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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.

A morphism of schemes is proper when it is a finite type morphism, a separated morphism, and a universally closed morphism. The valuative criterion for properness says, under the usual finite-type and Noetherian hypotheses, that $f:X\to Y$ is proper exactly when every commutative squarewith $R$ a discrete valuation ring and $K=\mathrm{Frac}R$ has a unique diagonal lift $\mathrm{Spec}R\to X$.

For ${\mathbb{P}}_k^2\to \mathrm{Spec}k$, a $K$-point is $[x_0:x_1:x_2]$ with not all $x_i$ zero. If $\pi$ is a uniformizer, multiply all coordinates by one power of $\pi$ so that ${\min }_{i}v(x_i)=0$. The resulting coordinates lie in $R$ and at least one is a unit, so they define an $R$-point extending the given $K$-point. If two extensions exist, on a chart where one coordinate is a unit their affine coordinate ratios agree in $K$ and therefore in the integral domain $R$; hence the extensions agree. This verifies existence and uniqueness directly.

A Weil divisor on $X$ is a finite formal sumover integral codimension-one closed subschemes $Z$. Since $X$ is regular in codimension one, the local ring at the generic point of each $Z$ is a discrete valuation ring. A nonzero rational function $f\in k(X{)}^{\times }$ therefore defines the principal divisor $\mathrm{div}(f)={\sum }_Z{v}_Z(f)[Z]$. The divisor class group is

Write $f:U={\mathbb{A}}_{\mathbb{C}}^1\setminus \{1\}\to {\mathbb{A}}_{\mathbb{C}}^1$ for $f(t)=t^2$. For any closed point $a\in \mathbb{C}$, the fiber is the zero-dimensional closed subscheme of $U$ cut out by $t^2-a$. It is finite over $\mathbb{C}$, and every finite morphism is proper. This includes $a=1$, whose fiber in $U$ consists only of $t=-1$.

The morphism itself fails the valuative criterion for properness. Take the discrete valuation ring $R=\mathbb{C}[[s]]$ with fraction field $K=\mathbb{C}((s))$. The $K$-point $t=1+s$ lies in $U(K)$ and lies over the $R$-point of the target defined byAny extension $\mathrm{Spec}R\to {\mathbb{A}}^1$ of that $K$-point must still be given by $t=1+s$, whose closed point maps to the deleted point $t=1$. It therefore cannot factor through $U$. The required lift does not exist, so $f$ is not proper. This is an instance of the fact that proper closed-point fibers do not imply properness.

For the discrete valuation $v$ corresponding to $|\cdot |$, the valuation ring and its maximal ideal areThe first is a subring of the field $K$, hence an integral domain. An element of ${\mathcal{O}}_K$ is a unit exactly when its valuation is zero, so every nonunit lies in $\mathfrak{m}$ and $\mathfrak{m}$ is the unique maximal ideal.

Choose a uniformizer $\pi$ with $v(\pi )=1$. If $I\ne 0$ is an ideal, the set of valuations of its nonzero elements has a least member $n$. Choose $x\in I$ with $v(x)=n$. Then $x=u{\pi }^n$ for a unit $u$, so $(x)=({\pi }^n)$. Every $y\in I$ has $v(y)\ge n$, hence $y/x\in {\mathcal{O}}_K$, and therefore $I=(x)$. Thus ${\mathcal{O}}_K$ is a discrete valuation ring, in particular a principal ideal domain.

For an algebraically closed field $k$, the strong Hilbert Nullstellensatz says that every ideal $J\subseteq k[x_1,\dots ,x_n]$ satisfiesIts weak form says that $V(J)=\varnothing$ exactly when $J=(1)$. Now let $S=\mathbb{C}[x_0,\dots ,x_n]$ and let $I\subseteq S$ be a homogeneous ideal. The affine zero set of $I$ is the affine cone over $V_{+}(I)$ together with the origin. Hence, when the projective zero set is empty, eitheror the affine zero set is exactly the origin. In the latter case the Hilbert Nullstellensatz givesthe irrelevant ideal of projective space. Equivalently, some power $(x_0,\dots ,x_n{)}^N$ is contained in $I$. These are precisely the alternatives in the Projective Nullstellensatz.

