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For a polynomial $f\in K[x]$ of degree $n$, its splitting field $L$ has $[L:K]\mid n!$, including in positive characteristic when the polynomial is inseparable. Prove this by induction on the degree. If $f$ is irreducible, adjoining a root gives degree $n$ and leaves a polynomial of degree $n-1$ to split, so the tower law for field extensions gives the divisibility. If $f=gh$ with positive degrees $r,s<n$, split $g$ first and then $h$ over the resulting field. The degree divides $r!s!$, which divides $(r+s)!$. This separate reducible case avoids the invalid step of assuming that the degree of an arbitrarily chosen irreducible factor must divide $n$.

Consider the constructible angle whose cosine is $5/6$ and sine is $\sqrt{11}/6$. If one fifth of this angle were constructible, its cosine $c$ would satisfy the displayed polynomial, by the identity $\cos 5\theta =16{\cos }^5\theta -20{\cos }^3\theta +5\cos \theta$. The identity follows by taking the real part of $(\cos \theta +i\sin \theta {)}^5$ and substituting ${\sin }^2\theta =1-{\cos }^2\theta$.

The polynomial is irreducible by Eisenstein's criterion at five, so $c$ has degree five over the rational numbers. The initial angle field $\mathbb{Q}(\sqrt{11})$ already has degree two, and adjoining any finite collection of constructible numbers gives total degree over the rational numbers a power of two. Such a field cannot contain the degree-five element $c$, by the tower law for field extensions. Hence this constructible angle cannot be divided into five equal angles by straightedge and compass construction. Some particular angles can be quintisected; the obstruction disproves a construction valid for all angles.

Doubling the volume of a unit cube requires constructing its new side length $\sqrt[3]{2}$. Its minimal polynomial $X^3-2$ is irreducible by Eisenstein's criterion at two, so this length has degree three. Every constructible number belongs to a finite tower of quadratic extensions, whose total degree is a power of two. The tower law for field extensions would force the degree three to divide that power of two, a contradiction. Thus no such straightedge and compass construction exists.

Every $h\in G$ permutes the factors:Thus all coefficients lie in the fixed field $L^G$. Let $E$ be the field they generate over $K$. ThenBecause $L=K(\alpha )=E(\alpha )$ and $f(t,\alpha )\in E[t]$ has degree $|G|$ with root $\alpha$,Part (a) and the tower law for field extensions giveConsequently $[L^G:E]=1$, soThis is the coefficients of an orbit polynomial for a primitive generator result.

The tower law for field extensions states that for finite field extensionsone hasTo prove it, let $(u_i{)}_{i=1}^m$ be a $K$-basis of $E$ and $(v_j{)}_{j=1}^n$ an $E$-basis of $L$. Every $x\in L$ can be written as ${\sum }_j{a}_j{v}_j$ with $a_j\in E$, and expanding each $a_j$ in the basis $(u_i)$ shows that the $mn$ products $u_i{v}_j$ span $L$ over $K$.

Ifthen independence of the $v_j$ over $E$ gives ${\sum }_i{c}_{ij}{u}_i=0$ for every $j$, and independence of the $u_i$ over $K$ gives every $c_{ij}=0$. Thus $(u_i{v}_j)$ is a $K$-basis of $L$, proving the formula.

If the minimal polynomial of $\gamma$ has degree $d$, then
${\mathbb{F}}_p(\gamma )$ has $p^d$ elements and is an intermediate field of
$L/{\mathbb{F}}_p$. The tower law for field extensions givesso $d\mid n$.

The multiplicative group of a finite field is cyclic. Choose a generator
$\gamma$ of $L^{\times }$, of order $p^n-1$. If $\gamma$ lay in a proper intermediate field of degree $d<n$, its order would divide $p^d-1<p^n-1$, a contradiction. Thus
${\mathbb{F}}_p(\gamma )=L$ and its minimal polynomial has degree $n$.

For arbitrary $r\ge 1$, let $E$ be the splitting field of
$X^{p^r}-X$ over ${\mathbb{F}}_p$. Its roots form a field: the Frobenius endomorphism shows they are closed under addition, subtraction, multiplication, and inversion. The derivative is $-1$, so there are exactly $p^r$ distinct roots. Thus this root field has order $p^r$. Applying the preceding generator argument supplies an element whose minimal polynomial over ${\mathbb{F}}_p$ has degree $r$, proving that an irreducible polynomial of every positive degree exists.

