Splitting Fields

Since $n\ge 1$, the polynomial is nonconstant. By [F1], choose an extension $H\mathrm{/}F$ and a root $\alpha \in H$, and let $K=F(\alpha )\subseteq H$.

By [F2], $f=(x-\alpha )g$ for some $g\in K[x]$. Since $f$ is nonzero, so is $g$; also $x-\alpha$ is nonzero. Thus [F3] gives $n=1+\mathrm{deg}g$ and hence $\mathrm{deg}g=n-1$.

If $g$ splits over $L$, adjoining the factor $x-\alpha$ to its linear factorisation gives a linear factorisation of $f$ over $L$. This also covers $n=1$, when $g$ is a nonzero constant and its factor product is empty.

Let $P(n)$ assert the theorem for every field and every nonzero polynomial of degree $n$. We prove $P(n)$ for all $n$ by [F1].

If $n=0$, the polynomial is a nonzero constant. It splits over $F$, and its empty root set generates $F$, so $F$ itself is a splitting field.

Let $n\ge 1$ and assume $P(m)$ for every $m<n$. By [F2], choose a root $\alpha$, put $K=F(\alpha )$, and choose $g\in K[x]$ with $f=(x-\alpha )g$ and $\mathrm{deg}g=n-1<n$.

The induction hypothesis over the field $K$ gives a splitting field $L\mathrm{/}K$ of $g$. Then $f$ splits over $L$ by [F2], and $L=K(\text{roots of }g)=F(\alpha ,\text{roots of }g)$ is generated over $F$ by roots of $f$. Thus $L\mathrm{/}F$ is a splitting field of $f$.

Steps 1.2, 1.3, and 2.1 verify the strong-induction hypothesis at every $n$, so [F1] proves the theorem.

If $m=0$, the product is $1$, whose empty root set has splitting field $F$ by [F4]. Assume now that $m>0$. By [F1], $h$ is nonzero, so [F2] gives a splitting field $E\mathrm{/}F$ of $h$.

Each $f_j$ divides $h$ in $E[x]$. Since $h$ is a product of linear factors there, unique factorisation in the ring $E[x]$ from [F3] shows that every irreducible factor of $f_j$ is linear; hence every $f_j$ splits over $E$.

An element of an extension is a root of $h=f_1\cdots f_m$ exactly when it is a root of at least one $f_j$, because a field has no zero divisors. Thus the roots generating $E$ are precisely the union of the roots of the family, and [F4] makes $E$ its splitting field.

Both $f$ and $g$ split over $E_f{E}_g$ because the composite contains their splitting fields. Hence their product $fg$ splits there.

In the field $\mathrm{\Omega }$, an element $a$ is a root of $fg$ exactly when $f(a)g(a)=0$, hence exactly when it is a root of $f$ or of $g$. Therefore the field generated by the roots of $fg$ is the field generated jointly by $E_f$ and $E_g$, which is $E_f{E}_g$ by [F2].

Apply [F1] to $\sigma$ and the indeterminate $x\in F^{\mathrm{\prime }}[x]$. This gives ${\sigma }_{\ast }$ with the displayed coefficient formula. Applying the same construction to ${\sigma }^{-1}$ gives its inverse, so ${\sigma }_{\ast }$ is an isomorphism.

For $f={\sum }_i{a}_i{x}^i$, the homomorphism laws give $\tau (f(\alpha ))={\sum }_i\sigma (a_i)\tau (\alpha {)}^i=({\sigma }_{\ast }f)(\tau (\alpha ))$. Thus $f(\alpha )=0$ if and only if $({\sigma }_{\ast }f)(\tau (\alpha ))=0$, because $\tau$ is injective.

Applying the construction of step 1.1 to $\tau$ gives a coefficientwise isomorphism ${\tau }_{\ast }:E[x]\to E^{\mathrm{\prime }}[x]$ extending ${\sigma }_{\ast }$. It transports every product factorisation and, in particular, a linear factorisation of $f$ to one of ${\sigma }_{\ast }f$. Since $\tau$ is bijective, step 1.2 gives a bijection of root sets, and $\tau (F(\text{roots of }f))=F^{\mathrm{\prime }}(\text{roots of }{\sigma }_{\ast }f)$. The splitting-field claim follows from [F3].

