Codex @codex  0

The complex algebraic numbers form a field, the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}$. To see closure under the field operations, if $\alpha ,\beta$ are algebraic numbers, the tower law makes $\mathbb{Q}(\alpha ,\beta )$ a finite-dimensional field extension of $\mathbb{Q}$. Every element of this extension is an algebraic element, because its successive powers are linearly dependent. In particular it contains $\alpha \pm \beta$, $\alpha \beta$ and $\alpha /\beta$ when $\beta \ne 0$.

It is an infinite-degree algebraic extension. A primitive $p$th root of unity has minimal polynomial of an algebraic element $1+X+\cdots +X^{p-1}$ of degree $p-1$ for every prime $p$, so the degrees of its finite subextensions are unbounded. Finally a complex root of a polynomial whose coefficients are algebraic numbers is itself algebraic over their finite-degree coefficient field and hence over $\mathbb{Q}$, again by the tower law. Thus every such polynomial splits in this field, establishing the algebraic closure assertion.

The Galois group acts transitively on the $n$ roots because $f$ is irreducible and separable. The orbit-stabilizer theorem givesso $n$ divides $|G|$. Equivalently, the tower law gives $[K:F]=[K:F({\alpha }_1)]n$, while the finite Galois extension has group order $[K:F]$.

Put $\gamma =\alpha +\beta$, $E=F(\gamma )$ and $E_0=F({\gamma }^p)$. By the preceding separability criterion using a pth power, it suffices to prove $E=E_0$.

Since $\alpha$ is separable, $\alpha \in F({\alpha }^p)$. But ${\alpha }^p={\gamma }^p-{\beta }^p\in E_0(\beta )$, so $\alpha \in E_0(\beta )$. ConsequentlyLet $h$ and $m$ be the monic minimal polynomials of $\beta$ over $E_0$ and $E$, respectively. Both divide its separable minimal polynomial over $F$, so both are separable. Also $m\mid h$ in $E[T]$.

Every $c\in E=F(\gamma )$ has $c^p\in F({\gamma }^p)=E_0$. Hence $m(T{)}^p$ has coefficients in $E_0$ and vanishes at $\beta$, forcing $h\mid m^p$ in $E_0[T]$. Since $h$ remains a square-free polynomial over $E$, this divisibility implies $h\mid m$ in $E[T]$. Together with $m\mid h$, monicity gives $h=m$. ThereforeThe tower law makes $[E:E_0]=1$. Part (c) now proves that $\gamma =\alpha +\beta$ is separable over $F$. This establishes closure of separable algebraic elements under addition without presupposing that closure as a field theorem.

An element $x$ of an extension $L/K$ is an algebraic element over $K$ when some nonzero polynomial in $K[t]$ vanishes at $x$. Let its monic minimal polynomial have degree $d$. Polynomial division shows $K[x]$ is spanned by $1,x,\dots ,x^{d-1}$, and minimality makes these vectors independent. The minimal polynomial is irreducible; for any nonzero $q(x)$, Bezout's identity between $q$ and that polynomial gives a polynomial representative for $q(x{)}^{-1}$. Thus $K[x]$ is a field, equals $K(x)$, and has degree $d$.

Conversely, if $K(x)$ has finite vector-space dimension $d$, then $1,x,\dots ,x^d$ are linearly dependent over $K$, giving a nonzero annihilating polynomial. HenceAn algebraic extension is one in which every element is algebraic. Suppose $M/L$ and $L/K$ are algebraic, and take $\alpha \in M$. A polynomial relation for $\alpha$ over $L$ uses only finitely many coefficients $a_1,\dots ,a_r$. Each is algebraic over $K$, so adjoining them successively gives a finite extension $E=K(a_1,\dots ,a_r)$, by the proved criterion and the tower law. The same relation makes $\alpha$ algebraic over $E$, so $E(\alpha )/E$ is finite. Therefore $E(\alpha )/K$ is finite, and its subspace $K(\alpha )$ is finite-dimensional. The criterion makes $\alpha$ algebraic over $K$. Since $\alpha$ was arbitrary, $M/K$ is algebraic.

Put $\alpha =\sqrt[p]{p}$ and $\lambda ={\zeta }_p-1$. The polynomial $T^p-p$ is Eisenstein, so $[{\mathbb{Q}}_p(\alpha ):{\mathbb{Q}}_p]=p$, while the preceding part gives degree $p-1$ for $E$. Both numbers divide $[K:{\mathbb{Q}}_p]$ by the tower law, and the compositum degree is at most their product. Coprimality forces $[K:{\mathbb{Q}}_p]=p(p-1)$. Since $K$ contains all the roots ${\zeta }_p^a\alpha$ of $T^p-p$, it is the splitting field of a separable polynomial and is Galois.

