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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.

For nilpotent elements $a,b$ of the commutative ring, choose positive integers $r,s$ with $a^r=b^s=0$. The binomial theorem giveseach term has either $j\ge r$ or $r+s-1-j\ge s$. Also $(-a{)}^r=0$, $0\in N$, and $(ta{)}^r=t^r{a}^r=0$ for every $t\in R$. Thus $N$ is an additive subgroup stable under multiplication by arbitrary ring elements:Commutativity is essential to the simple binomial argument; an arbitrary sum of nilpotent elements in a noncommutative ring need not be nilpotent.

A nilpotent element belongs to every prime ideal: if $a^r=0\in P$, repeated primality forces $a\in P$. Therefore $N\subseteq {\bigcap }_{P}P$.

Conversely, if $a\notin N$, none of its positive powers is zero. The multiplicative subset $S=\{1,a,a^2,\dots \}$ is disjoint from $(0)$. Part (b) supplies a prime ideal disjoint from $S$, hence not containing $a$. Thus $a$ is outside the intersection, proving the nilradical as the intersection of prime ideals identity:In the zero ring there are no prime ideals; the empty intersection is interpreted as the whole ring, which is also its nilradical. For nonzero rings this argument proves existence of a prime ideal whenever a nonnilpotent element is available.

Take the dual numbers $A=k[\epsilon ]/({\epsilon }^2)$ over any field $k$, let $B=k$, and let $\alpha$ send $\epsilon$ to zero. Its kernel is the nonzero ideal $(\epsilon )$, but both spectra of rings have exactly one point: every prime ideal contains the nilpotent element $\epsilon$, and the only prime of $A$ is $(\epsilon )$. Consequently the induced map is a homeomorphism for the Zariski topology, although it is not an isomorphism of schemes:More generally, quotienting by a nilpotent ideal gives a nilpotent thickening with unchanged underlying topology. The structure sheaf retains information that the topology alone cannot recover.

The Jacobson radical is $J(A)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals of $A$; equivalently, it is the intersection of the annihilators of all simple modules. It is a two-sided ideal. A projective module has the lifting property against every surjective R-module homomorphism. A finitely generated module is a projective module precisely when it is a direct summand of a finitely generated free module. An indecomposable module is nonzero and admits no direct sum decomposition into two nonzero submodules.

To calculate the top of an indecomposable projective module, use the right Artinian ring hypothesis, namely the descending chain condition on right ideals. The Hopkins-Levitzki theorem gives finite composition length of the right regular module, hence of its submodule $P$. Also $J=J(A)$ is a nilpotent ideal, and $A/J$ is a semisimple ring. Thus the quotient module $\overline{P}=P/PJ$ is a semisimple module; it is nonzero, since $P=PJ$ would imply $P=P{J}^m=0$ for sufficiently large $m$.

Suppose $\overline{P}$ were not a simple module. A nontrivial direct sum decomposition of this semisimple module would give an idempotent $\alpha \in {\mathrm{End}}_A(\overline{P})$ that is neither zero nor the identity. Writing $\pi :P\to \overline{P}$, the lifting property of the projective module $P$ gives $\beta \in {\mathrm{End}}_A(P)$ with $\pi \beta =\alpha \pi$. The Fitting lemma for an indecomposable module of finite composition length says that $\beta$ is either invertible or a nilpotent element. Its induced map $\alpha$ would then be invertible or a nilpotent element, respectively. Neither is possible for a nontrivial idempotent. Therefore $P/PJ(A)$ is simple. This argument does not assume that an embedded projective module automatically splits off from the ambient module.

A block of an Artinian algebra is a nonzero two-sided direct summand $Ae$ determined by a primitive central idempotent $e$: $e$ cannot be written as a sum of two nonzero orthogonal central idempotents. Its identity is $e$. For a finite-dimensional associative algebra, the blocks of an Artinian algebra give its unique decomposition as a finite product of indecomposable algebras, or equivalently as a direct sum of two-sided ideals.

For the block of S3 in characteristic three, put $A=k{S}_3$, $r=(123)$, $s=(12)$ and $a=r-1$. The group algebra has basis $r^i,r^{i}s$ for $0\le i<3$. In characteristic three,Consequently $I=aA$ is a two-sided nilpotent ideal, $I^3=0$, and $A/I\cong k{C}_2\cong k\times k$. A nilpotent ideal lies in the Jacobson radical, and a quotient that is a semisimple ring forces the reverse inclusion. HenceDefine orthogonal idempotents $e_{+}=(1+s)/2$ and $e_{-}=(1-s)/2$. They sum to one, so the right regular module decomposes asThe two summands are the indecomposable projectives of S3 in characteristic three. Each direct summand is a projective module, with basis $e_{\pm },e_{\pm }r,e_{\pm }{r}^2$. Its quotient module modulo multiplication by $J(A)$ is one-dimensional: the trivial representation for $e_{+}A$, and the sign representation for $e_{-}A$. Each is an indecomposable module, since two nonzero direct summands would each have nonzero quotient module modulo $J(A)$, contradicting its one-dimensional top. For additional detail, the successive factors of the radical series of a module are the trivial representation, sign representation, trivial representation for $e_{+}A$, and the sign representation, trivial representation, sign representation for $e_{-}A$. Indeed, modulo $J^2$ the relation $sa=-as+a^{2}s$ reverses the $s$-sign, whereas $s$ commutes with $a^2$.

