For integral solutions of x squared minus two equals y cubed, the equation implies $y\ge -1$, so $y+2>0$. If $y$ were even, then $x$ would be even and $x^2-2$ would be two or six modulo eight, whereas $y^3$ would be zero modulo eight. Thus $y$ and $x$ are odd. Also $3\nmid x$: otherwise $x^2-2\equiv 7(\mathrm{mod}9)$, which is not a cubic residue modulo nine. Reducing modulo three then gives $y\equiv 2(\mathrm{mod}3)$. ConsequentlyPut $\eta =\sqrt{-6}$ and $\alpha =x+(y+1)\eta$. Its field norm is $(y+2{)}^3$. The principal ideals $(\alpha )$ and $(\bar{\alpha })$ are coprime. Indeed a common prime ideal has underlying rational prime $p$ dividing $y+2$. The displayed gcd excludes two and three, so both two and $\eta$ are invertible modulo that prime ideal. The difference $\alpha -\bar{\alpha }=2(y+1)\eta$ therefore implies $p\mid y+1$, contradicting $p\mid y+2$.

Their product is $(y+2{)}^3$, so unique factorization of ideals in a number field and coprimality imply $(\alpha )=A^3$ for an integral ideal $A$. In the given ideal class group of order two, $[A{]}^3=1$ implies $[A]=1$. Hence $A=(u+v\eta )$ with integers $u,v$. Equality of principal ideals gives $\alpha =\epsilon (u+v\eta {)}^3$. The only units are $\pm 1$, and both are cubes, so the sign can be absorbed into $u,v$. Taking norms, and using $y+2>0$, givesThe coefficient of $\eta$ gives $y+1=3u^{2}v-6v^3$. Eliminating $y$ with the field norm identity yieldsThe denominator is nonzero for integral $v$. Polynomial division makes the integrality obstruction especially clear:If $u$ is integral, the reciprocal must be an integer, forcing $3v-1=\pm 1$. The only integral possibility is $v=0$. Then $u^2=1$, the field norm gives $y=-1$, and the real coefficient in the cube gives $x=u^3=\pm 1$. Both pairs directly satisfy the original equation. Therefore all integral solutions are

Choose a nonzero integral ideal $B$ whose class is $[A{]}^{-1}$; one can multiply the fractional inverse $A^{-1}$ by a suitable nonzero integer to obtain such a $B$. By the given bound choose $0\ne \beta \in B$ with $|N(\beta )|\le c_{K}N(B)$. Put $I_A=(\beta )B^{-1}$. Since $(\beta )\subseteq B$, multiplication by $B^{-1}$ makes $I_A$ an integral ideal. Its class is $[B{]}^{-1}=[A]$, and multiplicativity of ideal norms givesThis proves the class representative bound without accidentally constructing a representative of the inverse class instead.

To determine the Ideal class group of Q of square root minus thirty-four, write $\eta =\sqrt{-34}$. The given integral basis has discriminant $-136$, so the imaginary-quadratic bound is $4\sqrt{34}/\pi <8$. Every class therefore has a representative of ideal norm at most seven, and factoring such representatives shows that the ideal class group is generated by prime ideals of ideal norm at most seven.

Two ramifies: let $P_2=(2,\eta )$, with $P_2^2=(2)$. Three is inert because $X^2+34\equiv X^2+1$ is irreducible modulo three, so its prime ideal ideal norm is nine and supplies no small generator. At five and seven the roots are $\pm 1$, so the primes of ideal norm five and seven are conjugate pairs. Set $P_5=(5,\eta -1)$ and $P_7=(7,\eta +1)$. Their conjugates have inverse ideal classes. No new prime ideal has ideal norm four, since the only prime above two has residue degree one; six is not a prime power. Thus $[P_2],[P_5],[P_7]$ generate the ideal class group.

The element field norms $N(4+\eta )=50$ and $N(1+\eta )=35$, together with reduction at the split primes, giveThe ideal norm products verify both exponents. Consequently $[P_5{]}^2=[P_2]$ and $[P_7]=[P_5]$, so the whole group is generated by $[P_5]$ and its order divides four. The class $[P_2]$ is not trivial: a generator would be an element of the ring of integers $a+b\eta$ of field norm two, but $a^2+34b^2=2$ has no integer solution. Since $[P_2{]}^2=1$, its order is exactly two. Hence $[P_5]$ has order four and

The Dedekind factorization theorem says: if $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and the rational prime $p$ does not divide $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. Then the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, with integer lifts $f_i$, andThe index condition is part of the theorem and must be checked before applying it at a ramified prime.

