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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.

An algebraic field extension $E/K$ is separable if every $\alpha \in E$ has a separable minimal polynomial over $K$.

For an inseparable example, take $K={\mathbb{F}}_p(t)$ and $f(X)=X^p-t$. Eisenstein's criterion at the prime $t$ in ${\mathbb{F}}_p[t]$, followed by Gauss lemma for polynomials, proves irreducibility over its fraction field. Its derivative is zero, and in a splitting field it is $(X-t^{1/p}{)}^p$, so it is inseparable.

Now let $M/K$ be a separable splitting extension of a nonzero polynomial $f$. Every irreducible factor $g$ of $f$ has a root $\alpha \in M$, and $g$ is a scalar multiple of the minimal polynomial of $\alpha$. Separability of $M/K$ makes this polynomial separable. Therefore every irreducible factor of $f$ is separable, which proves the requested conclusion in the factorwise convention. The squarefree qualification under the alternative convention is explained in part (a).

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 $d$.

Set $K_r=K({\alpha }_1,\dots ,{\alpha }_r)$, beginning with $K_0=K$. Suppose $j:K_{r-1}\to L$ extends $i$. The minimal polynomial of ${\alpha }_r$ over $K_{r-1}$ divides its minimal polynomial over $K$. Its image under $j$ therefore divides $i(m_{{\alpha }_r})$, which splits into distinct linear factors in $L$. It consequently has exactly $[K_r:K_{r-1}]$ distinct roots in $L$. Part (i) supplies exactly this many extensions of $j$ to $K_r$.

Starting with the unique prescribed map on $K$, multiply these branching counts:For the final assertion, let $M$ be a splitting field of a separable polynomial and embed it in an algebraic closure $\Omega$. It is generated by roots whose minimal polynomials are separable and split in $\Omega$. Applying the just-proved count with $L=\Omega$ gives $[M:K]$ embeddings. Were $M/K$ inseparable, part (a) would give strictly fewer. Thus the splitting extension is separable. This works for the factorwise convention as well, since the generators' minimal polynomials divide the separable irreducible factors.

The cyclotomic polynomial isBy the allowed irreducibility, this is the minimal polynomial of $\xi$ and its degree is $\phi (n)$. The field $\mathbb{Q}(\xi )$ contains every $n$th root of unity and is the splitting field of $X^n-1$. This polynomial has no repeated roots in characteristic zero, so the extension is Galois.

An automorphism sends $\xi$ to another primitive root ${\xi }^a$, where $a$ is a unit modulo $n$. Conversely each such conjugate defines an embedding, and it is an automorphism of $\mathbb{Q}(\xi )$ because $a$ has an inverse modulo $n$. Composition corresponds to multiplication of exponents. Therefore

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.

The Kummer-Dedekind theorem states the following. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral and $p\nmid [{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor the minimal polynomial modulo $p$ as $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ into distinct monic irreducible factors. ThenThe index restriction is essential.

For the proposed ideal relation, reflexivity uses $\alpha =\beta =1$ and symmetry interchanges $\alpha ,\beta$. If $(\alpha )I_1=(\beta )I_2$ and $(\gamma )I_2=(\delta )I_3$, multiplication gives $(\alpha \gamma )I_1=(\beta \delta )I_3$. All multipliers remain nonzero. Thus it is an equivalence relation, namely equality of ideal classes.

Put $s=\sqrt{-2}$. The minimal polynomial is $x^2+2$ and the given full power basis makes every prime eligible for the theorem. Modulo $3$ its roots are $\pm 1$; modulo $5$ and $7$ it is irreducible; modulo $17$ its roots are $\pm 7$. ConsequentlyThe two inert ideals have norms $25$ and $49$; each split factor has norm equal to its rational prime.

For $p_1\equiv 1(\mathrm{mod}8)$, quadratic reciprocity and its supplementary laws give $(-2/p_1)=1$. Thus $x^2+2$ has two distinct roots modulo $p_1$, giving distinct prime ideals $P,P^{\prime }$ withThe given principal ideal domain property makes $P=(x+ys)$ for some integers $x,y$. Taking norms, which are positive here, gives $p_1=N(P)=x^2+2y^2$.

Let $\omega =(1+\sqrt{-3})/2$, whose minimal polynomial is $x^2-x+1$. For $p_2\equiv 1(\mathrm{mod}3)$, the cyclic group ${\mathbb{F}}_{p_2}^{\times }$ has a nontrivial cube root of unity $t$. Then $-t$ and $-t^2$ are distinct roots of $x^2-x+1$. By the Kummer-Dedekind theorem,Since the given ring of integers is a principal ideal domain, write $Q=(a+b\omega )$. Its norm isMultiplication by $4$ completes the square:Thus $x=2a+b$ and $y=b$ supply the required integers.

