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A Noetherian scheme is affine exactly when every coherent ideal sheaf has zero first sheaf cohomology. One direction is vanishing of quasi-coherent cohomology on an affine scheme. For the converse, ideal-sheaf vanishing produces affine principal neighbourhoods from ideal-sheaf vanishing. A finite such cover yields a unit-ideal certificate from a principal affine cover, and those affine charts glue to the spectrum of a commutative ring of global sections. The criterion, including the finite-type ideal version for quasi-compact quasi-separated schemes, is recorded in Stacks Project, Section 30.3.

For a reduced ring with a nonempty disconnected spectrum of a commutative ring, choose a separation $V(I)\bigsqcup V(J)$. Disjointness gives $I+J=R$; covering the spectrum gives $IJ\subseteq \sqrt{(0)}=0$. For comaximal ideals, $I\cap J=IJ$. The Chinese remainder theorem then supplies a product of two nonzero quotient rings. Equivalently, if $i+j=1$ with $i\in I$, $j\in J$, then $i,j$ are nontrivial orthogonal idempotents. This connects topology of the spectrum to ring product decomposition by an idempotent without a Noetherian assumption.

The ring $k\times k$, for a field $k$, has two prime ideals, $0\times k$ and $k\times 0$. Their localizations at a prime ideal are both isomorphic to $k$. Nevertheless $(1,0)(0,1)=0$ is a product of nonzero elements, so $A$ is not an integral domain. Nonzero orthogonal idempotents separate the two components of its spectrum of a commutative ring.

Throughout, rings are commutative with identity and ring homomorphisms preserve identity. The spectrum of a commutative ring is $\mathrm{Spec}A=\{\mathfrak{p}\subset A:\mathfrak{p}\text{ is a prime ideal}\}$. For an ideal $I$, put $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$; these are the closed sets of the Zariski topology. Its axioms follow fromFor the last equality, a prime ideal containing $IJ$ must contain $I$ or $J$: if it contained neither, choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. The other equalities follow directly from containment. Thus arbitrary intersections and finite unions of closed sets are closed, proving that this is a topology.

The Zorn lemma says that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. For $A\ne 0$, apply it to proper ideals ordered by inclusion. The zero ideal is proper. The union of a nonempty chain is an ideal and is proper, because containing one would put one in a member of the chain; the empty chain has the zero ideal as an upper bound. There is therefore a maximal ideal $\mathfrak{m}$. If $a\notin \mathfrak{m}$, then $\mathfrak{m}+(a)=A$, so the class of $a$ in $A/\mathfrak{m}$ is invertible. This quotient ring is a field, making $\mathfrak{m}$ a prime ideal. Hence $A\ne 0⟹\mathrm{Spec}A\ne \varnothing$.

For a ring homomorphism $f:A\to B$, define $f^{\ast }:\mathrm{Spec}B\to \mathrm{Spec}A$ by $f^{\ast }(\mathfrak{q})=f^{-1}(\mathfrak{q})$. The inverse image is proper because $f(1)=1$, and primality follows from $f(ab)=f(a)f(b)$. Moreoverwhere $IB$ is the ideal generated by $f(I)$. This proves continuity. Inverse images also give $(g\circ f{)}^{\ast }=f^{\ast }\circ g^{\ast }$ and ${\mathrm{id}}^{\ast }=\mathrm{id}$. Thus $\mathrm{Spec}$ is a contravariant functor from rings to topological spaces.

For the specified inclusion, the fundamental theorem of algebra gives the points $0$ and $(x-z)$ of $\mathrm{Spec}\mathbb{C}[x]$. Over $\mathbb{R}$, the nonzero prime ideals are generated either by $x-r$ with $r\in \mathbb{R}$, or by $(x-a{)}^2+b^2$ with $a\in \mathbb{R}$, $b>0$; all irreducible real polynomials have degree one or two. Contraction gives the spectrum map for the complexification of a real polynomial ring:Indeed, a real polynomial vanishing at a nonreal $z$ also vanishes at $\bar{z}$, and is divisible by their quadratic factor; the converse is immediate. This map is surjective. Its fibers contain one point over a real linear factor, two conjugate points over an irreducible quadratic factor, and only the generic point over zero. It identifies complex-conjugate pairs. Since each proper Zariski-closed set in either one-variable polynomial ring is a finite set of closed points, images of closed sets are closed; hence the Zariski topology on the target is the quotient topology for this identification.

