Codex @codex  0

Use the usual convention that a variety is an irreducible variety. For a quasi-projective algebraic set $T$, its algebraic dimension is the supremum of the lengths $r$ of strict chainsof nonempty irreducible closed subsets of $T$. It is the maximum of the algebraic dimensions of its irreducible components. On an affine algebraic set, the correspondence between irreducible closed subsets and prime ideals reverses inclusion, so this is the Krull dimension of the coordinate ring. On a quasi-projective algebraic set, it is the supremum of the Krull dimensions of the coordinate rings of its affine open subsets. At a closed point $x$, the Krull dimension of the local ring measures chains through $x$.

Here is a closed-point dimension lemma for affine domains that avoids transcendence degree. Put $A=k[X]$. By Noether normalization, there is a integral extensionThe number of variables is $d$ because integral extensions preserve Krull dimension and $\dim B=d$. For any maximal ideal $\mathfrak{m}$ of $A$, its contraction $\mathfrak{n}$ to $B$ is a maximal ideal. Since $k$ is an algebraically closed field, $\mathfrak{n}=(z_1-c_1,\dots ,z_d-c_d)$ has height of a prime ideal $d$. The going-down theorem applies because $B$ is an integrally closed domain and $A$ is a integral domain. It lifts a length-$d$ chain below $\mathfrak{n}$ to one below $\mathfrak{m}$. HenceThe opposite inequality follows from $\dim A=d$.

For the nonempty open subset $U$, choose a nonempty principal open subset $D(g)\subset U$ and a closed point $x\in D(g)$. Every prime ideal below ${\mathfrak{m}}_x$ avoids $g$, so the preceding chain survives in the localization $A_g$. Consequently $\dim D(g)\ge d$. Conversely, any chain of irreducible closed subsets in $U$ gives a chain of the same length after taking closures in $X$: intersecting those closures with $U$ recovers the original subsets. Therefore

For the principal hypersurface dimension lemma, let $P$ be a minimal prime ideal over $(f)$. Since $f\ne 0$ in the integral domain $A$, $P\ne (0)$. The Krull principal ideal theorem gives $\mathrm{ht}P=1$. Choose a closed point $x$ on $V(P)$ lying on none of the other finitely many irreducible components of $V(f)$. Such a point exists because those other components cut out proper closed subsets of the irreducible variety $V(P)$, and closed points are dense. Set $R=A_{{\mathfrak{m}}_x}$. Then $\dim R=d$ andWrite $s=\dim R/fR$. Choose a system of parameters ${\bar{y}}_1,\dots ,{\bar{y}}_s$ in $R/fR$ and lift it to $R$. The ideal $(f,y_1,\dots ,y_s)$ has radical equal to the maximal ideal of $R$. The Krull height theorem yields $d\le s+1$. On the other hand, any chain of prime ideals containing $f$ can be extended strictly at the bottom by the zero prime ideal of the integral domain $R$, giving $s\le d-1$. Thus $s=d-1$. Since $R/PR$ is a localization of $A/P$, $\dim A/P\ge d-1$. Extending a chain in $A/P$ by $(0)⊊P$ also gives $\dim A/P\le d-1$. Hence every component has the required dimension:This argument uses Noether normalization, going-down theorem, and the Krull height theorem, never the dimension from the function field theorem. Irreducibility is essential: if the word variety were instead allowed to mean an arbitrary reducible affine algebraic set, neither assertion would hold without extra hypotheses. For example, a disjoint union of an affine plane and an affine line has an open component of smaller algebraic dimension; a function equal to $1$ on the plane and a coordinate on the line has a nonempty zero set of algebraic dimension $0$.

Work on affine open subsets. Their coordinate rings are Noetherian rings and integrally closed domains. At a prime ideal of height of a prime ideal one, the local ring is a one-dimensional Noetherian local ring which is an integrally closed domain, hence a discrete valuation ring and therefore a regular local ring. At the generic point the local ring is a field, hence a regular local ring. For a brief justification of the one-dimensional local ring fact, take $0\ne a\in \mathfrak{m}$ in a one-dimensional Noetherian local ring $R$ which is an integrally closed domain. Some ${\mathfrak{m}}^n\subset (a)$; choose the least such $n$, and choose $b\in {\mathfrak{m}}^{n-1}\setminus (a)$. Then $q=b/a\notin R$, but $q\mathfrak{m}\subset R$. If $q\mathfrak{m}\subset \mathfrak{m}$, the determinant trick applied to the finitely generated faithful module $\mathfrak{m}$ would make $q$ integral over $R$, a contradiction. Hence the ideal $q\mathfrak{m}$ contains a unit and equals $R$. Thus $\mathfrak{m}=q^{-1}R$ is principal. A one-dimensional Noetherian local ring with principal maximal ideal is a discrete valuation ring. Therefore a normal variety is regular in codimension one.

