Let $\Omega$ be an algebraic closed field of characteristic $p$, then for any field $F$ of $q$ ($=p^a$) elements, $F$ can be embedded in $\Omega.$

I need above property to proof that every finite fields with the same order are isomorphic, but I don't know how to proof the above property.

Here is what I think: Every elements in $F$ satisfy the equation $x^q=x$. Now $x^q=x$ can be solved in $\Omega$, and its solutions form a subfield in $\Omega$. But I don't see why $F$ must isomorphic to such subfield.

So how to proof the property above?

BTW, it's known that there is an unique subfield of $q$ elements in $\Omega$, I don't know whether it helps or not.

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    $\begingroup$ Often $\Bbb{F}_q$ is constructed as a splitting field of $x^q-x$ over the prime field. A basic result from the theory of field extensions is that a splitting field is unique up to (a non-unique) isomorphism. Clearly you can construct a splitting field inside $\Omega$, so... $\endgroup$ – Jyrki Lahtonen Jan 18 '15 at 8:31
  • $\begingroup$ OK, I accept this answer, though I think it's a bit strange to use such result in this situation... $\endgroup$ – CYC Jan 18 '15 at 8:36
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    $\begingroup$ You do not need that property to prove two finite fields of the same order are isomorphic. None of the basic theorems about finite fields (existence, uniqueness up to isomorphism based on order, cyclicity of the nonzero elements, being perfect,...) require using algebraically closed fields. $\endgroup$ – KCd Jan 18 '15 at 8:37
  • $\begingroup$ It is a bit roundabout way of doing it. But quite economical, I think! $\endgroup$ – Jyrki Lahtonen Jan 18 '15 at 8:37
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    $\begingroup$ It's not economical if you think about the time invested in proving algebraic closures exist (even just for finite fields). $\endgroup$ – KCd Jan 18 '15 at 8:38

Lemma: Let $K/k$ be an algebraic field extensions and let $\phi: k \to C$ be a ring homomorphism where $C$ is an algebraically closed field. Then there exists a ring homomorphism $\sigma : K \to C$ which extends $\phi.$

Proof: Let $\mathcal F$ be the set of pairs $(F, f)$ where $k \subseteq F \subseteq K$ is a field and $f: F \to C$ is a ring homomorphism such that $f|_k = \phi.$ Since $(k, \phi ) \in \mathcal F,$ it is non-empty. Define a relation on $\mathcal F$ as follows: $(F, f) \leq (F', f') \Leftrightarrow F \subseteq F'$ and $f'|_F = f.$ This will give a partial order relation on $\mathcal F.$ Show that every chain has an upper bound and hence by Zorn's Lemma, it has a maximal element, say, $(F, \sigma).$ We claim that $F = K.$ If not, then there exists a $x \in K$ such that $x \notin F.$ Let $g(X) \in F[X]$ be the minimal polynomial of $x$ over $F.$ Note that $F(x) \cong F[X]/(g(X)).$ For any $p(X) \in F(X),$ let $p^{\sigma}(X)$ be a polynomial in $C[X]$ obtained by applying $\sigma$ to the coefficients of $p(X).$ Now $g^{\sigma}(X)$ has a root in $C,$ say $\alpha.$ Define a map $\psi : F[X] \to C$ by $p(X) \mapsto p^{\sigma}(\alpha).$ Then this amp induces a map $\overline{\psi}: F[X]/(g(X)) \to C.$ Using $F(x) \cong F[X]/(g(X))$ we see that the pair $(F(x), \overline{\psi}) \in \mathcal F$ and $(F, \sigma) \leq (F(x), \overline{\psi}),$ contradicting the maximality of $(F, \sigma).$ Hence $F = K.$

Now we use the above Lemma in this case. As you have already noted that $F$ is an algebraic extension of $\mathbb Z/ p\mathbb Z$ and we have an inclusion $\mathbb Z/ p\mathbb Z \to \Omega.$ So it has an extension $F \to \Omega.$

  • $\begingroup$ In the last line: why is $F \to \Omega$ an embedding? $\endgroup$ – No One Jul 18 at 3:45

I would give an economical argument:

  1. For every field $F$ of cardinality $q=p^d$ we have $x^q -x=0$ for all $x \in F$. (this is easy)

  2. For every $P \in \mathbb{F}_p[X]$ irreducible of degree $d$ we have $P \mid X^{q}-X$

Indeed, work in the field $F = \mathbb{F}_p[X]/P(X)$. The polynomial $P(X)$ has a root in $F$ which is also a root of $X^{q}-X$. So $P(X)$, $X^q-X$ have a common factor and therefore $P(X) \mid X^q - X$.

  1. Let $F'$ of degree $d'$, $F$ of degree $d$ and $d' \mid d$ . Then $F'$ imbeds into $F$.

Indeed: the extension $F'/\mathbb{F}_p$ is simple ( it's finite and separable) so $F'= \mathbb{F}_p[X]/Q(X)$.

From $1$. we have: $$X^q - X = \prod_{\alpha \in F} (X-\alpha)$$

However, from $2.$ we have

$Q(X) \mid (X^{q'}- X) \mid (X^q-X)$

It follows that $Q(X)$ has $d$ roots in $F$.

We are done.

Finding effectively a root of $Q(X)$ in $F$ might be a bit involved. Here is an example where $p=3$, $Q(X)=1+2 X + X^3$ and $F= \mathbb{F}_3[X]/(2 + X + 2 X^2 + X^4 + X^6)$. We need to find $a + b X + c X^2 + d X^3 + e X^4 + f X^5$ so that the remainder of $$1 + 2(a + b X + c X^2 + d X^3 + e X^4 + f X^5) + (a + b X + c X^2 + d X^3 + e X^4 + f X^5)^3 $$ after dividing by $2 + X + 2 X^2 + X^4 + X^6$ is $\ 0\!\!\! \mod 3$.

By brute force we find $3$ solutions

$$(a,b,c,d,e,f)= (0, 1, 0, 1, 2, 0),\ (1, 1, 0, 1, 2, 0), \ \text{or}\ (2, 1, 0, 1, 2, 0)$$

Therefore, we have a morphism of fields

$$\mathbb{F}_3[X]/(1+2X+X^3) \to \mathbb{F}_3[X]/(2 + X + 2 X^2 + X^4 + X^6)\\ X\mapsto X + X^3 + 2 X^4 $$


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