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The Dedekind factorization theorem says that, when a rational prime $p$ does not divide the index of an order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_K$, a factorizationof the minimal polynomial modulo $p$ gives

Let $K=\mathbb{Q}(\sqrt{d})$ with square-free $d$, and let $p$ be odd. The resulting splitting of rational primes in a quadratic field is:

Finally suppose $d>0$ and
$\alpha =A/C+(B/C)\sqrt{d}$ has norm $-1$, with integers
$A,B,C$ chosen coprime. ThenIf an odd $p\equiv 3(\mathrm{mod}4)$ ramified, then $p\mid d$. Reduction modulo $p$ gives
$A^2\equiv -C^2(\mathrm{mod}p)$. Since $-1$ is not a square modulo such a prime, $p\mid A,C$. Dividing the equation's divisibility shows $p\mid B$ as well (use that square-free $d$ has $p$-adic valuation one), contradicting coprimality. Hence no prime $p\equiv 3(\mathrm{mod}4)$ ramifies.

The automorphisms $1,\sigma ,\dots ,{\sigma }^{n-1}$ are distinct and hence linearly independent as maps $L\to L$ by the linear independence of distinct field embeddings. Therefore the operatoris not the zero map. Choose $\beta \in L$ for which $\alpha =T(\beta )\ne 0$. Reindexing the sum and using ${\sigma }^n=1$ givesThis is the Lagrange resolvent eigenvector for a cyclic field automorphism.

Let $F=L^{\sigma }$ be the fixed field. Since $\sigma ({\alpha }^n)={\alpha }^n$, the element $\alpha$ is a root ofIts orbit under $⟨\sigma ⟩$ isand these $n$ elements are distinct because $\alpha \ne 0$ and $\zeta$ is a primitive root of unity. Hence the minimal polynomial has at least $n$ distinct roots. It also divides the displayed degree-$n$ polynomial, so

Because $\mathrm{char}K=0$, the finite extension $M/K$ is separable. By the primitive element theorem, write $M=K(\theta )$. The minimal polynomial of $\theta$ over $K$ has a root in the algebraically closed field $L$, so sending $\theta$ to that root defines a $K$-field embedding $\iota :M\hookrightarrow L$.

The cyclic group $⟨\sigma ⟩$ acts faithfully on $L$ after replacing $d$ by the order of $\sigma$. The Artin fixed-field theorem givesand says that $L/K$ is a finite Galois extension with cyclic Galois group $⟨\sigma ⟩$. The image $E=\iota (M)$ is an intermediate field. Every subgroup of a cyclic group is normal, so the normal subextension criterion shows that $E/K$ is Galois, and its Galois group is a quotient of $⟨\sigma ⟩$, hence cyclic. Transporting this structure through $\iota$ proves that $M/K$ is Galois with cyclic Galois group. This is the finite extensions of the fixed field of a finite-order automorphism of an algebraically closed field theorem.

Algebraic closedness is necessary. Take $L=\mathbb{Q}$ and $\sigma =\mathrm{id}$, so $K=L^{\sigma }=\mathbb{Q}$, and let $M=\mathbb{Q}(\sqrt[3]{2})$. The extension has degree three but is not normal, since the two nonreal roots of $X^3-2$ do not belong to $M$. It is therefore not Galois.

Letbe a primitive polynomial. The Eisenstein criterion states that if a prime number $p$ satisfiesthen $F$ is an irreducible polynomial in $\mathbb{Z}[X]$, equivalently in $\mathbb{Q}[X]$ by Gauss lemma for polynomials.

To prove it, suppose that $F=GH$ with $G,H\in \mathbb{Z}[X]$ of positive degree. Primitivity and Gauss's lemma let us take such an integral factorisation if a factorisation over $\mathbb{Q}$ exists. Reducing modulo $p$ givesBecause $p$ does not divide the leading coefficient of $F$, neither factor loses degree on reduction. The polynomial ring ${\mathbb{F}}_p[X]$ is a unique factorization domain, so both reductions are monomials of positive degree. In particular, $p$ divides both constant terms $G(0)$ and $H(0)$. It follows that $p^2$ dividesa contradiction. This proves the criterion.

