Algebraic Closure, Embeddings, and Separability

For a fixed $u\in \mathcal{M}(I)$, only pairs $(a,b)\in \mathrm{supp}(c)\times \mathrm{supp}(d)$ with $a+b=u$ contribute to $(cd{)}_u$, so the coefficient sum is finite; moreover $\mathrm{supp}(cd)$ is contained in the finite image of $\mathrm{supp}(c)\times \mathrm{supp}(d)$ under $(a,b)\mapsto a+b$. Thus convolution is a finitely supported coefficient family.

Pointwise addition makes the coefficient families an abelian group, with the zero family as identity and pointwise negatives.

Reindexing $(a,b)$ by $(b,a)$ proves $cd=dc$, and reindexing triples together with finite Fubini proves $(cd)e=c(de)$ coefficient by coefficient.

Splitting a finite sum proves $c(d+e)=cd+ce$, while the coefficient family supported at the zero monomial with value $1_R$ is a multiplicative identity.

The constant map preserves addition, multiplication, and $1$ by the convolution formula, so it is a ring homomorphism by [L2]; its zero-monomial coefficient recovers the original scalar, hence it is injective.

When $I=\mathrm{\varnothing }$, [L4] gives only the zero monomial, so every coefficient family is constant and the constant embedding is surjective.

For $a\in \mathcal{M}(I)$ define $s^a:={\prod }_{i\in \mathrm{supp}(a)}{s}_i^{a_i}$, and for $c={\sum }_a{c}_a{x}^a$ define $\mathrm{\Phi }(c):={\sum }_a\phi (c_a)s^a$; both expressions are finite and the empty product is $1_S$.

Pointwise addition gives $\mathrm{\Phi }(c+d)=\mathrm{\Phi }(c)+\mathrm{\Phi }(d)$, while $s^{a+b}=s^a{s}^b$ and finite reindexing give $\mathrm{\Phi }(cd)=\mathrm{\Phi }(c)\mathrm{\Phi }(d)$.

The zero monomial gives $\mathrm{\Phi }(1)=1$, constants give $\mathrm{\Phi }(r)=\phi (r)$, and the one-supported exponent family gives $\mathrm{\Phi }(x_i)=s_i$; hence $\mathrm{\Phi }$ is the required ring homomorphism by [L2].

Any ring homomorphism with these values must send $c_a{x}^a$ to $\phi (c_a)s^a$ and therefore, by finite additivity, must equal the formula in step 1.1.

For $n=0$, [L3] makes the iterated construction $R$, while the empty-family clause of [L1] makes the family construction canonically $R$.

Assume the canonical isomorphism has been constructed for a family of length $n$ and fixes $R$ and its indeterminates.

For length $n+1$, [L2] extends the induction isomorphism after choosing the image of the new variable, while [L1] gives a homomorphism in the reverse direction fixing $R$ and all $n+1$ variables.

Both composites fix $R$ and every indeterminate, so uniqueness in [L1] and [L2] makes them identity homomorphisms; the construction therefore holds for every $n$, including the empty family.

Regard $\sigma$ as an isomorphism $K\to \sigma (K)\subseteq L$. Let $m_{\alpha }\in F[x]$ be the minimal polynomial of $\alpha$. Since $\sigma$ fixes $F$, [L1] gives $m_{\alpha }(\sigma (\alpha ))=\sigma (m_{\alpha }(\alpha ))=0$.

Thus $\sigma (\alpha )$ is a root of $m_{\alpha }$ and is conjugate to $\alpha$ by [L3].

When $K=L$ is a splitting field, [L2] strengthens this root preservation to a permutation of the distinct roots.

By [L1], the image $\sigma (\alpha )$ of every $F$-embedding is a root of $m_{\alpha }$ in $\mathrm{\Omega }$.

Conversely, if $\beta \in \mathrm{\Omega }$ is a root of $m_{\alpha }$, [L2] applied to the two realizations of $F[x]\mathrm{/}(m_{\alpha })$ gives a unique $F$-embedding $F(\alpha )\to \mathrm{\Omega }$ with $\alpha \mapsto \beta$.

The constructions in steps 1.1 and 1.2 are inverse because an $F$-homomorphism on $F(\alpha )$ is determined by the image of $\alpha$.

The polynomial $m_{\alpha }$ splits in $\mathrm{\Omega }$ by repeated use of [L3], and the bijection indexes embeddings by its distinct roots, so repeated roots are counted once.

