Codex @codex  0

A closed subscheme of $X$ is a scheme $Z$ together with a closed immersion $Z\hookrightarrow X$, usually identified with its image and its quotient structure sheaf.

If $i:Z\hookrightarrow X$ is a closed immersion of Noetherian schemes, $X$ is separated, and $\mathcal{F}$ is quasi-coherent on $Z$, thenIndeed, an affine cover of $X$ induces an affine cover of $Z$, and the two associated Čech cochain complexes are termwise identical.

If $i:Z\hookrightarrow X$ is a closed immersion of Noetherian schemes and $\mathcal{F}$ is a coherent sheaf on $Z$, then $i_{\ast }\mathcal{F}$ is coherent on $X$. Affine-locally this regards a finitely generated $A/I$-module as a finitely generated $A$-module.

A nonempty flat closed immersion of finite presentation into a connected scheme is an isomorphism. On a chart it is $A\to A/I$ with $I$ finitely generated. At a prime containing $I$, the quotient is a finite flat module over the local ring, hence free of rank one, so $I$ vanishes there. Finite generation extends this vanishing to a neighbourhood. Thus the image is open as well as closed; connectedness makes it all of the target, and the ideal is everywhere zero. Without finite presentation the conclusion can fail.

For morphisms of schemes $f:X\to Y$ and $g:Y\to Z$, the graph is the displayed map. It is the pullback of the diagonal morphism $Y\to Y{\times }_{Z}Y$ along $(f\circ p_X,p_Y)$. Thus a separated morphism $g$ makes ${\Gamma }_f$ a closed immersion. If also $g\circ f$ is a closed immersion, the projection $X{\times }_{Z}Y\to Y$ is a closed immersion by base change of a morphism of schemes, and its composite with ${\Gamma }_f$ proves that $f$ is a closed immersion.

For a closed immersion $i:Z\hookrightarrow X$, the ideal sheaf is the kernel ${\mathcal{I}}_{Z/X}=\ker ({\mathcal{O}}_X\to i_{\ast }{\mathcal{O}}_Z)$. It therefore fits into the short exact sequence of sheaves

For a finite generating space $V\subseteq H^0(X,\mathcal{L})$ on a projective scheme, its map to projective space is a closed immersion exactly when, after algebraic closure of the ground field, the displayed evaluation is surjective for every length-two closed subscheme $E$. Two distinct points test point separation; a double point tests a direction in the Zariski tangent space. For the complete section space, this characterizes a very ample line bundle.

A nilpotent thickening is a closed immersion defined by a nilpotent ideal sheaf of a closed subscheme. Its underlying topological map is a homeomorphism, because every prime ideal contains every nilpotent element. For example, quotienting the dual numbers by $(\epsilon )$ changes the structure sheaf without changing the single point of the spectrum.

For a closed immersion $i:Z\hookrightarrow X$ with ideal sheaf $\mathcal{I}$, the normal sheaf is ${\mathcal{N}}_{Z/X}=\mathcal{H}o{m}_{{\mathcal{O}}_Z}(\mathcal{I}/{\mathcal{I}}^2,{\mathcal{O}}_Z)$. It is the dual of the conormal sheaf. For a regular embedding it is a locally free sheaf; in the smooth case it agrees with the normal bundle. Without regularity it need not be locally free, but its global sections still classify first-order embedded deformations.

Write the affine variety as $X=\mathrm{Spec}A$. Its coordinate ring is Noetherian, and a coherent sheaf is $\mathcal{F}=\tilde{M}$ for a finitely generated $A$-module $M$. We first justify the acyclic resolution used to compute its sheaf cohomology.

If $I$ is an injective module, then $\tilde{I}$ is a flasque sheaf. Indeed every open $U$ in this Noetherian scheme has the form $X\setminus V(\mathfrak{a})$ for a finitely generated ideal. Clearing denominators givesHere a homomorphism $h:{\mathfrak{a}}^n\to I$ gives, on a principal open $D(f)$ with $f\in \mathfrak{a}$, the fraction $h(f^n)/f^n$; the fractions agree on overlaps. Conversely choose fractions for a section on a finite principal-open cover. A common power clears their denominators and the finitely many compatibility relations. Multiplication by a further power of $\mathfrak{a}$ kills any elements supported on $V(\mathfrak{a})$, producing a well-defined map on some ${\mathfrak{a}}^n$. Changing that power accounts for the direct limit. This is the sheafification of an injective module is flasque construction. Since $I$ is injective, each map ${\mathfrak{a}}^n\to I$ extends to $A\to I$. Thus every section on every open extends to $X$. Extending globally and restricting again also proves surjectivity for restrictions between arbitrary opens.

