Algebraic Closure, Embeddings, and Separability — Examples

By [L2], $A$ is a subfield of $\mathbb{R}$ containing $\mathbb{Q}$, and $A\mathrm{/}\mathbb{Q}$ is algebraic. Choose an algebraic closure $\mathrm{\Omega }\mathrm{/}\mathbb{Q}$ by [L1].

The polynomial $x^2+1$ has no root in the ordered field $A\subseteq \mathbb{R}$, since every square is nonnegative and $-1<0$ by [L4].

The identity on $\mathbb{Q}$ extends by [L3] to an embedding $A\to \mathrm{\Omega }$; denote its image by $A^{\mathrm{\prime }}$.

If $A^{\mathrm{\prime }}$ contained a root of $x^2+1$, its preimage under the isomorphism $A\to A^{\mathrm{\prime }}$ would be a root in $A$, contrary to step 1.2. The algebraically closed field $\mathrm{\Omega }$ does contain such a root, so $A^{\mathrm{\prime }}$ is proper in $\mathrm{\Omega }$ and is not algebraically closed. No use of $\mathbb{C}$ is required.

Every $a\in \mathrm{\Omega }$ is algebraic over ${\mathbb{F}}_p$ by [L6], so [L1] makes ${\mathbb{F}}_p(a)$ a finite field. Hence $\mathrm{\Omega }$ is the union of its finite subfields.

For $n\ge 1$, let $E_n$ be the roots in $\mathrm{\Omega }$ of $x^{p^n}-x$. This polynomial splits by [L6], and its derivative is $-1$, so [L7] gives exactly $p^n$ distinct roots. By [L2], the root set is closed under subtraction and multiplication, and it is closed under nonzero inverses; hence [L3] makes $E_n$ a subfield of order $p^n$. Any other subfield of that order consists entirely of roots by [L4], so it equals $E_n$.

If $a$ lies in a finite subfield of order $p^d$, choose $n\ge d$. Every element $b$ of that subfield satisfies $b^{p^d}=b$ by [L4]. Since $d$ divides $n!$, iterating Frobenius by [L2] gives $b^{p^{n!}}=b$, so the subfield lies in $E_{n!}$. The same argument shows $E_{n!}\subseteq E_{(n+1)!}$, and step 1.1 now shows that their nested union is all of $\mathrm{\Omega }$.

The irreducibles supplied by [L5] have roots in $\mathrm{\Omega }$ by [L6], and [L8] makes the generated simple subextensions have arbitrarily large finite degree. Therefore $\mathrm{\Omega }$ cannot be finite, while it is algebraic by [L6].

The polynomial $x^3-2$ is Eisenstein at $2$, so [L1] makes it the minimal polynomial of $r$. By [L3] its three roots in $\mathrm{\Omega }$ are distinct, and [L2] gives three $\mathbb{Q}$-embeddings of $\mathbb{Q}(r)$ into $\mathrm{\Omega }$.

The field $\mathbb{Q}(r)$ lies in $\mathbb{R}$. In an ordered field the map $x\mapsto x^3$ is strictly increasing, or directly $u^3-v^3=(u-v)(u^2+uv+v^2)$ with the second factor positive for $u\mathrm{\ne }v$; hence [L4] and [L5] make $r$ the only real root of $x^3-2$.

A $\mathbb{Q}$-automorphism of $\mathbb{Q}(r)$ must send $r$ to another root lying inside the same real field. Step 1.2 forces that image to be $r$, so the automorphism fixes the generator and is the identity.

Eisenstein at $2$, $3$, and $2$ makes $x^2-2$, $x^2-3$, and $x^2-6$ irreducible over $\mathbb{Q}$ by [L1]. Thus none of $\sqrt{2},\sqrt{3},\sqrt{6}$ is rational, and [L2] gives the basis $(1,\sqrt{2})$ of $\mathbb{Q}(\sqrt{2})\mathrm{/}\mathbb{Q}$.

If $\sqrt{3}=a+b\sqrt{2}$ with $a,b\in \mathbb{Q}$, squaring and using the basis gives $2ab=0$. If $b=0$, then $\sqrt{3}$ is rational; if $a=0$, then $2b$ is a rational square root of $6$. Both contradict step 1.1. Hence $x^2-3$ has no root in $\mathbb{Q}(\sqrt{2})$ and is irreducible there.

By [L2] the second tower step has degree two, and [L3] gives $[E:\mathbb{Q}]=4$.

The first square root has two distinct images by [L4]. Over each image of $\mathbb{Q}(\sqrt{2})$, the second square root has the two distinct images $\pm \sqrt{3}$; [L5] shows these fibres exhaust all extensions. Thus the four independent sign choices are exactly the four embeddings.

Suppose $t=(u\mathrm{/}v{)}^p$ with coprime nonzero $u,v\in {\mathbb{F}}_p[t]$, using [L1]. Then $u^p=t{v}^p$. In the UFD of [L2], the exponent of the irreducible factor $t$ on the left is divisible by $p$, while on the right it is congruent to $1$ modulo $p$, a contradiction. Thus $t$ is not a $p$th power in ${\mathbb{F}}_p(t)$.