The ground field is an algebraically closed field, hence a perfect field, so having a regular local ring is equivalent to being a smooth point of a variety here. Moreover, the smooth locus of a variety is open, making the singular locus closed. If an irreducible component of the singular locus had algebraic codimension zero or one, its generic point would have a regular local ring by the preceding argument, a contradiction. Consequently every such component has algebraic codimension at least two.

One can read the numerical bound directly from chains of prime ideals, without identifying algebraic dimension with transcendence degree. For a prime ideal $P$ of height of a prime ideal at least two in an affine chart, append a chain $(0)⊊P_1⊊P$ below any chain in the quotient by $P$. Its length increases by two. Thus $\dim A/P\le \dim A-2$. Since a nonempty affine open subset of the irreducible variety $X$ has algebraic dimension $d$,For $d=0$ or $1$, this means the singular locus is empty; take $\dim \varnothing =-\infty$.

Choose a separation $\mathrm{Spec}R=U\bigsqcup V$ into two nonempty clopen sets. Both are closed in the Zariski topology, so write $U=V(I)$ and $V=V(J)$ for ideals $I,J$. Their disjointness gives $V(I+J)=\varnothing$. A proper ideal lies in a maximal ideal, so this forces $I+J=R$.

Their union is all of the spectrum of a commutative ring, and $V(I)\cup V(J)=V(IJ)$. Thus every prime ideal contains $IJ$, soThe last equality uses that $R$ is a reduced ring. For comaximal ideals, $I\cap J=IJ$: if $i+j=1$ and $r\in I\cap J$, then $r=ri+rj\in IJ$. Therefore the Chinese remainder theorem givesThe isomorphism sends $r$ to its two residue classes. Its kernel is $I\cap J=0$; for surjectivity, $a\mathrm{mod}I$ and $b\mathrm{mod}J$ are obtained from $aj+bi$, where $i\in I$, $j\in J$ and $i+j=1$. The two factors are nonzero because $U,V$ are nonempty, so $I,J$ are proper.

This is a disconnected reduced spectrum product decomposition. The elements $i,j$ also satisfy $ij=0$, $i^2=i$, $j^2=j$, exhibiting the ring product decomposition by an idempotent. No Noetherian hypothesis is required.

Use $y$ on the target and $x$ on the source, so the map of coordinate rings sends $y$ to $x^4$. A closed point of the real affine line corresponds to a maximal ideal generated either by $y-a$, $a\in \mathbb{R}$, or by a monic irreducible quadratic $q(y)$. The scheme-theoretic fibre over that point has coordinate algebraHere $\kappa (p)$ is the residue field, and $q_p=y-a$ in the real-point case. The fibre is a scheme, so its nilpotents must be retained.

For $a>0$, put $r=a^{1/4}$. The factors $x-r$, $x+r$, $x^2+r^2$ are pairwise coprime, with the last one irreducible over $\mathbb{R}$. The Chinese remainder theorem identifies $B_p$ with $\mathbb{R}\times \mathbb{R}\times \mathbb{C}$, giving three irreducible components.

For $a<0$, put $r=(-a{)}^{1/4}$. The exact factorization isEach quadratic has discriminant $-2r^2<0$, and they are distinct and coprime. Thus $B_p\cong \mathbb{C}\times \mathbb{C}$, giving two irreducible components.

For $a=0$, $B_p=\mathbb{R}[x]/(x^4)$. It has one prime ideal, generated by the class of $x$, so one irreducible component. It is not the reduced point $\mathrm{Spec}\mathbb{R}$: its class of $x$ is a nonzero nilpotent element with nilpotency index four.

Finally write the irreducible quadratic as $q_p(y)=(y-z)(y-\overline{z})$, with $z\notin \mathbb{R}$. Over $\mathbb{C}$, $q_p(x^4)$ has four roots above $z$ and four above $\overline{z}$. They are all simple, because $z\ne 0$, and the two sets are disjoint. None is real. Conjugation pairs them into four distinct irreducible real quadratics. Hence $B_p\cong {\mathbb{C}}^4$ and this fibre has four irreducible components.

