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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.

The immersions and gluing. For $p>0$, the prime ideals of the localization ${\mathbb{Z}}_{(p)}$ are $(0)$ and $(p){\mathbb{Z}}_{(p)}$. Inverting $p$ gives ${\mathbb{Z}}_{(p)}[1/p]=\mathbb{Q}$, sois an open immersion sending the generic point to the generic point. For $p=0$, ${\mathbb{Z}}_{(0)}=\mathbb{Q}$, and take ${\psi }_0$ to be the identity. Identify each pair of these open subschemes using the identity of $\mathrm{Spec}\mathbb{Q}$. The identities compose to the identity on every triple overlap, so gluing of schemes along open subschemes applies. No finiteness of the family is required.

The resulting scheme $Y$ has one common generic point $\eta$ and a distinct closed point $y_p$ for every positive prime number $p$. Its affine open subschemes $U_p=\{\eta ,y_p\}$, for $p>0$, intersect pairwise in $U_0=\{\eta \}$. A subset $W\subseteq Y$ is open precisely when its intersection with every $U_p$ is open. In the two-point topological space $U_p$, the open subsets are $\varnothing$, $\{\eta \}$ and $U_p$. Consequentlywhere $S$ is any subset of the positive prime numbers, including the empty subset.

The canonical morphism. The ring homomorphisms $\mathbb{Z}\to {\mathbb{Z}}_{(p)}$ give morphisms of schemes $U_p\to \mathrm{Spec}\mathbb{Z}$ agreeing on the overlaps, hence a morphism of schemes $f:Y\to \mathrm{Spec}\mathbb{Z}$. By uniqueness of the structure morphism, this is the required map. Contraction of prime ideals givesThese are all points of $\mathrm{Spec}\mathbb{Z}$, each exactly once, so $f$ is a bijection. It is continuous because its restrictions to the open cover $(U_p)$ are continuous; explicitly, every nonempty open subset of $\mathrm{Spec}\mathbb{Z}$ is the complement of a finite collection of closed points, whose inverse image is one of the $W_S$.

Sections on every open. For any $S$, the sheaf gluing axiom on the cover of $W_S$ by $U_0$ and the $U_p$ with $p\in S$ identifieswhere the intersection over the empty set is $\mathbb{Q}$. Indeed, a section on $U_p$ is an element of ${\mathbb{Z}}_{(p)}$; restrictions to every overlap are its inclusion into $\mathbb{Q}$. Compatibility means that all these elements are the same rational number. In lowest terms $a/b$, this rational number lies in the intersection exactly when no prime number in $S$ divides $b$. Equivalently, this is the integer localization at a set of primes obtained by inverting all positive prime numbers outside $S$.

Now let $V\subseteq \mathrm{Spec}\mathbb{Z}$ be nonempty and open. Its finite complement consists of $(p_1),\dots ,(p_r)$; put $N=p_1\cdots p_r$, with $N=1$ for an empty complement. Then $V=D(N)$ andThe equality follows by the denominator criterion above. The map $f^{\#}$ is this same equality of subrings of $\mathbb{Q}$, so it is a ring isomorphism. Over the empty open subset both rings of sections are the zero ring. These identifications respect every restriction map. Thus the precise comparison of structure sheaves isHere “over every open set” refers to an open subset $V$ of the target and its inverse image. The preceding intersection formula additionally describes the sections on every open subset of $Y$, including ones whose images are not open in the target.

Topology and finiteness. The singleton $\{\eta \}=U_0$ is open in $Y$, but $\{(0)\}$ is not open in $\mathrm{Spec}\mathbb{Z}$: a nonempty open subset there contains all but finitely many of the infinitely many closed points. Therefore $f$ is not a homeomorphism, despite being a continuous bijection and giving the indicated isomorphism of structure sheaves after direct image.

The open cover $(U_p{)}_{p>0}$ of $Y$ has no finite subcover, because $y_q$ belongs only to $U_q$. Thus $Y$ is not a quasi-compact scheme, and part (a) rules out an affine scheme. Each ${\mathbb{Z}}_{(p)}$ is a Noetherian ring, being a localization of the Noetherian ring $\mathbb{Z}$. Hence the $U_p$ form an affine open cover by spectra of Noetherian rings, proving that $Y$ is a locally Noetherian scheme. A Noetherian scheme is a quasi-compact scheme, so

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 $U_i\cap U_j$.

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.

An inverse system of topological spaces over a directed set $I$ consists of spaces $X_i$ and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=1$ and $f_{ik}=f_{ij}{f}_{jk}$. Its inverse limit is a space $X$ with compatible continuous projections ${\pi }_i$ such that every compatible family of continuous maps $u_i:Y\to X_i$ factors uniquely through a continuous map $u:Y\to X$. To construct it, takewith the subspace topology inherited from the product topology. The map $u(y)=(u_i(y))$ has values in $X$ and is continuous by the definition of the product topology; its coordinates force uniqueness. If $X^{\prime }$ is another inverse limit, the two universal properties give continuous maps $X\to X^{\prime }$ and $X^{\prime }\to X$ whose composites have all the same projections as the identity. Uniqueness makes these composites the identity. The inverse limit exists and is unique up to a unique projection-preserving homeomorphism. An arbitrary inverse limit may be empty; nonemptiness requires additional hypotheses.

Use the usual Hausdorff convention for a topological group. An inverse limit of finite discrete groups is a closed subgroup of their product: each compatibility condition is closed because the finite factors are Hausdorff spaces. The Tychonoff theorem gives compactness, and distinct tuples are separated by a clopen coordinate cylinder, so the limit is totally disconnected. Coordinatewise multiplication and inversion are continuous, giving a profinite group.

