Structure sheaf of a scheme (${\mathcal{O}}_X$)

The structure sheaf ${\mathcal{O}}_X$ records the regular functions on every open subset of a scheme $X$. Its stalk ${\mathcal{O}}_{X,x}$ is the local ring of functions defined near $x$.

The sheaf of total quotient rings, or sheaf of meromorphic functions, of a scheme $X$ is constructed by locally allowing denominators that are nonzerodivisors. For an open set $U$, let $S(U)$ consist of sections of ${\mathcal{O}}_X(U)$ whose germ is a nonzerodivisor at every point of $U$. Sheafifying the presheafgives ${\mathcal{K}}_X$. The stalkwise regular-denominator condition ensures that restriction to smaller open sets is defined. Merely taking arbitrary fraction rings of section rings is not a substitute for this construction.^[ref]

For a locally Noetherian scheme, ${\mathcal{K}}_{X,x}$ is the total quotient ring of ${\mathcal{O}}_{X,x}$, and on an affine open $\mathrm{Spec}A$ its sections are the total quotient ring of $A$. These identifications require the stated hypothesis.^[ref]

For an integral scheme this recovers its field of rational functions locally. For $A=k[\epsilon ]/({\epsilon }^2)$, every nonzerodivisor is already a unit, so its total quotient ring is $A$ itself and retains a nonzero nilpotent. Total quotient rings therefore need not be fields or remove infinitesimal structure.

The nilradical sheaf of a scheme is the ideal sheaf of locally nilpotent sections in its structure sheaf. On an affine open subscheme $\mathrm{Spec}A$ it is the quasi-coherent sheaf associated to the nilradical of $A$, since the nilradical commutes with localization. Its quotient defines the reduced induced subscheme. On a Noetherian scheme it has a uniform nilpotence exponent: take nilpotence bounds on a finite affine open cover and use their maximum. On a general scheme, a locally nilpotent section need not have a single nilpotence exponent over a non-quasi-compact open subset.

A global idempotent on a scheme is zero or one in every local ring. Its one-locus is both open and closed. Conversely a clopen subset determines a global idempotent by gluing one on it and zero on its complement. Under this correspondence, decompositions into nonzero orthogonal idempotents correspond to disjoint clopen decompositions. When all connected components are open, the primitive idempotents correspond exactly to those components. Without orthogonality the assertion fails in characteristic two: $(1,0)=(1,1)+(0,1)$ in ${\mathbb{F}}_2\times {\mathbb{F}}_2$.

The units in the structure sheaf of a scheme form a multiplicative sheaf of abelian groups. On a smooth algebraic curve, a nonzero rational function is a unit at a point precisely when its discrete valuation there is zero. Its inclusion in the sheaf of nonzero rational functions on an irreducible variety relates divisor on an algebraic curve to the Picard group through the long exact sequence in sheaf cohomology.

A regular function on an open subset $U$ of a scheme $X$ is a section of ${\mathcal{O}}_X(U)$. On an affine variety it is locally a quotient of polynomial functions with a denominator that does not vanish.

On a variety, a regular function is a regular map to the affine line; the arbitrary-target regular-map concept is broader.

A global regular function is a global section of the structure sheaf of a scheme, defined on the entire scheme. These functions form its ring $H^0(X,{\mathcal{O}}_X)$. On an affine scheme this recovers the defining ring; on a positive-dimensional projective complete intersection it consists only of constants over the base field.

A regular function $f$ on an entire affine variety is a polynomial function, even if the variety is reducible. Put $J=\{g\in A(X):gf\in A(X)\}$, an ideal of its coordinate ring. Near a point $p$, write $f=a/b$ with $b$ nonzero, and choose a principal open subset $D(h)$ containing $p$ inside that neighbourhood. Then $(hb)f=ha$ on all of $X$: on $D(h)$ this is the fraction identity, and off $D(h)$ both sides vanish. Hence $hb\in J$ and $(hb)(p)\ne 0$. No point is a common zero of $J$, so the weak Hilbert Nullstellensatz implies $J=A(X)$. Thus $1\in J$ and $f\in A(X)$. This proof does not require a fraction field for a reducible coordinate ring.

Over an algebraically closed field, every regular function on a nonempty irreducible projective variety is constant. Regard the function as a morphism to ${\mathbb{P}}^1$ avoiding infinity. Its graph is closed; the closedness of projection from projective space makes its image closed in ${\mathbb{P}}^1$. A proper closed subset there is finite, and an irreducible finite image is a singleton. Irreducibility is the usual variety convention; on a disconnected projective algebraic set different components may have different constants.

For $W={\mathbb{A}}_k^2\setminus \{0\}$, restriction to $D(x)$ and $D(y)$ givesThe intersection is taken in $k(x,y)$ and the equality follows from unique factorization: a reduced denominator dividing powers of both $x$ and $y$ is a unit. The inclusion $W\hookrightarrow {\mathbb{A}}^2$ induces an isomorphism on global functions but is not an isomorphism, so $W$ is not an affine variety.