Let $V\subseteq {\mathbb{P}}^3$ be a smooth quadric surface. After a projective linear transformation, the Segre embedding identifiesChoose distinct points $p,q\in {\mathbb{P}}^1$. The two fibresare the promised disjoint curves on a smooth quadric surface: each is a smooth projective line, and $C_p\cap C_q=\varnothing$. By the Bézout theorem, any two nonempty curves in the projective plane intersect. An isomorphism of algebraic varieties preserves intersections and takes projective curves to projective curves, so these disjoint curves prove

Next let $W$ be a smooth projective curve and let $f:W⇢Y\subseteq {\mathbb{P}}^N$ be a rational map to a projective variety. At a point $P$ outside its initial domain, writeusing elements of the function field $\mathbb{C}(W)$. The local ring of a smooth algebraic curve at $P$ is a discrete valuation ring. If $m={\min }_i{v}_P(f_i)$, multiplying every coordinate by a function of valuation $-m$ makes all coordinates regular at $P$ and at least one a unit there. They therefore define a morphism near $P$. Repeating this at the finitely many missing points proves the extension of a rational map from a smooth projective curve.

Smoothness is essential. Let $\nu :{\mathbb{P}}^1\to C$ be the normalization of a nodal curve for a rational nodal cubic $C$. The inverse is a rational map $C⇢{\mathbb{P}}^1$ on the smooth locus. If it extended to a morphism $g:C\to {\mathbb{P}}^1$, then $g\circ \nu$ would agree with the identity on a dense Zariski-open subset of ${\mathbb{P}}^1$ and hence everywhere. But the two distinct points $a,b\in {\mathbb{P}}^1$ above the node would satisfya contradiction. This is the failure of rational-map extension on a singular curve.

Finally take the Fermat cubic curveand define the projective hypersurfaceBecause $F$ and $xyz$ have no common irreducible factor, the bihomogeneous polynomial $sF+txyz$ is irreducible, so $Z$ is an algebraic variety. Letbe the second coordinate projection. Every homogeneous cubic in three variables has a projective zero over $\mathbb{C}$, so every fibre is nonempty and $\pi$ is a surjective morphism. Over $p=[1:0]$ the fibre is the Fermat cubic curve, which is smooth and has genusOver $q=[0:1]$ the fibre isthe union of exactly three distinct projective lines, hence exactly three irreducible components. This is the cubic pencil with a triangular member required by the question.

A discrete valuation on a field $K$ is a surjective group homomorphismsatisfyingwhenever $x+y\ne 0$. Its discrete valuation ring is

One standard characterization defines a Dedekind domain as a Noetherian integrally closed domain of Krull dimension one. Equivalently, all its localizations at nonzero prime ideals are discrete valuation rings.

A prime Weil divisor is an integral closed subscheme of codimension one. If $Y$ has generic point ${\eta }_Y$, regularity in codimension one makes the local ring ${\mathcal{O}}_{X,{\eta }_Y}$ a discrete valuation ring with fraction field $K(X)$. Its normalized discrete valuationis the order of vanishing along $Y$: writing $f=u{\pi }^m$ for a unit $u$ and uniformizer $\pi$ gives ${\nu }_Y(f)=m$. Additivity of exponents makes this a group homomorphism.

One useful form of Hensel lemma is this: if $R$ is a complete discrete valuation ring, $f\in R[X]$, andthen there is a unique $a\in R$ such that $f(a)=0$ and $a\equiv a_1(\mathrm{mod}\pi )$. Indeed, after constructing $a_r$ with $f(a_r)\equiv 0(\mathrm{mod}{\pi }^r)$, choose the unique $t$ modulo $\pi$ for whichand put $a_{r+1}=a_r+{\pi }^{r}t$. The resulting Cauchy sequence converges by completeness, and the same first-order congruence proves uniqueness.

Apply this to $X^{p-1}-1$. Every nonzero class in ${\mathbb{F}}_p$ is a simple root, so it has a unique Teichmuller representative in ${\mathbb{Z}}_p$. These give all roots of unity of order prime to $p$. For odd $p$, the group $1+p{\mathbb{Z}}_p$ has no nontrivial torsion: if $u\ne 1$, then the binomial theorem gives $v_p(u^p-1)=v_p(u-1)+1$, which is incompatible with finite $p$-power order. HenceFor $p=2$, the subgroup $1+4{\mathbb{Z}}_2$ is torsion-free by the same argument, while $-1$ supplies the extra torsion element. Thus $\mu ({\mathbb{Q}}_2)=\{\pm 1\}\cong C_2$. This describes the roots of unity in a p-adic field for $K={\mathbb{Q}}_p$.