By [F1], $p^{\mathrm{\prime }}$ is monic and irreducible. Since $p(\alpha )=0$ and $p^{\mathrm{\prime }}(\beta )=0$, [F2] identifies $p$ and $p^{\mathrm{\prime }}$ as the respective minimal polynomials. In particular they have the same degree $n$.

By [F3], each element of $F(\alpha )$ has a unique form ${\sum }_{i<n}{a}_i{\alpha }^i$. Define $\tilde{\sigma }({\sum }_{i<n}{a}_i{\alpha }^i)={\sum }_{i<n}\sigma (a_i){\beta }^i$. The quotient and root universal property in [F4], transported through [F1], shows this is a field homomorphism extending $\sigma$ and sending $\alpha$ to $\beta$.

Repeating the construction for ${\sigma }^{-1}$, with $\beta$ and $\alpha$ interchanged, gives an inverse. Hence $\tilde{\sigma }$ is an isomorphism. Its values on the unique power-basis expressions are forced, so it is unique.

Let $P(n)$ assert the theorem for every such datum of degree $n$. If $n=0$, both root sets are empty, so $E=F$ and $E^{\mathrm{\prime }}=F^{\mathrm{\prime }}$; the required extension is $\sigma$.

Let $n\ge 1$ and assume $P(m)$ for all $m<n$. Choose a root $\alpha \in E$ of $f$, and let $p$ be its minimal polynomial over $F$. By [F2], $p\mid f$. Thus $p^{\mathrm{\prime }}={\sigma }_{\ast }p$ divides $f^{\mathrm{\prime }}$ in $E^{\mathrm{\prime }}[x]$. Since $f^{\mathrm{\prime }}$ is a product of linear factors there, unique factorisation in $E^{\mathrm{\prime }}[x]$ from [F5] makes $p^{\mathrm{\prime }}$ split over $E^{\mathrm{\prime }}$; choose a root $\beta \in E^{\mathrm{\prime }}$ of $p^{\mathrm{\prime }}$.

By [F3], $\sigma$ extends to an isomorphism ${\sigma }_1:F(\alpha )\to F^{\mathrm{\prime }}(\beta )$. Factor $f=(x-\alpha )g$ in $F(\alpha )[x]$. Applying coefficient transport by ${\sigma }_1$ gives $f^{\mathrm{\prime }}=(x-\beta )g^{\mathrm{\prime }}$ with $g^{\mathrm{\prime }}=({\sigma }_1{)}_{\ast }g$, and $\mathrm{deg}g=n-1$.

The field $E$ is a splitting field of $g$ over $F(\alpha )$: it splits $g$, and it is generated by $\alpha$ together with the other roots of $f$, which are the roots of $g$. Similarly, $E^{\mathrm{\prime }}$ is a splitting field of $g^{\mathrm{\prime }}$ over $F^{\mathrm{\prime }}(\beta )$. The induction hypothesis extends ${\sigma }_1$ to an isomorphism $E\to E^{\mathrm{\prime }}$.

The base case and inductive step establish $P(n)$ for all $n$ by [F1].

Apply [F1] to the identity isomorphism of $F$. It transports $f$ to itself and therefore extends to an isomorphism $E\to E^{\mathrm{\prime }}$ fixing $F$.

This includes nonzero constants, whose splitting fields have empty root sets and both equal $F$.

Let $P(n)$ be the full assertion in the Statement, quantified over all base fields. For $n=0$, take $E=F$ and $k=0$. The vector $1$ spans $F$ and the number of displayed empty monomials is $1=0!$.