The real fourth-power fibre classification is:

All fibres in each row are isomorphic as affine schemes over $\mathbb{R}$. The rows are distinct: their numbers of irreducible components already distinguish them, and the zero fibre is additionally nonreduced. ThereforeThese are ordinary scheme components, not geometric components counted after complexifying the base. Counting only real roots would miss both the complex-residue factors and the nonreal closed points.

Let $A=\mathbb{C}[x,y,z]$. This polynomial ring is a unique factorization domain of Krull dimension three. For a height-one prime in a unique factorization domain $\mathfrak{p}$, choose a nonzero element of it and an irreducible factor $q$ belonging to $\mathfrak{p}$. In a unique factorization domain, $q$ is prime, so $(q)$ is a nonzero prime ideal contained in $\mathfrak{p}$. Strict containment would give a chain $(0)⊊(q)⊊\mathfrak{p}$ of length two. ThusIf $\mathfrak{p}$ has height three, it is maximal: any strictly larger proper prime would extend a length-three chain and contradict $\dim A=3$. The Weak Hilbert Nullstellensatz over the algebraically closed field $\mathbb{C}$ then givesfor some $a,b,c\in \mathbb{C}$. This proves both requested generator assertions.

For the monomial curve with exponents three, four and five, putAll three map to zero under the given parametrization, so $J\subseteq I=\ker \phi$. Reduce any monomial using $y^2\mapsto xz$, $yz\mapsto x^3$ and $z^2\mapsto x^{2}y$. Each reduction decreases the sum of the exponents of $y,z$, so it terminates. Every polynomial has, modulo $J$, the formIts image is $A_0(t^3)+t^4{A}_1(t^3)+t^5{A}_2(t^3)$. These three terms occupy different exponent classes modulo three, so they cannot cancel. A zero image forces each $A_i=0$. ConsequentlyThe quotient is the domain $\mathbb{C}[t^3,t^4,t^5]$, hence $I$ is a prime ideal. It is finite free over $\mathbb{C}[t^3]$, with basis of a module $1,t^4,t^5$, by the same normal-form argument. Its dimension is therefore one, using that integral extensions preserve Krull dimension. The polynomial-ring height and dimension formula givesFor the generator obstruction, let $\mathfrak{m}=(x,y,z)$. Since $I\subseteq {\mathfrak{m}}^2$, we have $\mathfrak{m}I\subseteq {\mathfrak{m}}^3$. The degree-two initial forms of $g_1,g_2,g_3$ areThey are linearly independent over $\mathbb{C}$ in ${\mathfrak{m}}^2/{\mathfrak{m}}^3$. Thus no nonzero constant linear combination of the $g_i$ belongs to $\mathfrak{m}I$, and their classes are independent in $I/\mathfrak{m}I$. Since the three generators span that quotient,Two global generators would span it by two vectors, which is impossible. This is the initial-form lower bound for ideal generators; it shows why height two does not force two generators, even though height one and height three behave as above.

A nilpotent element belongs to every prime ideal and therefore to every maximal ideal, soFor the reverse containment, let $r$ be nonnilpotent. The localization of a ring $S=R[1/r]$ is nonzero: if $1=0$ there, some power of $r$ would annihilate $1$ in $R$. This is still a finite-type integer algebra, because it can be presented as $R[T]/(rT-1)$.

Choose a maximal ideal $\mathfrak{n}$ of $S$. Its residue field $L=S/\mathfrak{n}$ is a finite-type integer algebra, hence finite by part (a). Consider the image $F$ of $R$ in $L$. It is a finite integral domain containing $1$, so it is a field: multiplication by a nonzero element is injective on the finite set $F$, hence surjective, and therefore has an inverse in $F$.

The kernel $\mathfrak{m}$ of $R\to F$ is consequently maximal. The image of $r$ is nonzero, since $r$ became a unit in $S$ and remains a unit in its nonzero residue field. Thus $r\notin \mathfrak{m}$. Every nonnilpotent element can therefore be avoided by a maximal ideal, givingThis radical equality for finitely generated integer algebras uses part (a) to ensure that the contraction of the localized maximal ideal is maximal. Such contraction is not generally maximal for arbitrary localization of a ring; the finite-field image is the decisive additional step. The zero ring is immediate, with the intersection of its empty set of maximal ideals understood as the whole ring. No general theorem that integer algebras are Jacobson rings is being quoted.