By [L3], $x^p-t$ is irreducible.

Its derivative is zero, so [L4] makes it inseparable. In an extension containing $t^{1\mathrm{/}p}$, the characteristic-$p$ binomial identity gives $x^p-t=(x-t^{1\mathrm{/}p}{)}^p$.

Write $u=s^p$ and $v=t^p$. In the rational function field ${\mathbb{F}}_p(v)(u)$, the $u$-adic valuation of a $p$th power is divisible by $p$, so $u$ is not a $p$th power and $s\mathrm{\notin }F$. Likewise, in $F(s)={\mathbb{F}}_p(s)(v)$, the $v$-adic valuation shows that $v$ is not a $p$th power and $t\mathrm{\notin }F(s)$. These valuation statements follow from reduced fractions in the UFDs of [L1] and [L2]. Every element of $E$ has its $p$th power in $F$, so $(s,t)$ is a minimal generating family for an exponent-one purely inseparable extension. By [L3], $[E:F]=p^2$ and $\{s^i{t}^j:0\le i,j<p\}$ is an $F$-basis.

The base field $F$ is infinite because it contains the rational function field ${\mathbb{F}}_p(s^p)$ from [L1]. For each $c\in F$, put $u_c=s+ct$. Then $u_c^p=s^p+c^p{t}^p\in F$, while the basis in step 1.1 shows $u_c\mathrm{\notin }F$, so [L3] gives $[F(u_c):F]=p$.

If $c\mathrm{\ne }d$ and $F(u_c)=F(u_d)$, that common field contains $(u_c-u_d)\mathrm{/}(c-d)=t$ and then $s=u_c-ct$, so it equals $E$. This contradicts its degree $p$ against $[E:F]=p^2$. Hence the fields $F(u_c)$ are pairwise distinct.

There are therefore infinitely many intermediate fields, and [L4] says that the finite extension $E\mathrm{/}F$ is not simple. This refutes the stated universal claim.

The given algebraic closure contains a root $t_n$ of every $x^{p^n}-t$, and [L2] makes that root unique. Take $t_0=t$. Uniqueness gives $t_{n+1}^p=t_n$, so $F_n\subseteq F_{n+1}$.

A nested union of fields is a field, so $P$ is a field of characteristic $p$ containing $F_0={\mathbb{F}}_p(t)$. The distinct powers of the indeterminate $t$ show that this fraction field, and hence $P$, is infinite.

If $z=r(t_n)\mathrm{/}s(t_n)\in F_n$, replace $t_n$ by $t_{n+1}$ in the same rational expression. Since coefficients in ${\mathbb{F}}_p$ are fixed by Frobenius, [L2] shows that the resulting element of $F_{n+1}$ has $p$th power $z$. Thus Frobenius on $P$ is surjective.

By [L1], $P$ is perfect.

The polynomial in [L1] is an irreducible polynomial over a field which is not separable.

It is a counterexample to the universal claim, so the Statement is false.

Since every prime $p$ is greater than $1$, the two degrees in [L1] are unequal.

This finite extension refutes the universal equality.

If $\overline{{\mathbb{F}}_p}={\mathbb{F}}_p(\alpha )$ for one element $\alpha$, then $\alpha$ is algebraic and [L2] would make the extension finite.

This contradicts the infinitude in [L1]. Hence the algebraic extension $\overline{{\mathbb{F}}_p}\mathrm{/}{\mathbb{F}}_p$ is not simple, refuting the Statement.

Choose $\mathrm{\Omega }$ by [L1]. The polynomial $x^2+1$ is irreducible over $\mathbb{Q}$ and has a root $a\in \mathrm{\Omega }$ by [L5]. Its roots $a$ and $-a$ are distinct because $\mathbb{Q}$ has characteristic zero. By [L2], the assignment $a\mapsto -a$ gives a nonidentity $\mathbb{Q}$-embedding of $\mathbb{Q}(a)$ into $\mathrm{\Omega }$.

Extend that embedding across the algebraic extension $\mathrm{\Omega }\mathrm{/}\mathbb{Q}(a)$ using [L3]. Its image $E$ is algebraically closed because it is isomorphic to $\mathrm{\Omega }$. Every $b\in \mathrm{\Omega }$ is algebraic over $\mathbb{Q}\subseteq E$ by [L5], so [L6] gives a minimal polynomial over $E$; it has a root in $E$ and is therefore linear. Thus the resulting embedding $\tau :\mathrm{\Omega }\to \mathrm{\Omega }$ is surjective, hence is a nonidentity $\mathbb{Q}$-automorphism with $\tau (a)=-a$.

The identity and $\tau$ are distinct $\mathbb{Q}$-isomorphisms from the same algebraic closure to itself. Therefore uniqueness of the base-field isomorphism is false, although existence is true by [L4].