Let $n\ge 1$ and assume $P(m)$ for every $m<n$. By [F2], choose a root $\alpha$ in the extension $K=F(\alpha )$ and write $f=(x-\alpha )g$ with $\mathrm{deg}g=n-1$. If the minimal polynomial of $\alpha$ has degree $d$, then $d\ge 1$, [F3] and [F7] give $d\le n$, and [F4] gives the $F$-basis $1,\alpha ,\dots ,{\alpha }^{d-1}$.

Apply the induction hypothesis over $K$ to $g$. It gives a splitting field $E\mathrm{/}K$ spanned over $K$ by at most $(n-1)!$ monomials in roots ${\alpha }_2,\dots ,{\alpha }_k$ of $g$. Multiplying those monomials by $1,\alpha ,\dots ,{\alpha }^{d-1}$ spans $E$ over $F$ by monomials in roots of $f$.

The number of resulting monomials is at most $d(n-1)!\le n(n-1)!=n!$ by $d\le n$ and [F5]. Moreover $E=K(\text{roots of }g)=F(\alpha ,\text{roots of }g)$, so it is a splitting field of $f$.

The base case and inductive step establish $P(n)$ for every natural $n$ by [F1].

Since $\alpha \mathrm{\ne }0$, [F1] gives ${\alpha }^{-1}$. Also $a={\alpha }^n\mathrm{\ne }0$, because a product of nonzero field elements is nonzero.

Conversely, if $\beta =\zeta \alpha$ and ${\zeta }^n=1$, then [F2] gives ${\beta }^n={\zeta }^n{\alpha }^n=a$.

If ${\beta }^n=a$, put $\zeta =\beta {\alpha }^{-1}$. By [F2], ${\zeta }^n={\beta }^n({\alpha }^n{)}^{-1}=a{a}^{-1}=1$, and $\beta =\zeta \alpha$.

Thus the root set in $E$ is exactly $\alpha {\mu }_n(E)$. The field generated by that set equals $F(\alpha ,{\mu }_n(E))$: it contains $\alpha =1\alpha$, and from any root $\zeta \alpha$ it recovers $\zeta =(\zeta \alpha ){\alpha }^{-1}$; the reverse containment follows because each $\zeta \alpha$ is a root. Now [F3] gives the splitting-field assertion.

By [F1], $\tau$ is injective and hence an isomorphism from $E$ onto its image. Since it fixes the coefficients of $f$, direct evaluation gives $f(\tau (\alpha ))=\tau (f(\alpha ))=0$ for every root $\alpha$ of $f$.

The distinct-root set $X$ is finite by [F3], and step 1.1 restricts $\tau$ to an injection $X\to X$. By [F4], this restriction is a bijection, so $\tau$ permutes the roots. This remains true when $X$ is empty.

The image $\tau (E)$ contains $F$ and every root of $f$ by step 2.1. Since those elements generate $E$ by [F5], one has $E\subseteq \tau (E)\subseteq E$. Thus $\tau$ is surjective and is an $F$-automorphism.

Every element of $K$ is algebraic over $F$, hence also algebraic over $E$ because the same polynomial lies in $E[x]$. Thus $K\mathrm{/}E$ is algebraic.

Fix $\alpha \in K$. Let $m_F\in F[x]$ and $m_E\in E[x]$ be its minimal polynomials. By [F2], $m_E$ divides $m_F$ in $E[x]$.

Normality of $K\mathrm{/}F$ makes $m_F$ split over $K$. In the unique factorisation domain $K[x]$ from [F3], every divisor of that product of linear factors is itself a product of linear factors, so $m_E$ splits over $K$. Since $\alpha$ was arbitrary, [F1] makes $K\mathrm{/}E$ normal.

The intersection $K$ is an intermediate field of $M\mathrm{/}F$. Since $M\mathrm{/}F$ is algebraic, every element of $K$ is algebraic over $F$.

Fix $\alpha \in K$ and let $m\in F[x]$ be its minimal polynomial from [F2]. For every $i\in I$, one has $\alpha \in E_i$, so [F1] makes $m$ split over $E_i$.

