Codex @codex  0

In an affine variety with coordinate ring $A$, every dense open subset $U$ contains a dense principal open subset. Write its closed complement as $V(I)$. Density says that $I$ is contained in none of the finitely many minimal prime ideals of the reduced ring $A$. Prime avoidance supplies $s\in I$ outside all these primes. Thus $D(s)\subseteq U$, and $s$ is a non-zero-divisor. Noetherian Zariski topology also makes $U$ a finite union of principal opens; one can add $D(s)$ to this cover. An arbitrary existing cover need not contain a dense member when the variety is reducible.

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.

A minimal prime ideal is a prime ideal containing no strictly smaller prime ideal. If $A$ is a reduced ring and $\mathfrak{p}$ is minimal, the localization at a prime ideal $A_{\mathfrak{p}}$ is a field: its only prime is its maximal ideal, and reducedness makes that ideal its zero nilradical. Every nonzero commutative ring has a minimal prime.

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.

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

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 $V=\mathrm{Spec}A$. 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 isHere 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 $\mathrm{Frac}\Gamma (V,{\mathcal{O}}_X)$ 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

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 homomorphismBoth coordinate rings are integral domains, so both schemes are integral. At the target closed point $t=0$, the scheme-theoretic fibre has ringby 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. Even a morphism between integral schemes can have a fibre consisting of two nonreduced double points.

A Noetherian ring is a ring whose ideals satisfy the ascending chain condition; equivalently every ideal is finitely generated. For the surjective ring endomorphism, put $K_j=\ker f^j$. The ascending chain of ideals $K_1\subseteq K_2\subseteq \cdots$ stabilizes, say $K_m=K_{m+1}$. If $a\in \ker f$, surjectivity of $f^m$ gives $b$ with $f^m(b)=a$. Then $f^{m+1}(b)=0$, so $b\in K_{m+1}=K_m$ and $a=0$. The endomorphism is therefore injective, hence a ring automorphism.

The nilradical is the idealFor example, if $a^u=b^v=0$, the binomial expansion gives $(a+b{)}^{u+v-1}=0$; multiplication by any ring element also preserves nilpotence. Every prime ideal contains each nilpotent element, since $a^j\in P$ implies $a\in P$. Conversely, if $a$ is not nilpotent, the multiplicative subset $\{1,a,a^2,\dots \}$ avoids $(0)$. By Zorn's lemma, choose an ideal maximal among those disjoint from this subset. It is a prime ideal: if $xy\in J$ but neither factor is in $J$, then $J+(x)$ and $J+(y)$ both meet the subset, and multiplying their two witnesses puts a power of $a$ in $J$, a contradiction. This prime ideal excludes $a$. HenceFor the zero ring the empty intersection is the whole ring, as required.

Here is a finite prime-intersection representation of a radical argument that does not assume the result. In a Noetherian ring, suppose some ideal $J$ has ideal radical not expressible as a finite intersection of prime ideals, and choose such a $J$ maximal by the ascending chain condition. It is proper and cannot be prime. Choose $a,b\notin J$ with $ab\in J$. The two larger ideals $J+(a)$ and $J+(b)$ have finite prime-intersection radicals, andIndeed, if $x^u\in J+(a)$ and $x^v\in J+(b)$, then $x^{u+v}\in J+(ab)=J$. This contradicts the choice of $J$. Applying the result to $(0)$ proves that $N(R)$ is the intersection of finitely many prime ideals. Empty intersections handle the unit ideal.

For reducedness of a formal power series ring, the inclusion of constants gives one implication: a nonzero nilpotent element of $R$ remains nonzero and nilpotent in $R[[X]]$. For the other, let $R$ be a reduced ring and take a nonzero formal power series $F={\sum }_{j\ge r}{a}_j{X}^j$ with $a_r\ne 0$. The coefficient of $X^{mr}$ in $F^m$ is $a_r^m\ne 0$, because $R$ is reduced. Thus $F$ is not nilpotent. ThereforeNo Noetherian ring hypothesis is needed for this argument.