Conversely, let $G$ be a compact Hausdorff space and a totally disconnected topological group. We use the standard topological fact that a compact Hausdorff space that is totally disconnected has a basis of clopen sets. Given an identity neighbourhood, choose a clopen identity neighbourhood $U$ inside it. Compactness of $U$ and continuity of multiplication give a symmetric identity neighbourhood $V$ with $VU\subseteq U$: cover $U$ by finitely many neighbourhoods on which multiplication by a sufficiently small identity neighbourhood stays inside $U$, then intersect those identity neighbourhoods. Symmetry gives $vU=U$ for $v\in V$. Consequently $H=\{g:gU=U\}$ is a subgroup containing $V$, hence an open subgroup, and $H\subseteq U$ since $1\in U$. Its open cosets cover the compact group, so there are finitely many. The normal core $N={\bigcap }_{g\in G}gH{g}^{-1}$ is a finite intersection of open subgroups, hence an open normal subgroup contained in $U$. Thus the open normal subgroups form an identity-neighbourhood basis and have trivial intersection.

The canonical group homomorphism $G\to {\underleftarrow{\lim }}_{N}G/N$ is continuous and injective. For a compatible tuple of quotient elements, its inverse-image cosets in $G$ are closed and have the finite intersection property: a finite family is refined by the quotient for the intersection of its normal subgroups. Compactness gives an element in all those cosets, proving surjectivity. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The compact totally disconnected and inverse-limit definitions of a profinite group agree.

Index the finite nonempty sets $\mathcal{P}(N)$ by open normal subgroups, with the transition from $M$ to $N$ when $M\subseteq N$ given by image under $G/M\to G/N$. Images of Sylow subgroups under surjections of finite groups are Sylow subgroups. These transitions are surjective: if $S\le G/N$ is a Sylow subgroup, choose a Sylow subgroup in its full inverse image in $G/M$. It maps onto $S$, and its order has the full $p$-part of $|G/M|$, so it is a Sylow subgroup of $G/M$ too. The transitions respect composition, giving an inverse system of finite sets.

Its inverse limit is nonempty. Indeed, finitely many compatibility conditions can be satisfied by choosing one Sylow subgroup in a quotient refining all the involved quotients and taking its images; compactness of the product of the finite sets then gives a compatible family $(P_N{)}_N$. DefineThis is a closed subgroup of $G$. Its image in $G/N$ is exactly $P_N$: for a prescribed $s\in P_N$, each finite collection of conditions defining $P$ together with ${\pi }_N(g)=s$ is satisfiable in a common refining quotient, because the transitions on the chosen Sylow subgroups are surjective. The finite intersection property again supplies $g\in P$. Hence $P\cong {\underleftarrow{\lim }}_N{P}_N$ is a pro-p group and $PN/N=P_N$ for every $N$.

Finally, if a pro-p subgroup $Q$ contains $P$, its image in every finite quotient is a p-group containing the Sylow subgroup $P_N$, so that image equals $P_N$. Thus every $q\in Q$ belongs to every ${\pi }_N^{-1}(P_N)$, whence $Q\subseteq P$. The constructed $P$ is a maximal pro-p subgroup.

An essay arguing against the claim. A good definition can change which mathematical questions are visible. Nevertheless, I would argue that category theory derives its explanatory force from the interaction between definitions and theorems; there is no sound general ordering in which definitions contribute more than theorems.

The strongest case for the contrary view comes from universal properties. A product in a category is characterized by maps into it, rather than by a particular construction of ordered pairs. The same definition applies to sets, groups, and topological spaces. An adjunction goes further: its natural correspondence of arrows puts free constructions, quotient constructions, and several kinds of completion into a common language. Such definitions remove irrelevant choices and expose relationships that would be difficult to recognize from separate constructions.

But a universal-property definition does not supply an object. The definition of a product in a category says what a product would accomplish; an existence theorem says whether the category actually has one. Even uniqueness up to unique isomorphism is a short proof, not an extra clause to be silently supplied by the word “universal”. Given two products, their universal properties produce maps in both directions, and uniqueness forces the two composites to be identities. This elementary theorem makes the definition usable independently of a chosen construction.

The Yoneda lemma illustrates the distinction especially clearly. A representable functor is a definition, but the lemma proves that its natural transformations into any set-valued functor are determined by one element. Its proof evaluates at the identity and reconstructs the whole transformation by functoriality. Consequently the Yoneda embedding is full and faithful: an object's pattern of incoming arrows recovers its morphisms without loss. That is a substantive explanation of why the chosen definition is powerful. Merely calling a functor representable would not establish it.

Similarly, defining a monadic adjunction gives a precise meaning to recovering structured objects as algebras for a monad. It does not decide whether a given forgetful functor is monadic. The precise monadicity theorem translates that difficult classification into a usable test involving adjoint functors, reflected isomorphisms, and specific coequalizers. For compact Hausdorff spaces, the crucial step is proving that a set-split relation is closed. The definition of a monad alone does not contain this topological argument. The theorem connects the algebraic language to an independently meaningful class of spaces.

There is also a danger in praising definitions without hypotheses. A Cartesian closed category has exponential objects, but a category of structured objects need not be cartesian closed merely because underlying sets have function sets. The correct exponential in a functor category generally involves natural transformations out of representable functors, not pointwise function sets. Its existence and universal property require an argument. Theorems also reveal when tempting definitions cannot support the intended construction.

Definitions provide the vocabulary and select the structures worth studying; proofs establish existence, consequences, and limitations. Category theory makes this partnership unusually explicit because its definitions are often expressed through universal properties. Its distinctive achievement is to turn well-chosen definitions into reusable theorems, rather than to make theorems secondary.

An alternative essay arguing for the claim. In category theory, theorems are indispensable, but a carefully chosen definition often performs the greater act of organization. It identifies what many apparently unrelated constructions have in common and makes whole families of proofs possible. In that explanatory sense, definitions can be the subject's most influential contributions.

Consider a universal property. Describing a product in a category as an object representing pairs of incoming arrows replaces several different concrete constructions by one invariant pattern. The corresponding uniqueness proof is important but short; the lasting insight is that the same pattern is the right question in every category. Once the definition is available, it becomes meaningful to ask which functors preserve products, whether the products exist in a new category, and how they interact with other universal constructions.

An adjunction is an even more striking example. The existence of free groups, free modules, and a Stone-Čech compactification can be proved separately. The definition of an adjunction recognizes their common relation to forgetful functors and explains why arrows out of a free object are determined by arrows out of its generators. It also packages the relation as a monad. A large collection of isolated existence results becomes an instance of a shared structure, with reusable consequences about composition and categorical limits.

