic polynomial. An extension field $K$ of $F$ is a splitting field for $f$ if

  1. $f(X)=(X-a_1)(X-a_2)\dots(X-a_k)$ in $K[X]$ (the roots need not be distinct);
  2. $K=F(a_1,a_2,\dots,a_k)$

Theorem. If $K_1$ and $K_2$ are splitting fields of $f(X)\in F[X]$, then there exists a field isomorphism $\varphi\colon K_1\to K_2$ leaving $F$ pointwise fixed.

The proof is longish and can be found in any book on Galois theory, because it's a basic tool thereof.

Now consider the polynomial $X^{p^n}-X\in\mathbb{F}_p[X]$, where $\mathbb{F}_p$ is the $p$-element field (which is unique up to unique isomorphism).

Let $K$ be a splitting field of $f(X)$. Then $f(X)$ has $p^n$ distinct roots in $K$ (because the derivative of the polynomial is $-1$). On the other hand, the set of roots of $f(X)$ is a subfield of $K$: indeed, if $a,b$ are roots, then $$ (a+b)^{p^n}-(a+b)=a^{p^n}+b^{p^n}-a-b=0 $$ so $a+b$ is a root of $f$. Analogously $$ (ab)^{p^n}-ab=a^{p^n}b^{p^n}-ab=ab-ab=0 $$ and it's easy to check the reciprocals. Since also $0$ and $1$ are roots we're done.

Thus $K$ is the set of all roots of $f$ and therefore $|K|=p^n$.

Conversely, if $K$ is a field with $p^n$ elements, then the same argument as before shows that $X^{p^n}-X$ has $p^n$ distinct roots in $K$, so $K$ is a splitting field for $f(X)$.

Uniqueness up to isomorphism now follows from the theorem above.

egreg
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