For this cubic, $X^3-6$ is Eisenstein and has discriminant $-972$. We first verify ${\mathcal{O}}_K=\mathbb{Z}[\theta ]$. The square of the index divides 972, so only two and three can divide the index. If either prime $p$ divided it, the finite quotient group would contain an element of order $p$, giving an integral element $(a+b\theta +c{\theta }^2)/p$ with digits $0\le a,b,c<p$, not all zero. Its field norm isFor $p=2$, the numerator has valuation zero if $a=1$, valuation one if $a=0,b=1$, and valuation two if $a=b=0,c=1$. For $p=3$ the same three cases, according to the first nonzero digit, give valuations zero, one and two. None gives an integer field norm after dividing by $p^3$, contradicting integrality. Thus the index is one.

Multiplication by $a+b\theta$ in the power basis, or the product over the three conjugates, gives $\boxed{N(a+b\theta )=a^3+6b^3}$. DefineReduction of the cubic gives $X^3$ at two and three, $(X-1)(X^2+X+1)$ at five, and $(X-3)(X-5)(X-6)$ at seven. The quadratic factor at five is irreducible since its discriminant is two, a nonsquare modulo five. HenceThe norms are respectively $2,3,5,25,7,7,7$. Norms and reduction of the elements at those primes giveFor example their absolute element field norms are $6,5,2,21,70$; each indicated prime divides the element by substitution in its residue field, and the product of the indicated ideal norms already equals that absolute field norm. Thus no additional factor or exponent is possible.

In the ideal of the specified quotient every prime exponent cancels:Equality with the unit ideal implies that both $\mu$ and its inverse are integral, so $\boxed{\mu \text{ is a unit}}$. For the real root, $1<\theta <2$. The numerator is negative and has absolute value less than two, while the denominator is greater than five. Thus $-2/5<\mu <0$, proving $\boxed{\mu \ne \pm 1}$. Its field norm is $-1$, consistent with the ideal calculation.

For the basis $w_1,\dots ,w_d$, its discriminant isEquivalently it is the square of the determinant of the matrix of the $d$ embeddings applied to the basis vectors. It is nonzero because number fields are separable. An integral basis is a basis of the ring of integers as a free abelian group; in particular it is a rational basis of the field and expresses every algebraic integer uniquely with integer coefficients.

To prove existence, start with any rational basis and multiply each vector by a suitable nonzero integer to make it integral. This is possible by clearing coefficients in its monic rational minimal polynomial of an algebraic element and rescaling the variable. Let $M$ be the lattice spanned over $\mathbb{Z}$ by these integral vectors, and let $D\ne 0$ be its discriminant. For any algebraic integer $\alpha ={\sum }_j{c}_j{w}_j$, each trace $\mathrm{Tr}(\alpha w_i)$ is an integer. The resulting linear system has integral coefficient matrix with determinant $D$. Its adjugate matrix shows $c_j\in D^{-1}\mathbb{Z}$. ThusA subgroup of a finitely generated free abelian group is free and finitely generated. Here its rank is $d$, since it contains $M$ and is contained in a rank-$d$ lattice. Its free generators are the required integral basis.

For squarefree $d\ne 1$, the quadratic integral basis isThe source allows squarefree nonzero integers, so the degenerate value $d=1$ gives $K=\mathbb{Q}$ and basis $\{1\}$ instead of a two-element basis.

In the biquadratic field define the three automorphisms by changing respectively the sign of $\sqrt{r}$, the sign of $\sqrt{2}$, and both signs. Their actions areHence the three sums are $2a+2b\sqrt{2}$, $2a+2c\sqrt{r}$ and $2a+2d\sqrt{2r}$. If $\mu$ is integral, each sum is integral and lies in its displayed quadratic subfield. Their radicands are squarefree and respectively congruent to two, three and two modulo four. The quadratic basis formula therefore forces all their two coefficients to be integers, proving

A straightedge and compass construction uses finitely many steps drawing lines through known points and circles with known centres and constructible radii, and adjoining their intersection points. The coordinates obtained by line–line intersections require only field operations; line–circle and circle–circle intersections require at most a quadratic extension after eliminating a linear equation. Thus all coordinates lie in a tower of quadratic extensions of the initial coordinate field. The degree of any finite intermediate coordinate field containing that initial field divides a power of two and is therefore $\boxed{2^r}$ for some $r\ge 0$.