Put $\alpha =\sqrt[4]{7}$. The roots of $T^4-7$ are $\alpha ,i\alpha ,-\alpha ,-i\alpha$, so its splitting field isThe polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $7$, and $\mathbb{Q}(\alpha )\subset \mathbb{R}$ does not contain $i$. The tower law for field extensions therefore gives $[L:\mathbb{Q}]=8$.

Define automorphismsThey satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct, so they exhaust the Galois group. Hencethe dihedral group of order eight.

The Galois correspondence gives all intermediate fields. The full list, with a single generator for each field, isFor example, $\sqrt{7}+i$ has stabilizer $⟨r^2⟩$, and $(1+i)\alpha$ has stabilizer $⟨rs⟩$; the other entries follow similarly. Also $\alpha +i$ has trivial stabilizer, so its orbit has eight elements and the primitive element theorem conclusion $L=\mathbb{Q}(\alpha +i)$ holds explicitly.

Finally, the normal subextension criterion says that normal intermediate fields correspond exactly to normal subgroups. In $D_8$ these are $D_8$, $⟨r⟩$, the two index-two Klein four-groups, $⟨r^2⟩$, and $\{1\}$, which gives exactly the “yes” rows above. This is the subfields of the splitting field of x to the fourth minus seven lattice.

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.

First, the number of embeddings extending a prescribed base embedding is at most the extension degree. Prove this for any finite extension by choosing finitely many algebraic generators and applying part (i) successively. At each step the number of extensions is at most the degree of that simple step; multiplication and the tower law for field extensions give the total bound.

If $\tilde{K}/K$ is not separable, choose $\alpha$ with inseparable minimal polynomial of degree $e=[K(\alpha ):K]$. Its image under $i$ has strictly fewer than $e$ distinct roots in any target field: inseparability is preserved under the coefficient embedding. Thus there are $r<e$ embeddings of $K(\alpha )$ extending $i$. Each extends to $\tilde{K}$ in at most $[\tilde{K}:K(\alpha )]$ ways, giving at mostIf instead $i(m_{\alpha })$ does not split in $L$, it again has fewer than $e$ distinct roots in $L$, and exactly the same estimate applies. Hence either obstruction makes the embedding count strictly less than $d$.

Identify $\mathbb{C}$ with the Euclidean plane and start with the points $0,1$. A complex number is a constructible number if a finite sequence of drawing lines through known points and circles with known centres and known distances as radii, and selecting their intersections, produces that point.

Line-line intersections solve linear equations in the current coordinate field. Line-circle and circle-circle intersections require at most one quadratic equation after eliminating a coordinate. Hence all coordinates obtained lie in a finite tower of real quadratic extensions of $\mathbb{Q}$. If $\alpha =x+iy$ is constructible, choose a common such real field $F$ containing $x,y$. Adjoining $i$ if needed still gives an extension $F(i)/\mathbb{Q}$ of power-of-two degree. By the tower law for field extensions,A regular heptagon centred at $0$ with one vertex $1$ would have $\xi =e^{2\pi i/7}$ as another constructible vertex. But its cyclotomic degree is $\phi (7)=6$, which is not a power of two. Thus the regular heptagon is not constructible.

Let the monic minimal polynomial of $\alpha$ over $L$ have coefficients $a_0,\dots ,a_{r-1}$, and put $F=K(a_0,\dots ,a_{r-1})$. Each coefficient is algebraic over $K$, so adjoining these finitely many coefficients successively gives $[F:K]<\infty$. The same polynomial makes $\alpha$ algebraic over $F$, giving $[F(\alpha ):F]\le r$. The tower law for field extensions yields $[F(\alpha ):K]<\infty$. Its subfield $K(\alpha )$ therefore also has finite degree over $K$. This holds for every $\alpha \in M$, proving $M/K$ is algebraic.

A splitting field of a nonconstant polynomial $f$ over a field $K$ is an field extension $L$ in which $f$ factors into linear factors and which is generated over $K$ by those roots. We prove existence and the divisibility simultaneously by induction on $n=\mathrm{deg}f$; constant nonzero polynomials have splitting field $K$.

If $f$ is reducible, write $f=gh$ with positive degrees $r,s<n$. By induction first split $g$, obtaining $E/K$ of degree dividing $r!$, then split $h$ over $E$, obtaining $L/E$ of degree dividing $s!$. The tower law for field extensions gives $[L:K]\mid r!s!\mid (r+s)!$. All adjoined elements are roots of $f$, so $L$ has the required minimality.