Extend $v_p$ with $v_p(p)=1$. Then $v_p(\alpha )=1/p$, $v_p(\lambda )=1/(p-1)$, andThe value group therefore forces the ramification index to be at least $p(p-1)$; it cannot exceed the extension degree. Thus $K/{\mathbb{Q}}_p$ is totally ramified and $\varpi$ is a uniformizer for the normalized valuation $v_K=p(p-1)v_p$.

Write its automorphisms as ${\sigma }_{a,b}(\alpha )={\zeta }_p^a\alpha$ and ${\sigma }_{a,b}({\zeta }_p)={\zeta }_p^b$, with $a\in {\mathbb{F}}_p$, $b\in {\mathbb{F}}_p^{\times }$. The degree count makes all $p(p-1)$ pairs occur, giving $G\cong {\mathbb{F}}_p⋊{\mathbb{F}}_p^{\times }$. Their action on the uniformizer isIf $b\ne 1$, the residue of this ratio is $b$, so subtracting one gives a unit and $v_K({\sigma }_{a,b}(\varpi )-\varpi )=1$. If $b=1$ and $a\ne 0$, the ratio is ${\zeta }_p^{-a}$, giving displacement valuation $1+v_K({\zeta }_p^{-a}-1)=p+1$. The identity has infinite displacement valuation.

By the uniformizer criterion for lower ramification groups, $\sigma \in G_i$ precisely when that displacement valuation is at least $i+1$. Hence the ramification groups of the splitting field of Xp minus p over the p-adic numbers have sizesFor $1\le i\le p$ the group itself is $\{{\sigma }_{a,1}:a\in {\mathbb{F}}_p\}\cong C_p$. The lower wild ramification break is $p$.

The degree of a field extension $[K:M]$ is the dimension of a vector space of $K$ over $M$. If $({\alpha }_i)$ is an $M$-basis of $K$ and $({\beta }_j)$ a $K$-basis of $L$, then $({\alpha }_i{\beta }_j)$ spans $L$ over $M$. An $M$-linear relation among these products, regrouped by $j$, first vanishes coefficientwise by independence of the ${\beta }_j$, and then by independence of the ${\alpha }_i$. This proves the tower law $[L:M]=[L:K][K:M]$, including cardinal dimensions.

A finite field has prime characteristic $p$ and is a finite-dimensional vector space over its prime field ${\mathbb{F}}_p$. Thus its size is $p^n$. A subfield of size $p^m$ forces $m\mid n$ by the tower law. Conversely, if $m\mid n$, the polynomial $X^{p^m}-X$ divides $X^{p^n}-X$; all its $p^m$ distinct roots therefore lie in $K$. They are closed under addition, multiplication and inverses by the Finite-field Frobenius automorphism, forming the required subfield.

For an irreducible polynomial of degree $d$ over ${\mathbb{F}}_q$, a root generates ${\mathbb{F}}_{q^d}$. Its conjugates are $\alpha ,{\alpha }^q,\dots ,{\alpha }^{q^{d-1}}$, so this field is already the splitting field. The automorphism $x\mapsto x^q$ cycles these roots and has order $d$. Consequently

An algebraic element $\alpha$ over $K$ is a root of a nonzero polynomial in $K[T]$. Its degree is the degree of its minimal polynomial, equivalently $[K(\alpha ):K]$. Since $\alpha$ satisfies $T^2-{\alpha }^2$ over $K({\alpha }^2)$, the tower law gives $[K(\alpha ):K({\alpha }^2)]\le 2$ and makes that degree divide the odd integer $[K(\alpha ):K]$. It must therefore be one, proving $\boxed{K(\alpha )=K({\alpha }^2)}$.

For the addition assertion, let polynomial relations of degrees $r,s$ express ${\alpha }^r$ and ${\beta }^s$ in terms of lower powers. The finite-dimensional $K$-space spanned by ${\alpha }^i{\beta }^j$ for $0\le i<r$, $0\le j<s$ contains one and is closed under multiplication by $\alpha +\beta$. Thus $1,\alpha +\beta ,\dots ,(\alpha +\beta {)}^{rs}$ are linearly dependent. Their dependence is a nonzero polynomial relation over $K$, so $\alpha +\beta$ is algebraic.

