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TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

For a discrete valuation ring $A$,Let $\pi$ be a uniformizer, so its unique nonzero maximal ideal is $\mathfrak{m}=(\pi )$. We use the defining description of a discrete valuation ring: it is a non-field local ring that is a principal ideal domain, and every nonzero element is $u{\pi }^n$ for a unit $u$ and an integer $n\ge 0$.

Every nonzero proper ideal is therefore $({\pi }^n)$ for some $n\ge 1$. Such an ideal is a prime ideal only when $n=1$: for $n\ge 2$, $\pi {\pi }^{n-1}\in ({\pi }^n)$ while neither factor belongs to $({\pi }^n)$. The ideal $(\pi )$ is prime because it is maximal. Also $(0)$ is prime because $A$ is an integral domain. Thusand the longest strict chain has length one.

The original PDF asks about a discrete valuation ring; the local TeX transcription's phrase “discrete variable $X$” is an extraction error.

For a product in a category of prevarieties, the projections $p_X,p_Y$ must induce a bijectionfor every prevariety $T$. If $Z$ and $Z^{\prime }$ both have this property, their projection pairs determine morphisms $Z\to Z^{\prime }$ and $Z^{\prime }\to Z$. Each composite has the same projections as the identity, so uniqueness in the universal property makes both composites identities. This gives the unique isomorphism respecting the projections.

Here the classical convention is that a prevariety is locally an affine variety, with a finite affine cover, and is irreducible; an algebraic variety is a separated prevariety, meaning that its diagonal morphism has closed image. The same separation argument works in the convention that also permits reducible varieties. For separated $X,Y$, under the rearrangement $(X\times Y{)}^2\cong X^2\times Y^2$, the diagonal isIt is closed, proving separation. For completeness, irreducibility in the classical convention follows from the affine calculation below: the products of affine charts are irreducible, and their pairwise intersections are nonempty since nonempty open subsets of each irreducible factor meet. An open cover by irreducible open sets with pairwise nonempty intersections has irreducible union. Thus the product is an algebraic variety.

To establish the coordinate ring of a product of affine varieties, write $A=k[X]=k[x_1,\dots ,x_r]/J$ and $B=k[Y]=k[y_1,\dots ,y_s]/L$. The algebrahas zero set $X\times Y$. To see directly that $(J,L)$ is prime when $X,Y$ are irreducible, take nonzero $F,G\in A{\otimes }_{k}B$. Express $F={\sum }_i{a}_i\otimes b_i$ with the $b_i$ linearly independent over $k$. The set of $x\in X$ for which $F_x={\sum }_i{a}_i(x)b_i$ is nonzero is a nonempty open set: it is the union of the nonvanishing loci of the $a_i$, and at least one $a_i$ is nonzero. The Hilbert Nullstellensatz guarantees that such a function cannot vanish at every point. The analogous set for $G$ is also nonempty; their intersection contains a point $x$ by irreducibility. Since $B$ is an integral domain, $F_x{G}_x\ne 0$, whence $FG\ne 0$. Thus $A{\otimes }_{k}B$ is an integral domain, and the Hilbert Nullstellensatz identifies $(J,L)$ with the defining ideal of its zero set. The resulting affine variety has the universal property: algebra maps from $A{\otimes }_{k}B$ to $\Gamma (T,{\mathcal{O}}_T)$ correspond to pairs of algebra maps from $A,B$, and these correspond to morphisms into $X,Y$. This dictionary holds for arbitrary $T$, since coordinate functions give the morphisms on affine charts and agree on overlaps. Uniqueness of the product now proves

For the final part put $T=A{\otimes }_{k}A$ and $\mu (x\otimes y)=xy$. Restriction to the diagonal morphism is $\mu$, so $I=\ker \mu$. Write $\delta (a)=1\otimes a-a\otimes 1$. The elements $\delta (a)$ generate $I$: quotienting by them identifies the two copies of $A$ and gives precisely the multiplication quotient $T/I\cong A$. The map $D(a)=[\delta (a)]$ is $k$-linear, vanishes on $k$, and takes values in $I/I^2$. Sinceit obeys $D(ab)=aD(b)+bD(a)$ for the stated first-factor action. Hence it is a derivation of an algebra. The universal property of the Module of Kähler differentials supplies an $A$-linear map $\theta :{\Omega }_{A/k}^1\to I/I^2$ with $\theta (da)=D(a)$.

