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For a commutative $k$-algebra $A$, let $I$ be the kernel of multiplication $\mu :A{\otimes }_{k}A\to A$. The formula $a\mapsto [1\otimes a-a\otimes 1]$ is a derivation of an algebra into the quotient module $I/I^2$, with $A$ acting through the first factor. It induces an isomorphism from the module of Kähler differentials. An inverse is induced by $q(x\otimes y)=xdy$: the identity $q(rs)=\mu (r)q(s)+\mu (s)q(r)$ shows $q(I^2)=0$. For $z={\sum }_i{x}_i\otimes y_i\in I$, the equality ${\sum }_i{x}_i{y}_i=0$ gives $[z]={\sum }_i{x}_i[1\otimes y_i-y_i\otimes 1]$, proving that the two maps are inverse. For an affine variety, $I$ is the defining ideal of its diagonal morphism.

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.

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.

A quasi-coherent sheaf is a sheaf of modules which locally has a presentation by free module sheaves:where the index sets may be infinite. Equivalently, on affine neighbourhoods it is an affine module sheaf. A coherent sheaf is locally finitely generated, with the additional requirement that the kernel of every map from a finite free module sheaf is locally finitely generated. On a Noetherian scheme, these conditions are equivalent to being locally associated with a finitely generated module.

For an affine variety $X$ with coordinate ring $A$ and an $A$-module $M$, setRestriction on principal open subsets is the natural module localization map. On an arbitrary open subset, sections are families of these stalk values locally represented by a single fraction from a localization. This constructs the affine module sheaf; its global sections are $M$. Choose a free-module presentation $A^{(J)}\to A^{(I)}\to M\to 0$. Exactness of localization and its compatibility with direct sums yieldThus $\tilde{M}$ is a quasi-coherent sheaf, without a finite-generation assumption.

We now prove coherence of an affine module sheaf. If $M$ is finitely generated, take a surjection $A^r\to M$. Since $A$ is a Noetherian ring, its kernel is finitely generated, so $M$ has a finite presentation of a module. More generally, on every principal affine open, the kernel of a map $A_a^s\to M_a$ is a submodule of a finite free module over the Noetherian ring $A_a$, and is finitely generated. On sheafifying, this verifies both parts of the definition of a coherent sheaf.

Conversely, if $\tilde{M}$ is coherent, choose a finite principal-open cover $X={\bigcup }_{i}D(a_i)$ on which it has finitely many generators. Write those local generators as $m_{ij}/a_i^{r_{ij}}$. Their numerators generate a finite submodule $N={\sum }_{i,j}A{m}_{ij}$ of $M$, and $N_{a_i}=M_{a_i}$ for every $i$. To check that $M/N=0$, let $q\in M/N$. Each equality after localization gives an exponent $r_i$ such that $a_i^{r_i}q=0$. The ideal generated by the $a_i^{r_i}$ is the unit ideal: otherwise a maximal ideal containing all of them would give a point outside the principal-open cover. A linear combination of these powers equal to $1$ consequently annihilates $q$. Every $q$ is zero, proving

Next consider a morphism $\varphi :{\mathcal{G}}_1\to {\mathcal{G}}_2$ between quasi-coherent sheaves. On an affine open $V$ of the target variety, use the assumed affine module-sheaf equivalence to write ${\mathcal{G}}_j{|}_V=\widetilde{M_j}$, with $M_j=\Gamma (V,{\mathcal{G}}_j)$. The sheaf morphism is the sheafification of the module map $\psi =\Gamma (V,\varphi )$. Indeed, its action on a section $m/a^r$ of a principal open is forced by module-linearity and restriction to be $\psi (m)/a^r$. Therefore exactness of localization givesThese identities are compatible with restriction, so $(\ker \varphi ){|}_V=\widetilde{\ker \psi }$. Since this holds on an affine cover, the kernel sheaf is quasi-coherent on every variety. This proof of kernels of quasi-coherent sheaf morphisms does not assert that taking global sections is right exact.

For direct image along a morphism of affine varieties, write $Y$ with coordinate ring $A$, $X$ with coordinate ring $B$, and let $\alpha :A\to B$ be the induced ring homomorphism. If $\mathcal{F}=\tilde{M}$ on $X$, regard $M$ as an $A$-module by restriction of scalars. For $a\in A$,The last equality identifies localization with the scalar action of $a$. It identifies the direct image sheaf with the affine module sheaf of the restricted module, so $f_{\ast }\mathcal{F}$ is quasi-coherent.

