Theorem 1.6.1 (Kronecker) If $k$ is a field and $f(x)\in k[x]$, there exists an extension field $K/k$ with $f$ a product of linear polynomials in $K[x]$.

Proof.

The proof is by induction on $\deg(f)$.

• If $\deg(f)=1$, then $f$ is linear and we can choose $K=k$.

• If $\deg(f)>1$, write $f=pg$, where $p(x),g(x)\in k[x]$ and $p$ is irreducible.

  Let $F=k/(p(x))$, then $F$ containing $k$ and a root $z$ of $p$.

  Hence, in $F[x]$, there is $h(x)$ with $p=(x-z)h$, and so $f=(x-z)hg$.

  By induction, there is a field $K$ containing $F$ (and hence $k$) so that $hg$, and hence $f$, is a product of linear factors in $K[x]$. $\square$

Definition 1.6.1 If $K/k$ is an extension field and $f(x)\in k[x]$ is nonconstant, then $f$ splits over $K$ if $f(x)=a(x-z_1)\cdots(x-z_n)$, where $z_1,\dots,z_n\in K$ and $a\in k$. An extension field $E/k$ is called a splitting field of $f$ over $k$ if splits over $E$, but $f$ does not split over any proper subfield of $E$.

Corollary 1.6.1 If $k$ is a field and $f(x)\in k[x]$, then a splitting field of $f$ over $k$ exists.

Proof.

By Kronecker's Theorem, there is an extension field $K/k$ such that $f$ splits in $K[x]$; say, $f(x)=a(x-\alpha_1)\cdots(x-\alpha_n)$. The subfield $E=k(\alpha_1,\dots,\alpha_n)$ of $K$ is a splitting field of $f$ over $k$. $\square$

A splitting field of $f(x)\in k[x]$ is a smallest extension field $E/k$ containing all the roots of $f$. We say "a" splitting field instead of "the" splitting field because it is not obvious whether an two splitting fields of $f$ over $k$ are isomorphic (they are). Analysis of this technical point will not only prove uniqueness of splitting fields, it will enable us to prove that any two finite fields with the same number of elements are isomorphic.

Example 1.6.1 Let $f(x)=x^n-1\in k[x]$ for some field $k$, and let $E/k$ be a splitting field. Because every finite subgroup of the multiplicative group $k^\times$ is cyclic, the set of all $n$th roots of identity in $E$ is a cyclic group. Then let $\omega$ be a generator of it. It follows that $k(\omega)=E$ is a splitting field of $f$.

Proposition 1.6.1 Let $p$ be prime, then let $k$ be a field. If $f(x)=x^p-c\in k[x]$ and $\alpha$ is a $p$th root of $c$ (in some splitting field), then either $f$ is irreducible in $k[x]$ or $c$ has a $p$th root in $k$. In either case, if $k$ contains the $p$th roots of identity, then $k(\alpha)$ is a splitting field of $f$.

Theorem 1.6.2 (Galois) If $p$ is a prime and $n$ is a positive integer, then there exists a field having exactly $p^n$ elements.

Proof.

Write $q=p^n$, and consider the polynomial $g(x)=x^q-x\in \mathbb Z_p[x]$.

By Kronecker's Theorem, there is an extension field $K/\mathbb Z_p$ with $g$ a product of linear factors in $K[x]$. Define
$$E=\{\alpha\in K;g(\alpha)=0\};
$$
that is, $E$ is the set of all the roots of $g$. Since the derivative $g'(x)=qx^{q-1}-1=p^nx^{q-1}-1=-1$, we have $\gcd(g,g')=1$. Therefore, all the roots of $g$ are distinct; that is, $E$ has exactly $q=p^n$ elements.

The theorem will follow if $E$ is a subfield of $K$. Of course, $1\in E$. If $a,b\in E$, then $a^q=a$ and $b^q=b$. Therefore, $(ab)^q=a^qb^q=ab$, and $ab\in E$. $(a-b)^q=a^q-b^q=a-b$, so that $a-b\in E$. Finally, if $a\ne 0$, $a^{q-1}=1$ gives $a^{q-2}$ is the inverse of $a$. $\square$

After this, we write $\mathbb F_{p^n}$ as a finite field having $p^n$ elements.