For a prime $p$, translate the geometric sum by one:Its leading coefficient is one, every other coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. It is therefore Eisenstein at $p$. Translation $X\mapsto X+1$ is an automorphism of $\mathbb{Z}[X]$, soThis is the geometric-sum irreducibility criterion.

The evaluation homomorphismhas image $\mathbb{Z}[\zeta ]$ and contains $(f)$ in its kernel. Since $f$ is monic, division by $f$ in $\mathbb{Z}[X]$ writes every $G$ as $G=Qf+R$ with $\mathrm{deg}R<\mathrm{deg}f$. If $G(\zeta )=0$, then $R(\zeta )=0$; the irreducibility of $f$ says that $f$ is the minimal polynomial of $\zeta$, so $R=0$. Thus $\ker ({\mathrm{ev}}_{\zeta })=(f)$, and the first isomorphism theorem for rings gives

Now take $p=3$. Then ${\zeta }^2+\zeta +1=0$, and the Eisenstein integers areComplex conjugation sends $\zeta$ to ${\zeta }^2$, so the field norm isGiven $z=x+y\zeta \in \mathbb{C}$, choose integers $m,n$ with $|x-m|,|y-n|\le \frac{1}{2}$. For $q=m+n\zeta$,For $\alpha ,\beta \in \mathbb{Z}[\zeta ]$ with $\beta \ne 0$, apply this to $z=\alpha /\beta$ and put $r=\alpha -q\beta$. ThenHence the norm is a Euclidean function, proving that $\mathbb{Z}[\zeta ]$ is a Euclidean domain. This is the Euclidean norm on the Eisenstein integers.

Finally suppose $A\in {\mathrm{GL}}_n(\mathbb{Z})$ satisfies $A^2+A+I=0$. Make the free abelian group $M={\mathbb{Z}}^n$ into a $\mathbb{Z}[\zeta ]$-module by definingThis is well-defined precisely because $A$ obeys the same polynomial relation as $\zeta$. The module is finitely generated. It is also torsion-free module: if $0\ne \alpha \in \mathbb{Z}[\zeta ]$ and $\alpha v=0$, multiplication by the conjugate of $\alpha$ gives $N(\alpha )v=0$, and the additive group ${\mathbb{Z}}^n$ has no nonzero integer torsion.

A Euclidean domain is a principal ideal domain, and the structure theorem for finitely generated modules over a principal ideal domain says that a finitely generated torsion-free module over one is free. Thusfor some $m$. Since $\mathbb{Z}[\zeta ]$ has free module basis $1,\zeta$ over $\mathbb{Z}$, comparison of abelian ranks gives $n=2m$. There can consequently be no such matrix when $n$ is odd.

If $n=2m$, choose a $\mathbb{Z}[\zeta ]$-basis $v_1,\dots ,v_m$. Thenis a $\mathbb{Z}$-basis, and $A^2{v}_i=-v_i-A{v}_i$. In this basis the matrix of $A$ is a direct sum of $m$ copies ofEvery admissible matrix is therefore conjugate in ${\mathrm{GL}}_n(\mathbb{Z})$ to $C^{\oplus m}$. Hence there is exactly one conjugacy class for even $n$, and none for odd $n$. This is the classification of integral matrices satisfying the third cyclotomic polynomial.

A finite extension $L/K$ is normal when every irreducible polynomial in $K[T]$ having one root in $L$ splits completely over $L$.

Suppose first that $L/K$ is normal. Choose ${\alpha }_1,\dots ,{\alpha }_m$ with $L=K({\alpha }_1,\dots ,{\alpha }_m)$, and let $f_j$ be the minimal polynomial of ${\alpha }_j$ over $K$. Every $f_j$ splits in $L$ by normality, and $L$ is generated by their roots because it is already generated by the particular roots ${\alpha }_j$. Thus $L$ is the splitting field of $f_1\cdots f_m$.

Conversely, suppose that $L$ is the splitting field of $f\in K[T]$. Every $K$-embedding $\sigma :L\to \overline{K}$ permutes the roots of $f$, so $\sigma (L)=L$. By the embedding characterization of normality, $L/K$ is normal. This proves the finite normal extension as a splitting field criterion.