An element is algebraic over $F$ when some nonzero polynomial in $F[x]$ vanishes at it. Its minimal polynomial is the monic polynomial of least positive degree with this property; scaling a least-degree nonzero vanishing polynomial makes it monic.

If $m_1,m_2$ are two such polynomials, they have the same least degree. Their difference has smaller degree because both are monic, and it also vanishes at $a$. Minimality forces the difference to be zero. Hence the monic minimal polynomial is unique.

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.

The degree $[K:F]$ is the dimension of $K$ as a vector space over $F$. The tower theorem states that for intermediate fields $F\subseteq E\subseteq K$ in a finite extension,If the degree is prime, one factor must be one, forcing $E=K$ or $E=F$; no strictly intermediate field exists.

Let $K$ have 32 elements. Its prime subfield is ${\mathbb{F}}_2$ and $[K:{\mathbb{F}}_2]=5$. Every element outside that subfield has minimal polynomial of degree five, by the tower theorem. Also every member of $K$ satisfies $x^{32}=x$: nonzero elements form a group of order 31. Hence $x^{32}-x$ splits over $K$, with distinct roots because its derivative is $-1=1$ in characteristic two.

Every irreducible degree-five polynomial over ${\mathbb{F}}_2$ divides $x^{32}-x$. Indeed its quotient field has 32 elements, and its residue class of $x$ satisfies that equation. Thus every such polynomial splits into five distinct roots in $K$. Conversely each of the 30 elements outside ${\mathbb{F}}_2$ has one such minimal polynomial, and different minimal polynomials have disjoint root sets. Consequently the number isThere is no extra leading-coefficient factor, since every nonzero leading coefficient over ${\mathbb{F}}_2$ is one.

The polynomial identityshows that the roots are the ninth roots of unity which are not third roots. These are precisely the primitive ninth roots of unity, all powers of ${\zeta }_9=e^{2\pi i/9}$. Thus the splitting field is $\mathbb{Q}({\zeta }_9)$. The supplied irreducibility statement makes the displayed degree-six polynomial the minimal polynomial of ${\zeta }_9$, giving

The notation $K=F(\delta )$ means that $K$ is the smallest subfield containing both $F$ and $\delta$, namely the simple field extension generated by $\delta$. Its elements can be expressed as rational functions in $\delta$ with coefficients in $F$ and nonzero denominator.

Write $d={\delta }^2\in F$. The element $\delta$ is a root of $X^2-d$, so its minimal polynomial over $F$ has degree at most two. Since $\delta \notin F$, that degree cannot be one. The degree of the simple field extension equals the degree of the minimal polynomial, givingIn particular $1,\delta$ form an $F$-basis, so every element is uniquely $a+b\delta$. This argument also works in characteristic two; separability is not needed for this part.

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.

A primitive seventh root generates all seventh roots, so the splitting field is $\mathbb{Q}({\zeta }_7)$. Its minimal polynomial is the given irreducible cyclotomic polynomial $1+x+\cdots +x^6$. Therefore $\boxed{[K:\mathbb{Q}]=6}$.

For a simple field extension, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis when the minimal polynomial has degree $d$: reduction modulo that polynomial spans the extension, and a relation of smaller degree would contradict minimality. Hence $[F(\lambda ):F]=d$. The field is $K$, whose dimension is $n$, giving $d=n$.

The characteristic of a field is the least positive integer $r$ with $r{1}_K=0$, or zero if none exists. In a finite field a positive such integer exists because the sequence $0,1_K,2\cdot 1_K,\dots$ must repeat.

The prime subfield is the intersection of all subfields of $K$, equivalently the subfield generated by $1_K$. In characteristic $p$ it is the copy of ${\mathbb{F}}_p$ formed by the multiples of $1_K$. Every field automorphism preserves $1_K$ and addition, so it fixes every element of this subfield.

Finally, once the order-$n$ Frobenius automorphism below is established, all its $n$ powers are distinct. Any automorphism fixes $F$ and is determined by its value at the generator $\lambda$. That value must be a root of the degree-$n$ minimal polynomial of $\lambda$, so there are at most $n$ automorphisms. Hence

The cubic has no root in ${\mathbb{F}}_2$, since its values at $0$ and $1$ are both $1$. A cubic over a field is irreducible precisely when it has no root. Consequentlyis a field extension of degree $3$, with eight elements. The Frobenius endomorphism $x\mapsto x^2$ takes a root of a polynomial over ${\mathbb{F}}_2$ to another root, so the roots are $\alpha ,{\alpha }^2,{\alpha }^4$. They are distinct: equality between any two would make the Frobenius endomorphism orbit have length less than $3$, contradicting the minimal polynomial of $\alpha$. They all lie in $K$, and one of them generates $K$. Hence