For the source's notation this assumes $K\supseteq K_0$, as is customary when previously constructed points are retained. If only the coordinates of a final subset are called $K$, replace it by the compositum $K_{0}K$; otherwise the expression $[K:K_0]$ need not be defined. The power-of-two condition is necessary and is not being asserted as sufficient for an arbitrary algebraic number.

Starting from a unit cube, doubling its volume would construct a length $\sqrt[3]{2}$. Its polynomial $X^3-2$ is Eisenstein at two, so its degree over $\mathbb{Q}$ is three, which cannot divide a power of two. Thus doubling the cube is impossible by ruler and compass.

Set $c=\cos \theta$, $s=\sin \theta$. The real part of $(c+is{)}^5$ is $c^5-10c^3{s}^2+5c{s}^4$. Replacing $s^2$ by $1-c^2$ givesChoose the given angle to have cosine $5/6$. It is constructible, since its unit circle point has coordinates $(5/6,\sqrt{11}/6)$. If a fifth of that angle were constructible and had cosine $c$, the identity would implyThis polynomial is Eisenstein at five: all nonleading coefficients are divisible by five and the constant term is not divisible by 25, while 96 is not divisible by five. Hence $[\mathbb{Q}(c):\mathbb{Q}]=5$. But the initial angle field is $\mathbb{Q}(\sqrt{11})$, itself of degree two, so adjoining any finite collection of constructible coordinates to it would still give total degree over $\mathbb{Q}$ a power of two. It could not contain a degree-five element. Thus general angle quintisection is impossible with straightedge and compass.

A finite extension is Galois when it is both normal and separable; over $\mathbb{Q}$ separability is automatic, so it is equivalently a splitting field of a polynomial with distinct roots. Its Galois group consists of the field automorphisms fixing $\mathbb{Q}$. If $G=\mathrm{Gal}(L/\mathbb{Q})$, the Fundamental theorem of Galois theory gives mutually inverse inclusion-reversing bijectionsThey satisfy $[L:L^H]=|H|$ and $[L^H:\mathbb{Q}]=[G:H]$. The intermediate extension $K/\mathbb{Q}$ is Galois exactly when $H=\mathrm{Gal}(L/K)$ is normal, and then $\boxed{\mathrm{Gal}(K/\mathbb{Q})\cong G/H}$ via restriction.

Every permutation of the four roots permutes the three unordered partitions into two pairs and hence permutes the three ${\beta }_j$. Thus each coefficient of $g(X)={\prod }_{j=1}^3(X-{\beta }_j)$ is a symmetric polynomial in the four roots. The Fundamental theorem of symmetric polynomials expresses it in the elementary symmetric functions, which are rational coefficients of $f$. Hence $g\in \mathbb{Q}[X]$. Each product ${\alpha }_i{\alpha }_j$, $i<j$, occurs exactly twice in ${\beta }_1+{\beta }_2+{\beta }_3$, so that sum is $2q$. Therefore the quadratic coefficient is $\boxed{-2q}$.

For the given cubic write $g(X)=(X-2{)}^3+12$. The translated cubic $Y^3+12$ is Eisenstein at three, hence irreducible. It has one real root and a nonreal conjugate pair. Its Galois group is transitive on three roots, so its order is divisible by three; complex conjugation is a transposition. A subgroup of $S_3$ with both a three-cycle and a transposition is all of $S_3$. HenceLet $G=\mathrm{Gal}(L/\mathbb{Q})\le S_4$. Irreducibility of the quartic makes the root action transitive, so four divides $|G|$. The cubic splitting field $K$ lies inside $L$ by the definitions of the ${\beta }_j$, and restriction onto its Galois group gives a quotient of order six. Thus six also divides $|G|$. Since $|G|$ divides 24, it is 12 or 24. If it were 12, the supplied index-two result would give $G=A_4$. The restriction kernel would then be a normal subgroup of $A_4$ of index six, contrary to the other supplied result. Thus

Every automorphism fixes zero and one, and preserves sums, products, negatives and multiplicative inverses of nonzero elements. Consequently elements fixed by every member of $G$ are closed under these operations and form a field. Artin fixed-field theorem gives $\boxed{[L:L^G]=|G|}$ for a finite automorphism group.