Because $p\equiv 11(\mathrm{mod}12)$, we have $p\equiv 3(\mathrm{mod}4)$ andFor $\theta =(1+\sqrt{-p})/2$, the minimal polynomial isSince $p\equiv 2(\mathrm{mod}3)$, reduction modulo $3$ gives $f(T)\equiv T(T-1)$. Its two roots are distinct, sowith distinct prime ideals of norm $3$. Thus $3$ splits completely, as recorded in splitting of three in Q of square root minus p.

The ideal class group ${\mathrm{Cl}}_K$ is the group of nonzero fractional ideals of ${\mathcal{O}}_K$ modulo the subgroup of principal fractional ideals. Saying that ${\mathfrak{p}}_1$ has order $n$ means that $n$ is the least positive integer for which ${\mathfrak{p}}_1^n$ is principal.

Suppose for contradiction that $n$ is odd and ${\mathfrak{p}}_1$ has order $n$. Then ${\mathfrak{p}}_1^n=(\alpha )$ for some $\alpha \in {\mathcal{O}}_K$, and taking norms givesWrite $\alpha =(a+b\sqrt{-p})/2$, where $a,b\in \mathbb{Z}$ have the same parity. ThenIf $b\ne 0$, the right-hand side is at least $p$, contradicting $p>3^{n+2}>4\cdot 3^n$. Hence $b=0$, so $\alpha$ is a rational algebraic integer and therefore an integer. But then $3^n=N(\alpha )={\alpha }^2$, impossible when $n$ is odd. This proves the odd-order obstruction for a split prime in an imaginary quadratic field and shows that the order cannot equal $n$.

The Liouville approximation theorem states that if $\alpha$ is an irrational algebraic number of degree $d\ge 2$, then there is a constant $c(\alpha )>0$ such that every reduced rational $p/q$ satisfies

Let $P\in \mathbb{Z}[X]$ be the minimal polynomial of $\alpha$, of degree $d$. Since $P$ is irreducible of degree greater than one, $P(p/q)\ne 0$. Moreover,and henceOn the compact interval $I=[\alpha -1,\alpha +1]$, putIf $p/q\in I$, the mean value theorem gives a point $\xi$ between $\alpha$ and $p/q$ such thatThus $|\alpha -p/q|\ge M^{-1}{q}^{-d}$. If $p/q\notin I$, then $|\alpha -p/q|>1\ge q^{-d}$. Taking anyproves the stated strict inequality.

Since $\alpha$ generates $K^{\times }$, it cannot lie in a proper subfield, so its minimal polynomial over ${\mathbb{F}}_2$ has degree $d$. Write it asThe identity $P(\alpha )=0$ and characteristic two giveMultiplying by ${\alpha }^n$ and applying the linear map $T$ yieldsThis is a binary linear recurrence of order at most $d$, so $(x_n)$ is the output of a linear-feedback shift register of length at most $d$.

An element $x\in L$ is separable over $K$ when its minimal polynomial $m_x(T)\in K[T]$ has distinct roots in a splitting field.

Because $m_x$ is irreducible, it is separable exactly whenIf $m_x^{\prime }\ne 0$, then $\mathrm{deg}{m}_x^{\prime }<\mathrm{deg}{m}_x$, so irreducibility makes this greatest common divisor equal to one. If $m_x^{\prime }=0$, the zero-derivative criterion givesfor some $g\in K[T]$, and $m_x$ is inseparable. Therefore

The Jordan normal form theorem says that every complex square matrix is similar to a direct sum of Jordan blocks, uniquely up to their order. For an eigenvalue $\lambda$, if its blocks have sizes $n_1,\dots ,n_r$, then its algebraic multiplicity and geometric multiplicity areand its contribution to the minimal polynomial is $(t-\lambda {)}^{{\max }_j{n}_j}$. Thuswhere $s_{\lambda }$ is the largest $\lambda$-block size.

