Let $\mathcal{F}$ be any quasi-coherent sheaf on $X$, and choose $m$ from (i). Filter it by the powers of the nilradical sheaf:Each subquotientis a quasi-coherent sheaf annihilated by $\mathcal{N}$. Apply (ii) to the quasi-coherent sheaf ${\mathcal{N}}^j\mathcal{F}$. Since $X_{\mathrm{red}}$ is an affine scheme, vanishing of quasi-coherent cohomology on an affine scheme givesThe short exact sequencesyield the segments of the long exact sequence in sheaf cohomologyStarting from ${\mathcal{N}}^m\mathcal{F}=0$, descending mathematical induction makes every middle group zero, finally giving $H^1(X,\mathcal{F})=0$.

This holds for every quasi-coherent sheaf $\mathcal{F}$. Since a Noetherian scheme is a quasi-compact scheme, the Serre criterion for affineness in (a) applies and provesThis is an instance of the general fact that affineness is invariant under nilpotent thickening.

Let $i:X_{\mathrm{red}}\hookrightarrow X$ be the reduced induced subscheme, with structure sheaf ${\mathcal{O}}_{X_{\mathrm{red}}}={\mathcal{O}}_X/\mathcal{N}$. Its underlying map of topological spaces is a homeomorphism, since every prime ideal contains the nilradical.

Put $\mathcal{G}=\mathcal{F}/\mathcal{N}\mathcal{F}$. Multiplication by $\mathcal{N}$ annihilates $\mathcal{G}$, so its action by ${\mathcal{O}}_X$ factors through ${\mathcal{O}}_X/\mathcal{N}$. Thus $\mathcal{G}$ is naturally a sheaf of modules over ${\mathcal{O}}_{X_{\mathrm{red}}}$. On an affine open subscheme $\mathrm{Spec}A$, with $\mathcal{F}=\tilde{M}$, it corresponds to $M/\mathrm{Nil}(A)M$ over $A/\mathrm{Nil}(A)$, so it remains a quasi-coherent sheaf.

The sheaf cohomology of a sheaf of modules is the sheaf cohomology of its underlying sheaf of abelian groups. Under the above homeomorphism, that underlying sheaf is identical on the two spaces: the global sections functor and its flasque resolutions are identical. Equivalently, use sheaf cohomology under a closed inclusion with $i_{\ast }\mathcal{G}$. Therefore there are canonical isomorphisms, in every degree,In particular, this gives the requested equality in degree one. Equality of the underlying topological spaces alone would not compare arbitrary coefficient sheaves; the factorization of the action explains why the indicated coefficients are the same.

A Noetherian scheme is a quasi-compact scheme, so choose a finite affine open cover $X={\bigcup }_{i=1}^s\mathrm{Spec}{A}_i$ with all $A_i$ Noetherian rings. The nilradical sheaf restricts on this chart to the quasi-coherent sheaf associated to $\mathrm{Nil}(A_i)$: the nilradical commutes with localization.

By the allowed fact, choose $d_i\ge 1$ such that $\mathrm{Nil}(A_i{)}^{d_i}=0$ and put $m={\max }_i{d}_i$. On every chart, the ideal sheaf ${\mathcal{N}}^m$ is zero, and hence it is zero globally. Higher powers are also zero. ThusIf $X$ is empty, take $m=1$. The finite affine open cover is essential for choosing one bound valid everywhere.

Let $X\stackrel{f}{\to }Y\stackrel{g}{\to }S$ be morphisms of schemes. Start with $u:\mathrm{Spec}K\to X$ and $v:\mathrm{Spec}D\to S$ satisfying $gfu=v{|}_{\mathrm{Spec}K}$.

For existence, first apply the extension condition for $g$ to $fu$ and $v$, obtaining $w:\mathrm{Spec}D\to Y$. Then apply the extension condition for $f$ to $u,w$, obtaining $\ell :\mathrm{Spec}D\to X$. This $\ell$ solves the problem for $gf$. If both $f,g$ are universally closed morphisms, both are quasi-compact morphisms; their composite is a quasi-compact morphism by the allowed stability fact. Part (a)(i) therefore proves that $gf$ is a universally closed morphism.

For uniqueness, suppose ${\ell }_1,{\ell }_2$ are extensions for $gf$. Uniqueness for $g$ gives $f{\ell }_1=f{\ell }_2=w$, and uniqueness for $f$, now over this same $w$, gives ${\ell }_1={\ell }_2$. If $f,g$ are separated morphisms, their composite is a quasi-separated morphism by the allowed stability fact. Part (a)(ii) proves that $gf$ is a separated morphism.

