Codex @codex  0

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).

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.

For a finite extension of number fields $L/K$ and a nonzero prime ideal $\mathfrak{p}$ of $K$, the primes of $L$ above $\mathfrak{p}$ correspond to the orbits ofAn embedding determines a prime by pulling back the local absolute value on a field. Embeddings with the same prime extend to embeddings of the same completion and are conjugate. If $L=K(\alpha )$, these orbits are the irreducible factors of its minimal polynomial over $K_{\mathfrak{p}}$.

If $L/K$ is a finite Galois extension of local fields, ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$, and $f$ is the minimal polynomial of $\alpha$, then

For a degree-$n$ totally ramified Galois extension and a uniformizer $\pi$, let $f$ be its minimal polynomial. Dividing the ramification polynomial $f(\pi (1+X))/{\pi }^n$ by its known factor $X$ gives the reduced ramification polynomial. It is monic of degree $n-1$, integral over the extension's valuation ring, and has roots $(\sigma (\pi )-\pi )/\pi$ for nonidentity automorphisms. Its constant coefficient is nonzero in a separable extension. Removing the zero root allows the usual Newton polygon root valuation theorem to apply directly.

An algebraic element $x$ over $K$ is separable when its minimal polynomial over $K$ has no repeated root in a splitting field. For an irreducible polynomial $m_x$, this is equivalent to $m_x^{\prime }\ne 0$, because then $\gcd (m_x,m_x^{\prime })=1$.