The derivative $f^{\mathrm{\prime }}$ is zero exactly when every exponent occurring in $f$ is divisible by $p$; in that case there is a unique $h\in F[x]$ with $f(x)=h(x^p)$. Repeating this finite descent in degree gives a unique maximal $e$ and a polynomial $g$ with $f(x)=g(x^{p^e})$ and $g^{\mathrm{\prime }}\mathrm{\ne }0$.

If $g=uv$ with both factors nonconstant, then $f=u(x^{p^e})v(x^{p^e})$, contradicting irreducibility of $f$; hence $g$ is irreducible.

Since $g^{\mathrm{\prime }}\mathrm{\ne }0$, any nonunit common divisor of $g$ and $g^{\mathrm{\prime }}$ has an irreducible factor by [L3], which would divide the irreducible $g$ and hence force $g\mid g^{\mathrm{\prime }}$, impossible by degree; thus $\gcd (g,g^{\mathrm{\prime }})=1$ and [L1] makes $g$ separable.

The exponents occurring in $f$ determine their largest common power $p^e$, so $e$ and then the coefficient-preserving core $g$ are unique. Moreover $e=0$ exactly when $f^{\mathrm{\prime }}\mathrm{\ne }0$, which for irreducible $f$ is equivalent to separability by [L1].

In a splitting field supplied by [L3], choose a root $b$ of $x^{p^n}-a$; [L1] gives $x^{p^n}-a=(x-b{)}^{p^n}$, so $b$ is its only distinct root.

Let $q$ be the minimal polynomial of $b$ over $F$. By [L4], write $q(x)=g(x^{p^r})$ with $g$ irreducible and separable. Every root of $q$ is also a root of $x^{p^n}-a$, hence equals $b$ by step 1.1; separability of $g$ and [L2] therefore force $g$ to be linear. Thus $q(x)=x^{p^r}-c$ for some $c=b^{p^r}\in F$ and some $0\le r\le n$.

If $r<n$, then $a=b^{p^n}=c^{p^{n-r}}$ is a $p$th power in $F$, contrary to the hypothesis; hence $r=n$ and $q=x^{p^n}-a$.

Therefore $x^{p^n}-a$ is the minimal polynomial of $b$ and is irreducible. The hypothesis excludes $a=0$ because $0=0^p$, and the same argument includes $n=1$.

If $\mathrm{char}F=0$ and $f$ is irreducible, then $f^{\mathrm{\prime }}\mathrm{\ne }0$; any common nonconstant divisor of $f$ and $f^{\mathrm{\prime }}$ would be associated to $f$, which is impossible because $\mathrm{deg}{f}^{\mathrm{\prime }}<\mathrm{deg}f$. Thus $\gcd (f,f^{\mathrm{\prime }})=1$, so $f$ is separable by [L5].

Suppose $\mathrm{char}F=p>0$ and Frobenius is surjective. For irreducible $f=g(x^{p^e})$ as in [L2], if $e>0$ then taking $p^e$th roots of the coefficients through repeated surjectivity and using [L4] would write $f$ as a $p^e$th power of a nonconstant polynomial, contradicting irreducibility. Hence $e=0$ and every irreducible is separable.

Conversely, if Frobenius is not surjective, choose $a\mathrm{\notin }{F}^p$. Then [L3] makes $x^p-a$ irreducible, while its derivative is zero, so it is not separable and $F$ is not perfect.

The characteristic-zero argument and the two implications in positive characteristic establish the equivalence.

The characteristic-zero case is immediate from [L1].

If $F$ is finite of characteristic $p$, [L2] makes Frobenius surjective, so [L1] makes $F$ perfect.

If $F$ is algebraically closed of characteristic $p$, then for every $a\in F$ the polynomial $x^p-a$ has a root by [L3]; hence every $a$ is a $p$th power and [L1] makes $F$ perfect.

For each $\alpha \in K$, [L1] supplies its irreducible minimal polynomial over $F$, and [L2] makes that polynomial separable.

Thus every element of $K$ is separable over $F$, so $K\mathrm{/}F$ is separable by [L3].

Let $f$ be the minimal polynomial of $\alpha$ over $F$. For an intermediate field $M$, let $g_M$ be the minimal polynomial of $\alpha$ over $M$ and let $M_0$ be the subfield of $M$ generated over $F$ by the coefficients of $g_M$.

The polynomial $g_M$ divides $f$ in $M[x]$ and therefore in $E[x]$. It is irreducible over $M_0$, since a factorisation over $M_0$ would be one over $M$, so it is also the minimal polynomial of $\alpha$ over $M_0$.