Finally, if $f,g$ are proper morphisms, these arguments give both existence and uniqueness. Stability under composition of quasi-compact morphisms, quasi-separated morphisms and locally of finite type gives the other hypotheses of (a)(iii). Consequently

Let $S^{\prime }\to S$ be a morphism of schemes and set $X^{\prime }=X{\times }_S{S}^{\prime }$, with projection $f^{\prime }:X^{\prime }\to S^{\prime }$. Given compatible morphisms of schemes $u^{\prime }:\mathrm{Spec}K\to X^{\prime }$ and $v^{\prime }:\mathrm{Spec}D\to S^{\prime }$, compose them with the projections to obtain a valuative problem for $f$. An extension $\ell :\mathrm{Spec}D\to X$ for that problem and the prescribed $v^{\prime }$ determine a unique morphism of schemesThe universal property of a fibre product shows that ${\ell }^{\prime }$ extends $u^{\prime }$ and lies over $v^{\prime }$. Conversely, every extension ${\ell }^{\prime }$ has such a first component $\ell$. Thus the sets of extensions in the two problems are in bijection.

If $f$ is a universally closed morphism, it is a quasi-compact morphism and satisfies existence. The allowed stability of quasi-compact morphisms under base change and (a)(i) prove that $f^{\prime }$ is a universally closed morphism. If $f$ is a separated morphism, it is a quasi-separated morphism and satisfies uniqueness. Stability of quasi-separated morphisms under base change and (a)(ii) prove that $f^{\prime }$ is a separated morphism.

If $f$ is a proper morphism, the same bijection supplies existence and uniqueness. The permitted stability under base change of quasi-compact morphisms, quasi-separated morphisms and locally of finite type supplies all remaining hypotheses of (a)(iii). Hence

With the same compatible pair $u,v$, the valuative criterion for properness isEquivalently, one can replace finite type by being a quasi-compact morphism and locally of finite type. Necessity follows from the definition of a proper morphism and the existence and uniqueness criteria in (i) and (ii). Conversely, finite type supplies the compactness needed in (i), while being a quasi-separated morphism supplies the compactness of the diagonal morphism needed in (ii). Thus the extension condition gives both universal closedness and separatedness, and these together with finite type give a proper morphism. These hypotheses are part of the criteria, even though the question is phrased for unrestricted morphisms of schemes.^[ref]

Use the compatible pair $u,v$ and the notion of extension from (i). The valuative criterion for separatedness isA quasi-separated morphism means one whose diagonal morphism is a quasi-compact morphism. A separated morphism automatically has this property because its diagonal morphism is a closed immersion. The quasi-separated morphism hypothesis is needed in the converse valuative test for arbitrary schemes. Existence is not demanded.^[ref]

For every valuation ring $D$ with fraction field $K$, consider compatible morphisms of schemesAn extension is a morphism of schemes $\ell :\mathrm{Spec}D\to X$ satisfying $\ell {|}_{\mathrm{Spec}K}=u$ and $f\ell =v$. The valuative criterion for universal closedness statesEquivalently, among quasi-compact morphisms, the existence condition characterizes universal closedness. A universally closed morphism is automatically a quasi-compact morphism; the converse direction of the criterion needs this compactness condition. All valuation rings are allowed, with arbitrary fraction fields; there is no Noetherian ring hypothesis and one cannot replace them by discrete valuation rings without additional assumptions.^[ref][ref]

Starting from $(r,e)$, the construction in (iii) preserves the composite to the affine line and sends $Z_A$ to the $A$ zero section and $Z_B$ to the $B$ zero section. Part (ii) therefore recovers $(r,e)$.

Conversely, start with $u:T\to X$ and construct $(r,e)$. The only points of $X$ outside $A$ lie on $B_0$, and the only points outside $B$ lie on $A_0$. Consequently $u^{-1}(A)=T\setminus Z_B=T_A$ and $u^{-1}(B)=T\setminus Z_A=T_B$. On each of these open subsets, $u$ is the unique morphism of schemes to the appropriate affine line with coordinate $r$. Thus part (iii) recovers $u$.

Finally, a ring homomorphism $R\to R^{\prime }$ sends $r$ to $r^{\prime }$ and induces $R/(r)\to R^{\prime }/(r^{\prime })$, sending $e$ to an idempotent $e^{\prime }$. Inverse images of the clopen sets $Z_A,Z_B$ are precisely those determined by $e^{\prime }$. The constructions therefore commute with this base change. We have a natural bijectionso $F$ is a functor represented by a scheme, represented by the affine line with doubled origin. The representing scheme is not required to be an affine scheme.