Corollary 1.6.2 For every prime $p$ and every integer $n\ge 1$, there exists an irreducible polynomial $g(x)\in \mathbb F_p[x]$ of degree $n$. In fact, if $\alpha$ is a primitive element of $\mathbb F_{p^n}$, then its minimal polynomial $g(x)=\operatorname {irr}(\alpha,\mathbb F_p)$ has degree $n$.

Proof.

Let $E/\mathbb F_p$ be an extension field with $p^n$ elements, and let $\alpha \in E$ be a primitive element.

Let $g(x)=\operatorname{irr}(\alpha,\mathbb F_p)\in \mathbb F_p[x]$, then $g(x)$ is irreducible.

If $\deg(g)=d$, the Proposition 1.5.6 gives $[\mathbb F_p[x]/(g):\mathbb F_p]=d$; and $\mathbb F_p[x]/(g)\cong \mathbb F_p(\alpha)=E$.

Therefoer $d=[E:\mathbb F_p]=n$. $\square$

Lemma 1.6.1 Let $\varphi:k\to k'$ be an isomorphism of fields, and let $\varphi_*:k[x]\to k'[x]$ be the ring isomorphism of the Corollary 1.2.1:
$$\varphi_*:g(x)=a_0+a_1x+\cdots+a_nx^n\mapsto g'(x)=\varphi(a_0)+\varphi(a_1)x+\cdots+\varphi(a_n)x^n.
$$
Let $f(x)\in K[x]$ and $f'(x)=\varphi_*(f)\in k'[x]$, If $E$ is a splitting field of $f$ over $k$ and $E'$ is a splitting field of $f'$ over $k'$, then there is an isomorphism $\Phi:E\to E'$ extending $\varphi$:
$$%https://darknmt.github.io/res/xypic-editor/#eyJub2RlcyI6W3sicG9zaXRpb24iOlswLDBdLCJ2YWx1ZSI6IkUifSx7InBvc2l0aW9uIjpbMSwwXSwidmFsdWUiOiJFJyJ9LHsicG9zaXRpb24iOlswLDFdLCJ2YWx1ZSI6ImsifSx7InBvc2l0aW9uIjpbMSwxXSwidmFsdWUiOiJrJyJ9XSwiZWRnZXMiOlt7ImZyb20iOjIsInRvIjozLCJ2YWx1ZSI6IlxcdmFycGhpIiwibGFiZWxQb3NpdGlvbiI6InJpZ2h0In0seyJmcm9tIjowLCJ0byI6MiwiaGVhZCI6Im5vbmUifSx7ImZyb20iOjEsInRvIjozLCJoZWFkIjoibm9uZSJ9LHsiZnJvbSI6MCwidG8iOjEsImxpbmUiOiJkYXNoZWQiLCJ2YWx1ZSI6IlxcUGhpIn1dfQ==
\xymatrix{
E \ar@{-}[d] \ar@{-->}[r]^{\Phi} & E' \ar@{-}[d] \\
k \ar@{->}[r]_{\varphi} & k'
}
$$
Theorem 1.6.3 If $k$ is a field and $f(x)\in k[x]$, then any two splitting fields of $f$ over $k$ are isomorphic via an isomorphism that fixes $k$ pointwise.

Corollary 1.6.3 (Moore) Any two finite fields having exactly $p^n$ elements are isomorphic.

Proof.

If $E$ is a field with $q=p^n$ elements, then Lagrange's Theorem applied to the multiplicative group $E^\times$ shows that $a^{q-1}=1,\forall a\in E^\times$.

So ever element of $E$ is a root of $f(x)=x^q-x\in \mathbb F_p[x]$, and so $E$ is a splitting field of $f$ over $\mathbb F_p$. $\square$

Finite fields are often called Galois fields in honor of their discoverer. In light of Corollary 1.6.3, we nay speak of the field with $q$ elements, where $q=p^n$ is a power of a prime $p$, and we denote it by
$$\mathbb F_q.
$$