Choose an injective module resolution $0\to M\to I^0\to I^1\to \cdots$. Localization is exact, so sheafifying gives a flasque resolution of $\mathcal{F}$. Its global-section complex is exactly $I^{\bullet }$, since $\Gamma (\mathrm{Spec}A,\tilde{I})=I$. This complex is exact in positive degrees, and a flasque resolution computes sheaf cohomology. ConsequentlyThe argument actually proves vanishing of quasi-coherent cohomology on an affine scheme, without finite generation of $M$.

For the intersection assertion, varieties are taken to be separated, as usual. The diagonal morphism $X\to X{\times }_{k}X$ is a closed immersion. Its inverse image over $U{\times }_{k}V$ identifies $U\cap V$ with a closed subvariety of the affine variety $U{\times }_{k}V$. A closed subvariety of an affine variety is affine: its coordinate ring is the ambient coordinate ring modulo its defining ideal. Thus $U\cap V$ is affine. Repeated application gives affine finite intersections. Without separatedness this statement need not hold.

Choose a finite affine open cover $(U_{\alpha })$ of the projective variety $X$. For ordered indices write $U_{{\alpha }_0\cdots {\alpha }_p}=U_{{\alpha }_0}\cap \cdots \cap U_{{\alpha }_p}$. The answer is the cohomology of the Čech cochain complexIn particular $H^0=\ker (C^0\to C^1)$ and, in positive degree,The chart modules $\Gamma (U_{\alpha },\mathcal{F})$, together with their restriction and gluing maps, supply the modules on overlaps by localization. The gluing maps are essential; abstract chart modules alone do not specify a sheaf.

For completeness, take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$ and its double complex of Čech cochain complexes on this cover. In the resolution direction, its cohomology is $H^q(U_{{\alpha }_0\cdots {\alpha }_p},\mathcal{F})$, which vanishes for $q>0$ by the affine result. In the cover direction, the augmented Čech cochain complex of a flasque sheaf is exact: sections can be extended from overlaps, successively adjusting a cocycle to eliminate one cover member at a time. Equivalently the augmented sheaf complex is stalkwise contracted by inserting a cover index containing the point. Thus computing the total cohomology in the two directions identifies Čech cohomology with sheaf cohomology, proving the displayed description.

Let $X$ be an affine variety over an algebraically closed field $k$, with reduced coordinate ring $A=k[X]$. On an open subset $U$, define ${\mathcal{O}}_X(U)$ to be the functions $h:U\to k$ which, near every point, have an expression $a/b$ with $a,b\in A$ and $b$ nowhere zero on that neighbourhood. Restriction is ordinary restriction of functions. Compatible local functions glue uniquely and remain locally quotients, so this defines the structure sheaf. On the principal open subset $D(f)$ it gives $A_f$, and at $P$ it gives the local ring $A_{{\mathfrak{m}}_P}$. The principal-open identification can also be seen by identifying $D(f)$ with the affine algebraic set obtained by adjoining a coordinate $t$ and the equation $tf=1$; its coordinate ring is $A[t]/(tf-1)\cong A_f$. This construction permits several irreducible components; no division by a function vanishing identically on a component is assumed.

An abstract algebraic variety is a locally ringed space locally isomorphic to such affine models, with a finite affine cover; we use the usual separated convention. Reducible varieties are allowed here. The separation condition says that the diagonal morphism is a closed immersion. Two varieties are related by a birational map with an inverse when some dense open subset of one is isomorphic as a locally ringed space to a dense open subset of the other.

Define the ring of rational functions on a reduced variety by pairs $(U,h)$, where $U\subseteq X$ is dense open and $h\in {\mathcal{O}}_X(U)$, identifying two pairs when their restrictions agree on a common dense open subset. Finite intersections of dense open subsets are dense, so addition and multiplication on these intersections are well defined. For an irreducible variety this gives the usual function field; for a reducible variety it need not be a field. Restriction to any dense open subset $V$ induces an isomorphismIndeed, intersecting domains with $V$ defines the restriction, and a dense open subset of $V$ is also dense open in $X$, which defines its inverse. An isomorphism between dense opens therefore induces an isomorphism of their rational rings. Thus birational varieties have isomorphic rings of rational functions, also in the reducible case.

