= Solution
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 $V(N)=\operatorname{Spec}A$, so the spectrum is irreducible. Combining the two criteria gives
$$
\boxed{A\text{ is an integral domain}
\iff\operatorname{Spec}A\text{ is reduced and irreducible}.}
$$
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> $U=\operatorname{Spec}A$ and $V=\operatorname{Spec}B$ inside them. Their disjoint union is itself an <affine open subscheme> $\operatorname{Spec}(A\times B)$. Since $A,B$ are nonzero, $(1,0)(0,1)=0$ contradicts the domain condition. Therefore
$$
\boxed{X\text{ is integral}\iff X\text{ is reduced and irreducible}.}
$$
The <generic point> of an <integral scheme> $X$ is the unique point $\eta$ with $\overline{\{\eta\}}=X$. For existence, take a nonempty <affine open subscheme> $V=\operatorname{Spec}A$. Its point $(0)$ has closure containing $V$, which is dense in $X$, so its closure in $X$ is all of $X$. For uniqueness, both proposed <generic points> lie in every nonempty <open subset>, hence in $V$, where the only dense point is $(0)$. The <function field> is
$$
K(X)=\mathcal O_{X,\eta}=\operatorname{Frac}(A).
$$
Here the <stalk> description shows that the <field of fractions> is independent of the choice of nonempty <affine open subscheme>.
Every nonempty <open subscheme> $U$ contains $\eta$, so taking a <germ> there defines $\Gamma(U,\mathcal O_X)\to K(X)$. If a section has zero germ, restrict it to any nonempty <affine open subscheme> $V\subseteq U$. Its image in $\operatorname{Frac}\Gamma(V,\mathcal O_X)$ is zero. Since this <coordinate ring> is an <integral domain>, the section is zero on $V$. Such affines cover $U$, and the <sheaf gluing axiom> makes the section zero on $U$. Thus the <generic-point embedding of regular functions> is
$$
\boxed{\Gamma(U,\mathcal O_X)\hookrightarrow K(X).}
$$
For a <nonreduced reducible fibre between integral schemes>, take both source and target to be the <affine line> over $\mathbb C$, and use the <ring homomorphism>
$$
\mathbb C[t]\longrightarrow\mathbb C[x],\qquad t\longmapsto x^2(x-1)^2.
$$
Both <coordinate rings> are <integral domains>, so both <schemes> are integral. At the target <closed point> $t=0$, the <scheme-theoretic fibre> has ring
$$
\mathbb C[x]/\bigl(x^2(x-1)^2\bigr)
\cong\mathbb C[x]/(x^2)\times\mathbb C[x]/((x-1)^2),
$$
by the <Chinese remainder theorem>. Its underlying space has two distinct <closed points>, so it is reducible. The class of $x(x-1)$ is nonzero but has square zero, so it is not reduced. \b[Even a morphism between integral schemes can have a fibre consisting of two nonreduced double points.]
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