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A ramified prime makes the finite algebra ${\mathcal{O}}_K/p{\mathcal{O}}_K$ have a nonzero nilpotent element. Multiplication by its product with any other element is nilpotent and has trace zero, so the reduced trace pairing is degenerate. Therefore the field discriminant is divisible by $p$. This proves finiteness of the ramified rational primes without requiring explicit prime factorization in the field.

The dual-number ring over a field $k$ consists of elements $a+b\epsilon$ with ${\epsilon }^2=0$. The element $\epsilon$ is a nonzero nilpotent element; its ideal is the ring's only prime ideal. Hence its affine scheme has the same one-point topology as $\mathrm{Spec}k$ while its structure sheaf differs.

Over a perfect field $k$ of characteristic $p$, the scheme-theoretic fiber of the Absolute Frobenius morphism of ${\mathbb{A}}_k^1$ over a rational point $a$ isIts underlying space is one point, but its coordinate ring has a nonzero nilpotent element, so the fibre is a length-$p$ nonreduced scheme.

An irreducible scheme is one whose underlying topological space is a nonempty irreducible topological space. It can have nilpotent elements in its structure sheaf; irreducibility concerns only the topology. A reduced scheme is an integral scheme precisely when it is irreducible.

For a finite CW complex, every element in the kernel of the rank map in topological K-theory is a nilpotent element. One proof adjoins one positive-dimensional cell at a time: the restriction kernel is square-zero by the relative product in topological K-theory. For connected spaces the rank-zero kernel is Reduced topological K-theory; for disconnected spaces the kernel of restriction to just one basepoint can contain nonzero idempotents.

A nilpotent thickening is a closed immersion defined by a nilpotent ideal sheaf of a closed subscheme. Its underlying topological map is a homeomorphism, because every prime ideal contains every nilpotent element. For example, quotienting the dual numbers by $(\epsilon )$ changes the structure sheaf without changing the single point of the spectrum.

The nilradical of a commutative ring is its ideal of nilpotent elements, equivalently the radical of an ideal applied to its zero ideal.

For every commutative ring $R$,Every nilpotent element lies in every prime ideal. If $a$ is not nilpotent, an ideal maximal among those disjoint from $\{1,a,a^2,\dots \}$ is a prime ideal avoiding $a$, by the Zorn lemma and the maximality argument.

A scheme is nonreduced when its structure sheaf contains a nonzero nilpotent element. For an affine scheme $\mathrm{Spec}(A/I)$, this is equivalent to the ideal $I$ not being radical.

Use $y$ on the target and $x$ on the source, so the map of coordinate rings sends $y$ to $x^4$. A closed point of the real affine line corresponds to a maximal ideal generated either by $y-a$, $a\in \mathbb{R}$, or by a monic irreducible quadratic $q(y)$. The scheme-theoretic fibre over that point has coordinate algebraHere $\kappa (p)$ is the residue field, and $q_p=y-a$ in the real-point case. The fibre is a scheme, so its nilpotents must be retained.

For $a>0$, put $r=a^{1/4}$. The factors $x-r$, $x+r$, $x^2+r^2$ are pairwise coprime, with the last one irreducible over $\mathbb{R}$. The Chinese remainder theorem identifies $B_p$ with $\mathbb{R}\times \mathbb{R}\times \mathbb{C}$, giving three irreducible components.

For $a<0$, put $r=(-a{)}^{1/4}$. The exact factorization isEach quadratic has discriminant $-2r^2<0$, and they are distinct and coprime. Thus $B_p\cong \mathbb{C}\times \mathbb{C}$, giving two irreducible components.

For $a=0$, $B_p=\mathbb{R}[x]/(x^4)$. It has one prime ideal, generated by the class of $x$, so one irreducible component. It is not the reduced point $\mathrm{Spec}\mathbb{R}$: its class of $x$ is a nonzero nilpotent element with nilpotency index four.