Here $\mathbb{Q}(t)$ is the fraction field of $\mathbb{Q}[t]$, with $t$ transcendental; its elements are fractions of rational-coefficient polynomials, with nonzero denominator and the usual equality by cross-multiplication. Substitution of $1-t$ or $1/t$ defines an automorphism, each its own inverse. With $\sigma \tau$ meaning first apply $\tau$ to a field element and then $\sigma$, its image on $t$ is $1/(1-t)$. Iterating gives $(t-1)/t$ and then $t$, so its order is three. The relations reduce every word to at most six possibilities, and the six resulting images areThey are distinct rational functions, so the group has six elements. Its faithful permutation action on $\{0,1,\infty \}$ identifies it with $\boxed{S_3}$.

For the first fixed field put $u=t(1-t)$. It is fixed by $\sigma$, and $t$ satisfies $X^2-X+u=0$ over $\mathbb{Q}(u)$. The degree is exactly two because the nontrivial automorphism $\sigma$ fixes that field. The fixed-field degree is also two, soFor the inversion put $v=t+1/t$. It is fixed and $t$ satisfies $X^2-vX+1=0$. The same degree argument givesThe source requests polynomial generators in both cases. The second requirement is impossible. If a nonconstant polynomial $p(t)=a_d{t}^d+\cdots +a_0$ were fixed by inversion, $p(t)=p(1/t)$ would equate a Laurent polynomial with highest exponent $d>0$ to one with no positive exponent, a contradiction. Thus the only polynomial invariants of inversion are constants, which cannot generate its nonconstant fixed field. The fixed field of inversion in a rational function field therefore has the second displayed rational-function generator.

For an irreducible polynomial $f\in K[X]$, take $L=K[X]/(f)$ and let $\alpha$ be the class of $X$. The quotient is a field, is generated by $\alpha$ over $K$, and satisfies $f(\alpha )=0$. Its degree is $\boxed{[L:K]=\mathrm{deg}f}$, with basis $1,\alpha ,\dots ,{\alpha }^{\mathrm{deg}f-1}$.

A splitting field is a field generated over $K$ by all roots of the polynomial and in which the polynomial is a product of linear factors. Prove existence and the factorial bound together by induction on the degree. Constant nonzero polynomials and linear polynomials require no extension. For a reducible degree-$n$ polynomial write $g=hk$ with positive degrees $r,n-r$. Split $h$ by induction, then split $k$ over that field. The resulting field is generated by all roots of $g$, and its degree over $K$ is at most $r!(n-r)!\le n!$. If $g$ is irreducible, adjoin one root in degree $n$, divide out its linear factor, and split the degree-$(n-1)$ quotient over that extension. The tower formula gives degree at most $n(n-1)!=n!$. This also works in inseparable characteristic and proves the required bound without assuming Galoisness.

For the sextic with constant term three, Eisenstein's criterion at three gives degree six. Choose a root $\alpha$, so ${\alpha }^3$ is a square root of $-3$. Thensatisfies ${\zeta }^2-\zeta +1=0$, and is a primitive sixth root of unity: ${\zeta }^3=-1$ and $\zeta \ne -1$. The six distinct roots are $\alpha {\zeta }^j$ for $0\le j<6$, all in $\mathbb{Q}(\alpha )$. Hence the simple extension is itself the splitting field and $\boxed{[L:\mathbb{Q}]=6}$.

For the constant term two, Eisenstein at two gives $[\mathbb{Q}(\gamma ):\mathbb{Q}]=6$ for a root $\gamma$. Its splitting field is $\mathbb{Q}(\gamma ,{\zeta }_6)$. The supplied noncontainment of a root of $X^2+3$ means that adjoining ${\zeta }_6=(1+\sqrt{-3})/2$ has degree two. Thus

Work in ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-d}]$, since $d\equiv 2(\mathrm{mod}4)$. First $x$ must be odd: if it were even, coprimality would make $a$ odd, so $x^2+a^{2}d\equiv 2(\mathrm{mod}4)$, which cannot be an odd power $y^p$ for $p\ge 3$. Thus $y$ is odd too. Also no odd prime $\ell$ can divide both $x$ and $d$. Such a prime does not divide $a$, and square-freeness of $d$ gives $v_{\ell }(x^2+a^{2}d)=1$, whereas $v_{\ell }(y^p)$ is divisible by $p$. Hence $\gcd (x,d)=1$.