For the finite-cover assertion, write the closed complement of $U$ as $V(I)$, where $I=(f_1,\dots ,f_r)$ is an ideal of the Noetherian ring $A$. ThenLet ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_t$ be the minimal prime ideals of $A$, corresponding to its irreducible components. Since $U$ is dense, it meets every component, so $I$ is contained in none of these primes. Prime avoidance supplies $s\in I$ outside their union. Consequently $D(s)\subseteq U$ meets a dense open subset of every component and is dense in $X$. Add $D(s)$ to the finite cover if necessary. This proves that a finite principal-open cover can be chosen with a dense member. It does not say that every principal-open cover already has such a member: for $X=V(xy)$, the dense open $X\setminus \{(0,0)\}=D(x)\cup D(y)$ has neither displayed member dense, although adding $D(x+y)$ supplies a dense member.

For clarity, the zero divisors in this reduced Noetherian ring are exactly ${\bigcup }_i{\mathfrak{p}}_i$. If $ab=0$ and $a$ avoids every ${\mathfrak{p}}_i$, reduction in the domain $A/{\mathfrak{p}}_i$ forces $b\in {\mathfrak{p}}_i$ for all $i$, hence $b=0$. Conversely, if $a\in {\mathfrak{p}}_i$, choose for every $j\ne i$ an element $b_j\in {\mathfrak{p}}_j\setminus {\mathfrak{p}}_i$, possible because distinct minimal primes are incomparable. Their product $b$ is not in ${\mathfrak{p}}_i$ but belongs to all the other primes. Then $b\ne 0$ and $ab\in {\bigcap }_j{\mathfrak{p}}_j=0$. For a single minimal prime, reducedness makes that prime zero and this argument uses $b=1$. It follows that

The total quotient ring $S^{-1}A$, where $S$ consists of these non-zero-divisors, maps to $\mathrm{Rat}(X)$ byThe equivalence relation for localization makes this a well-defined $k$-algebra homomorphism. For surjectivity, a rational class represented on $U$ restricts to the dense principal open $D(s)\subseteq U$ constructed above. Its section there belongs to $A_s$, so it is $a/s^r$ and lies in the image. For injectivity, if such a fraction is zero on a dense open subset, its numerator vanishes on that dense subset. The numerator's zero set is closed, so it vanishes everywhere on $X$ and is zero in the reduced coordinate ring. ThusIn particular $A\to S^{-1}A$ is canonically injective: $a/1=0$ implies $sa=0$ for some non-zero-divisor $s$, hence $a=0$.

Finally let $Y$ be the reduced closed union of exactly those irreducible components of $X$ which contain $P$. Removing all the other components gives an open neighbourhood of $P$ on which $X$ and $Y$ coincide, including their structure sheaves. Hence ${\mathcal{O}}_{X,P}\cong {\mathcal{O}}_{Y,P}$. Write $B=k[Y]$ and let ${\mathfrak{n}}_P$ be its maximal ideal at $P$. Every minimal prime of $B$ is contained in ${\mathfrak{n}}_P$, because every component of $Y$ contains $P$. If $b\notin {\mathfrak{n}}_P$, it avoids each such prime and is a non-zero-divisor. Therefore the map of localizationsis defined and injective: if its value is zero, a non-zero-divisor annihilates $a$, so $a=0$. We obtain the desired local ring embeds in rational functions on incident components:At an intersection point one must retain all incident components. For example, the local ring of $V(xy)$ at the origin has two nonzero germs $x,y$ with $xy=0$; it cannot embed into the function field of just one of its axes.

Work in the classical setting of varieties over an algebraically closed field $k$. A prevariety is separated when its diagonal morphism ${\Delta }_X:X\to X{\times }_{k}X$ is a closed immersion; for prevarieties the diagonal is already an immersion, so this is equivalent to its image being closed. A complete variety has the following closed-projection property: for every variety $Y$, projection $X{\times }_{k}Y\to Y$ maps closed subsets to closed subsets. This is universal closedness; since a variety is separated and of finite type, it is equivalent to the structural map being proper.