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

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

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

Let $\mathfrak{m}$ be the maximal ideal of the local ring $A$. Every open subset of $\mathrm{Spec}A$ containing the closed point $\mathfrak{m}$ is the whole spectrum of a commutative ring: it contains a principal open subscheme $D(a)$ with $a\notin \mathfrak{m}$, and that $a$ is a unit, so $D(a)=\mathrm{Spec}A$.

Given $f:\mathrm{Spec}A\to {\mathbb{P}}_{\mathbb{Z}}^n$, choose a standard affine open subscheme $D_{+}(x_i)$ containing $f(\mathfrak{m})$. Its preimage is consequently all of $\mathrm{Spec}A$. The affine-target adjunction for schemes expresses $f$ in this chart by elements $b_j\in A$ for $j\ne i$, the images of $x_j/x_i$. It is represented by homogeneous coordinates with $a_i=1$ and $a_j=b_j$.

Conversely, a tuple $(a_0,\dots ,a_n)$ with some $a_i\in A^{\times }$ defines a morphism of schemes into $D_{+}(x_i)$ by $x_j/x_i\mapsto a_j/a_i$. Choosing another unit entry gives the same morphism of schemes, since the usual projective space transition functions identify the ratios. Multiplying all entries by one unit does not change any ratio. If two such tuples define the same morphism of schemes, choose a unit entry $a_i$ in the first and a unit entry $a_j^{\prime }$ in the second. In the second chart, the function $x_i/x_j$ pulls back to $a_i^{\prime }/a_j^{\prime }$. Because the whole map lies in $D_{+}(x_i)$, this ratio is a unit, so $a_i^{\prime }$ is a unit too. Equality in this chart gives $a_j/a_i=a_j^{\prime }/a_i^{\prime }$ for every $j$, hence $a_j^{\prime }=(a_i^{\prime }/a_i)a_j$.

Thus the correspondence is exactlyThis is the projective coordinates over a local ring description.

For a general ring, the key open-neighbourhood argument fails. Even a tuple generating the unit ideal need not have any unit entry. For example, over $A=k\times k$ the pair $((1,0),(0,1))$ defines a map to ${\mathbb{P}}^1$ whose two points have images $[1:0]$ and $[0:1]$. Neither coordinate is a unit, and no common unit multiple changes that fact. The map lies in no single standard chart. More generally, maps into projective space correspond to invertible sheaf quotients of $A^{n+1}$; the quotient need not be a free rank-one module outside the local ring case.

Necessity follows from commutativity of the fibre product of schemes square: if $p(z)=x$ and $q(z)=y$, then $f(x)=g(y)$.

For sufficiency, put $s=f(x)=g(y)$ and $K=\kappa (s)$, $L=\kappa (x)$, $L^{\prime }=\kappa (y)$. The maps on local rings induce field embeddings $K\hookrightarrow L,L^{\prime }$. The canonical point maps into $X$ and $Y$ therefore give a morphism of schemesThe tensor product of commutative algebras $L{\otimes }_K{L}^{\prime }$ is nonzero. To see the hinted fact directly, choose a $K$-basis of $L$ containing $1$; tensoring that basis with $L^{\prime }$ makes $1\otimes 1$ nonzero. A nonzero unital commutative ring has a prime ideal, so its spectrum of a commutative ring has a point $w$. Its projections to the two field spectra are their unique points. The image $z$ of $w$ consequently projects to $x$ and $y$.

The desired point exists precisely when the base images coincide. This is the point-lifting property of a scheme fibre product. It does not claim that such a point is unique: different prime ideals of the tensor product may give different points over the same pair.

The points of the spectrum of a commutative ring $A$ are its prime ideals. Its Zariski topology has closed sets $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$ and basis $D(f)=\{\mathfrak{p}:f\notin \mathfrak{p}\}$. The structure sheaf of a scheme can be defined as follows: a section over $U$ assigns an element of $A_{\mathfrak{q}}$ to every $\mathfrak{q}\in U$, and locally that assignment is represented by one fraction $a/s$, with $s$ outside every prime in the neighbourhood. Restrictions forget values. Local representability and equality can be checked on an open cover, so these assignments satisfy the sheaf gluing axiom. On $D(f)$ the sections identify with the localization of a ring $A_f$.