A valuable definition also has to distinguish structures rather than merely rename them. An abelian category is not just an additive category with categorical kernels and categorical cokernels: normality of monomorphisms and epimorphisms matters. The additive example with truncated primary torsion shows why those clauses belong in the definition. The definition therefore preserves exactly the structure on which theorems about images and exact sequences depend.

This view does not make proofs dispensable. The Yoneda lemma justifies the use of representable functors, and the precise monadicity theorem tests when a proposed algebraic description is valid. Rather, it argues that the best categorical definitions have unusually broad explanatory reach: they select the invariant structures to which those proofs apply. If “matter more” means providing the reusable framework in which many theorems become instances of one idea, category-theoretic definitions have a strong claim to that role.

If an open set contains an even integer, invariance under every even translation makes it contain all even integers. Similarly, containing an odd integer makes it contain all odd integers. Conversely each parity class is invariant under those translations. Thus the proposed open sets are exactlyThey form a partition topology. Both the empty and whole sets are included, and any union or finite intersection is again a union of parity classes. Hence this is a topological space. It is not the discrete topology: no singleton is open.

For a sheaf of abelian groups $\mathcal{F}$ on a topological space $X$, putThese arbitrary families of germs form a sheaf, and restrictions are projections. Extending a family by zero proves that $C(\mathcal{F})$ is flasque. Sending a section to all its germs gives an injective sheaf map $\mathcal{F}\to C(\mathcal{F})$, since a section with every germ zero vanishes locally and hence globally.

Set $Q^{-1}=\mathcal{F}$, $I^j=C(Q^{j-1})$ and $Q^j=\mathrm{coker}(Q^{j-1}\to I^j)$, using the sheaf cokernel. Compose $I^j\to Q^j\to I^{j+1}$ to obtain the Godement resolutionIt is exact by construction, with flasque terms. Define sheaf cohomology by the cohomology of its global-section complex:The resolution principle for sheaf cohomology identifies this with the computation using any flasque resolution.

The elementary input for acyclicity is the flasque-kernel section-lifting lemma: in a short exact sequence with flasque kernel, sections of the quotient lift over every open subset. Consequently a quotient of two flasque sheaves is flasque, because a quotient section lifts and that lift extends. If $\mathcal{F}$ itself is flasque, induction through $0\to Q^{j-1}\to I^j\to Q^j\to 0$ makes every $Q^j$ flasque. The same lifting lemma makes these sequences exact on global sections. Hence the global-section complex has no positive-degree cohomology, provingThis deduction uses the lifting lemma, not the desired cohomological vanishing as an assumption.

For the rational-section construction, add two representative sections after restricting to the intersection of their dense domains. Multiply by a representative rational function in the same way. Finite intersections of dense open sets are dense, and further restricting both representatives does not change their resulting class. The module axioms hold on their common domain. Thus rational sections form a module over $\mathrm{Rat}(X)$; irreducibility is not needed for this module statement. When $X$ is irreducible, $\mathrm{Rat}(X)$ is the function field $K=k(X)$.

Suppose now that $\mathcal{F}$ is locally free and $X$ irreducible. A germ at $P$ is represented on a nonempty open neighbourhood, which is dense, so it defines a rational section. Two representatives of the germ agree on a neighbourhood of $P$ and therefore give the same class. If the class is zero, trivialize $\mathcal{F}$ on a neighbourhood of $P$. Its coefficient functions vanish on a dense open subset; regular functions on a reduced irreducible variety with this property vanish identically. The germ is therefore zero. This proves the stalk inclusion into rational sections of a locally free sheaf:Local freeness matters: a nonzero torsion germ can vanish on a dense open set.

On an irreducible algebraic curve, every proper closed subset is finite, and every open subset is quasi-compact because the Zariski space is Noetherian. Let $R=\mathrm{Rat}(\mathcal{F})$. The constant sheaf $\mathcal{R}(\mathcal{F})$ has sections $R$ on every nonempty open set: such opens are irreducible and hence connected. DefineRestriction drops the components outside the smaller open set. For a compatible family over an open cover of $U$, take a finite subcover by quasi-compactness. The union of the finite supports of these finitely many sections is finite. Their agreeing components therefore glue to a unique finite-support tuple on $U$. This verifies the sheaf gluing axiom, including arbitrary covers, rather than merely asserting that a presheaf direct sum is always a sheaf.

A rational section is regular on some dense open set, whose complement on a curve is finite. Its classes $[s{]}_P\in R/{\mathcal{F}}_P$ therefore have finite support, defining a sheaf map $\mathcal{R}(\mathcal{F})\to \mathcal{P}(\mathcal{F})$. At $P$, the first sheaf has stalk $R$. The second has stalk $R/{\mathcal{F}}_P$: any finite tuple can be restricted to a neighbourhood omitting all its other support points, and its $P$-component is unaffected by such restriction. The resulting sequence on stalks isExactness of sheaves can be tested on stalks, so this proves the rational principal-parts resolution on an algebraic curveBoth right-hand sheaves are flasque: restrictions for the constant rational-section sheaf are identities between nonempty opens, and restrictions for the rational principal-parts sheaf are projections, with extension by zero. ConsequentlySurjectivity on stalks has not been confused with surjectivity on global sections.

For an explicit failure of global surjectivity take $X={\mathbb{P}}^1$ and $\mathcal{F}=\mathcal{O}(-2)$. Use coordinate $z=X_1/X_0$ and frame $e_0=X_0^{-2}$ on the finite chart. At infinity use $w=1/z$ and $e_{\infty }=X_1^{-2}=z^{-2}{e}_0$. Prescribe the rational principal part of $z^{-1}{e}_0$ at zero and zero at every other point. If a rational section $r(z)e_0$ realized this tuple, then $r-z^{-1}$ would be regular at zero and $r$ would be regular at every other finite point. Thus $r=z^{-1}+p(z)$ with $p\in k[z]$. Regularity at infinity would requireto be regular at $w=0$. Its $w^{-1}$ term cannot cancel with any term of the polynomial contribution, whose powers are at most $-2$. This is impossible. Hence the global rational-principal-part map need not be surjective.