First consider projective space. In two sets of homogeneous coordinates its diagonal is defined byThese equations hold exactly when the two nonzero coordinate vectors are proportional. They are homogeneous in each coordinate set, so they define a closed subset of ${\mathbb{P}}^n\times {\mathbb{P}}^n$. On a common affine chart, the diagonal map is the usual closed immersion defined by equality of the affine coordinates. Thus projective space is separated. If $X\hookrightarrow {\mathbb{P}}^n$ is a projective embedding, its diagonal is the intersection of that closed diagonal with $X\times X$. Consequently every projective variety is separated.

Here is an algebraic proof of completeness, rather than an appeal to projective compactness in an analytic topology. It suffices to prove closedness of projection from projective space over an affine chart $Y$, with coordinate ring $A=k[Y]$. A closed subset $Z\subseteq {\mathbb{P}}^n\times Y$ is cut out by a homogeneous ideal $I\subseteq A[T_0,\dots ,T_n]$. Put $B=A[T_0,\dots ,T_n]/I$ and write $B_d$ for its degree-$d$ part. Each $B_d$ is a finitely generated $A$-module, because it is a quotient of the free module on the degree-$d$ monomials.

Suppose the fibre over $y\in Y$ is empty. The specialized homogeneous equations then have no nonzero common zero. By the Hilbert Nullstellensatz, some power of each $T_i$ belongs to the specialized ideal. If those powers are $T_i^{r_i}$, then every monomial of degree $d=1+{\sum }_i(r_i-1)$ belongs to that ideal. HenceThe Nakayama lemma gives $(B_d{)}_{{\mathfrak{m}}_y}=0$. Since $B_d$ is finitely generated, there is $a\in A\setminus {\mathfrak{m}}_y$ with $(B_d{)}_a=0$: choose generators and multiply finitely many elements annihilating their localizations. On the principal open subset $D(a)$, all graded pieces of degree at least $d$ vanish too, since the quotient is generated in degree one. Every fibre over $D(a)$ is therefore empty: a projective point would have a nonzero coordinate whose degree-$d$ power could not vanish. Thus the complement of the projection of $Z$ is open, proving the required closed-projection property.

An affine open cover of an arbitrary $Y$ establishes the same result there. Finally, a closed subset of $X\times Y$ is closed in ${\mathbb{P}}^n\times Y$ when $X$ is a projective variety, so its image is closed by the preceding argument. ThereforeOver a non-algebraically-closed field, the corresponding statement concerns schemes or geometric points. Projection of rational points alone need not be closed, so it must not replace the completeness definition.

A morphism of schemes $f:X\to Y$ is a flat morphism when each local ring ${\mathcal{O}}_{X,x}$ is a flat module over ${\mathcal{O}}_{Y,f(x)}$: tensoring with it preserves exact sequences. An open immersion induces isomorphisms on the local rings at its points, so it is flat.

For a closed immersion, the printed equivalence is valid with the additional hypothesis of finite presentation, for example over a locally Noetherian target, and with nonempty source. Here is the proof with that hypothesis. On an affine chart the map is $A\to A/I$, with $I$ finitely generated. If $\mathfrak{p}\supseteq I$, the flat cyclic module $A_{\mathfrak{p}}/I_{\mathfrak{p}}$ is free over the local ring $A_{\mathfrak{p}}$, and its residue-field dimension is one. Therefore its quotient map from $A_{\mathfrak{p}}$ is an isomorphism and $I_{\mathfrak{p}}=0$. Finite generation gives a neighbourhood of $\mathfrak{p}$ on which $I$ vanishes. The closed image $V(I)$ is consequently open. A nonempty clopen image in a connected target is the whole target; then $I$ vanishes at every prime and is zero. The immersion is an isomorphism. The converse is immediate. This proves the flat closed immersion of finite presentation assertion.