Let $R$ be a discrete valuation ring with uniformizer $\pi$, fraction field $K$, and normalized discrete valuation $\nu :K^{\times }\to \mathbb{Z}$. A map $\mathrm{Spec}K\to {\mathbb{P}}_{\mathbb{Z}}^n$ is a projective pointwith at least one nonzero coordinate. Put $m={\min }_{a_i\ne 0}\nu (a_i)$ and set $b_i={\pi }^{-m}{a}_i$. Then every $b_i$ lies in $R$, and at least one $b_j$ is a unit.

On the standard affine chart $U_j$ of projective space, the ratios $b_i/b_j$ all lie in $R$. They therefore define a map $\mathrm{Spec}R\to U_j\subseteq {\mathbb{P}}_{\mathbb{Z}}^n$ whose generic restriction is the original point. This proves existence in the DVR case. The assumed separatedness of ${\mathbb{P}}_{\mathbb{Z}}^n\to \mathrm{Spec}\mathbb{Z}$, through the valuative criterion for separatedness, gives uniqueness. Thus the morphism satisfies the requested DVR form of the valuative criterion for properness.

One form of Hensel lemma is: if $R$ is a complete discrete valuation ring, $f\in R[X]$, and $f(a_1)\equiv 0(\mathrm{mod}\pi )$ while $f^{\prime }(a_1)\not\equiv 0(\mathrm{mod}\pi )$, then there is a unique $\alpha \in R$ with $f(\alpha )=0$ and $\alpha \equiv a_1(\mathrm{mod}\pi )$.

Inductively, if $f(a_n)\equiv 0(\mathrm{mod}{\pi }^n)$, choose $t$ modulo $\pi$ so thatand put $a_{n+1}=a_n+{\pi }^{n}t$. The unit $f^{\prime }(a_n)$ makes $t$ unique. The resulting sequence is Cauchy, so completeness gives a root $\alpha$. Applying the same first-order congruence to two roots proves uniqueness.

A p-modular system consists of a characteristic-zero field $K$, a discrete valuation ring $\mathcal{O}\subset K$, and its residue field $k$ of characteristic $p$. It is a splitting system for a finite group when both $K$ and $k$ split the relevant group representations.

Let $R$ be a complete discrete valuation ring of residue characteristic of a field $p$, with maximal ideal $\mathfrak{m}$, and let $F$ be a one-dimensional formal group law over $R$. Its multiplication series is $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the invertible morphism criterion for formal group laws constructs its inverse with coefficients in $R$. Both formal power series converge on $\mathfrak{m}$, since their coefficients are integral and the powers of an element of $\mathfrak{m}$ tend to zero. Their compositions are the identity there. Thus multiplication by $n$ is bijective, including when $p=2$ and $n$ is odd. Every finite-order element of $F(\mathfrak{m})$ consequently has order of a group element equal to a prime power with prime $p$.

In a complete discrete valuation ring of residue characteristic $p$, the power map $u\mapsto u^m$ is a bijection on $1+\mathfrak{m}$ when $p\nmid m$. For a principal unit $a$, the Hensel lemma lifts the simple residue root one of $T^m-a$ uniquely. This can absorb principal-unit factors in tame radical descriptions of local fields.

The maximal ideal of an integrally closed Noetherian local domain of Krull dimension one is principal. Hence every such local ring is a discrete valuation ring.

On a normal integral Noetherian scheme, a nonzero rational function defines the displayed Weil divisor, with the sum over codimension-one prime divisors. The local ring at each such divisor is a discrete valuation ring, so the order is well defined. The divisor class group is the group of Weil divisors modulo principal Weil divisors. On a smooth curve this agrees with a principal divisor on an algebraic curve.

A uniformizer of a discrete valuation ring $R$ is a generator $\pi$ of its unique nonzero maximal ideal. Every nonzero $x$ in the fraction field has a unique expression $x=u{\pi }^n$ with $u$ a unit and $n\in \mathbb{Z}$.