For an affine curve it is always surjective, and here is a direct construction. Write $X=\mathrm{Spec}A$, $K=\mathrm{Frac}A$ and $\mathcal{F}=\tilde{M}$. By the affine module sheaf construction, $R=M{\otimes }_{A}K=:M_K$ and ${\mathcal{F}}_P=M_{{\mathfrak{m}}_P}$. Choose representatives $v_P\in M_K$ for a prescribed finite tuple, and choose one nonzero $a\in A$ with $a{v}_P\in M$ for all of them. The zero set $T=V(a)$ is finite. The ring $A/(a)$ is Artinian, since it is Noetherian of dimension zero, and its Artinian decomposition into local factors givesPrescribe the class of $a{v}_P$ in the indicated component for each requested point in $T$, and zero in the other components. A requested point outside $T$ already has zero principal part because $a$ is a unit there. Lift the tuple to $w\in M$. Then $s=w/a\in M_K$ satisfies $s-v_P\in M_{{\mathfrak{m}}_P}$ at every requested point in $T$, and is regular at every other point of $T$. Outside $T$ it is regular because $a$ is invertible. It therefore realizes exactly the prescribed tuple. This proves affine principal-parts interpolation for a locally free sheaf and givesNo nonsingularity assumption was used; the construction applies to singular irreducible affine curves as well.

Let $J:\mathbf{CompHaus}\hookrightarrow \mathbf{Top}$ be the inclusion. Both categories are locally small categories. By the Tychonoff theorem, small products in a category of compact Hausdorff spaces are compact Hausdorff, with their ordinary product topology. The equalizer of two continuous maps between compact Hausdorff spaces is closed in the source: it is the inverse image of the closed diagonal of the target. Hence it is again compact Hausdorff. The usual construction of small categorical limits from products and equalizers proves that $\mathbf{CompHaus}$ is a complete category, and that $J$ preserves its small limits as limits in $\mathbf{Top}$.

We verify the solution-set condition for an arbitrary topological space $X$. Put $\kappa =|X|$ and $\theta =2^{2^{\kappa }}$. Given a continuous map $f:X\to K$ with $K$ compact Hausdorff, letThis closed subspace is compact Hausdorff, and $f(X)$ is dense in it. The allowed cardinal estimate therefore gives $|Y|\le \theta$. The map $f$ factors through the continuous map $X\to Y$ and the subspace inclusion $Y\hookrightarrow K$.

Choose a fixed set $S$ of cardinality $\theta$. Consider all compact Hausdorff topologies on subsets of $S$, together with all continuous maps from $X$ into the resulting spaces. There are only a set of such topologies and functions. Every $Y$ above is homeomorphic to one of these spaces after relabelling its underlying set. Consequently every $X\to JK$ factors through a member of this set of maps. This proves the solution-set condition, including $X=\varnothing$.

The general adjoint functor theorem now gives the compact Hausdorff reflectionIts unit ${\eta }_X:X\to J\beta X$ is universal for continuous maps from $X$ to compact Hausdorff spaces. For arbitrary $X$, this is a reflection, and the unit is not required to be a topological embedding.

The Absolute Frobenius morphism is the identity on the underlying topological space and acts on the structure sheaf by $a\mapsto a^p$. On an affine scheme it is induced by the ring homomorphism $A\to A$, $a\mapsto a^p$. This is additive in characteristic $p$, and the inverse image of a prime ideal $\mathfrak{p}$ is again $\mathfrak{p}$, since $a^p\in \mathfrak{p}$ exactly when $a\in \mathfrak{p}$. These affine maps agree under localization and hence glue.

A perfect ring of characteristic p is one whose Frobenius endomorphism is bijective. Since the Absolute Frobenius morphism fixes points, its map on the stalk at $x$ is precisely the Frobenius endomorphism of ${\mathcal{O}}_{X,x}$. A morphism of sheaves is an isomorphism exactly when it is an isomorphism on all stalks. ThereforeIn the reverse direction the inverse sheaf map, together with the identity on points, gives the inverse morphism of locally ringed spaces. This is perfection in the Frobenius sense, not the unrelated homological use of the word perfect.

For the function field assertions below, interpret the normal schemes as integral, as is implicit in writing a single field $k(X)$ or $k(Y)$. General normal schemes need not be irreducible; without this interpretation those fields are not defined globally as single fields. Set $K=k(Y)\subseteq L=k(X)$. For an affine open $V=\mathrm{Spec}A$ in $Y$, finiteness gives $f^{-1}V=\mathrm{Spec}B$, with $A\subseteq B$ finite and both rings normal domains. In fact $B$ is the integral closure of $A$ in $L$: any element of $L$ integral over $A$ is also integral over $B$ and therefore belongs to $B$. This local description yields the two forms of Frobenius factorization through a finite normal cover.

A topological space is connected if it cannot be written as the union of two disjoint nonempty open subsets. Equivalently its only subsets that are both open and closed are the empty set and the whole space. For a subspace topology, open means relatively open within that subspace. A path-connected space is connected, since a separation would pull back along a path to a separation of an interval.

A proper filter on a set $X$ is a nonempty family $\mathcal{F}\subseteq \mathcal{P}(X)$ that excludes $\varnothing$, is closed under finite intersections, and is upward closed under inclusion. In particular $X\in \mathcal{F}$. In a topological space, convergence of a filter $\mathcal{F}\to x$ means that every neighbourhood of $x$ belongs to $\mathcal{F}$. An ultrafilter is a proper filter maximal under inclusion; equivalently, it contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.

To justify the equivalence, if $A\notin \mathcal{U}$ and every member of $\mathcal{U}$ met $A$, adjoining $A$ would generate a larger proper filter. Maximality therefore supplies $F\in \mathcal{U}$ with $F\cap A=\varnothing$, and upward closure gives $X\setminus A\in \mathcal{U}$. Both complementary sets cannot belong to a proper filter. Conversely, a filter with this dichotomy cannot acquire any new set without also containing its complement, which would give an empty intersection.

The ultrafilter characterization of compact Hausdorff spaces, stated without proof, is: a topological space is compact and Hausdorff exactly when every ultrafilter on it converges to a unique point.