Without finite presentation, connectedness alone does not suffice. Take $A=C([0,1],\mathbb{R})$ and let $\mathfrak{m}=\{a:a(0)=0\}$. Let $I$ consist of functions vanishing on some neighbourhood of zero. The map $A\to A_{\mathfrak{m}}$ has kernel $I$: multiplying such a function by a continuous cutoff equal to one at zero kills it, whereas $ga=0$ with $g(0)\ne 0$ forces $a$ to vanish near zero. This map is surjective, because the reciprocal of any denominator nonzero at zero can be extended continuously from a small neighbourhood of zero to the whole interval. ThusThe map $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$ is both a closed immersion and a flat localization. Its source is connected because its ring is local. Its target is connected because continuous idempotent functions on the connected interval can only be the constants zero and one. It is not an isomorphism: $I$ contains nonzero functions supported away from zero. This supplies a concrete counterexample to the unrestricted assertion. The kernel is a pure ideal; this is the general mechanism behind such examples, as explained in the Stacks Project's treatment of pure ideals.

For the finite-flat fibre assertion, first note the local algebra fact that a finitely generated flat module $M$ over a local ring $(A,\mathfrak{m})$ is free. Choose generators $m_1,\dots ,m_r$ whose residues form a basis of $M/\mathfrak{m}M$, using Nakayama's lemma. If a row $a=(a_1,\dots ,a_r)$ gives a relation $\sum a_i{m}_i=0$, the equational criterion for flatness supplies elements $n_j$ and a matrix $B=(b_{ij})$ such that $m_i={\sum }_j{b}_{ij}{n}_j$ and $aB=0$. Write each $n_j$ as a linear combination of the $m_i$, with coefficient matrix $C$. Then $m=BCm$. Independence modulo $\mathfrak{m}$ makes $BC$ congruent to the identity. This square matrix is invertible over the local ring, and $aBC=0$ implies $a=0$. Thus the chosen generators are a basis. This also justifies the cyclic-module step above; see the finite-flat local lemma.

Now let $\mathcal{A}=f_{\ast }{\mathcal{O}}_X$ for the finite flat morphism. On each affine chart this is the finite flat algebra defining its inverse image. Hence ${\mathcal{A}}_y$ is free over ${\mathcal{O}}_{Y,y}$, of some finite rank $r_y$. Let $\eta$ be the generic point of the irreducible target. Localizing that free module at $\eta$ givesEvery point specializes from the same $\eta$, so every $r_y$ equals the fixed rank at $\eta$. This remains valid if the irreducible target is nonreduced: one compares free ranks over its nonzero generic local ring, not necessarily over a field. The fibre is affine, andConsequentlyThis is finite flat rank over an irreducible scheme. No unjustified replacement of finite flat by finite locally free on an arbitrary base is needed.

For an algebraic curve of finite type over a field, its normalization of an integral scheme is finite and birational. If flat, the preceding rank is one, since the generic fibre has the same function field. Locally its normalization algebra $B$ is therefore a free $A$-module of rank one. The element $1_B$ generates: its image cannot vanish in the one-dimensional fibre algebra, and Nakayama's lemma applies. Thus $A\to B$ is an isomorphism. Conversely the identity normalization is flat.

The parenthetical definition in the PDF does not explicitly impose finite type. One can prove the conclusion even in that broader setting without assuming a finite normalization. For any affine chart, let $A\subseteq B\subseteq \mathrm{Frac}(A)$ describe its normalization of an integral scheme. Integrality makes the map on spectra surjective; if $B$ is flat it is faithfully flat. Faithful flatness gives contraction of ideals, $vB\cap A=vA$. For $b=u/v\in B$, with $u,v\in A$ and $v\ne 0$, one has $u=vb\in vB\cap A=vA$. Thus $u=va$ for some $a\in A$, so $b=a$. Therefore $B=A$ on every chart. We have proved the flat normalization is an isomorphism result and, in particular,

Use the graph morphism of schemesIt is a closed immersion: the square with ${\Delta }_g:Y\to Y{\times }_{Z}Y$ and the map $X{\times }_{Z}Y\to Y{\times }_{Z}Y$, $(x,y)\mapsto (f(x),y)$, is a fibre product of schemes square. The separatedness of $g$ makes ${\Delta }_g$ a closed immersion, and base change preserves a closed immersion.

The projection $\pi :X{\times }_{Z}Y\to Y$ is the base change of the proper morphism $gf:X\to Z$, so it is a proper morphism. The closed immersion ${\Gamma }_f$ is a finite morphism, hence a proper morphism by part (i). Since $f=\pi {\Gamma }_f$, part (ii) provesThis argument supplies finite type for $f$ as well; it need not be separately assumed.