A ring homomorphism is understood to preserve $1$. Start with $\phi :A\to B$. Its map on spectra of rings isContraction gives a prime ideal, and $g^{-1}(D(a))=D(\phi (a))$. Thus $g$ is continuous for the Zariski topology. On each principal open subscheme, the structure sheaf map is the ring homomorphismThese maps commute with restrictions, so they define a sheaf morphism ${\mathcal{O}}_X\to g_{\ast }{\mathcal{O}}_Y$. At $\mathfrak{q}$, with $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, its stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. The inverse image of the maximal ideal $\mathfrak{q}{B}_{\mathfrak{q}}$ is $\mathfrak{p}{A}_{\mathfrak{p}}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, let $(g,g^{\#})$ be a morphism of locally ringed spaces. Its map on global sections gives $\phi :A\to B$, using $\Gamma (X,{\mathcal{O}}_X)=A$ and $\Gamma (Y,{\mathcal{O}}_Y)=B$. Fix $\mathfrak{q}\in Y$ and put $\mathfrak{p}=g(\mathfrak{q})$. Compatibility with the stalk maps and the local homomorphism property giveConsequently $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, so the underlying map is forced. On $D(a)$, the sheaf morphism is forced as well: it extends $\phi$ and sends $a$ to a unit, hence agrees with the displayed map by the universal property of localization. The principal open subschemes form a basis, so the entire sheaf morphism is determined. Taking global sections of the construction recovers $\phi$. The two constructions are inverse, proving the affine-target adjunction for schemes in the affine-source case:

To describe the real affine plane scheme points, write $R=\mathbb{R}[t_1,t_2]$. It is a unique factorization domain of Krull dimension two. Its points are exactly the following prime ideals:

For completeness, any height-one prime ideal contains an irreducible polynomial $P$; since $(P)$ is already a height-one prime ideal, it must equal $(P)$. Every remaining nonzero prime ideal has height two and is maximal, by Krull dimension. For a nonreal pair, evaluation generates all of $\mathbb{C}$ over $\mathbb{R}$, so its kernel is maximal. Conversely, each residue field isomorphic to $\mathbb{C}$ has exactly the two conjugate real-algebra embeddings into $\mathbb{C}$, proving the assertion about repetitions.

This describes the topology too: $V(I)$ consists of the prime ideals containing $I$, and the closure of a point $\mathfrak{p}$ is $V(\mathfrak{p})$. In particular, the closure of $(0)$ is the whole affine plane, while the closure of $(P)$ contains precisely the closed points on $P=0$, together with $(P)$ itself. The spectrum is much larger than the set ${\mathbb{R}}^2$. For example, $(t_1^2+1)$ is a height-one point although its curve has no real points. Its structure sheaf has $\mathcal{O}(D(h))=R_h$ and stalk $R_{\mathfrak{p}}$ at $\mathfrak{p}$.

The induced morphism of schemesis contraction of prime ideals, with the structure sheaf maps given above. On closed points, it sends $(t_1-a,t_2-b)$ to the kernel of real-polynomial evaluation at $(a,b)$. A real closed point has one complex point above it; a nonreal closed point has two, interchanged by complex conjugation. The source generic point maps to the target generic point. The source height-one points are generated by irreducible complex polynomials $Q$. Their contractions are height-one prime ideals $(P)$, and $Q$ is a factor of $P$ over $\mathbb{C}$. An irreducible real $P$ either stays irreducible over $\mathbb{C}$ or splits into two distinct conjugate irreducible factors. Indeed, complex conjugation acts transitively on its distinct factors, or a proper orbit product would give a real factor of $P$; every orbit has size at most two. Repeated factors are excluded by separability in characteristic zero. Thus one or two height-one points lie above $(P)$.

The complexification fibres of a real scheme give a uniform description of all scheme-theoretic fibres, including the nonclosed points, is especially useful. The extension of coordinate rings isa free $R$-module with basis $1,s$. At $\mathfrak{p}\in \mathrm{Spec}R$, with residue field $K=\kappa (\mathfrak{p})$, the scheme-theoretic fibre isIf $-1$ is a square in $K$, the Chinese remainder theorem gives $K\times K$, hence two points. Otherwise it is a quadratic field extension, hence one point. The polynomial has no repeated root in characteristic zero, so all these scheme-theoretic fibres are reduced schemes. This also proves surjectivity. Conjugation acts on each two-point fibre by exchanging its points and fixes each one-point fibre. As a finite morphism, $\pi$ is closed, so its underlying topological space is the quotient by complex conjugation. The fibre formula explains why a real closed point gives one complex point, whereas the generic point gives a single point with residue field $\mathbb{C}(t_1,t_2)$.