For the given matrix,Moreover,HenceThe eigenvalue $-1$ therefore has one Jordan block of size two, so

For a ring homomorphism $A\to B$, the module of Kähler differentials ${\Omega }_{B/A}$ is the $B$-module generated by symbols $db$, subject toEquivalently, it represents $A$-linear derivations:For $A\to B\to C$, the Transitivity exact sequence for Kähler differentials isIf $B\to C=B/I$ is surjective, the Conormal exact sequence for Kähler differentials is

Let $L/K$ be a finite field extension. By the primitive element theorem, its maximal separable field extension $L_s/K$ is simple, and transitivity reduces the calculation to a simple algebraic extension. If $L=K(\alpha )$ with minimal polynomial $m_{\alpha }$, thenThus a separable simple extension has zero differentials. Conversely, if $L/K$ is not separable, the purely inseparable part has a generator whose minimal polynomial has zero formal derivative in positive characteristic, producing a nonzero differential. Hence

Now let $\mathrm{char}K=p>0$. If $L=K(a)$, $a^p\notin K$, and $a^{p^2}\in K$, then the minimal polynomial is $T^{p^2}-a^{p^2}$ and has zero derivative, soIf $L=K(a,b)$, where $a\notin K$, $b\notin K(a)$, and $a^p,b^p\in K$, then $[K(a):K]=[L:K(a)]=p$. Both defining equations have zero derivative, and

For the three morphisms, compute the Sheaf of relative Kähler differentials on coordinate rings.

If $\mathrm{char}k\ne 2$, this is $k[y]/(y)dy$ and its support is the origin $V(y)=V(x,y)$. If $\mathrm{char}k=2$, it is free of rank one and its support is all of $X$.

Its support is the origin $V(x)$ in every characteristic; the module is $k[x]/(x)$ unless $\mathrm{char}k=2$, when it is $k[x]/(x^2)$.

Its support is the entire component $V(x)\subset X$.

For a finite extension $L/K$ of local fields, let $e=e(L/K)$ be its ramification index and $f=f(L/K)=[k_L:k_K]$ its residue-field degree. An unramified extension has $e=1$. A totally ramified extension has $e=[L:K]$, equivalently $f=1$. A tamely ramified extension has separable residue extension and ramification index coprime to the residue characteristic.

Let $\bar{\theta }$ generate the finite extension $k_L/k_K$, and let $\bar{g}$ be its minimal polynomial. Lift $\bar{g}$ to a monic $g\in {\mathcal{O}}_K[X]$ and choose any lift $t\in {\mathcal{O}}_L$ of $\bar{\theta }$. Since finite fields are perfect fields, ${\bar{g}}^{\prime }(\bar{\theta })\ne 0$. The simple-root form of Hensel lemma, applied inside $L$, gives $\theta \in {\mathcal{O}}_L$ withSet $K_0=K(\theta )$. Its residue field contains $k_K(\bar{\theta })=k_L$, soThe equation $g(\theta )=0$ gives the reverse inequality. Thus $[K_0:K]=f$, its residue-field degree is $f$, and $e(K_0/K)=1$; hence $K_0/K$ is unramified. Since $k_{K_0}=k_L$, the extension $L/K_0$ has residue-field degree one and is totally ramified. This constructs the maximal unramified subextension of a local field extension.

The minimal polynomial $m_{\varphi }$ is the unique monic polynomial of least positive degree with $m_{\varphi }(\varphi )=0$. Uniqueness follows because division of one monic annihilator by another would produce a lower-degree annihilator. Cayley–Hamilton says that the characteristic polynomial annihilates $\varphi$; division then shows

The map $\theta :\mathbb{C}[x]\to V$, $p\mapsto p(\varphi )a_1$, is surjective because $1,x,\dots ,x^{n-1}$ map to the given basis. Its kernel is generated by the minimal polynomial and the quotient has dimension $n$, so $\mathrm{deg}{m}_{\varphi }=n$. The relation for $\varphi (a_n)$ gives

Make $V$ a $k[x]$-module by $xv=\phi (v)$. The structure theorem over the Euclidean domain $k[x]$ decomposes its torsion module aswhere the $p_{\alpha }$ are monic irreducibles. Each summand is indecomposable: its submodules form a chain, so two nonzero submodules cannot form a direct sum. This is the desired decomposition into invariant indecomposable subspaces.