Put $\alpha =x+a\sqrt{-d}$. Suppose a prime ideal $P$ divides both $(\alpha )$ and $(\overline{\alpha })$, and let $\ell$ be the rational prime below it. It then contains $2x$ and $2a\sqrt{-d}$. The case $\ell =2$ is impossible because $N((\alpha ))=y^p$ is odd. If $\ell$ is odd, the first inclusion gives $\ell \mid x$, and $\gcd (a,x)=1$ makes $a$ nonzero in the residue field. The second inclusion therefore gives $\sqrt{-d}\in P$, so $\ell \mid d$, contradicting the previous paragraph. Thus $(\alpha )$ and $(\overline{\alpha })$ are coprime ideals. In particular they generate the unit ideal, the precise meaning of coprimality here; no unique factorization of elements is assumed.

Their product is $(y{)}^p$. Unique prime ideal factorization and disjoint prime support imply $(\alpha )=A^p$ for an integral ideal $A$. In the ideal class group, $[A{]}^p=1$. Since $p$ does not divide the class number, Lagrange's theorem forces $[A]=1$. Write $A=(u+v\sqrt{-d})$. Equality of the principal ideals gives $\alpha =\epsilon (u+v\sqrt{-d}{)}^p$. The only units are $\pm 1$, since a unit satisfies $u^2+d{v}^2=1$ and $d\ge 2$. Both signs are $p$th powers for odd $p$, so they can be absorbed into the generator:Apply the class-number obstruction from a ramified odd exponent. For the equation with $a=1$ and $p\mid d$, compare the coefficients of $\sqrt{-d}$ in this identity. The coefficient on the right isIts $j=1$ term is divisible by $p$, and every $j\ge 3$ term contains a factor $d$, so the whole coefficient is divisible by $p$. The coefficient on the left is one, a contradiction. Thus there are no positive integer solutions under the stated class-number and divisibility assumptions.

Finally,The integer $174=2\cdot 3\cdot 29$ is square-free, is two modulo four, and is divisible by three. If its imaginary quadratic class number were not divisible by three, the preceding result with $p=3$ would prohibit this solution. Consequently

Minkowski convex body theorem states that a centrally symmetric convex measurable subset of ${\mathbb{R}}^n$ with volume greater than $2^n$ times the covolume of a full Euclidean lattice contains a nonzero lattice point. The equivalent closed-boundary form follows by compactness for compact convex bodies.

The set $S_t$ is centrally symmetric. It is convex because for $0\le \lambda \le 1$, the triangle inequality gives $\Vert \lambda u+(1-\lambda )v{\Vert }_1\le \lambda \Vert u{\Vert }_1+(1-\lambda )\Vert v{\Vert }_1\le t$. It is a diamond of area $2t^2$. For the embedded nonzero integral ideal $I$, the given covolume is $2N(I)\sqrt{d}$. Hence Minkowski convex body theorem applies whenFor the resulting nonzero $\alpha \in I$, write $x={\sigma }_1(\alpha )$, $y={\sigma }_2(\alpha )$. ThenDecrease $t$ to $2\sqrt{N(I)\sqrt{d}}$. A fixed slightly larger diamond contains only finitely many lattice points, so one nonzero $\alpha$ recurs along a sequence of such $t$ values. Passing to the limit provesGiven a target ideal class $C$, choose an integral ideal $I$ in $C^{-1}$. Since $\alpha \in I$, the ideal $J=(\alpha )I^{-1}$ is integral, belongs to $C$, and has ideal normFor $K=\mathbb{Q}(\sqrt{14})$, every ideal class thus has an integral representative of norm at most $\sqrt{14}<4$. The only possibilities are norms one, two or three. The prime two ramifies, giving the unique ideal of norm two $P_2=(2,\sqrt{14})$. The element $4+\sqrt{14}$ has field norm $16-14=2$, so $(4+\sqrt{14})=P_2$, making it principal. The prime three is inert because $X^2-14\equiv X^2+1(\mathrm{mod}3)$ is an irreducible polynomial; its prime ideal has norm nine, so there is no ideal of norm three. Every possible small representative is principal, proving class number one for Q square root of fourteen. Therefore