If $f$ is irreducible, take $E=K[x]/(f)$ and let $\alpha$ be the image of $x$. Then $[E:K]=n$ and $f=(x-\alpha )h$ in $E[x]$. Apply induction to $h$, of degree $n-1$, to obtain $[L:E]\mid (n-1)!$. ThereforeThis proof also covers inseparable polynomials; it does not incorrectly identify extension degree with automorphism count in that case.

Since $K$ is an $F$-vector subspace of the finite-dimensional vector space $L$, it has a finite $F$-basis $u_1,\dots ,u_m$. Any finite $F$-spanning set for $L$ also spans $L$ over $K$, so choose a finite $K$-basis $v_1,\dots ,v_r$ of $L$.

Every $\ell \in L$ can be expanded first in the $v_j$ and then in the $u_i$, showing that the products $u_i{v}_j$ span $L$ over $F$. If ${\sum }_{i,j}{c}_{ij}{u}_i{v}_j=0$ with $c_{ij}\in F$, independence of the $v_j$ over $K$ gives ${\sum }_i{c}_{ij}{u}_i=0$ for every $j$; independence of the $u_i$ then gives every $c_{ij}=0$. The products form an $F$-basis, proving the tower law for field extensions:

Put $E=F(a,b)$. The tower law for field extensions shows that both $m$ and $n$ divide $[E:F]$. Since $m,n$ are coprime, their product divides $[E:F]$, so $[E:F]\ge mn$.

The minimal polynomial of $b$ over $F$ has degree $n$ and is still an annihilating polynomial over $F(a)$. The minimal polynomial over $F(a)$ divides it, soThe upper and lower bounds agree:Coprimality is the step forcing the lower bound to equal the product; without it the two generated fields can overlap substantially.

Eisenstein criterion says that a primitive integer polynomial is irreducible over $\mathbb{Q}$ if some prime divides all coefficients except the leading one, does not divide the leading coefficient, and its square does not divide the constant coefficient. The passage from primitive integer factorization to rational factorization is Gauss lemma for polynomials.

Translate the variable:This is Eisenstein at two. Translation is an automorphism of $\mathbb{Q}[x]$, so $x^8+1$ itself is irreducible.

Set $\zeta =e^{i\pi /8}$ and $c=\cos (\pi /8)$. Since $\zeta$ is a root of that irreducible degree-eight polynomial, $[\mathbb{Q}(\zeta ):\mathbb{Q}]=8$. Also $c=(\zeta +{\zeta }^{-1})/2$, so $\mathbb{Q}(c)\subseteq \mathbb{Q}(\zeta )$ and $\zeta$ solves $T^2-2cT+1=0$. The extension degree is at most two. It is not one, since every element of $\mathbb{Q}(c)$ is real and $\zeta$ is not. The tower law for field extensions now gives

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.

Induct through the finitely many construction steps. Initially all available coordinates lie in $\mathbb{Q}$. If coordinates obtained so far lie in a real field $F$, the line through two of its points has coefficients in $F$. A circle with centre in $F^2$ and radius a distance between two such points has equationwhose coefficients lie in $F$, since the squared distance $r^2$ lies in $F$; the radius itself need not lie there.

Part (b) therefore puts every new intersection in $F$ or a real quadratic extension of $F$. If a step produces several points, adjoin their coordinates successively, omitting extensions that add nothing. A finite construction ends in a real field $K$ containing the desired coordinates, with a tower of genuine quadratic extensionsThe tower law for field extensions yieldsA construction involving no nontrivial extension has $n=0$ and degree one.

Eisenstein's criterion says that an integer polynomial with leading coefficient not divisible by a prime $\ell$, all other coefficients divisible by $\ell$, and constant coefficient not divisible by ${\ell }^2$ is irreducible over the rationals, provided it is primitive; the usual Gauss's lemma supplies the passage from integer to rational factors.

Apply it with $\ell =2$ toAll nonleading coefficients are even and the constant is not divisible by four. A translation of the variable preserves reducibility, so $x^8+1$ is irreducible over $\mathbb{Q}$.

Let $\zeta =e^{i\pi /8}$ and $c=\cos (\pi /8)$. The minimal polynomial of $\zeta$ is $x^8+1$, so $[\mathbb{Q}(\zeta ):\mathbb{Q}]=8$. Since $2c=\zeta +{\zeta }^{-1}$, the number $\zeta$ satisfies $t^2-2ct+1=0$ over $\mathbb{Q}(c)$. This real field cannot contain the nonreal $\zeta$, so that extension has degree exactly two. The tower law for field extensions gives