Define the $k$-linear map $q:T\to {\Omega }_{A/k}^1$ by $q(x\otimes y)=xdy$. It is $A$-linear for the first-factor action and, by the Leibniz rule, satisfiesTherefore $q(I^2)=0$ and restriction induces $\psi :I/I^2\to {\Omega }_{A/k}^1$. We have $\psi \theta (da)=da$, so $\psi \theta$ is the identity because the $da$ generate the Module of Kähler differentials. Conversely, if $z={\sum }_i{x}_i\otimes y_i\in I$, then ${\sum }_i{x}_i{y}_i=0$ andThus the conormal module of the diagonal gives

Suppose first that $\overline{f(X)}=Y$. If $g\in \ker f^{\ast }$, then $g\circ f=0$, so the zero set of the regular function $g$ contains $f(X)$ and hence its closure $Y$. Thus $g=0$ in the coordinate ring $k[Y]$, and $f^{\ast }$ is injective.

Conversely, if $\overline{f(X)}$ were a proper closed subset of $Y$, its vanishing ideal in $k[Y]$ would contain a nonzero regular function $g$. Then $g$ vanishes on $f(X)$, so $f^{\ast }(g)=0$, contradicting injectivity. ThereforeThis is the coordinate-ring criterion for a dominant morphism of affine varieties.

For an irreducible affine variety, define dimension byIrreducibility makes both coordinate rings integral domains. The injective map $f^{\ast }:k[Y]\hookrightarrow k[X]$ extends to an embedding of function fieldsAn algebraically independent family in $k(Y)$ remains algebraically independent in $k(X)$, so the dimension of an irreducible affine variety by transcendence degree gives

Let the torsion submodule of the module over a ring $M$ beThis is a submodule. Indeed, if $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs\ne 0$ because the scalar ring is an integral domain, and $rs(m+n)=0$; scalar multiples and additive inverses are handled similarly.

Take the quotient module $M_0=M/T(M)$ and let $q:M\to M_0$ be the quotient R-module homomorphism. It is torsion-free: if $0\ne r\in R$ and $r(m+T(M))=0$, then $rm\in T(M)$, so $srm=0$ for some $s\ne 0$. Since $sr\ne 0$, this puts $m$ in $T(M)$ and therefore $m+T(M)=0$.

Now let $N$ be torsion-free and let $f:M\to N$ be an $R$-module homomorphism. If $m\in T(M)$ and $rm=0$ for $r\ne 0$, then $rf(m)=f(rm)=0$, so torsion-freeness gives $f(m)=0$. Hence $T(M)\subseteq \ker f$, and the universal property of a quotient module gives a unique homomorphismwith $\boxed{f=f_0\circ q}$. Thus $M/T(M)$ is the maximal torsion-free quotient of $M$.

The set $S=1+I$ is a multiplicative subset, since $(1+a)(1+b)=1+(a+b+ab)$. The localization of a ring $S^{-1}R$ consists of fractions $r/s$, whereSimilarly, the localization of a module $S^{-1}M$ consists of fractions $m/s$, with the analogous equivalence relation. The canonical maps areThe Localization of a Noetherian ring is Noetherian. Since a finitely generated module over a Noetherian ring is a Noetherian module, localizing a finite generating set shows that $S^{-1}M$ is Noetherian over $S^{-1}R$.

The prime ideal correspondence for localization identifies $\mathrm{Spec}{S}^{-1}R$ with the primes $P\in \mathrm{Spec}R$ disjoint from $S$. Hereso the spectrum of localization away from one plus an ideal is

An element $m$ maps to zero exactly when $(1+a)m=0$ for some $a\in I$. Such an equation gives $m=(-a{)}^{j}m\in I^{j}M$ for every $j$, proving one inclusion. Conversely, suppose $m\in {\bigcap }_{j\ge 1}{I}^{j}M$. The Artin-Rees lemma applied to $Rm\subseteq M$ says that for some $c$,Taking $n=c+1$ gives $m\in I(Rm)$, so $m=am$ for some $a\in I$. Then $(1-a)m=0$ and $1-a\in S$. Thus

For failure without Noetherianity, takeThe strict chain $(t)⊊(t^{1/2})⊊(t^{1/4})⊊\cdots$ shows that $R$ is not Noetherian. Since $t^q=(t^{q/2}{)}^2$, one has $I^2=I$, and hence ${\bigcap }_j{I}^j=I\ne 0$. But $R$ is an integral domain, so its localization map into $(1+I{)}^{-1}R$ is injective. This is the non-Noetherian failure of the intersection formula for localization.