Finally write the irreducible quadratic as $q_p(y)=(y-z)(y-\overline{z})$, with $z\notin \mathbb{R}$. Over $\mathbb{C}$, $q_p(x^4)$ has four roots above $z$ and four above $\overline{z}$. They are all simple, because $z\ne 0$, and the two sets are disjoint. None is real. Conjugation pairs them into four distinct irreducible real quadratics. Hence $B_p\cong {\mathbb{C}}^4$ and this fibre has four irreducible components.

The real fourth-power fibre classification is:

All fibres in each row are isomorphic as affine schemes over $\mathbb{R}$. The rows are distinct: their numbers of irreducible components already distinguish them, and the zero fibre is additionally nonreduced. ThereforeThese are ordinary scheme components, not geometric components counted after complexifying the base. Counting only real roots would miss both the complex-residue factors and the nonreal closed points.

A nilpotent element belongs to every prime ideal and therefore to every maximal ideal, soFor the reverse containment, let $r$ be nonnilpotent. The localization of a ring $S=R[1/r]$ is nonzero: if $1=0$ there, some power of $r$ would annihilate $1$ in $R$. This is still a finite-type integer algebra, because it can be presented as $R[T]/(rT-1)$.

Choose a maximal ideal $\mathfrak{n}$ of $S$. Its residue field $L=S/\mathfrak{n}$ is a finite-type integer algebra, hence finite by part (a). Consider the image $F$ of $R$ in $L$. It is a finite integral domain containing $1$, so it is a field: multiplication by a nonzero element is injective on the finite set $F$, hence surjective, and therefore has an inverse in $F$.

The kernel $\mathfrak{m}$ of $R\to F$ is consequently maximal. The image of $r$ is nonzero, since $r$ became a unit in $S$ and remains a unit in its nonzero residue field. Thus $r\notin \mathfrak{m}$. Every nonnilpotent element can therefore be avoided by a maximal ideal, givingThis radical equality for finitely generated integer algebras uses part (a) to ensure that the contraction of the localized maximal ideal is maximal. Such contraction is not generally maximal for arbitrary localization of a ring; the finite-field image is the decisive additional step. The zero ring is immediate, with the intersection of its empty set of maximal ideals understood as the whole ring. No general theorem that integer algebras are Jacobson rings is being quoted.

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.

The Noether normalization lemma is valid over every field $k$, including finite fields: a nonzero finitely generated algebra $C$ contains elements $y_1,\dots ,y_d$ with algebraic independence over $k$ such that $C$ is finite as a module over the polynomial ring $k[y_1,\dots ,y_d]$. We prove it without an infinite-field hypothesis.

Write $C=k[a_1,\dots ,a_n]$ and induct on $n$. If the generators have algebraic independence, they already give a polynomial ring and the assertion is immediate. Otherwise choose a nonzero relation $F(a_1,\dots ,a_n)=0$. Choose an integer $M$ larger than every exponent in the monomials of $F$, and setIn $F(Y_1+T^M,Y_2+T^{M^2},\dots ,Y_{n-1}+T^{M^{n-1}},T)$, the largest power of $T$ has a nonzero coefficient in $k$. Indeed, a monomial with exponents $(e_1,\dots ,e_n)$ contributes top weight $e_n+{\sum }_{i<n}{e}_i{M}^i$. These weights are distinct by uniqueness of base-$M$ expansion, so only one monomial contributes the highest power. After rescaling, the substituted relation is monic in $T$.

Consequently $a_n$ is an integral element over $C^{\prime }=k[b_1,\dots ,b_{n-1}]$, and $C=C^{\prime }[a_n]$ is finite over $C^{\prime }$. Apply the induction hypothesis to $C^{\prime }$ and compose the finite module extensions. This proves Noether normalization by weighted substitutions. The induction reaches $n=0$, where $C=k$. For an integral domain $C$, taking fraction fields makes the resulting extension finite algebraic, so $d$ is the transcendence degree of $\mathrm{Frac}(C)/k$. More generally $d=\dim C$: integral extensions preserve Krull dimension, and a polynomial algebra in $d$ variables has Krull dimension $d$.