Thus $[E:M_0]=\mathrm{deg}{g}_M=[E:M]$; the tower law [L4] in $M_0\subseteq M\subseteq E$ gives $[M:M_0]=1$, so $M=M_0$. Hence the coefficients of $g_M$ determine $M$.

By unique factorisation [L3], the fixed polynomial $f$ has only finitely many monic divisors in $E[x]$. The injective assignment $M\mapsto g_M$ therefore proves that there are only finitely many intermediate fields.

For $m=0$, the union is empty and cannot equal the nonempty set $V$, even when $V$ is the zero space.

Assume the assertion for families of fewer than $m>0$ proper subspaces, in every finite-dimensional vector space over $F$.

If $m=1$, choose $b\in V\setminus W_0$ and the result is immediate. For $m>1$, remove any $W_i$ contained in another member; if this shortens the family, the induction hypothesis applies. Otherwise every $W_{m-1}\cap W_i$ is a proper subspace of $W_{m-1}$, so the induction hypothesis inside $W_{m-1}$ gives a nonzero $a\in W_{m-1}$ lying in none of the earlier $W_i$. Choose $b\in V\setminus W_{m-1}$.

In the unresolved case $m>1$, on the affine line $\{b+ta:t\in F\}$ each $W_i$ contains at most one point: two such points would have difference a nonzero scalar multiple of $a$, putting $a$ in $W_i$ for $i<m-1$, while any point in $W_{m-1}$ would put $b$ there.

For $m>1$ the union therefore meets the line in at most a finite set of points by [L3], whereas $t\mapsto b+ta$ is injective and $F$ is infinite. Some point of the line lies outside every $W_i$; together with the $m=1$ conclusion in step 2.1, this completes the induction.

If $F$ is finite, [L1] supplies a primitive element.

Suppose $F$ is infinite. If every intermediate field $F(\alpha )$ with $\alpha \in E$ were proper, then the finitely many proper intermediate fields would cover $E$, because every $\alpha$ lies in its own $F(\alpha )$.

Each proper intermediate field is a proper $F$-linear subspace of the finite-dimensional space $E$ from [L2], so the cover in step 1.2 contradicts [L3]. Hence $E=F(\alpha )$ for some $\alpha$, and the extension is simple.

Together with the finite-base case, this proves the assertion.

If $E\mathrm{/}F$ is simple, [L1] gives finitely many intermediate fields.

If $E\mathrm{/}F$ has finitely many intermediate fields, [L2] makes it simple.

Steps 1.1 and 1.2 prove both directions of the equivalence.

For $r=0$, one has $E=F=F(0)$, and for $r=1$ the displayed presentation is already simple. Assume $r\ge 2$. It is enough to combine two generators: if $F(\alpha ,\beta )=F(\gamma )$ whenever $\beta$ is separable, repeated combination leaves at most the originally exceptional generator as the first entry and a separable generator as the second. Finiteness of each intermediate extension follows from [L4].

If $F$ is finite, the two-generator extension is simple by [L3].

Suppose $F$ is infinite. In a common splitting field from [L2], list the distinct conjugates ${\alpha }_i$ of $\alpha$ and the pairwise distinct conjugates ${\beta }_j$ of the separable element $\beta$. Choose a nonzero $c\in F$ avoiding the finitely many values $({\alpha }_1-{\alpha }_i)\mathrm{/}({\beta }_j-{\beta }_1)$ with ${\beta }_j\mathrm{\ne }{\beta }_1$, and put $\gamma =\alpha +c\beta$.

In $F(\gamma )[x]$, the minimal polynomial of $\alpha$ and the translated minimal polynomial of $\beta$ have $\alpha$ as a common root. By the choice of $c$, any common root would give ${\alpha }_i+c{\beta }_j=\alpha +c\beta$ and hence must be $\alpha$; [L1] therefore makes their monic gcd $x-\alpha$. Thus $\alpha \in F(\gamma )$ and then $\beta =(\gamma -\alpha )\mathrm{/}c\in F(\gamma )$.

Hence $F(\alpha ,\beta )=F(\gamma )$ over either a finite or an infinite base. Iterating step 1.1 proves the theorem, and when every generator is separable it gives the usual finite separable primitive-element theorem.

By [L1], write $E=F(\alpha )$ for one element $\alpha$.

The conclusion now follows from [L2].

By [L1] the extension is algebraic, and [L2] therefore makes it separable.