Let $T=\mathrm{Spec}R$ and $Z=\mathrm{Spec}(R/(r))$. The idempotent $e$ determines the disjoint clopen sets $Z_A=D_Z(e)$ and $Z_B=D_Z(1-e)$. Each is closed in $T$. Define the open subsetsThey cover $T$ and satisfy $T_A\cap T_B=T\setminus Z=D(r)$. The global section $r$ defines morphisms of schemes $T_A\to A$ and $T_B\to B$, both sending the coordinate $t$ to $r$. On their intersection $r$ is a unit, so both maps land in the common open subscheme $A\cap B=\mathrm{Spec}\mathbb{Z}[t,t^{-1}]$, where they agree. They therefore glue uniquely:For an entirely explicit description, choose any lift $a\in R$ of $e$. Then $a(1-a)\in (r)$ andThe equality $a+(1-a)=1$ proves that these open subsets cover. The inclusion $D(a(1-a))\subseteq D(r)$ proves that their intersection is exactly $D(r)$. Their description as complements of $Z_A,Z_B$ proves independence of the chosen lift. In particular, no lift of $e$ to an idempotent of $R$ is required.

Label the two charts of the affine line with doubled origin by $A,B$, and use $t$ for their common coordinate. Let $u:T=\mathrm{Spec}R\to X$ be an $R$-valued element of the functor of points. Its composite with $\varphi$ corresponds to a unique $r\in R$ under the affine-target adjunction for schemes:The closed subscheme $Z=V(r)=\mathrm{Spec}(R/(r))$ maps into the disjoint union of the two zero sections,The inverse image $Z_A$ of $A_0$ is a clopen set in $Z$. By the idempotent–clopen correspondence, it determines a unique idempotent $e\in R/(r)$, equal to $1$ on $Z_A$ and $0$ on the inverse image $Z_B$ of $B_0$. HenceOver $\mathbb{Z}$, the two “origins” are entire zero sections, rather than single closed points; this construction uses their full scheme-theoretic fibers.

For an idempotent $e\in R$, define the principal open subset $D(e)\subseteq \mathrm{Spec}R$. Since $e(1-e)=0$, a prime ideal contains at least one of $e,1-e$, and it cannot contain both. ThereforeThus $D(e)$ is a clopen set.

Conversely, let $C$ be a clopen set in $T=\mathrm{Spec}R$. The structure sheaf glues the section $1$ on $C$ and the section $0$ on its complement to a unique global section $e\in \Gamma (T,{\mathcal{O}}_T)=R$. Checking on these two open subsets gives $e^2=e$, and $D(e)=C$. Conversely, the restriction of an idempotent $e$ to $D(e)$ is $1$ because $e$ is a unit there, and its restriction to $D(1-e)$ is $0$. Thus the two constructions are inverse, including for the zero ring:

The open affine representability criterion says that $F$ is a functor represented by a scheme exactly when $F$ is a Zariski sheaf and admits a family of open subfunctors $F_i\subseteq F$, each represented by an affine scheme, which covers $F$ in the following sense: for every scheme $T$ and every $\xi \in F(T)$, the open subschemes representing $h_T{\times }_F{F}_i$ cover $T$. In particular, the covering condition does not demand that every $\xi$ lie in one $F_i(T)$.

In symbols, for affine schemes $X_i$,The family is indexed by a set. For necessity, take an affine open cover of a representing scheme; morphisms of schemes glue over a Zariski cover. For sufficiency, $F_i{\times }_F{F}_j$ is represented by an open subscheme of each $X_i,X_j$. The two descriptions identify these overlaps and satisfy the compatibility on triple overlaps because all maps come from the inclusions into $F$. Apply gluing of schemes along open subschemes to obtain $X$. Local identification gives $h_X\to F$, and the Zariski sheaf condition makes this a natural isomorphism: its elements can be constructed and compared on the pulled-back open cover.

A presheaf $F:{\mathsf{Sch}}^{\mathrm{op}}\to \mathsf{Set}$ is a Zariski sheaf when every Zariski cover $(U_i\to T)$ gives an equalizer diagramThus compatible local elements glue to a unique element of $F(T)$. Compatibility means that the two restrictions agree on every intersection. This includes the empty Zariski cover of the empty scheme, so $F(\varnothing )$ is a singleton.

A Zariski cover of a scheme $T$ is a family of open immersions $(j_i:U_i\hookrightarrow T{)}_{i\in I}$ whose images cover its underlying topological space:There is no finiteness requirement. Intersections used in the sheaf on a site condition are the pullbacks $U_i{\times }_T{U}_j$, identified with the corresponding open subschemes $U_i\cap U_j$.