A useful bridge to the Hilbert Nullstellensatz is the Zariski lemma. If a field $E$ is a finitely generated algebra over $k$, normalization makes it finite and integral over a polynomial ring $P=k[y_1,\dots ,y_d]$. A subring over which a field is integral is a field: for nonzero $a\in P$, an integral equation for $a^{-1}\in E$, multiplied by $a^{r-1}$, expresses $a^{-1}$ as an element of $P$. Therefore $P$ must be a field. A polynomial ring in a positive number of variables is not a field, since a variable has no polynomial inverse. Hence $d=0$ and $E/k$ is a finite field extension. This proves the Zariski lemma.

Now let $k$ be an algebraically closed field. The Weak Hilbert Nullstellensatz says that every maximal ideal of $k[X_1,\dots ,X_n]$ is uniquely of the formTo prove it, the residue field $k[X]/\mathfrak{m}$ is a field generated as a $k$-algebra by the images of the variables. The Zariski lemma makes it finite algebraic over $k$, and algebraic closedness makes it $k$. Thus each $X_i$ has an image $a_i\in k$. The evaluation map has kernel the displayed ideal: subtracting the constant value of a polynomial expresses its difference as a combination of $X_i-a_i$. That kernel is maximal and contained in $\mathfrak{m}$, so equality holds. Conversely every evaluation kernel is maximal because its quotient is $k$. Uniqueness follows from the variable images. Every proper ideal is contained in a maximal ideal, so it has a common zero; equivalently, an ideal with no common zero is the whole ring.

For an ideal $I\subseteq k[X_1,\dots ,X_n]$, let $V(I)$ be its common-zero set and let $I(V(I))$ be all polynomials vanishing on that set. The Strong Hilbert Nullstellensatz statesThe inclusion $\sqrt{I}\subseteq I(V(I))$ follows because a field has no nonzero nilpotent elements. For the other inclusion, take $f$ vanishing on $V(I)$, with $f\ne 0$, and form the Rabinowitsch trick idealIt has no common zero: at a zero of $I$ the second generator has value one. The Weak Hilbert Nullstellensatz in $n+1$ variables gives $J=(1)$. Hence a finite identity has the form $1={\sum }_j{h}_j(X,T)g_j(X)+h(X,T)(1-Tf)$, with $g_j\in I$. Substitute $T=f^{-1}$ in the localization of a ring $k[X{]}_f$. Clearing the finitely many powers of $f$ occurring in denominators gives $f^r\in I$ for some $r$, so $f\in \sqrt{I}$. The case $f=0$ is immediate. This completes all three proofs. The algebraically closed hypothesis belongs to the two forms of the Hilbert Nullstellensatz; it was not needed for the Noether normalization lemma or the Zariski lemma.

All rings below are nonzero and commutative with identity; the zero ring is trivially Artinian. A primary ideal is proper. An ideal $Q$ is primary precisely when every zero divisor of $R/Q$ is a nilpotent element. Its radical of an ideal $\mathfrak{p}=\sqrt{Q}$ is prime: if $ab\in \sqrt{Q}$ and $a\notin \sqrt{Q}$, apply the primary property to $(ab{)}^r\in Q$ to obtain a power of $b$ in $Q$. In a Noetherian ring, finite generation of $\mathfrak{p}$ gives ${\mathfrak{p}}^s\subseteq Q$ for some $s$. For generators $u_i$ with $u_i^{s_i}\in Q$, one can take $s=1+{\sum }_i(s_i-1)$.

We first prove existence of primary decomposition. The ascending chain condition implies that every proper ideal is a finite intersection of irreducible ideals. Otherwise choose a maximal counterexample $I$ under inclusion. It is not irreducible, so $I=J\cap K$ with both $J,K$ strictly larger; their finite irreducible decompositions give one for $I$, a contradiction.