For a finite set partition $X=A_1\cup \cdots \cup A_n$, if no cell belonged to $\mathcal{U}$, the ultrafilter dichotomy would put all $X\setminus A_j$ in $\mathcal{U}$. Their finite intersection is empty, a contradiction. Thus some $A_j\in \mathcal{U}$, and at most one cell can belong because distinct cells are disjoint. The converse is: a proper filter on a set meeting every finite partition by containing one cell is an ultrafilter. Apply the partition hypothesis to $X=A\cup (X\setminus A)$ to obtain the dichotomy, then the maximality argument above. Empty cells can be omitted, and the cases $A=\varnothing$ or $A=X$ are automatic.

For a function $f:X\to Y$, the image of a filter isSince preimages preserve inclusions and finite intersections, this is a proper filter on a set $Y$. Equivalently, it is the filter generated by $f(F)$ for $F\in \mathcal{F}$: if $f^{-1}(B)\in \mathcal{F}$, then its image is contained in $B$, while $f(F)\subseteq B$ implies $F\subseteq f^{-1}(B)$. For an ultrafilter $\mathcal{U}$, the identity $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$ shows that $f_{\ast }\mathcal{U}$ contains exactly one of $B,Y\setminus B$. Therefore the pushforward filter of an ultrafilter is an ultrafilter.

Now let $x\in {\ell }^{\infty }$, the l-infinity sequence space, and put $a_x(n)=n^{-1}{\sum }_{j=1}^n{x}_j$. All these Cesàro means lie in the closed disk $K_x=\{z:|z|\le \Vert x{\Vert }_{\infty }\}$, a compact Hausdorff space. The pushforward filter $a_{x\ast }\mathcal{U}$ contains $K_x$; restricting it to that disk and applying the ultrafilter theorem gives a unique limitFor real sequences the disk can be replaced by a closed interval. This is the ultrafilter construction of a Banach limit.

The identities $a_{\alpha x+\beta y}(n)=\alpha a_x(n)+\beta a_y(n)$ give linearity of $l$. Explicitly, the two sets on which $|a_x(n)-l(x)|<\delta$ and $|a_y(n)-l(y)|<\delta$ both belong to $\mathcal{U}$, hence so does their intersection. On that intersection the difference from $\alpha l(x)+\beta l(y)$ is at most $(|\alpha |+|\beta |)\delta$, proving $l(\alpha x+\beta y)=\alpha l(x)+\beta l(y)$. If $x_n\ge 0$, its means lie in $[0,\Vert x{\Vert }_{\infty }]$, so $l(x)\ge 0$. Thus $l$ is a positive linear functional. The bound $|l(x)|\le \Vert x{\Vert }_{\infty }$ gives $\Vert l\Vert \le 1$, while the constant-one sequence satisfies $l(1)=1$. Hence $\boxed{\Vert l\Vert =1}$.

Because $\mathcal{U}$ contains no finite sets, its dichotomy puts every cofinite set in $\mathcal{U}$. If $x_n\to L$ ordinarily, the Cesàro theorem for convergent sequences follows directly by splitting the sum: for a fixed $N$ with $|x_j-L|<\epsilon$ for $j>N$,The finite head tends to zero, so $a_x(n)\to L$ ordinarily. Every sufficiently long tail belongs to $\mathcal{U}$, making its ultralimit also $L$. Thus $\boxed{x_n\to L⟹l(x)=L}$.

Finally, for the left shift on bounded sequences $S$,Taking ultralimits along the same cofinite-containing ultrafilter gives $\boxed{l(Sx)=l(x)}$. This establishes every defining property of a Banach limit.

A Hausdorff space is a topological space in which any two distinct points have disjoint open neighborhoods. Let $K$ be a compact subset of a Hausdorff space and $x\notin K$. For each $y\in K$, choose disjoint open neighborhoods $U_y$ of $y$ and $W_y$ of $x$. The $U_y$ cover $K$, so compactness gives a finite subcover $U_{y_1},\dots ,U_{y_m}$. Their paired intersection $W={\bigcap }_{r=1}^m{W}_{y_r}$ is an open neighborhood of $x$ disjoint from $K$. Thus every point of its complement has an open neighborhood in the complement, proving every compact subset of a Hausdorff space is closed. The empty set case is immediate.

In the cocountable topology on an uncountable set, any two nonempty open sets intersect: their combined complements are countable and cannot exhaust $X$. Thus the space is not Hausdorff.

Nevertheless every compact subset is closed here. In fact only finite subsets are compact. To prove this, suppose $K$ is infinite and choose distinct points $d_1,d_2,\dots \in K$. The open setsform an increasing open cover of $X$, and hence of $K$. Any finite selection is contained in $U_N$ for its largest index $N$, which misses $d_N\in K$. So $K$ is not compact. Finite sets are compact in any topological space, and in this cocountable topology they are closed because their complements are open. This proves the claimed behavior of compact subsets of a cocountable space.

A topological group is a group equipped with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous maps. The product uses the product topology. Translation and inversion are therefore homeomorphisms. Hausdorffness is sometimes included in the definition; the two open-subgroup arguments below do not require it.

For the inverse system, take a directed set $I$, topological spaces $X_i$, and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=\mathrm{id}$ and $f_{ik}=f_{ij}{f}_{jk}$. An inverse limit is a space $X$ with compatible continuous maps ${\pi }_i:X\to X_i$ such that every compatible family $q_i:Y\to X_i$ factors uniquely as $q_i={\pi }_{i}q$ through a continuous map $q:Y\to X$. Existence follows from the concrete constructionwith the subspace topology of the product topology. The only possible map $q$ is $q(y)=(q_i(y){)}_i$. Compatibility puts its image in $X$, and coordinatewise continuity makes it continuous. This proves the universal property of an inverse limit, even when $X$ is empty. Given two inverse limits, their universal properties supply maps each way; the composites have the same coordinates as the identity, so uniqueness makes them identities. The inverse limit exists and is unique up to the unique homeomorphism respecting its projections.

A profinite group can be defined as an inverse limit of finite discrete groups, or as a compact Hausdorff space that is a totally disconnected space, equipped with continuous group operations. For the first implication, use the Tychonoff theorem. The compatibility equalities define a closed subgroup of the compact Hausdorff product. Two distinct points are distinguished by a finite discrete coordinate, so no connected subset contains both. Consequently the inverse limit is a compact Hausdorff totally disconnected topological group.

