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The non-Archimedean part of the Ostrowski theorem says that every nontrivial Non-Archimedean absolute value on $\mathbb{Q}$ is equivalent to the p-adic absolute value for a unique prime $p$.

Indeed, $|n|\le 1$ for every integer $n$. Nontriviality gives a prime $p$ with $|p|<1$, and there cannot be two such primes because the Bezout identity would make $1$ a sum of two terms of absolute value below one. If $a$ is coprime to $p$, another Bezout identity shows $|a|=1$. Consequentlyfor every $x\in {\mathbb{Q}}^{\times }$, which is a positive real power of $|x{|}_p$.

More generally, the Non-Archimedean absolute values on a number field $K$ are indexed, up to equivalence, by the nonzero prime ideals $\mathfrak{p}\subset {\mathcal{O}}_K$. The value attached to $\mathfrak{p}$ isTo prove completeness of the list, restrict an absolute value to $\mathbb{Q}$ and obtain a rational prime $p$. Its valuation ring contains ${\mathcal{O}}_K$ away from a unique prime above $p$; equivalently, its centeris a nonzero prime ideal. Since ${\mathcal{O}}_{K,\mathfrak{p}}$ is a discrete valuation ring, every $x\in K^{\times }$ is a unit times a power of a uniformizer, so the given value is equivalent to the displayed $\mathfrak{p}$-adic value. An absolute value trivial on $\mathbb{Q}$ is trivial on the algebraic extension $K$, so no further cases occur.

The minimal polynomial of $\beta$ isand part (b) showed that ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$. Modulo $3$,The quadratic has no root in ${\mathbb{F}}_3$, so it is irreducible. The Dedekind factorization theorem therefore givesThe two factors are distinct prime ideal, with norms $3$ and $3^2=9$, respectively.

Modulo $2$,and the quadratic is irreducible over ${\mathbb{F}}_2$. HenceThis is a product of two distinct proper prime ideals, so $(2)$ is not prime. These decompositions are collected in prime ideals above two and three in the cubic field of discriminant minus 307.

Suppose $xy\in J$ and $x\notin J$. Then the colon ideal $(J:x)$ properly contains $J$ because it contains $y$. If also $y\notin J$, then $J+(x)$ properly contains $J$. The maximality of $J$ in the family makes both larger ideals lie outside the family. Applying the stated closure condition with $I=J$ and $a=x$ would imply $J$ lies outside the family, a contradiction. Hence $x\in J$ or $y\in J$, and $J$ is a prime ideal. This argument is the Prime ideal principle for an Oka family.

An irreducible complex analytic hypersurface in a complex manifold $X$ is a closed irreducible analytic subset of pure complex codimension one. A local defining function of a complex analytic hypersurface $Y$ at $x$ is a holomorphic function $f$ on a neighbourhood $U$ such thatThe necessary local algebra is that the stalk ${\mathcal{O}}_{X,x}$ is a regular local ring, hence a unique factorization domain, and that the local branches of a hypersurface germ determine finitely many height-one prime ideals. Each is principal; the product of their generators gives $f$, and removing repeated factors makes it reduced. This also covers a globally irreducible hypersurface that has several local branches at a singular point.

A divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{n}_{Y}Y$ of irreducible analytic hypersurfaces with integer coefficients. On a sufficiently small $U_{\alpha }$, local defining functions give a meromorphic equation $f_{\alpha }$ for $D$. The quotients $f_{\alpha }/f_{\beta }$ are nowhere-zero holomorphic functions. Gluing frames byproduces the holomorphic line bundle associated to a divisor $[D]$, and $f_{\alpha }{e}_{\alpha }$ gives its canonical meromorphic section with divisor $D$.

The Euler sequence on complex projective spaceimplies $\det T{\mathbb{CP}}^n\cong \mathcal{O}(n+1)$. Taking the dual determinant yields the canonical bundle of complex projective spacewhere $H$ is a hyperplane divisor.