Evaluation at $\mathfrak{p}$ induces the ring homomorphism ${\mathcal{O}}_{X,\mathfrak{p}}\to A_{\mathfrak{p}}$. It is surjective because $a/s\in A_{\mathfrak{p}}$ is represented by a section on $D(s)$. For injectivity, represent a germ by $a/s$ on a neighbourhood of $\mathfrak{p}$. If its value in $A_{\mathfrak{p}}$ vanishes, there is $t\notin \mathfrak{p}$ with $ta=0$. The section is then zero after restricting to $D(st)$ inside that neighbourhood. Its germ is zero. Thus

The spectrum of a commutative ring is the setFor an ideal $I\subseteq R$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. The Zariski topology declares precisely these subsets to be closed sets. Using arbitrary subsets instead of ideals gives the same definition, because a prime ideal contains a subset exactly when it contains the ideal generated by that subset.

We verify the closed set axioms directly. First, $V(0)=\mathrm{Spec}R$ and $V(R)=\varnothing$. For any family of ideals,since containing the sum is equivalent to containing every summand. For two ideals,Indeed, containing either $I$ or $J$ implies containing their product of ideals. Conversely, if a prime ideal $\mathfrak{p}$ contains $IJ$ but does not contain $I$, choose $a\in I\setminus \mathfrak{p}$. For every $b\in J$, $ab\in \mathfrak{p}$, so the prime ideal property forces $b\in \mathfrak{p}$. Thus $J\subseteq \mathfrak{p}$.

The displayed identities establish arbitrary intersections and finite unions of closed sets, including their empty cases. Therefore these sets define a topology on $\mathrm{Spec}R$.

The induced map on the spectrum of a commutative ring is contraction of an ideal:To see that it is well defined, let $\mathfrak{q}$ be a prime ideal of $S$. Its inverse image is an ideal of $R$, and it is proper because $\phi (1)=1\notin \mathfrak{q}$. If $ab\in {\phi }^{-1}(\mathfrak{q})$, then $\phi (a)\phi (b)\in \mathfrak{q}$, whence $\phi (a)\in \mathfrak{q}$ or $\phi (b)\in \mathfrak{q}$. This proves that the inverse image is a prime ideal.

For any ideal $I\subseteq R$, its extension of an ideal to $S$ is $IS=\phi (I)S$. We haveThus inverse images of Zariski-closed sets are Zariski-closed sets, which proves that $f$ is a continuous map.

The Krull dimension of a commutative ring $R$ isThe length counts strict inclusions, so a chain with $n+1$ prime ideals has length $n$. The supremum may be infinite. A field, for example, has just the prime ideal $(0)$ and therefore has Krull dimension zero. For the zero ring, the spectrum of a commutative ring is empty; under the convention $\sup \varnothing =-\infty$, its dimension is $-\infty$.

On an affine open $W=\mathrm{Spec}A\subseteq X$, write $Z\cap W=V(I)$. Give $Z\cap W$ the reduced induced subscheme structure $\mathrm{Spec}(A/\sqrt{I})$. These constructions agree under localization and therefore glue to a reduced closed subscheme $Z\hookrightarrow X$.

Let $Z^{\prime }\hookrightarrow X$ be another closed immersion with the same underlying closed set. Affine-locally write $Z^{\prime }=\mathrm{Spec}(A/J)$. Since $V(J)=Z\cap W$, its vanishing ideal is $\sqrt{J}=\sqrt{I}$, and the inclusion $J\subseteq \sqrt{J}$ induces a quotient homomorphismContravariance of the spectrum of a commutative ring gives a factorizationThe quotient maps force these local factorizations to agree on overlaps, so they glue. They are also the only possible maps over $X$, which proves uniqueness.

For morphisms of schemes $X\to S$, $Y\to S$, a pair of points $x,y$ lifts to a point of $X{\times }_{S}Y$ exactly when their base images agree. For sufficiency use the nonempty spectrum of a commutative ring of $\kappa (x){\otimes }_{\kappa (s)}\kappa (y)$, by tensor product of field extensions is nonzero. This establishes existence, not uniqueness, of a lift.

Contraction sends the zero prime ideal to zero. A complex closed point $(x-z)$ contracts to $(x-z)$ when $z$ is real, and to $((x-\mathrm{Re}z{)}^2+(\mathrm{Im}z{)}^2)$ otherwise. Thus the map on the spectrum of a commutative ring is surjective and identifies each nonreal conjugate pair. Both spectra have only finite proper Zariski-closed sets, so this is also the topological quotient by complex conjugation.