The characteristic polynomial iswhereas the minimal polynomial is the least common multiple

Suppose first that $L/K$ is a totally ramified extension of degree $n$, and let $\alpha$ be a uniformizer of $L$. If the valuation on $L$ is normalized by $v_L(\alpha )=1$, then $v_L(K^{\times })=n\mathbb{Z}$. The value group of $K(\alpha )$ already contains both $n\mathbb{Z}$ and $1$, so its ramification index over $K$ is at least $n$. Hence $[K(\alpha ):K]\ge n$, and therefore $L=K(\alpha )$.

Letbe the minimal polynomial of $\alpha$. Every conjugate of $\alpha$ has positive valuation, so each $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverwhich means $v_K(a_0)=1$. Thus $f$ is an Eisenstein polynomial.

Conversely, if $\alpha$ is a root of an Eisenstein polynomial of degree $n$, the Eisenstein criterion makes that polynomial irreducible and its Newton polygon gives $v_L(\alpha )=1/n$ when $v_K$ is normalized. Consequently $e(K(\alpha )/K)\ge n$; equality with the field degree forces $e=n$ and residue-field degree one. Thus $K(\alpha )/K$ is totally ramified.

For the general result, choose logarithms of nonzero algebraic numbers ${\alpha }_1,\dots ,{\alpha }_s$ and algebraic coefficients ${\beta }_0,\dots ,{\beta }_s$, and putLetwhere $f_{{\alpha }_j}$ is the minimal polynomial and its $H$ is the naive polynomial height, and letThe general lower bound for a linear form in logarithms states that, if $\Lambda \ne 0$, thenwhere the effective constant $C$ depends only on $s$ and the degree of the number field generated by all the data.

For the improved homogeneous result, take $b_1,\dots ,b_s\in \mathbb{Z}$ andWith the same $A_j$, after ordering the terms setThe Baker lower bound for a homogeneous linear form in logarithms givesBoth constants are effective. The division by $\log A_s$ in $B^{\ast }$ is the improvement that matters when ${\alpha }_s$ has variable height.

For coprime integers $p,q$ with $q>0$,This follows either directly from the real and p-adic absolute values, or from the height-Mahler measure formula applied to the primitive minimal polynomial $qX-p$.

Suppose $P$ and $Q$ were linearly dependent over $\mathbb{Q}$. Then $P=cQ$ for some $c\in \mathbb{Q}$, with the zero-polynomial cases included. Since $d\ge 3$, $c+\alpha \ne 0$, and thereforeIf this has multiplicity $t$ at $\alpha$, then the minimal polynomial of $\alpha$ to the power $t$ divides $Q$ in $\mathbb{Q}[X]$. Hence $dt\le \mathrm{deg}Q\le n$.

The construction in part (b) givesWhen $\kappa <1$, this exceeds $n/d$ for all sufficiently large $n$ depending only on $\kappa$ and $d$, contradicting $dt\le n$. Thus $P,Q$ are linearly independent.

Squaring and eliminating the inner radical givesThus $\alpha ,\beta$ are roots ofand $\gamma ,\delta$ are roots ofBoth polynomials are irreducible by the Eisenstein criterion at $3$, so these are the four minimal polynomials over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{3})$, the relevant radicands areThe norms of $a$ and $b$ to $\mathbb{Q}$ equal $6$, so neither can be a square in $K$. Also $(u+v\sqrt{3}{)}^2=6$ with $u,v\in \mathbb{Q}$ would force $uv=0$, followed by either $u^2=6$ or $v^2=2$, both impossible. Thus $6$ is not a square either.

Similarly, over $K^{\prime }=\mathbb{Q}(\sqrt{6})$ the radicands $3+\sqrt{6}$ and $3-\sqrt{6}$ have nonsquare norm $3$, while $(u+v\sqrt{6}{)}^2=3$ would force either $u^2=3$ or $v^2=1/2$. Hence $3$ is not a square in $K^{\prime }$. Both extensions therefore satisfy all the hypotheses of part (b).