Dedekind factorization theorem states that if $K=\mathbb{Q}(\theta )$ with integral $\theta$ and $p$ does not divide $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, then a factorization $\overline{f}={\prod }_i{\overline{f}}_i^{e_i}$ of its minimal polynomial of an algebraic element into distinct monic irreducible polynomials over ${\mathbb{F}}_p$ givesThe factors are distinct prime ideals; the residue degrees and ramification indices satisfy ${\sum }_i{e}_i\mathrm{deg}{f}_i=[K:\mathbb{Q}]$. Unique prime ideal factorization holds in the ring of integers.

For a square-free integer $d\equiv 1(\mathrm{mod}4)$, ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-d}]$, so the index condition is automatic. If $p_i\mid d$, the polynomial $X^2+d$ reduces to $X^2$ modulo $p_i$. HenceTherefore $d{\mathcal{O}}_K=({\prod }_i{P}_i{)}^2=(\sqrt{-d}{)}^2$. By unique prime ideal factorization, ${\prod }_i{P}_i=(\sqrt{-d})$.

Let $S$ be a nonempty proper subset and put $m={\prod }_{i\in S}{p}_i$. For $d>1$, $1<m<d$ and $m$ is a square-free integer. If ${\prod }_{i\in S}{P}_i=(u+v\sqrt{-d})$, its ideal norm givesThe inequality $m<d$ forces $v=0$, leaving $m=u^2$, impossible for a square-free integer greater than one. Thus every such subproduct is nonprincipal.

These are ramified prime classes for discriminant minus four d. The ideal classes $[P_i]$ have order dividing two. Their product is the identity, and the preceding proof rules out every other nonempty subset relation. The homomorphism ${\mathbb{F}}_2^k\to \mathrm{Cl}(K)$ defined by these classes consequently has kernel exactly $\{0,(1,\dots ,1)\}$. Its image is thereforeThe boundary case $d=1$ has $k=0$ and makes the printed exponent $k-1$ meaningless; its field is $\mathbb{Q}(i)$ with trivial ideal class group. The subgroup statement is understood for $d>1$, while the prime-factor and proper-subset statements are empty for $d=1$.

For the $n$ distinct complex embeddings ${\sigma }_1,\dots ,{\sigma }_n$ of the number field, define the discriminant of elements of a number field byThe equality follows by multiplying the transpose of the embedding matrix by itself. Each field trace is rational: it is also the trace of the rational matrix of multiplication by $w_i{w}_j$ on $K$. Thus the determinant lies in $\mathbb{Q}$. If the $w_i$ are algebraic integers, the products and their conjugates are algebraic integers, and so are their sums. A rational algebraic integer is an integer, hence all trace entries and their determinant are integers.

Write $u_j={\sum }_i{a}_{ij}{v}_i$. The inclusion of the generated abelian groups makes $A=(a_{ij})$ an integer matrix. Its embedding matrix is the embedding matrix of the $v_i$ multiplied by $A$. Taking squared determinants givesFor independent algebraic integers $v_i$, choose an integral basis ${\omega }_i$ of ${\mathcal{O}}_K$. The same formula gives $D(v)=[{\mathcal{O}}_K:{\sum }_i\mathbb{Z}{v}_i{]}^{2}D(\omega )$. If $D(v)$ is nonzero and a square-free integer, the integer index must be one, proving that the $v_i$ form an integral basis.

Now put $\beta =3\alpha +4$. Since $\mathbb{Q}(\beta )=\mathbb{Q}(\alpha )$ has degree three, the monic polynomialis its minimal polynomial of an algebraic element. The field norm is the negative of its constant term:For the discriminant calculation, $f^{\prime }(\alpha )=3{\alpha }^2+4\alpha$. In the basis $1,\alpha ,{\alpha }^2$, using ${\alpha }^3=-2{\alpha }^2-1$, multiplication by this element has matrixThus the given derivative-norm formula gives $D(1,\alpha ,{\alpha }^2)=-59$. Since $\alpha$ satisfies a monic polynomial with integer coefficients, $1,\alpha ,{\alpha }^2$ are algebraic integers; the fact that a squarefree discriminant certifies an integral basis proves

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.