For the converse, use the topological result that a compact Hausdorff totally disconnected space has a basis of clopen sets. Choose a clopen set $U$ containing $1$ inside any specified identity neighbourhood. For each $u\in U$, continuity gives an identity neighbourhood $V_u$ and an open neighbourhood $W_u$ of $u$ with $V_u{W}_u\subseteq U$. Finitely many $W_u$ cover the compact set $U$. Intersect their $V_u$ and its inverse to obtain a symmetric identity neighbourhood $V$ with $VU\subseteq U$. Symmetry also gives $V^{-1}U\subseteq U$, and hence $vU=U$ for every $v\in V$. The setis a subgroup containing $V$, so it is open. It is contained in $U$, since $1\in U$ and $h=h\cdot 1\in hU=U$. By part (ii), $H$ has finite index. Its normal core of a subgroup is a finite intersection of conjugates of $H$, hence an open normal subgroup $N$ contained in $U$. This proves the open normal subgroup basis of a profinite group.

The natural group homomorphismis injective because the intersection of these $N$ is $\{1\}$. For a compatible family of cosets, any finitely many have nonempty intersection: their intersection subgroup is another indexing subgroup, and its prescribed coset lies in each. The finite intersection property and compactness give a point belonging to all the cosets, proving surjectivity. Finally a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The two definitions of profinite group are equivalent.

A locally ringed space is a topological space $X$ equipped with a sheaf of commutative rings ${\mathcal{O}}_X$ whose stalk ${\mathcal{O}}_{X,x}$ at every point is a local ring. A morphism of locally ringed spaces $(f,f^{\#}):(Y,{\mathcal{O}}_Y)\to (X,{\mathcal{O}}_X)$ consists of a continuous map $f:Y\to X$ and a morphism of sheaves of rings $f^{\#}:{\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$, with every induced stalk map ${\mathcal{O}}_{X,f(y)}\to {\mathcal{O}}_{Y,y}$ a local homomorphism: the inverse image of the target maximal ideal is the source maximal ideal. Here $f_{\ast }{\mathcal{O}}_Y$ is the direct image sheaf, with sections ${\mathcal{O}}_Y(f^{-1}U)$ on $U$.

Let $\varphi :A\to B$ be a ring homomorphism. Define the map between the spectra of rings byThe inverse image is a prime ideal, and $f^{-1}(D(a))=D(\varphi (a))$. Thus $f$ is a continuous map, because the principal open subsets form a basis. On these principal open subsets, define the structure sheaf map using the localizationsThese maps commute with restrictions and consequently give a morphism of sheaves. At $\mathfrak{q}$, putting $\mathfrak{p}={\varphi }^{-1}\mathfrak{q}$, the stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. An element $b/s$ belongs to its inverse image of $\mathfrak{q}{B}_{\mathfrak{q}}$ exactly when $b\in \mathfrak{p}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, take a morphism of locally ringed spaces $f:\mathrm{Spec}B\to \mathrm{Spec}A$. Its map on global sections gives a ring homomorphismLocality of the stalk map implies, for every $a\in A$,Hence the point map must be $\mathfrak{q}\mapsto {\varphi }^{-1}\mathfrak{q}$. On $D(a)$, its structure sheaf map must send $b/1$ to $\varphi (b)/1$ and the inverse of $a$ to the inverse of $\varphi (a)$. Thus it is precisely the localization map constructed above. Agreement on this basis proves agreement of the morphisms of sheaves. In particular, the inducing ring homomorphism exists and is unique.

If $\varphi$ is surjective, write $B=A/I$. The correspondence between prime ideals of $A/I$ and prime ideals of $A$ containing $I$ identifies $\mathrm{Spec}B$ homeomorphically with the closed subset $V(I)$: the principal open subsets correspond on both sides. Moreover, the stalk maps are the surjections $A_{\mathfrak{p}}\to (A/I{)}_{\mathfrak{p}/I}$, so ${\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$ is surjective.

For the converse, it is important not to assume that global sections preserve an arbitrary surjection of sheaves. Here the affine module-sheaf equivalence supplies the extra information: on every $D(a)$,Thus $f_{\ast }{\mathcal{O}}_Y$ is the quasi-coherent sheaf associated with the $A$-module $B$, and the cokernel sheaf of ${\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$ is associated with the module $C=\mathrm{coker}(A\to B)$, since localization of a module is exact. Surjectivity of the morphism of sheaves gives $C_{\mathfrak{p}}=0$ for every prime ideal $\mathfrak{p}$. This implies $C=0$: if $0\ne c\in C$, its proper annihilator lies in a maximal ideal $\mathfrak{m}$, and then $c/1\ne 0$ in $C_{\mathfrak{m}}$. Therefore $A\to B$ is surjective. We have proved the closed immersion criterion for affine schemes:

A proper morphism of schemes is a separated morphism of finite type which is a universally closed morphism: after every base change, the resulting map of underlying topological spaces takes closed subsets to closed subsets.

The valuative criterion for properness, in the Noetherian setting used here, says that a finite type morphism $f:X\to Y$ is a proper morphism if and only if the following existence and uniqueness condition holds. For every valuation ring $V$, its fraction field $K$, and compatible morphisms of schemes $a:\mathrm{Spec}K\to X$ and $b:\mathrm{Spec}V\to Y$ with $f\circ a=b{|}_{\mathrm{Spec}K}$, there is exactly one morphism of schemes $h:\mathrm{Spec}V\to X$ such that $h{|}_{\mathrm{Spec}K}=a$ and $f\circ h=b$. All valuation rings are allowed; no discreteness assumption is needed in this formulation. The following arguments also use that each of the three defining conditions for a proper morphism is preserved by base change.

The affine space $X={\mathbb{A}}_k^n$ is an integral scheme, hence its underlying topological space is irreducible. Every nonempty open subset of an irreducible topological space is irreducible and in particular connected. A section of the constant sheaf with value the function field $K=k(t_1,\dots ,t_n)$ is a locally constant function, so on any such open subset it is constant. Thus its sections are $K$ on each nonempty open subset and zero on the empty set. All restriction maps are either identities on $K$ or maps onto zero.