An irreducible ideal in a Noetherian ring is primary. Pass to the quotient and suppose zero is irreducible. If $xy=0$ with $x\ne 0$, the ascending chain of annihilators of $y^j$ stabilizes, say at $j=r$. Any element of $(x)\cap (y^r)$ can be written $z=ax=b{y}^r$. Then $yz=0$, so $b\in \mathrm{Ann}(y^{r+1})=\mathrm{Ann}(y^r)$ and $z=0$. Irreducibility of zero implies $(y^r)=0$, since $(x)\ne 0$. Thus $y$ is nilpotent. This proves irreducible ideals are primary in Noetherian rings, and hence the Lasker–Noether theorem.

An intersection of finitely many primary ideals with the same radical $\mathfrak{p}$ is again primary: if $ab$ belongs to all of them and $b\notin \mathfrak{p}$, the primary property forces $a$ into every component. Combining such components and removing redundant ones yields a minimal primary decompositionThe radicals are uniquely determined, but the components need not all be unique. Here is a proof identifying the invariant radicals as the associated primes of a module $R/I$.

Every nonzero module over a Noetherian ring has a nonzero element with a prime annihilator: maximize the annihilator of a nonzero element using the ascending chain condition. If $abm=0$ and $bm\ne 0$, maximality gives $\mathrm{Ann}(bm)=\mathrm{Ann}(m)$ and hence $am=0$. This is the maximal annihilator of a module element is prime argument. For a $\mathfrak{p}$-primary quotient $R/Q$, the radical of the annihilator of every nonzero element is $\mathfrak{p}$: it is contained in $\mathfrak{p}$ by the primary property and contains a power of $\mathfrak{p}$ because ${\mathfrak{p}}^s\subseteq Q$. Thus the only possible associated prime is $\mathfrak{p}$, and it does occur.

The diagonal injection $R/I\hookrightarrow {⨁}_{i}R/Q_i$ shows that every associated prime of $R/I$ is one of the ${\mathfrak{p}}_i$. Indeed, for an element whose annihilator $P$ is prime, that annihilator is the intersection of the finitely many component annihilators, all containing $P$. Their product is contained in $P$, so primality forces one of them to equal $P$; its component element is nonzero and has associated prime ${\mathfrak{p}}_i$. Conversely, irredundancy gives $b\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$. The nonzero cyclic submodule generated by $b+I$ embeds in $R/Q_i$. It has an associated prime, necessarily ${\mathfrak{p}}_i$, which is then an associated prime of $R/I$. We have proved the first uniqueness theorem for primary decomposition:Also the zero divisors on $R/I$ are exactly ${\bigcup }_i{\mathfrak{p}}_i$. For a scalar killing a nonzero element, extend its element annihilator to a maximal element annihilator containing it; the preceding argument gives an associated prime containing that scalar. The converse is immediate from the definition of an associated prime.

The minimal members of the set $\{{\mathfrak{p}}_i\}$ are the isolated primes of a primary decomposition. They are exactly the primes minimal over $I$: if a prime contains ${\bigcap }_i{\mathfrak{p}}_i$, it contains one ${\mathfrak{p}}_i$ by the product argument. A component belonging to an isolated prime is unique. Localize at ${\mathfrak{p}}_i$. Every other component becomes the whole ring, since its radical contains an element outside ${\mathfrak{p}}_i$ whose power lies in that component. A ${\mathfrak{p}}_i$-primary ideal contracts unchanged from this localization, because $sa\in Q_i$ with $s\notin {\mathfrak{p}}_i$ implies $a\in Q_i$. Hence the second uniqueness theorem for primary decomposition givesAn embedded primary component can vary. For example, in $k[x,y]$,Both second components are $(x,y)$-primary, since their quotients are dual-number algebras. For the second equality, reducing modulo $(x^2,y+x)$ makes $y=-x$; an element $xg$ vanishes exactly when $g$ has zero constant term, giving $(x^2,xy)$. The isolated component is $(x)$ and the embedded associated prime is $(x,y)$.

