Embed finite field in algebraic closed field

Question

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.

Answer 1

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 $F \to \Omega.$

Answer 2

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 $P(X) \mid X^q - X$.

3. 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

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