The constant sheaf is therefore a flasque sheaf. It has its natural ${\mathcal{O}}_X$-module structure because a regular function on a nonempty open subset embeds in the function field and acts by multiplication. Part (i) now givesThis is an instance of constant sheaves on irreducible spaces are flasque; no restriction on $n$ is needed.

The general adjoint functor theorem says that, for a functor $U:\mathcal{A}\to \mathcal{B}$ between locally small categories, with $\mathcal{A}$ a complete category, $U$ has a left adjoint if and only if it preserves small limits and satisfies the solution-set condition. The solution-set condition means that for each $B\in \mathcal{B}$ there is a set of arrows $b_i:B\to U{A}_i$ such that every $b:B\to UA$ factors as $U(a)b_i$ for some $i$ and some $a:A_i\to A$.

Here is the smallness argument behind the theorem. A complete category that is locally small and has a weakly initial set $(W_i)$ has an initial object. Form the product in a category $W={\prod }_i{W}_i$. It is weakly initial, since a projection $W\to W_i$ followed by a chosen arrow $W_i\to X$ gives an arrow $W\to X$. Take the simultaneous equalizer $e:E\to W$ of all endomorphisms of $W$ with $1_W$; the family is a set by local smallness. The object $E$ is also weakly initial. For parallel arrows $a,b:E\rightrightarrows X$, take their equalizer $j:Y\to E$ and a map $t:W\to Y$. Since $ejt$ is an endomorphism of $W$, the defining property of $e$ gives $ejte=e$. Cancellation of the monomorphism $e$ gives $jte=1_E$. Thus the monomorphism $j$ has a right inverse and is an isomorphism, so $a=b$. Existence of arrows out of $E$ was already established, proving the initial-object lemma for complete categories with a weakly initial set.

If $U$ preserves small categorical limits, the comma category $(B\downarrow U)$ is complete: take the limiting object in $\mathcal{A}$, and use preservation to assemble the arrows from $B$. It is locally small, and the solution-set condition provides a weakly initial set. Its initial object is an arrow ${\eta }_B:B\to ULB$ with a natural bijectionFor a morphism $v:B\to B^{\prime }$, define $Lv$ by $U(Lv){\eta }_B={\eta }_{B^{\prime }}v$. Uniqueness proves functoriality and naturality, giving a left adjoint $L⊣U$. Conversely, a right adjoint preserves categorical limits, because applying the adjunction converts a limiting cone over a diagram into the corresponding limiting cone of hom-sets. Its adjunction unit at $B$ is a singleton solution set. This proves both directions of the general adjoint functor theorem.

Apply it to the inclusion $J:\mathbf{KHaus}\hookrightarrow \mathbf{Top}$, where $\mathbf{KHaus}$ is the full subcategory of compact Hausdorff spaces. A small product in a category of compact Hausdorff spaces is again compact Hausdorff by the Tychonoff theorem. An equalizer of two continuous functions into a Hausdorff space is closed, hence compact Hausdorff. Products and equalizers construct all small categorical limits, and the inclusion preserves these constructions. Both categories are locally small.

Fix a topological space $X$ and a continuous function $f:X\to K$ into a compact Hausdorff space. The closure $K_f=\overline{f(X)}$ is compact Hausdorff and has a dense subset of cardinality at most $|X|$. The permitted cardinal estimate givesFor $X=\varnothing$, use $K_f=\varnothing$. Choose a set of representatives of all compact Hausdorff topologies on underlying sets of cardinality at most this bound, and all continuous functions from $X$ to those representatives. Every $f$ factors through one of them, using the inclusion $K_f\hookrightarrow K$. This is the solution-set condition. Consequently the inclusion of compact Hausdorff spaces has a left adjoint, the compact Hausdorff reflection:The universal arrow $X\to RX$ has dense image: factoring it through its closed image and applying uniqueness supplies an inverse to that image inclusion. It need not be an embedding when $X$ is not sufficiently separated; the assertion is a compact Hausdorff reflection, rather than a claim that every topological space embeds in a compact Hausdorff space.

For completeness, the alternative Special adjoint functor theorem can be proved from the same initial-object argument. Its limit form assumes that $\mathcal{A}$ is complete, locally small and well-powered, with a small cogenerating family $(Q_i{)}_{i\in I}$; then $U:\mathcal{A}\to \mathcal{B}$, with $\mathcal{B}$ locally small, has a left adjoint precisely when it preserves small limits. To prove sufficiency, start with $b:B\to UA$ and intersect all subobjects $M\hookrightarrow A$ through which $b$ factors after applying $U$. Well-poweredness makes this a small intersection of subobjects, and preservation of limits gives a smallest supporting object $(M,b_M)$. For maps $s,t:M\rightrightarrows Q_i$, equality $U(s)b_M=U(t)b_M$ makes their equalizer another supporting subobject. Minimality forces that equalizer to be invertible, so $s=t$. Thereforeis an injective function. The evaluation embedding into cogenerator products embeds $M$ in ${\prod }_i{Q}_i^{S_i}$, where $S_i$ is the realized subset of the fixed set $\mathcal{B}(B,U{Q}_i)$. There are only a set of choices of these subsets, only a set of subobjects of each resulting product, and only a set of arrows from $B$ to each $U$-image. They give a solution set, so the already proved general adjoint functor theorem applies. Necessity is again preservation of limits by a right adjoint. For $\mathbf{KHaus}$, monomorphisms are embeddings onto closed subspaces, and $[0,1]$ is a coseparator because continuous functions to it separate points. Thus the Special adjoint functor theorem gives the same reflection directly.

For the covariant category in this part, put $\mathcal{E}=[\mathcal{C},\mathbf{Set}]$ and $r_c=\mathcal{C}(c,-)$. Its global-sections functor is $\Gamma =\mathrm{Hom}(1,-)$, the limit of the diagram. If $\Gamma$ is also an inverse image, it is a left adjoint and preserves all small colimits.