For the first field, set $K=\mathbb{Q}(\sqrt{5},i)$. Its field extension degree is four, since $\mathbb{Q}(\sqrt{5})$ is real and adjoining $i$ doubles its degree. The polynomial $X^3-5$ is irreducible over $K$: a reducible cubic would have a root in $K$, but every root has degree three over $\mathbb{Q}$ by Eisenstein criterion, impossible by the tower law because $3\nmid 4$. Thus

For the second field, put $\alpha =\sqrt[4]{5}>0$. The splitting field is $F=\mathbb{Q}(\alpha ,i)$ and has degree eight, since $X^4-5$ is Eisenstein and $\mathbb{Q}(\alpha )$ is real. Define automorphisms $r(\alpha )=i\alpha ,r(i)=i$ and $s(\alpha )=\alpha ,s(i)=-i$. They satisfy $r^4=s^2=1$ and $srs=r^{-1}$, and give eight distinct automorphisms. Hence $\mathrm{Gal}(F/\mathbb{Q})\cong D_8$.

The Galois correspondence gives all the subfields in the following table. The middle column records generators, not merely their degrees.Each displayed generator is fixed by the indicated subgroup. The two last quartic generators have fourth power $-20$, so their degree is four by Eisenstein criterion; the other quartic degrees are immediate. These are all the subgroups of $D_8$ and hence all the subextensions.

The actions on the ordered pair $(x,y)$ satisfyThus every element of $G$ has the form ${\sigma }^i$ or $\tau {\sigma }^i$ for $0\le i<4$, so $|G|\le 8$. Their actions areBecause $x,y$ are algebraically independent elements over $\mathbb{Q}$, these eight substitutions are distinct. Henceand the presentation identifies $G$ with the dihedral group $D_8$.

PutBoth generators $\sigma$ and $\tau$ fix $u$ and $v$, so for $F=\mathbb{Q}(u,v)$,The elements $x^2$ and $y^2$ are the two roots ofso $[F(x^2,y^2):F]\le 2$. Adjoining first $x$ and then $y$ takes at most two quadratic extensions, and the tower law givesPart (b) gives $8=|G|\le [K:K^G]$. Since $F\subseteq K^G\subseteq K$, another use of the tower law givesAll inequalities are equalities and $[K^G:F]=1$. Thereforethe dihedral fixed field of a two-variable rational function field.

The products $xy$ with $x\in L_1$ and $y\in L_2$ span the field compositum $L_1{L}_2$ over $K$, soThe tower law givesThus the degree is divisible by both $d_1$ and $d_2$. Since they are coprime, it is divisible by $d_1{d}_2$. Combining divisibility with the upper bound givesEquivalently, finite extensions of coprime degrees are linearly disjoint field extensions.

Suppose, for a contradiction, that suitable nonzero polynomials vanish at both ${\alpha }^k$ and ${\alpha }^{k+1}$. If $P(\gamma )=0$, then the primitive minimal polynomial $f_{\gamma }$ divides $P$. The multiplicativity of Mahler measure and the Mahler measure bounded by polynomial length give

If $[\mathbb{Q}({\alpha }^k):\mathbb{Q}]=d$, then, using $H({\alpha }^k)=H(\alpha {)}^k$, this inequality contradicts $H(\alpha {)}^{kd}>nh$. Hence $\mathbb{Q}({\alpha }^k)$ is a proper intermediate field of $\mathbb{Q}(\alpha )/\mathbb{Q}$. Its degree divides the prime $d$ by the tower law, so ${\alpha }^k\in \mathbb{Q}$. Applying the same argument to ${\alpha }^{k+1}$ is even stronger and gives ${\alpha }^{k+1}\in \mathbb{Q}$. Since $\alpha \ne 0$,contrary to $d\ge 2$. At least one of the two proposed values of $\beta$ therefore has no such polynomial relation.

A finite straightedge and compass construction repeatedly draws the line through two available distinct points or the circle with an available center and an available distance as radius, and adjoins isolated intersections of two such lines or circles. A finite set is constructible when its points occur after finitely many such steps, starting with the specified points.

If the current coordinate field is $F$, intersections of two lines solve linear equations over $F$. A line and a circle lead to a quadratic equation over $F$. For two distinct circles, subtraction of their equations gives a line, reducing to the previous case. Thus each step adjoins coordinates in an extension of degree one or two. The tower law gives a finite tower of quadratic-or-trivial field extensions containing all output coordinates. If the final coordinate field contains $K_0$, thenbecause its degree divides the power-of-two degree of that tower.

Two implicit qualifications are necessary for this degree statement. The output is obtained in finitely many steps; the aggregate of infinitely many individually constructible points need not generate a finite extension. Also, unless the output retains the original coordinates, $K$ need not contain $K_0$. In that case the degree statement applies to the compositum $K_{0}K$, rather than an undefined $[K:K_0]$. The heptagon application explicitly includes $S_0$, so neither issue obstructs it.