Suppose $Y$ is affine and $X={\bigcup }_{i=1}^r{U}_i$ is a finite affine cover. Since a variety is separated, every $U_i\cap U_j$ is affine: it is identified with the inverse image of the closed diagonal morphism inside the affine product $U_i{\times }_k{U}_j$. Write $g_i=f{|}_{U_i}$ and $g_{ij}=f{|}_{U_i\cap U_j}$. Form the morphism of sheaves of modulesBoth sums are finite, and every summand is quasi-coherent by the affine calculation. For every open $V\subseteq Y$, the sheaf gluing axiom on $f^{-1}(V)$ identifies the kernel of this map on sections with $\mathcal{F}(f^{-1}(V))$. HenceThe kernel result proved above establishes quasi-coherence.

Finally, for arbitrary $Y$, take an affine neighbourhood $V\subseteq Y$ and apply the preceding argument to $f^{-1}(V)\to V$. The open subset $f^{-1}(V)$ is again a variety and has a finite affine cover, because varieties have Noetherian Zariski topology. The restriction of $f_{\ast }\mathcal{F}$ to $V$ is exactly the direct image for this restricted morphism. ThusThe finiteness and separation conditions are the relevant hypotheses, not affineness of the original source or target. More generally, quasi-coherence of direct image under a quasi-compact quasi-separated morphism is valid. If one permits nonseparated finite-type prevarieties, each overlap in the argument has a finite affine cover; replacing its summand by the sum of direct images from these charts gives the same kernel, since agreement can be tested on a cover. This also proves the assertion in that convention. It does not extend by this argument to arbitrary spaces with an unrestricted infinite cover.

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.

The composite of two finite type morphisms is of finite type: on suitable affine open subsets, take the algebra generators of the first algebra together with the images of the algebra generators of the second. To check separatedness, factor the diagonal morphism of the composite asThe first arrow is a closed immersion, and the second is the base change of the closed immersion ${\Delta }_g:Y\to Y{\times }_{Z}Y$. Hence their composite is a closed immersion.

Finally, after any base change $Z^{\prime }\to Z$, the composite is $X^{\prime }\stackrel{f^{\prime }}{\to }{Y}^{\prime }\stackrel{g^{\prime }}{\to }{Z}^{\prime }$. Both arrows take closed subsets to closed subsets, so their composite does as well. This is universal closedness. All three conditions hold, and therefore

A finite morphism is a finite type morphism. It is also a separated morphism: on an affine scheme chart $\mathrm{Spec}B\to \mathrm{Spec}A$, its diagonal morphism is represented by the surjective multiplication ring homomorphism $B{\otimes }_{A}B\to B$, and is therefore a closed immersion by Question 1.

We prove the lifting condition in the valuative criterion for properness. Choose an affine open subset of $Y$ containing the image of the closed point of $\mathrm{Spec}V$. Every point of $\mathrm{Spec}V$ is a generalization of its closed point, so the whole map factors through this affine open subset. Its inverse image under the finite morphism is $\mathrm{Spec}B$, with $B$ a finitely generated $A$-module. Every $b\in B$ is integral over $A$: apply the determinant trick to multiplication by $b$ on a finite set of module generators to obtain a monic polynomial which annihilates all generators and hence annihilates $1$.

In the given valuative square, we have compatible ring homomorphisms $A\to V$ and $B\to K$. The image in $K$ of each $b\in B$ satisfies a monic polynomial over $V$. To see directly why it belongs to $V$, suppose $x\in K\setminus V$. The defining property of a valuation ring gives $x^{-1}\in V$, and it lies in the maximal ideal since it cannot be a unit. A monic equation for $x$, multiplied by $x^{-n}$, would then express $1$ as a sum of elements in that maximal ideal, a contradiction. This proves that valuation rings are integrally closed and that $B\to K$ factors through $B\to V$. The factorization is unique because $V\hookrightarrow K$ is injective. Thus the lift exists uniquely, proving

A morphism of schemes $f:X\to Y$ is a separated morphism if its diagonal morphismis a closed immersion. The diagonal morphism is induced by the pair of identity maps on $X$. Thus separatedness is a condition relative to the given target $Y$, not a requirement that $X$ be absolutely separated.

Since $g$ is a separated morphism, the diagonal morphism ${\Delta }_g:Y\to Y{\times }_{Z}Y$ is a closed immersion. The squareis cartesian. Indeed, on test schemes its pullback condition is $(f(x),y)=(t,t)$, forcing $y=t=f(x)$, and leaving exactly the choice of $x$. The graph morphism of schemes ${\Gamma }_f$ is therefore a base change of a morphism of schemes of ${\Delta }_g$. Part (c) shows that ${\Gamma }_f$ is a closed immersion.

Now assume additionally that $h=g\circ f:X\to Z$ is a closed immersion. The projection $p_Y:X{\times }_{Z}Y\to Y$ is its base change of a morphism of schemes along $g$, so is a closed immersion by part (c). Sincepart (b) shows that $f$ is a closed immersion. This covers the further assumption in the final unheaded paragraph.