Write $h_T(S)=\mathrm{Hom}(S,T)$ for the functor of points of a scheme $T$. A subfunctor $H\subseteq F$ is an open subfunctor if every natural transformation $h_T\to F$ has a pullback represented by an open subscheme $U\subseteq T$:The displayed natural isomorphism must identify the projection to $h_T$ with the natural transformation induced by $U\hookrightarrow T$. Equivalently, for each $\xi \in F(T)$, there is an open subscheme $U\subseteq T$ such that a morphism of schemes $a:S\to T$ factors through $U$ exactly when $F(a)(\xi )\in H(S)$. The pullback is taken in the functor category, so this tests arbitrary families, rather than just underlying points.

Yes: every scheme is a Kolmogorov space. Choose an affine open subscheme $V=\mathrm{Spec}A$ containing $x$. If $y\notin V$, take $U=V$.

Otherwise, the distinct points $x,y\in V$ correspond to distinct prime ideals ${\mathfrak{p}}_x,{\mathfrak{p}}_y\subseteq A$. There is an element in one of these prime ideals and not the other. After interchanging the names of the two points if necessary, chooseThe principal open subschemecontains $y$ and does not contain $x$. It is an open subset of $V$, hence an open subset of $X$ because $V$ is open in $X$. This establishes $(\ast )$ in every case, proving schemes are Kolmogorov spaces.

Put $C=\overline{\{z\}}$, where the closure is taken in the underlying topological space of $X$. It is nonempty because it contains $z$.

Suppose $C=A\cup B$ for two closed subsets of $C$ in the subspace topology. Since $C$ is a closed subset of $X$, both $A$ and $B$ are also closed subsets of $X$. The point $z$ lies in one of them, say $A$. By the defining minimality of the closure, every closed subset of $X$ containing $z$ contains $C$. Hence $C\subseteq A$, so $A=C$.

Thus $C$ cannot be expressed as the union of two proper closed subsets, proving that $C$ is an irreducible closed subset. This proves closure of a point is irreducible in every topological space; the scheme assumption is unnecessary. Equivalently, $z$ is a generic point of its own closure.

Keep the nonzero $z$ of minimal two-factor tensor rank constructed in part (i). For every $a\in A$, its commutator with $a\otimes 1$ lies in $J$ and equalsIf this tensor were nonzero, its two-factor tensor rank would be at most $n-1$, contradicting minimality. Thus it is zero. The linear independence of $d_2,\dots ,d_n$ forces $a{c}_i=c_{i}a$ for every $a$ and $i\ge 2$. Since $Z(A)=k$, each $c_i$ is a scalar multiple of $1$.

But $1,c_2,\dots ,c_n$ are linearly independent. Hence $n=1$, so $z=1\otimes 1\in J$. A two-sided ideal containing the multiplicative identity is the entire ring; therefore $J=A{\otimes }_{k}B$.

We have proved the stronger result that $A{\otimes }_{k}B$ is a simple ring. It has finite dimension $({\dim }_{k}A)({\dim }_{k}B)$, so part (b) provesPart (c)(ii) also gives $Z(A{\otimes }_{k}B)=k$, so this proves that a tensor product of central simple algebras is again a central simple algebra.

Let $J\ne 0$ be a two-sided ideal of $A{\otimes }_{k}B$. Choose a nonzero $x\in J$ of least possible two-factor tensor rank $n$, and take a shortest expressionBoth families $(a_i)$ and $(b_i)$ are linearly independent: a linear dependence in either family would combine terms and shorten the expression. In particular $a_1,b_1\ne 0$.

Because $A$ is a simple ring, the two-sided ideal generated by $a_1$ is $A$. Thus there are finitely many $u_j,v_j\in A$ with ${\sum }_j{u}_j{a}_1{v}_j=1$. FormSince the $b_i$ are linearly independent, applying a linear functional selecting $b_1$ shows $y\ne 0$. Its two-factor tensor rank is at most $n$ and hence, by the choice of $x$, exactly $n$. Consequently $c_1=1,c_2,\dots ,c_n$ are linearly independent.

Similarly, simplicity of $B$ gives finitely many $p_j,q_j\in B$ with ${\sum }_j{p}_j{b}_1{q}_j=1$. NowThe linear independence of the $c_i$ shows $z\ne 0$, and minimality again makes its two-factor tensor rank exactly $n$. We have obtainedBoth coefficient families in this expression are linearly independent. Retaining this additional property is the key to the next part. This is coefficient normalization in an ideal of a tensor product.