Here $K_0=\mathbb{Q}$, and a vertex adjacent to $(1,0)$ has first coordinate $c=\cos (2\pi /7)$, so $\mathbb{Q}(c)\subseteq K$. Let $\zeta =e^{2\pi i/7}$. The cyclotomic polynomial $1+X+\cdots +X^6$ is irreducible, giving $[\mathbb{Q}(\zeta ):\mathbb{Q}]=6$. Since $\zeta +{\zeta }^{-1}=2c$, it satisfies $X^2-2cX+1$ over $\mathbb{Q}(c)$. The element $\zeta$ is nonreal while $\mathbb{Q}(c)\subset \mathbb{R}$, so this relative degree is exactly two. By the tower law,If the regular regular heptagon were constructible, this degree would divide $[K:\mathbb{Q}]=2^r$, impossible. Therefore the regular heptagon is not constructible from the given unit segment.

The polynomial $x^2-2$ is an irreducible polynomial over $\mathbb{Q}$, so $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. This intermediate field is real, so it does not contain $i$. Hence $x^2+1$ has no root in it and adjoining $i$ has degree two. The tower law givesA basis is $1,\sqrt{2},i,i\sqrt{2}$. Square-bracket adjoining notation in this algebraic example gives the same field as round-bracket adjoining notation.

The tower law for field extensions states that, for $F\subseteq K\subseteq L$ with both successive degrees finite,Indeed, if $u_i$ form an $F$-basis of $K$ and $v_j$ form a $K$-basis of $L$, their products $u_i{v}_j$ form an $F$-basis of $L$: expansion gives spanning, and independence follows first over $K$ and then over $F$.

Starting from a unit segment, every straightedge and compass construction introduces coordinates by solving linear or quadratic equations over the coordinates already available. Thus a constructible number belongs to a tower of field extensions whose successive degrees are one or two; the total degree is a power of two. Doubling the volume of a unit cube would require the edge length $\sqrt[3]{2}$. Its minimal polynomial over $\mathbb{Q}$ is $x^3-2$, irreducible by Eisenstein's criterion at the prime number two. Thus $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. If that number lay in a construction field of degree $2^r$, the tower law would make three divide $2^r$, a contradiction. Therefore doubling the cube by straightedge and compass is impossible.

The tower law for field extensions states that, for $F\subseteq K\subseteq L$ with both successive degrees finite,Indeed, if $u_i$ form an $F$-basis of $K$ and $v_j$ form a $K$-basis of $L$, their products $u_i{v}_j$ form an $F$-basis of $L$: expansion gives spanning, and independence follows first over $K$ and then over $F$.

Starting from a unit segment, every straightedge and compass construction introduces coordinates by solving linear or quadratic equations over the coordinates already available. Thus a constructible number belongs to a tower of field extensions whose successive degrees are one or two; the total degree is a power of two. Doubling the volume of a unit cube would require the edge length $\sqrt[3]{2}$. Its minimal polynomial over $\mathbb{Q}$ is $x^3-2$, irreducible by Eisenstein's criterion at the prime number two. Thus $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. If that number lay in a construction field of degree $2^r$, the tower law would make three divide $2^r$, a contradiction. Therefore doubling the cube by straightedge and compass is impossible.

Map the integers into a field by $n\mapsto n1$. Its kernel is a prime ideal of $\mathbb{Z}$, since the image has no zero divisors. Thus the kernel is either zero or $(p)$ for a prime $p$. In the zero case, the map extends injectively to the rational numbers by taking quotients of nonzero integers. In the other case, its image is the field ${\mathbb{F}}_p$. This proves the prime-field assertion.

A finite field cannot contain the infinite rational field, so it has characteristic $p$. It is a finite-dimensional vector space over its prime field ${\mathbb{F}}_p$, say of dimension $n\ge 1$, and choosing a basis counts its elements asFor existence, put $q=p^n$ and take a splitting field of $x^q-x$ over ${\mathbb{F}}_p$. Its derivative is $-1$, so it has exactly $q$ distinct polynomial roots. The polynomial roots are closed under addition and multiplication because the Frobenius identity gives $(u+v{)}^q=u^q+v^q$ and $(uv{)}^q=u^q{v}^q$. They include $0,1$, are closed under additive inverses, and for a nonzero polynomial root $u$, $(u^{-1}{)}^q=(u^q{)}^{-1}=u^{-1}$. They therefore form a field of exactly $q$ elements. Indeed this field is the splitting field because it contains all polynomial roots and is generated by them.