The center of an associative algebra is spanned by the conjugacy class sums $1$, $r+r^2$, and $(1+r+r^2)s$. Since $r+r^2=-1+a^2$ and $1+r+r^2=a^2$, this center of an associative algebra isIts vector subspace $k{a}^2\oplus k{a}^{2}s$ is a square-zero ideal. If $d1+n$ is a central idempotent, then $d^2=d$ and $(2d-1)n=0$. Thus $d=0$ or $1$ and $n=0$. There is no nontrivial central idempotent, so the whole group algebra is its single block. The two three-dimensional projective modules above are a decomposition of the regular module, not two blocks of an Artinian algebra.

We first establish the nilradical as the intersection of prime ideals:Every nilpotent element belongs to every prime ideal. In the other direction, if $a$ is not nilpotent, the multiplicative subset $T=\{1,a,a^2,\dots \}$ avoids zero. By the Zorn lemma, there is an ideal $J$ maximal among those disjoint from $T$. It is a prime ideal: if $uv\in J$ and neither $u$ nor $v$ belongs to $J$, then $J+(u)$ and $J+(v)$ each meet $T$. Multiplying elements in these two intersections gives an element of $T$ lying in $J$, a contradiction. Since $a\notin J$, this proves the identity. The identity also holds for the zero ring, with the intersection over an empty family interpreted as the whole ring.

A Zariski-closed set $V_R(I)$ contains the image of $f$ exactly whenThe last equality follows because $\phi (r)$ is a nilpotent element exactly when $r^n\in \ker \phi$ for some $n\ge 1$. Consequently the smallest Zariski-closed set containing the image, namely its closure, isThis is all of $\mathrm{Spec}R$ precisely when the kernel of a ring homomorphism is contained in every prime ideal of $R$. Hence

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

False. In the polynomial ring $R=k[x,y]$ over a field, takeThe ideal $P$ is a maximal ideal, with quotient ring $R/P\cong k$. AlsoIn this last ring, an element with nonzero constant term is a unit, and an element with zero constant term is a multiple of $y$ and has square zero. Thus every zero divisor is a nilpotent element. This is equivalent to $I$ being a primary ideal: a zero product in the quotient ring with one nonzero factor forces the other factor to be nilpotent. Its radical of an ideal is $P$, so $I$ is $P$-primary.

However, $y\notin I$, so $I\ne P$. Moreover $x\in I$ while $x\notin P^2$, since every polynomial in $P^2$ has no term of total degree one. For every $n\ge 2$, $P^n\subseteq P^2$, so $I\ne P^n$. Hence

True. An irreducible ideal $I$ is a proper ideal such that $I=J\cap K$ forces $I=J$ or $I=K$. Passing to the quotient ring $A=R/I$ turns this into the assertion that zero cannot be the intersection of two nonzero ideals. Also $A$ is Noetherian: ascending chains of ideals lift to ascending chains of ideals containing $I$ in $R$.

Suppose $ab=0$ in $A$, and suppose $b$ is not a nilpotent element. The ascending chain condition makes the chain of annihilatorsstabilize. Choose $n\ge 1$ with $\mathrm{Ann}(b^n)=\mathrm{Ann}(b^{n+1})$. We claimIf $z=ua=v{b}^n$ lies in this intersection, then $bz=uab=0$, so $v{b}^{n+1}=0$. The equality of annihilators implies $v{b}^n=0$, hence $z=0$. Since $b$ is not nilpotent, $(b^n)\ne 0$. Irreducibility of zero therefore forces $(a)=0$, or $a=0$.

We have proved that $ab=0$ and $a\ne 0$ force $b$ to be a nilpotent element. This says precisely that zero is a primary ideal in $A$, and therefore that $I$ is a primary ideal in $R$.

If its radical of an ideal is to be denoted by a prime ideal, this requires no extra hypothesis: for any primary ideal $I$, $uv\in \sqrt{I}$ and $u\notin \sqrt{I}$ imply $u^n{v}^n\in I$ with $u^n\notin I$, so some power of $v$ lies in $I$. Thus $v\in \sqrt{I}$.

All Topological K-theory groups below are complex and are indexed modulo two using Bott periodicity. For $n\ge 1$, Complex K-theory of a sphere givesFor an even-dimensional sphere, choose a generator $u$ of Reduced topological K-theory. Its square vanishes: the Chern character sends $u$ to a top-degree cohomology class, whose square is zero, and the Chern character is injective on this torsion-free abelian group. Thus the ring answers areThe exceptional zero-dimensional sphere consists of two points: $K^0(S^0)=\mathbb{Z}\times \mathbb{Z}$ with coordinatewise multiplication and $K^{-1}(S^0)=0$. In particular its reduced generator is idempotent rather than square-zero.