The relevance to Artinian rings is particularly sharp in Krull dimension zero. Suppose $R$ is Noetherian and all its primes are maximal. Apply a minimal primary decomposition to zero. Its radicals are distinct maximal ideals, so the components are pairwise comaximal ideals: each contains a power of its radical, and expanding $1=(a+b{)}^N$ for $a+b=1$ shows that powers of comaximal ideals remain comaximal. The Chinese remainder theorem gives the Artinian decomposition into local factorsEach factor has one maximal ideal, and that ideal is nilpotent. Its finite radical filtration has successive layers finitely generated over its residue field, hence of finite vector-space dimension. It follows that the factor, and therefore $R$, has a finite composition series of a module, so $R$ is Artinian. This proves the Noetherian dimension-zero criterion for an Artinian ring in this direction. Applied to $R/I$, it says that the quotient is Artinian exactly when all primes over $I$ are maximal. In that case no primary component is embedded, so all components are unique.

For completeness, the converse does not require initially assuming Noetherianity. In an Artinian ring, an Artinian domain is a field: stabilization of $(a^j)$ and cancellation gives an inverse to every nonzero $a$. Thus every prime is maximal. There are finitely many maximal ideals, since infinitely many distinct ones would give strictly descending finite intersections; comaximality guarantees strictness by the Chinese remainder theorem. Its nilradical $N$ is nilpotent. Indeed, its powers stabilize at $J=N^r$, with $J^2=J$. If $J\ne 0$, choose an ideal $C$ minimal among those satisfying $JC\ne 0$. Then $JC=C$, since $J(JC)=JC\ne 0$. Choose $x\in C$ with $Jx\ne 0$; minimality also gives $Rx=C$. Therefore $x=ax$ for some $a\in J\subseteq N$. But $1-a$ is a unit, contradicting $x\ne 0$. Hence $J=0$.

Now $R/N$ is a finite product of residue fields. Each $N^j/N^{j+1}$ is an Artinian module over that product, and therefore a finite direct sum of finite-dimensional vector spaces: an infinite-dimensional vector space would have a strictly descending chain of subspaces. The finite filtration by powers of $N$ makes $R$ a module of finite composition length, in particular Noetherian. Combining both directions yieldsThus primary decomposition separates the local pieces of a zero-dimensional ring, while their nilpotent maximal ideals record the multiplicities that the reduced set of primes alone does not detect.

Take $\mathcal{A}$ to be the category of finitely presented commutative rings with identity, and put $\mathcal{C}={\mathcal{A}}^{\mathrm{op}}$. In its presheaf topos, the tautological ring $U(A)=A$ is generic. The additional domain axioms are coherent sequents, so they are imposed by a quotient-theory coverage on $\mathcal{C}$.

Concretely, declare the zero ring covered by the empty family. For every finitely presented $A$ and elements $a,b\in A$ with $ab=0$, declare the two opposite quotient arrows associated withto be a covering family at $A$. These quotient rings are finitely presented. Pullback and transitivity generate a Grothendieck topology $J$ from these families. The empty cover forbids $0=1$; the two quotient covers make every zero product locally have a zero factor. Conversely, any internal integral domain satisfies exactly the continuity conditions prescribed by these generating covers. Thusand its generic domain is the associated sheaf $K=a_{J}U$, with the ring operations transported through the left-exact sheaf reflector.

This coverage is not standard, meaning not all representables are sheaves; in modern terminology it is not subcanonical. For an explicit obstruction, use $A=\mathbb{Z}/4\mathbb{Z}$ and $a=b=2$. The two quotient arrows are the same map $A\to \mathbb{Z}/2\mathbb{Z}$, so their generated sieve is a singleton cover. Consider the representable on $\mathcal{C}$ corresponding to $R=\mathbb{Z}[t]$:Its distinct sections $t\mapsto 0$ and $t\mapsto 2$ become equal after restriction to $\mathbb{Z}/2\mathbb{Z}$. Hence this representable is not even separated for $J$. The nilpotent element has to disappear in the generic domain, which explains this failure of standardness.