The hypotheses on the homogeneous polynomial $p$ say thatis a smooth projective hypersurface of degree $k$, so its divisor line bundle is $[V]\cong \mathcal{O}(kH)$. The Adjunction formula givesThis is the canonical bundle of a smooth projective hypersurface.

Now fix an isomorphism $\Phi :[P]\to [Q]$ and regard $s_P$ and ${\Phi }^{-1}{s}_Q$ as holomorphic sections of the same holomorphic line bundle. They have no common zero because $P\ne Q$. Their homogeneous coordinates therefore define a well-defined holomorphic mapIn a local frame, the quotientis a meromorphic function with divisor $(f)=P-Q$. Thus $F^{-1}(0)=P$ and $F^{-1}(\infty )=Q$, both with multiplicity one. The degree of a holomorphic map $F$ is therefore one. A nonconstant degree-one holomorphic map between compact connected Riemann surfaces is a biholomorphism, so the displayed map is biholomorphic.

The Ostrowski theorem says that every nontrivial absolute value on a field defined on $\mathbb{Q}$ is equivalent either to the usual absolute value or to $|\cdot {|}_p$ for a unique prime $p$.

Let an absolute value on the number field $K$ extend $|\cdot {|}_p$. Its valuation ring determinesa prime ideal satisfying $\mathfrak{p}\cap \mathbb{Z}=(p)$. Conversely, each prime $\mathfrak{p}$ above $p$ defines the normalized absolute valuewhere $e_{\mathfrak{p}}=v_{\mathfrak{p}}(p)$. It restricts to $|\cdot {|}_p$. The correspondence between extensions and primes follows either from the valuation ring or from local factorization and extended absolute values; distinct primes give inequivalent valuations. Thus these $|\cdot {|}_{\mathfrak{p}}$ are exactly the extensions, up to equivalence.

For the tensor-product assertion, choose a primitive element of a field extension $\alpha$ for $K/\mathbb{Q}$, with minimal polynomial $f$. Because number fields are separable, over ${\mathbb{Q}}_p$ it factors into distinct irreduciblesindexed by the primes $\mathfrak{p}\mid p$. The Chinese remainder theorem givesThe $i$th factor is the completion of a number field at a prime ideal $K_{{\mathfrak{p}}_i}$. Under these identifications the isomorphism is the natural diagonal map $x\otimes a\mapsto (ax{)}_{\mathfrak{p}}$, proving the p-adic tensor decomposition of a number field

Let $K$ be a number field. Its places consist of its real embeddings, conjugate pairs of complex embeddings, and the finite places associated with nonzero prime ideals $\mathfrak{p}$ of its ring of integers of a number field. At a real or complex place use the usual absolute value. If $\mathfrak{p}$ lies over the prime number $p$ with ramification index $e_{\mathfrak{p}}$, normalize its absolute value byThese normalizations extend the standard absolute values on $\mathbb{Q}$.

Write $d_v=[K_v:{\mathbb{Q}}_v]$ for the local degree of a place. Thus $d_v$ is $1$ at a real place, $2$ at a complex place, and $e_v{f}_v$ at a finite place. The Absolute multiplicative Weil height isThe product formula shows that this is unchanged when $K$ is replaced by a larger number field containing $\alpha$.

No. Take the Noetherian ring $R=k[x]$, the finitely generated $R$-moduleand $N=0$. Its annihilator of a module iswhich is a prime ideal and hence a primary ideal. But with $r=x$ and $m=(0,\overline{1})$ we have $m\ne 0$ and $rm=0$, while $x^{k}M\ne 0$ for every $k$ because the free summand survives. Thus $N$ is not a primary submodule.

For a prime ideal $P$ of a polynomial ring over a field, the quotient's Krull dimension is the transcendence degree of its fraction field, and its height is $n$ minus that degree. This dimension theorem identifies algebraic codimension with the length of prime chains. It must not be applied indiscriminately to every Noetherian ring without its polynomial-ring hypotheses.