The height of a prime ideal $P$ is

The Krull principal ideal theorem states that if $R$ is Noetherian and $P$ is minimal among the primes containing a proper principal ideal $(a)$, thenSuppose otherwise that $P_0⊊Q⊊P$. Quotient by $P_0$ and localize at $P$; it is enough to consider a Noetherian local domain whose maximal ideal $P$ is the only prime containing $(a)$ and which has $0⊊Q⊊P$.

For $n\ge 1$, putThis is a $Q$-primary ideal. Since $R/(a)$ is a zero-dimensional Noetherian ring, it is Artinian, and the descending chain $(Q^{(n)}+(a))/(a)$ eventually stabilizes. Thus, for all sufficiently large $n$, every $q_n\in Q^{(n)}$ can be writtenNow $ra\in Q^{(n)}$, while $a\notin Q$; $Q$-primaryness gives $r\in Q^{(n)}$. HenceThe Nakayama lemma applied to $Q^{(n)}/Q^{(n+1)}$ gives $Q^{(n)}=Q^{(n+1)}$. Localizing at $Q$ makes all sufficiently large powers of the nonzero maximal ideal $Q{R}_Q$ equal. A nonzero element of the stable power then belongs to ${\bigcap }_n{Q}^n{R}_Q$, contradicting the Krull intersection theorem. This proves the theorem.

Now let $R$ be a Noetherian integral domain. If $R$ is a unique factorization domain and $P$ has height one, choose $0\ne x\in P$ and an irreducible factor $p$ of $x$. In a UFD, $p$ is prime, soHeight one forces $P=(p)$.

Conversely, suppose every height-one prime is principal. Noetherianity makes $R$ an atomic domain. Given an irreducible element $p$, choose a prime $P$ minimal over $(p)$. The principal ideal theorem gives $\mathrm{ht}P=1$, so $P=(q)$ by hypothesis. Since $p=qr$ and $p$ is irreducible, $r$ is a unit; hence $(p)=P$ is prime. Thus every irreducible is a prime element, and an atomic domain with this property is a UFD. Therefore

Suppose first that the affine variety $V$ is irreducible. If $fg\in I(V)$, thenBoth pieces are Zariski closed in $V$. Irreducibility forces one to equal $V$, so either $f\in I(V)$ or $g\in I(V)$. Thus $I(V)$ is a prime ideal.

Conversely, suppose that $I(V)$ is prime and that $V=A\cup B$ with $A,B$ closed in $V$. If both were proper, the Hilbert Nullstellensatz would giveChoose $f\in I(A)\setminus I(V)$ and $g\in I(B)\setminus I(V)$. Their product vanishes on $A\cup B=V$, so $fg\in I(V)$, contradicting primality. Hence one piece equals $V$. This proves the criterion of irreducibility and prime vanishing ideal:

For the semicubical parabolaconsiderCertainly $(y^2-x^3)\subseteq \ker \theta$. Divide any polynomial by the monic polynomial $y^2-x^3$ as a polynomial in $y$; its remainder is $a(x)y+b(x)$. If this remainder lies in $\ker \theta$, thenThe first summand contains only odd powers of $t$ and the second only even powers, so both vanish and $a=b=0$. ThereforeThe latter is a subring of the integral domain $k[t]$, hence is a domain. Thus $(y^2-x^3)$ is prime and

An affine variety of dimension $d$ is smooth at $p$ when its Zariski tangent space has dimension $d$; equivalently, the Jacobian criterion requires the Jacobian of its defining ideal to have rank $n-d$. For this plane curve, put $F(x,y)=y^2-x^3$. Its gradient isAt $(0,0)\in C$ the gradient vanishes, so the tangent space is all of $k^2$ and has dimension two rather than the curve dimension one. Hence the result on the coordinate ring and singularity of the semicubical parabola gives

Assume that $p$ is a prime element. If $p=ab$, then $p$ divides $ab$, so $p$ divides $a$ or $b$. If $p\mid a$, write $a=pc$; cancellation in the integral domain gives $1=bc$, so $b$ is a unit. The other case is symmetric. Hence every prime element is irreducible.

Every field is an integral domain, so if $R/(p)$ is a field then it is an integral domain.

PutSinceis an integral domain, $(x,z)$ is a prime ideal. Moreover $\dim R=3$ and $\dim R/(x,z)=2$, so its height of an ideal is one. Therefore $D=V(x,z)$ is a prime Weil divisor.