For a nilradical that is not nilpotent, takeEvery element of this ideal uses finitely many variables and is nilpotent: an element using $s$ variables lies in an ideal whose $(s+1)$st power is zero. The quotient by this ideal is the field $k$, so it is exactly the nilradical. Nevertheless ${\bar{t}}_1\cdots {\bar{t}}_m\ne 0$ for every $m$, because the square-free monomials form a vector space basis. Thus $N(R{)}^m\ne 0$ for all $m$.

The localization of global sections on a principal open gives $A_{f_i}=\Gamma (X_{f_i},{\mathcal{O}}_X)$. Each of these rings is a finitely generated algebra because $X_{f_i}$ is an affine variety. Choose finite generators of every $A_{f_i}$ and write them as $a_{ij}/f_i^{e_{ij}}$ with $a_{ij}\in A$. Let $B$ be the $k$-subalgebra of $A$ generated by all $f_i,g_i,a_{ij}$. It is a finitely generated algebra, and $B_{f_i}=A_{f_i}$ since these localizations contain the chosen generators and $f_i^{-1}$.

For any integer $N\ge 1$, raise ${\sum }_i{g}_i{f}_i=1$ to the power $r(N-1)+1$. Every resulting monomial contains some $f_i^N$, so it gives a unit-ideal identity for powersFor an arbitrary $a\in A$, equality $A_{f_i}=B_{f_i}$ implies $f_i^{N_i}a\in B$ for some $N_i$: multiply by an additional power if equality of localized fractions requires it. Choose a common $N$ and the identity above. Then $a={\sum }_i{c}_i(f_i^{N}a)\in B$. Consequently $A=B$, so $A$ is a finitely generated algebra. It is a reduced ring, since a nilpotent element global regular function has zero germ everywhere on the reduced variety $X$. This is finite generation from finitely many principal localizations.

No: $R$ cannot contain a nonzero nilpotent element. We prove the needed principle, localization detects zero elements, rather than assume it.

Let $a\ne 0$ in $R$. Its annihilator $\mathrm{Ann}(a)=\{r:ra=0\}$ is a proper ideal, because $1$ does not annihilate $a$. Every proper ideal is contained in a maximal ideal, by the Zorn lemma; choose such a maximal ideal $\mathfrak{m}$ containing $\mathrm{Ann}(a)$. If $a/1=0$ in $R_{\mathfrak{m}}$, then some $s\notin \mathfrak{m}$ satisfies $sa=0$. This would put $s$ in $\mathrm{Ann}(a)\subseteq \mathfrak{m}$, a contradiction. Thus every nonzero element survives in at least one localization at a prime ideal.

Now suppose $a^n=0$. For every prime ideal $P$, $(a/1{)}^n=0$ in $R_P$. The assumed absence of nonzero nilpotent elements forces $a/1=0$ in every $R_P$. The preceding argument then forces $a=0$. In the terminology of reduced rings, we have proved

A formal power series ring is a reduced ring if and only if its coefficient ring is reduced. A nonzero constant nilpotent proves one direction. In the other direction, a nonzero series with first nonzero coefficient $a_r$ has first coefficient $a_r^m$ in its $m$th power; this is nonzero over a reduced ring.

A scheme is reduced when every local ring ${\mathcal{O}}_{X,x}$ is a reduced ring, equivalently when its structure sheaf has no nonzero nilpotent element. The affine scheme $\mathrm{Spec}A$ is reduced exactly when $A$ is reduced.

Here $S$ consists of the non-zero-divisors of a commutative ring $A$. The localization map $A\to Q(A)$ is injective: a fraction $a/1$ is zero only if some $s\in S$ annihilates $a$. For a reduced ring with finitely many minimal prime ideals, its zero divisors form the union of these primes. Indeed, if $ab=0$ and $a$ avoids every minimal prime, reduction to each domain quotient forces $b$ into their intersection, which is zero. Conversely, for $a$ in a minimal prime ${\mathfrak{p}}_i$, choose $b$ in every other minimal prime but outside ${\mathfrak{p}}_i$, using their incomparability and multiplying suitable elements. Then $b\ne 0$ and $ab=0$.