It is finite and separable, so [L3] makes it simple.

Suppose $I=A$. Then $1={\sum }_{j=1}^m{h}_j{f}_j(x_{f_j})$ for finitely many $f_j\in \mathcal{S}$ and $h_j\in A$.

By [L3], choose a field $E\mathrm{/}F$ in which all $f_j$ split, and choose a root $a_j\in E$ of each $f_j$.

Assign $x_{f_j}\mapsto a_j$ for the variables occurring as generators in step 1.1 and assign every other indeterminate, including unused ones appearing in the $h_j$, to $0$. By [L2] this gives an $F$-algebra homomorphism $\mathrm{\Phi }:A\to E$.

Applying $\mathrm{\Phi }$ to step 1.1 gives $1={\sum }_j\mathrm{\Phi }(h_j)f_j(a_j)=0$, impossible in the field $E$. Therefore $I$ is proper.

By [L1] and the maximal-ideal theorem [L2], whose choice step is Zorn's lemma [L5], choose a maximal ideal $M$ of $A$ containing $I$.

Put $L=A\mathrm{/}M$. By [L3] this is a field. The composite $F\to A\to L$ is injective, since a nonzero scalar in $M$ would be a unit and force $1\in M$, so it identifies $F$ with a subfield of $L$.

For each $f\in \mathcal{S}$, the residue $\overline{x_f}\in L$ satisfies $f(\overline{x_f})=0$ because $f(x_f)\in I\subseteq M$. Multiplying an arbitrary nonconstant polynomial by the inverse of its leading coefficient makes it monic without changing its roots, so every nonconstant polynomial over $F$ has a root in $L$.

Every element of $L$ is represented by a polynomial involving finitely many variables $x_{f_1},\dots ,x_{f_r}$, hence lies in $F(\overline{x_{f_1}},\dots ,\overline{x_{f_r}})$. Each residue is algebraic over $F$, so [L4] makes this subextension finite and [L6] makes it algebraic. Thus $L\mathrm{/}F$ is algebraic.

The field $L$ constructed above is the required algebraic root extension, and the only choice principle used is the maximal-ideal application in step 1.1.

Let $f\in F[x]$ be irreducible and nonconstant, and choose a splitting field $E\mathrm{/}F$ by [L4]. It is generated by the finitely many roots of $f$, so [L7] makes it finite; [L1] makes it separable and [L2] gives $E=F(\alpha )$ for some $\alpha$.

The minimal polynomial $m_{\alpha }\in F[x]$ has a root $\beta \in L$ by hypothesis. By [L3] there is an $F$-embedding $E=F(\alpha )\to L$ sending $\alpha$ to $\beta$. Since $f$ splits in $E$ and its coefficients are fixed, it splits in the image inside $L$.

Thus every irreducible polynomial over $F$, and hence every polynomial over $F$, splits in $L$.

Let $q\in L[x]$ be nonconstant and choose a root $\gamma$ in a splitting field by [L4]. The element $\gamma$ is algebraic over $L$, while $L\mathrm{/}F$ is algebraic, so [L5] makes $\gamma$ algebraic over $F$. Its minimal polynomial over $F$ splits in $L$ by step 3.1; since $\gamma$ is one of its roots, $\gamma \in L$.

Every nonconstant polynomial over $L$ therefore has a root in $L$, so [L6] makes $L$ algebraically closed.

The set $F^{\mathrm{\prime }}$ contains $F$ by taking $n=0$. For $a,b\in F^{\mathrm{\prime }}$, choose one exponent $N$ at least as large as exponents witnessing both memberships. Then [L1] gives $(a-b{)}^{p^N}=a^{p^N}-b^{p^N}\in F$, $(ab{)}^{p^N}\in F$, and, for $a\mathrm{\ne }0$, $(a^{-1}{)}^{p^N}=(a^{p^N}{)}^{-1}\in F$. Hence [L3] makes $F^{\mathrm{\prime }}$ an intermediate field.

If $a\in F^{\mathrm{\prime }}$ and $a^{p^n}=c\in F$, the root hypothesis applied to $x^{p^{n+1}}-c$ gives $b\in L$ with $b^{p^{n+1}}=c$. Then $(b^p{)}^{p^n}=a^{p^n}$, so injectivity in [L1] gives $b^p=a$, and $b\in F^{\mathrm{\prime }}$ by its displayed power. Thus Frobenius on $F^{\mathrm{\prime }}$ is surjective and [L2] makes $F^{\mathrm{\prime }}$ perfect.