Localizing at $x$ eliminates $y$ and givesa unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(R)$ is generated by the height-one primes containing $x$. Sincethese are $D=V(x,z)$ and $E=V(x,w)$. Both occur with multiplicity one in the principal divisorThe units of $R_x\cong k[z,w][x,x^{-1}]$ are exactly $c{x}^n$ with $c\in k^{\ast }$ and $n\in \mathbb{Z}$. Consequently the only relation supplied by localization is $[D]+[E]=0$, andThis is the divisor class group of the three-dimensional affine quadric cone.

Write $A=k[x,y]$, $X=\mathrm{Spec}A$, and $U=X\setminus Z$. Since $A$ is an integral domain, no nonzero global section is supported only at the origin, soThe punctured affine plane has the affine cover $U=D(x)\cup D(y)$. Its Čech cochain complex for the structure sheaf isConsequently

Because $X$ is affine, the higher sheaf cohomology of ${\mathcal{O}}_X$ vanishes. The long exact sequence for local cohomology therefore givesand $H_Z^i(X,{\mathcal{O}}_X)=0$ for $i\ge 3$. The nonzero group has the $k$-basisThis computes the local cohomology of the affine plane supported at the origin in every degree.

Repeated formal addition givesfor positive $n$, and the same formulas extend to all $n\in \mathbb{Z}$ using the formal inverse.

Now let $R$ be a field of characteristic $p>0$ and let $h:{\hat{\mathbb{G}}}_a\to {\hat{\mathbb{G}}}_m$ be a homomorphism. Compatibility with multiplication by $p$ givesThe left side is $h(0)=0$, while the Frobenius identity gives the right side asThe formal power series ring over a field is an integral domain, so $h(X{)}^p=0$ implies $h=0$. Hence there are no nonzero homomorphisms from the formal additive group to the formal multiplicative group.

Subtracting the two defining equations givesConsequently the variety isEach member is irreducible. Indeed, after imposing its linear equation, its coordinate ring isThe invertible linear substitutionturns this intowhich is an integral domain. The two curves are distinct and neither contains the other, so they are precisely the irreducible components of two intersecting complex hyperbolas.

The zero element is torsion. If $m,n\in T$, choose nonzero $r,s\in R$ with $rm=0$ and $sn=0$. Since $R$ is an integral domain, $rs\ne 0$, andThus $m+n\in T$. If $a\in R$, then $r(am)=a(rm)=0$, so $am\in T$. Hence $T$ is closed under addition and scalar multiplication and is an submodule of $M$.

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.

Let $\mathcal{F}$ be the proper nonprincipal ideals. A chain in $\mathcal{F}$ has a nonprincipal union: if its union were $(a)$, then $a$ would belong to one member of the chain, forcing that member to equal the union and be principal. The union is also proper. Thus Zorn lemma gives a maximal member whenever $\mathcal{F}$ is nonempty.

The family satisfies the condition from part (a). Indeed, supposeWrite $a=bd$ and $b=i+ra$, where $i\in I$. Every $x=bs\in I$ has $as=dx\in I$, so $s\in (c)$ and $I\subseteq (bc)$. Conversely $ac\in I$ because $c\in (I:a)$, while $ic\in I$ because $i\in I$. Thus $bc=ic+rac\in I$, proving $(bc)\subseteq I$. Hence $I=(bc)$ is principal.

If a nonprincipal ideal existed, part (a) would therefore produce a nonprincipal prime ideal, contrary to the hypothesis. Every ideal is principal, so the integral domain is a principal ideal domain.

For a point $p\in {\mathbb{A}}_k^1$, let $a$ be the image of $t$ in its residue field $\kappa (p)$. The scheme-theoretic fiber isIf $p\ne (t)$, then $a\ne 0$. The quadratic polynomial $xy-a{z}^2$ is irreducible. Indeed, after setting $z=0$, any hypothetical linear factors must restrict, up to nonzero scalars, to $x$ and $y$; comparing the $xz$ and $yz$ coefficients then forces both $z$ coefficients to vanish, contradicting the nonzero $z^2$ coefficient. Hence its homogeneous coordinate ring is an integral domain, so $Y_p$ is an integral scheme.

At the origin $p=(t)$, the fiber is $V_{+}(xy)$, the union of the two distinct projective lines $x=0$ and $y=0$, and is therefore not irreducible. It is nevertheless a reduced scheme because the ideal $(xy)=(x)\cap (y)$ equals its radical. Every other fiber is integral and hence reduced. Thus the fiber is integral exactly away from the origin, and it is reduced at every point of ${\mathbb{A}}_k^1$.

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.