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

The coincidence locus of two scheme morphisms is the fibre productThe diagonal morphism ${\Delta }_Y$ is a locally closed immersion, and this property is stable under base change of a morphism of schemes, so $E\hookrightarrow X$ is locally closed. The universal property of a fibre product says that a morphism $T\to X$ factors through $E$ exactly when $f{|}_T=g{|}_T$. Thus $E$ is the largest locally closed subscheme on which they coincide. If $Y$ is separated, ${\Delta }_Y$ is a closed immersion, and its base change $E\hookrightarrow X$ is closed.

A morphism of schemes is proper when it is of finite type, separated, and universally closed.

Each property is local on the target. Finite type is local by its affine definition. The restrictions of the diagonal morphism of $\phi$ over the open sets $U_i$ are closed immersions; since being a closed subset is local on an open cover, the diagonal itself is a closed immersion, so $\phi$ is separated. Finally, after any base change $Y^{\prime }\to Y$, the inverse images $U_i^{\prime }=Y^{\prime }{\times }_Y{U}_i$ cover $Y^{\prime }$. For every closed $C\subseteq X{\times }_Y{Y}^{\prime }$, its image has closed intersection with every $U_i^{\prime }$ because the restricted base-changed morphism is closed. The image is therefore closed in $Y^{\prime }$. Thus $\phi$ is universally closed and hence proper. This proves that properness is local on the target.

A group scheme over $k$ is a $k$-scheme $G$ equipped with multiplication $m:G{\times }_{k}G\to G$, identity $e:\mathrm{Spec}k\to G$, and inversion $j:G\to G$ satisfying the associativity, identity, and inverse diagrams.

ConsiderThe diagonal morphism is ${\Delta }_{G/k}=q^{-1}(e)$. For a finite-type $k$-scheme the rational identity point is closed, so its inverse image is closed. Thus the diagonal is a closed immersion and every such group scheme over a field is a separated scheme.

In characteristic $p>0$, the infinitesimal additive groupis a nonreduced group scheme. Its comultiplication is $t\mapsto t\otimes 1+1\otimes t$, which is well defined because $(t\otimes 1+1\otimes t{)}^p=t^p\otimes 1+1\otimes t^p$.

A morphism $f:X\to S$ is a separated morphism when its diagonal morphismis a closed immersion.

For the requested example, take the affine plane with doubled origin: glue two copies $U,V\cong {\mathbb{A}}_k^2$ by the identity onThe opens $U$ and $V$ are affine, while their intersection is the punctured affine plane, which is not affine. Indeed, its regular functions are still $k[x,y]$; if it were affine, the canonical map to $\mathrm{Spec}k[x,y]={\mathbb{A}}_k^2$ would be an isomorphism, contrary to the missing origin. The resulting scheme is not separated: in a separated scheme, the intersection of two affine opens is the inverse image of the closed diagonal inside their affine product and is therefore affine.

Fix $x\in X$ and choose affine opens $x\in U\subseteq X$ and $f(x)\in W\subseteq S$ with $f(U)\subseteq W$. The open subset $U{\times }_{W}U$ of $X{\times }_{S}X$ contains $\Delta (x)$. If $U=\mathrm{Spec}A$ and $W=\mathrm{Spec}R$, its diagonal is induced by the surjectionso it is a closed immersion. Hence every diagonal morphism is locally a closed immersion into an open subset, and therefore is a locally closed immersion.

For ${\mathbb{A}}_k^1$, the product is ${\mathbb{A}}_k^2=\mathrm{Spec}k[x,y]$ and the diagonal is $V(x-y)$. Its complement is the principal affine open

Take instead the separated scheme $X={\mathbb{A}}_k^2$. The complement of its diagonal in $X{\times }_{k}X\cong {\mathbb{A}}_k^4$ is ${\mathbb{A}}_k^4\setminus {\mathbb{A}}_k^2$, where the removed diagonal has codimension two. Its global functions still form the polynomial ring of ${\mathbb{A}}_k^4$. Were the complement affine, its canonical morphism to $\mathrm{Spec}$ of this ring would identify it with all of ${\mathbb{A}}_k^4$, which is impossible. Thus this diagonal complement is not affine.

In the classical convention over an algebraically closed field $k$, a prevariety is an irreducible topological space with a sheaf of rings which has a finite open cover by affine varieties, with compatible local identifications. A prevariety need not be separated. A classical algebraic variety is a separated prevariety: its diagonal morphism has closed image. Some authors allow reducible prevarieties and varieties; the separation criterion and product construction remain the same under that convention.