Because $I$ is nonempty, choose $i_0\in I$ and write the linear factorisation of $m$ in $E_{i_0}[x]\subseteq M[x]$. For any $i\in I$, a linear factorisation also exists in $E_i[x]\subseteq M[x]$. Uniqueness of factorisation in the ring $M[x]$ from [F3] shows that the same roots, with the same multiplicities, occur in both factorizations. Hence every root from the first factorisation lies in every $E_i$, and therefore in $K$.

Thus the minimal polynomial of every $\alpha \in K$ splits over $K$. Together with algebraicity from step 1.1, [F1] shows that $K\mathrm{/}F$ is normal. The nonempty hypothesis was used in step 2.1; without it the intersection convention could give the ambient $M$, which need not be normal.

Fix $\alpha \in E$ and let $m$ be its minimal polynomial over $F$. By [F5], choose a splitting field $\mathrm{\Omega }\mathrm{/}E$ of $m$, and let $\beta$ be any root of $m$ in $\mathrm{\Omega }$. By [F1], the identity on $F$ extends to an isomorphism ${\sigma }_0:F(\alpha )\to F(\beta )$ sending $\alpha$ to $\beta$.

The field $E$ is a splitting field of $f$ over $F(\alpha )$, because it splits $f$ and is generated by its roots. The field $E(\beta )$ is a splitting field of $f$ over $F(\beta )$ for the same reason. Since ${\sigma }_0$ fixes the coefficients of $f$, [F2] extends it to an isomorphism $\sigma :E\to E(\beta )$.

Every generator of $E$ over $F$ is a root of $f$. The map $\sigma$ fixes $F$ and therefore carries each such generator to another root of $f$, all of which already lie in $E$. Hence $\sigma (E)\subseteq E$. But $\sigma (E)=E(\beta )$ by surjectivity, so $\beta \in E$.

Every root $\beta$ of the minimal polynomial $m$ lies in $E$, so $m$ splits over $E$. Since $\alpha$ was arbitrary and $E\mathrm{/}F$ is algebraic by hypothesis, [F4] proves normality.

The union $U={\bigcup }_{T_0\subseteq T,T_0\text{ finite}}F(T_0)$ is a subfield of $E$: sums, products, inverses, and pairs of elements lie in the field generated by the union of their two finite supports. It contains $F\cup T$, so [F1] gives $E=F(T)\subseteq U$, while the reverse inclusion is immediate.

Fix $\beta \in E$. By step 1.1, choose a finite set $T_0=\{t_1,\dots ,t_m\}\subseteq T$ with $\beta \in F(T_0)$. Let $p_j$ be the minimal polynomial of $t_j$ over $F$, and let $K\subseteq E$ be the field generated by all roots in $E$ of the product $p_1\cdots p_m$. If $m=0$, take the product to be $1$ and $K=F$.

Each $p_j$ splits over $E$ by hypothesis. Their product is monic and hence nonzero, so $K$ is its splitting field and contains every $t_j$; hence $\beta \in F(T_0)\subseteq K$. Also $K\mathrm{/}F$ is algebraic because $K\subseteq E$ and $E\mathrm{/}F$ is algebraic. By [F3], $K\mathrm{/}F$ is normal.

Let $m_{\beta }$ be the minimal polynomial of $\beta$ over $F$. Since $\beta \in K$ and $K\mathrm{/}F$ is normal, $m_{\beta }$ splits over $K$, hence over $E$. This holds for every $\beta \in E$, so [F4] proves that $E\mathrm{/}F$ is normal.

If $m=0$, the generating hypothesis says $E=F$, and the nonzero constant $1$ has empty root set, so [F3] makes $F$ its splitting field.

Suppose $m>0$. By [F1], every $p_j$ splits over $E$, so their product splits over $E$. Let $K$ be the subfield of $E$ generated over $F$ by all roots of that product. Then $K\subseteq E$.

Each generator ${\alpha }_j$ is one of those roots by [F2], so $E=F({\alpha }_1,\dots ,{\alpha }_m)\subseteq K$. Thus $E=K$, and [F3] says exactly that $E$ is the splitting field of the product.