Put $\theta =\sqrt{p+\sqrt{p}}$. Then $({\theta }^2-p{)}^2=p$, so $\theta$ is a root ofThis polynomial satisfies Eisenstein's criterion at $p$: every nonleading coefficient is divisible by $p$, while the constant coefficient $p(p-1)$ is not divisible by $p^2$. ThereforeNow suppose $p=n^2+1$, taking $n>0$ without loss of generality. Let $\beta =\sqrt{p-\sqrt{p}}$. Since $\theta \beta =n\sqrt{p}$ and $\sqrt{p}={\theta }^2-p$,Consequently all four distinct roots $\pm \theta ,\pm \beta$ lie in this field. This gives a cyclic quartic field from a nested quadratic radical. It is a splitting field in characteristic zero, hence a normal extension and a Galois extension.

There is a Galois automorphism $\sigma$ sending $\theta$ to $\beta$: a root image defines an embedding, and normality makes it an automorphism of the splitting field. It sends $\sqrt{p}={\theta }^2-p$ to $-\sqrt{p}$, soThus $\theta \mapsto \beta \mapsto -\theta \mapsto -\beta \mapsto \theta$ is a four-cycle. The Galois group has order four and contains an element of order four, givingFor the requested lower degrees, $(q,r)=(3,4)$ gives $\sqrt{q+\sqrt{r}}=\sqrt{5}$ and degree two; $(q,r)=(3,1)$ gives $2$ and degree one. Both pairs use positive integers.

An algebraic element is a separable algebraic element over $K$ if its minimal polynomial of an algebraic element has distinct roots in a splitting field. A field extension is a separable extension if all its elements are separable algebraic elements.

The formal derivative is additive and satisfies $D(X^m)=m{X}^{m-1}$. For monomials, $D(X^r{X}^s)=(r+s)X^{r+s-1}=X^{r}D(X^s)+X^{s}D(X^r)$. Distributing over the finitely many coefficients proves the product rule $D(fg)=fDg+gDf$ over any field. If $f=(X-\alpha {)}^{2}h$ in an extension, the rule gives $Df=2(X-\alpha )h+(X-\alpha {)}^{2}Dh$, so $Df(\alpha )=0$, including in characteristic two. If $f,Df$ were relatively prime in $K[X]$, the Bézout's identity for polynomials would give $uf+vDf=1$, whose evaluation at $\alpha$ is impossible. Hence they are not relatively prime.

In characteristic zero, the formal derivative of a positive-degree irreducible polynomial $m$ is nonzero and has smaller degree. Thus $\gcd (m,Dm)=1$, since an irreducible polynomial has no proper nonconstant factor. By the previous argument, $m$ has no repeated root. Every element of a finite field extension is algebraic, so the extension is a separable extension.

For $K={\mathbb{F}}_p(t)$, let ${\alpha }^p=t$. The characteristic-$p$ identity gives $X^p-t=(X-\alpha {)}^p$, so $K(\alpha )$ is already a splitting field. It is a proper extension by the allowed no-root fact. More precisely, the minimal polynomial of an algebraic element $\alpha$ divides $X^p-t$ and, over $K(\alpha )$, equals $(X-\alpha {)}^r$ for some $1\le r\le p$. If $r<p$, its next-to-leading coefficient is $-r\alpha \in K$, forcing $\alpha \in K$ because $r\ne 0$ in ${\mathbb{F}}_p$, a contradiction. Hence its minimal polynomial of an algebraic element is $X^p-t$, with its one root repeated $p$ times. The extension is therefore inseparable, of degree $p$.

By contrast, every finite field extension $L/{\mathbb{F}}_p$ is a separable extension. If $[L:{\mathbb{F}}_p]=n$, then $|L|=p^n$. Lagrange's theorem in its nonzero multiplicative group gives $a^{p^n}=a$ for all $a\in L$. Thus every element is a root of $X^{p^n}-X$, whose formal derivative is $-1$; it has no repeated roots. Every minimal polynomial of an algebraic element divides this separable polynomial, proving the assertion.