For the Euler characteristic identity, induct over the cells of a finite CW complex. The starting CW complex consisting of $v$ vertices has $K^0\cong {\mathbb{Z}}^v$, $K^{-1}=0$, and $\chi =v$. If $X$ is obtained from $A$ by attaching one $d$-dimensional cell, the quotient is $S^d$, and the K-theory six-term exact sequence isThese abelian groups are finitely generated by induction. Tensoring the exact sequence with $\mathbb{Q}$ preserves exactness. The alternating sum of dimensions in a cyclic six-term exact sequence is zero, soTherefore

There is a genuine convention issue in the nilpotence assertion. With the usual definition of Reduced topological K-theory as the kernel of restriction to one basepoint, the assertion needs $X$ to be connected. For $X=S^0$ based at its first point, $(0,1)$ lies in ${\tilde{K}}^0(X)$ and satisfies $(0,1{)}^m=(0,1)$ for every $m\ge 1$. Thus it is not a nilpotent element. For an arbitrary finite CW complex, the correct assertion uses the rank map in topological K-theory on every connected component:For a connected space, $I(X)={\tilde{K}}^0(X)$.

Here is an induction proving the corrected assertion. On the vertices, $I(X)=0$. For a positive-dimensional cell attachment $A\subset X$, the ideal $J=\ker (K^0(X)\to K^0(A))$ is square-zero. Indeed, lift two elements of $J$ to relative topological K-theory $K^0(X,A)$. Their product is induced by the reduced diagonalThis map is null-homotopic, since $d<2d$ and $S^{2d}$ is $(2d-1)$-connected. The relative product in topological K-theory is therefore zero, giving $J^2=0$. If $a\in I(X)$, its restriction belongs to $I(A)$, so induction gives $(a{|}_A{)}^m=0$ for some $m$. Then $a^m\in J$ and $a^{2m}=0$. We have proved every Topological K-theory class of rank zero on every connected component is a nilpotent element, and hence the requested result for connected $X$.

The first isomorphism theorem for rings states that a ring homomorphism $\varphi :R\to S$ induces an isomorphismBecause $J$ is an ideal, sums and products of elements of $R+J$ remain in $R+J$, so it is a subring. The surjective homomorphismhas kernel $R\cap J$. The theorem givesEvaluation at $-2$ givesso this quotient is a field of characteristic zero. The other quotient isIt has characteristic $3$ but is not a field, since the nonzero class of $X-1$ is nilpotent.

Two nonisomorphic punctual schemes areTheir underlying spaces each have one point, but the second has a nonzero nilpotent element and the first is reduced.

After translating their common support to the origin, punctual closed subschemes of ${\mathbb{A}}_k^1$ correspond to ideals of $k[t]$ whose radical is $(t)$. Since $k[t]$ is a principal ideal domain, each such ideal is $(t^r)$ for a unique $r\ge 1$. Its coordinate ring has basis $1,t,\dots ,t^{r-1}$ and hence dimension $r$. Equal dimensions force equal exponents, so $Z$ and $Z^{\prime }$ are in fact the same closed subscheme after the common coordinate choice, and in particular are isomorphic.

In ${\mathbb{A}}_k^2$, the idealsdefine distinct punctual closed subschemes supported at the origin. Both quotient rings have dimension two over $k$, with bases $1,y$ and $1,x$ respectively. Thus they have equal-dimensional global-section spaces despite being distinct embedded closed subschemes.

Pass to $A=R/I$. It is enough to prove that the zero ideal of $A[t]$ is primary. The zero ideal of $A$ is primary, so every zero divisor of $A$ is nilpotent element.

Suppose $fg=0$ in $A[t]$ with $g\ne 0$. By McCoy theorem, some nonzero $a\in A$ satisfies $af=0$. Hence every coefficient of $f$ is a zero divisor and therefore nilpotent. There are only finitely many coefficients, so the ideal they generate is nilpotent; consequently some power of $f$ is zero. This proves that $(0)$ is primary in $A[t]$, and the coefficientwise quotient of a polynomial ringshows that $I^e$ is primary.

Write $X=\mathrm{Spec}B$, $Y=\mathrm{Spec}A$, and let the morphism correspond to a homomorphism $\phi :A\to B$. Put $I=\ker \phi$ andThe injection $A/I\hookrightarrow B$ gives $X\to Z$, and the quotient $A\to A/I$ gives a closed immersion $Z\hookrightarrow Y$, so $f$ factors through $Z$.

If $f$ factors through another closed subscheme $Z^{\prime }=\mathrm{Spec}(A/J)$, then $J\subseteq \ker \phi =I$. The resulting quotient $A/J\to A/I$ induces the unique factorization $Z\to Z^{\prime }\to Y$. Thus $Z$ is the scheme-theoretic image.

It remains to identify its underlying set. A principal open subscheme $D(a)\subseteq \mathrm{Spec}A$ misses $f(X)$ exactly when $D(\phi (a))$ is empty, equivalently when $\phi (a)$ is a nilpotent element. Hence the ideal of functions vanishing set-theoretically on $f(X)$ has radical $\sqrt{\ker \phi }$. The closure is thereforeas required.