A strictly increasing finite sequence of prime ideals. Its length is the number of strict inclusions, so the displayed chain has length $s$. The Krull dimension of a ring is the supremum of such lengths; the height of a prime ideal restricts to chains ending at that prime.

A prime filtration of a module over a commutative ring is a finite increasing chain from zero to the module with each nonzero successive quotient module isomorphic to $R/P$ for a prime ideal $P$. Every finitely generated module over a Noetherian ring has one. In each nonzero quotient, a maximal annihilator of a module element is prime supplies a cyclic submodule $R/P$; lift it and repeat. The ascending chain condition for a Noetherian module makes the process terminate. This is not necessarily a composition series: the prime ideals need not be maximal.

In a Noetherian ring, an ideal $I$ disjoint from a nonempty multiplicative subset $S$ extends to an ideal $P$ maximal among those disjoint from $S$, by the ascending chain condition. It is proper. If $ab\in P$ but neither factor belongs to $P$, both $P+(a)$ and $P+(b)$ meet $S$. Multiplying such representatives puts an element of $S$ in $P$, a contradiction. Thus $P$ is a prime ideal. For a general commutative ring the same argument follows after applying the Zorn lemma to disjoint ideals; the union of a chain is disjoint from $S$.

Every nonzero ideal of the ring of integers of a number field factors uniquely as a finite product of nonzero prime ideals. Thus each $x\in K^{\times }$ has integer valuations ${\mathrm{ord}}_{\mathfrak{p}}(x)$, almost all zero, determined by the factorization of its principal ideal.

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 $K_{\mathfrak{p}}$.

A commutative Artinian ring has finitely many prime ideals, and every one is a maximal ideal. If $\mathfrak{p}$ is prime, the Artinian domain $R/\mathfrak{p}$ is a field: stabilization of $(x)\supseteq (x^2)\supseteq \cdots$ gives $x^n=x^{n+1}a$, and cancellation gives $xa=1$. Infinitely many distinct maximal ideals ${\mathfrak{m}}_i$ would give a strictly descending chainbecause distinct maximal ideals are comaximal.

For a nonzero ideal $I$ in a Dedekind domain, $v_P(I)$ is the exponent of the nonzero prime ideal $P$ in the unique factorization of ideals in a Dedekind domain. Then $I\subseteq J$ if and only if $v_P(I)\ge v_P(J)$ for every $P$, and $v_P(I+J)=\min (v_P(I),v_P(J))$.

A prime-power ideal has the form ${\mathfrak{p}}^m$ for a prime ideal $\mathfrak{p}$ and positive integer $m$.

For an irreducible variety which is an affine variety of dimension $d$ and a nonzero nonunit regular function $f$, each irreducible component of $V(f)$ has dimension $d-1$. Choose a closed point on just the component in question. In its local ring $R$, $\sqrt{(f)}$ is that component's prime ideal. A system of parameters of $R/(f)$, lifted and supplemented by $f$, generates a maximal-primary ideal; the Krull height theorem gives $\dim R/(f)\ge d-1$. Extending any chain above $(f)$ by the zero prime of the domain gives the reverse inequality. This proof does not identify dimension with transcendence degree.

The points of the real affine plane as a scheme are its prime ideals: the zero ideal, the principal ideals of irreducible real polynomials, and its maximal ideals. The latter have residue field $\mathbb{R}$ or $\mathbb{C}$, by the Zariski lemma and the real closed field property. They correspond respectively to real coordinate pairs and conjugate pairs of nonreal complex coordinate pairs. Thus the scheme contains both nonclosed generic points and closed points absent from the real locus. Its topology is the Zariski topology, and its structure sheaf has sections $\mathbb{R}[x,y{]}_h$ on each principal open subscheme $D(h)$.

The complement of a multiplicative subset is a union of prime ideals exactly when the subset is saturated. If $a\notin S$, saturation ensures $(a)\cap S=\varnothing$; extend this to a prime ideal avoiding a multiplicative subset. Conversely, a product in $S$ cannot have a factor in a prime ideal contained in the complement.