Consider the canonical pointwise-surjective mapAt $d$, the summand $r_d(d)$ contains $1_d$, so $q$ is an epimorphism. In a presheaf topos epimorphisms are regular, and a colimit-preserving functor preserves their coequalizer presentations. Thus $\Gamma q$ is surjective. Since $\Gamma$ preserves coproducts, some summand has a global element, yielding a section $s:1\to r_c$ of its unique map $r_c\to 1$. The constant singleton functor is a retract of a covariant representable.

Conversely, if $1$ is such a retract in a category, $\mathrm{Hom}(1,-)$ is a retract of $\mathrm{Hom}(r_c,-)$, which is evaluation at $c$ by the Yoneda lemma. Evaluation preserves colimits pointwise. The colimit comparison for a retract functor is a retract of the evaluation comparison; a retract of an isomorphism is an isomorphism, so $\Gamma$ preserves all small colimits. It also preserves finite limits because it is representable. A right adjoint can be constructed byFor $h:d\to d^{\prime }$, precomposition gives $r_{d^{\prime }}\to r_d$, and applying $\Gamma$ and then $\mathrm{Set}(-,S)$ gives $R(S)(d)\to R(S)(d^{\prime })$. The presentation of every functor as a colimit of representables gives $\mathrm{Hom}(F,R(S))\cong \mathrm{Set}(\Gamma F,S)$. Hence $\Gamma ⊣R$, so $\Gamma$ is an inverse image and the topos is local.

The retract condition has a precise interpretation as the initial object criterion for a covariant local topos. Write the section as a natural family $u_d:c\to d$. Naturality says $h{u}_d=u_{d^{\prime }}$ for every $h:d\to d^{\prime }$. Put $e=u_c$. Then $e^2=e$ and every $f:c\to d$ satisfies $fe=u_d$. In the idempotent completion of $\mathcal{C}$, the object $(c,e)$ is initial: an arrow from it to $(d,k)$ must satisfy $v=kve$, and the identities force $v=u_d$; this arrow exists because $k{u}_d=u_d$ and $u_{d}e=u_d$.

Conversely an initial object $(c,e)$ in the idempotent completion gives that natural family and the retract. ThereforeIf $\mathcal{C}$ is already idempotent-complete, this is equivalent to an initial object in $\mathcal{C}$ itself, and $\Gamma$ is evaluation there. The word is initial, not terminal, because this question uses covariant functors.

One cannot omit idempotent completion for a general small category. Take the one-object category of the monoid $\{1,0\}$ with absorbing zero. It has no initial object because its endomorphism set has two elements, but the natural family $u=0$ satisfies $m0=0$ for every $m$. Its covariant functor category is local. Concretely, a functor is a set with an idempotent operator, and its global sections are the fixed points of that operator.

For a topological space $X$, there is an equally explicit criterion. If $\mathbf{Sh}(X)$ is local, apply $\Gamma$ to the epimorphism ${\coprod }_i{U}_i\to 1$ associated with any open cover $X={\bigcup }_i{U}_i$, regarding the opens as subterminal sheaves. Since $\Gamma U_i$ is a singleton exactly when $U_i=X$, and otherwise empty, coproduct and epimorphism preservation force some $U_i=X$.

Thus every open cover of $X$ contains $X$ itself. Conversely this open cover criterion for a local sheaf topos condition is equivalent to existence of a point $x$ whose only open neighborhood is $X$: the union of all proper open sets cannot be all of $X$, so choose $x$ outside it. Such a point plainly forces any cover to contain $X$. Empty spaces fail the condition.

At this point, the stalk functor $F\mapsto F_x$ is simply $F(X)=\Gamma F$, since there is only one open neighborhood. Its right adjoint is the skyscraper sheaf of setsA morphism $F\to R(S)$ is determined exactly by its map $F(X)\to S$, because all proper opens omit $x$. Hence $\Gamma ⊣R$, and $\Gamma$ preserves finite limits, proving locality. ConsequentlyIn a $T_0$ space this distinguished point is unique and closed. In a $T_1$ space the condition forces $X$ to be a singleton. Indistinguishable points in a non-$T_0$ space can all have this property; uniqueness of an actual point is not part of the general criterion.

An irreducible scheme is a nonempty scheme whose underlying topological space cannot be expressed as the union of two proper closed subsets. Equivalently, any two nonempty open subsets meet. A reduced scheme is one whose local rings have no nonzero nilpotent elements. Equivalently, every affine open subscheme has a reduced ring of regular functions. We may define an integral scheme as a nonempty scheme for which the coordinate ring of every nonempty affine open subscheme is an integral domain. We shall show that this is equivalent to being reduced and irreducible. The nonempty convention matters: the empty scheme is reduced but is not irreducible or integral.

For a commutative ring $A$, the spectrum of a ring is reduced exactly when $A$ is a reduced ring. One direction follows since localization preserves reducedness. Conversely, if $a$ is a nonzero nilpotent element, choose a prime ideal containing its proper annihilator. Then $a/1$ cannot vanish at that localization, contradicting reducedness of its local ring.

The spectrum of a ring is irreducible exactly when its nilradical $N=\sqrt{(0)}$ is a prime ideal. To see the essential implication directly, if $ab\in N$, then $D(a)\cap D(b)=D(ab)$ is empty. Irreducibility forces $D(a)$ or $D(b)$ to be empty, hence $a\in N$ or $b\in N$. Also $N$ is proper because the spectrum is nonempty. Conversely, if $N$ is a prime ideal, the point $N$ has closure $V(N)=\mathrm{Spec}A$, so the spectrum is irreducible. Combining the two criteria gives

Now suppose $X$ is reduced and irreducible. Every nonempty affine open subscheme is also reduced and irreducible: irreducibility passes to nonempty open subsets, because their nonempty open subsets are nonempty opens of $X$. The affine criterion makes each coordinate ring an integral domain, so $X$ is integral. Conversely, suppose all its nonempty affine coordinate rings are integral domains. Their localizations show that $X$ is reduced. If $X$ were reducible, there would be disjoint nonempty open subsets; choose nonempty affine open subschemes $U=\mathrm{Spec}A$ and $V=\mathrm{Spec}B$ inside them. Their disjoint union is itself an affine open subscheme $\mathrm{Spec}(A\times B)$. Since $A,B$ are nonzero, $(1,0)(0,1)=0$ contradicts the domain condition. Therefore