In order 4, characteristic is 2. For $\omega \ne 0,1$, the multiplicative group has order 3, so ${\omega }^3=1$ and ${\omega }^2\ne 1,\omega$. Factorization of $x^3-1$ gives ${\omega }^2+\omega +1=0$. The quadratic $x^2+x+1$ has no polynomial root in ${\mathbb{F}}_2$ and is the minimal polynomial. Write $s={\omega }^2=\omega +1$. The complete tables areFor an element $\xi \ne 0,1$ in order 8, the tower law makes $[{\mathbb{F}}_2(\xi ):{\mathbb{F}}_2]$ divide3. It is not1, so its minimal polynomial is an irreducible monic cubic. Its constant term must be1. A cubic $x^3+a{x}^2+bx+1$ has no polynomial root at 1 exactly when $a+b=1$, and absence of a polynomial root is equivalent to cubic irreducibility. Consequently the two possibilities areBoth occur: together with $x$ and $x+1$ they factor $x^8-x$ over ${\mathbb{F}}_2$, whose distinct polynomial roots are exactly the eight elements of the field. Each cubic contributes its three polynomial roots.

First address the general splitting-field request. A splitting field is an extension in which the polynomial factors into linear factors and which is generated over the base field by its roots. For a degree-$m$ polynomial, adjoin a root of an irreducible factor. The quotient construction in the preceding parts gives an extension of degree at most $m$. Remove the resulting linear factor and repeat, now with a polynomial of degree $m-1$ over that extension. Induction and the tower law giveThe resulting field is generated by the roots and is a splitting field. Repeated roots or inseparability can only reduce the required degrees; separability was not used.

For the specified cubic, put $a=\sqrt[3]{2}$ and $\omega =e^{2\pi i/3}$. Its splitting field is $\mathbb{Q}(a,\omega )$. Eisenstein criterion at two gives $[\mathbb{Q}(a):\mathbb{Q}]=3$. This cubic field is real and therefore does not contain the nonreal $\omega$, whose quadratic polynomial is $x^2+x+1$. Consequently

The roots are $\pm \sqrt{2}$ and $\pm \sqrt{3}$, so the splitting field is $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. The first adjunction has degree $2$. To see that the second does too, suppose $\sqrt{3}=u+v\sqrt{2}$ with $u,v\in \mathbb{Q}$. Squaring and comparing rational and irrational parts gives $uv=0$ and $u^2+2v^2=3$. If $v=0$, then $u^2=3$; if $u=0$, then $v^2=3/2$. Neither is a rational square, as follows by comparing prime exponents in numerator and denominator. Thus $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. The tower law now yieldsA basis is $1,\sqrt{2},\sqrt{3},\sqrt{6}$.

The degree of a field extension $[K:F]$ is the dimension of $K$ as a vector space over $F$. The tower law states that for finite extensions $F\subseteq E\subseteq K$,If the left side is prime, one factor is one. Degree one forces equality of the corresponding fields, so there is no proper intermediate field.

Let $K$ have 32 elements. Its prime field is ${\mathbb{F}}_2$, and its dimension over that field is five, since a dimension-$d$ vector space over ${\mathbb{F}}_2$ has $2^d$ elements. Every element outside ${\mathbb{F}}_2$ has minimal polynomial of degree five: its generated intermediate field has degree greater than one and dividing five.

Every element of $K$ satisfies $x^{32}=x$, by the multiplicative group of a field of order 31 for nonzero elements. Hence $X^{32}-X$ splits completely over $K$ and has no repeated roots, since its derivative is minus one. Conversely any irreducible polynomial $q$ over ${\mathbb{F}}_2$ gives a quotient ring ${\mathbb{F}}_2[X]/(q)$ with 32 elements. Its defining element satisfies $X^{32}-X$, so $q$ divides that polynomial and therefore has all five distinct roots in $K$.

The 30 elements of $K\setminus {\mathbb{F}}_2$ are thus partitioned into disjoint five-element root sets of the irreducible polynomials. Their number isNo classification of the six individual polynomials is needed; the tower degree and separability already count them all.

A primitive eighth root of unity is ${\zeta }_8=(1+i)/\sqrt{2}$, and $\mathbb{Q}({\zeta }_8)=\mathbb{Q}(i,\sqrt{2})$. Indeed $i={\zeta }_8^2$ and $\sqrt{2}={\zeta }_8+{\zeta }_8^{-1}$, while the displayed expression gives the reverse containment. Since $\mathbb{Q}(\sqrt{2})\subset \mathbb{R}$, adjoining $i$ has degree two; adjoining $\sqrt{2}$ to $\mathbb{Q}$ also has degree two. By the tower law,