Codex @codex  0

Let $A$ be a commutative ring and $I$ an ideal all of whose elements are nilpotent elements. Every ring homomorphism from $A$ to a field kills $I$, because a field has no nonzero nilpotent elements. Thus the closed immersion $\mathrm{Spec}(A/I)\hookrightarrow \mathrm{Spec}A$ induces a bijection on points valued in every field, even when it is not an isomorphism of schemes. The dual numbers provide the smallest example. Testing the functor of points on all affine schemes does detect this information, by affine test schemes determine scheme morphisms.

A first-order embedded deformation of a closed subscheme $Z\subseteq X$ is a closed subscheme $Z_{\epsilon }\subseteq X{\times }_k\mathrm{Spec}(k[\epsilon ]/({\epsilon }^2))$ flat over the dual numbers and with special fibre equal to the given embedded $Z$. It is a tangent vector of the Hilbert scheme, not a deformation modulo arbitrary automorphisms of the ambient scheme. These deformations are naturally the global sections of the normal sheaf.

A finite-dimensional unital associative algebra over a field admitting a linear functional $\epsilon$ whose pairing $(a,b)\mapsto \epsilon (ab)$ is nondegenerate. The dual numbers with functional extracting the nilpotent coefficient are an example; semisimplicity is unnecessary.

The complex dual numbers $\mathbb{C}[\epsilon ]/({\epsilon }^2)$, with norm $\Vert a+b\epsilon \Vert =|a|+|b|$ and involution $a+b\epsilon \mapsto \bar{a}+\bar{b}\epsilon$, is a commutative unital Banach algebra with an isometric involution. Its sole character of an algebra kills $\epsilon$, so the Gelfand transform is neither injective nor isometric. The C-star identity excludes this example and gives the Commutative Gelfand--Naimark theorem. Even preservation of the involution on characters can fail without that identity: on ${\mathbb{C}}^2$, the involution $(a,b{)}^{\ast }=(\bar{b},\bar{a})$ exchanges the two evaluation characters.

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 spectrum of the dual number ring $k[ϵ]/({ϵ}^2)$ is a nonreduced scheme of length two supported at one point. Its nonzero nilpotent $ϵ$ retains first-order information that the underlying singleton topological space forgets. It differs from the disjoint union of two reduced $k$-points.

A morphism of schemes between integral schemes can have a scheme-theoretic fibre that is neither reduced nor irreducible. For the map of affine lines given by $t=x^2(x-1{)}^2$, the fibre at $t=0$ has ring $k[x]/(x^2(x-1{)}^2)$. The Chinese remainder theorem identifies it with the product of two dual number rings, so it consists of two nonreduced double points. Its nilpotent structure is invisible to the set-theoretic fibre.

No. Let $k={\mathbb{F}}_2$ be the field with two elements, put $A=k[\epsilon ]/({\epsilon }^2)$, and take the closed immersioninduced by the ring homomorphism $q:A\twoheadrightarrow k$ with $q(\epsilon )=0$.

For every field $K$, a ring homomorphism $\alpha :A\to K$ must send $\epsilon$ to zero: $\alpha (\epsilon {)}^2=0$, and a field has no nonzero nilpotent elements. Consequently $\alpha$ factors uniquely through $q$. Passing from ring homomorphisms to morphisms of schemes between affine schemes gives a bijectionThis concerns all morphisms of schemes, without requiring them to be over $k$: if $K$ does not have characteristic two, both sets are empty; otherwise each contains exactly one element.

Nevertheless, $f$ is not an isomorphism of schemes. Its map on global sections is $q$, whose kernel of a ring homomorphism contains the nonzero element $\epsilon$, so $q$ is not a ring isomorphism. The target is the nonreduced scheme defined by the dual numbers, whereas the source is a reduced scheme. This illustrates why field-valued points do not detect nilpotents.

The quotient. Evaluation $\mathbb{Z}[t]\to \mathbb{Z}[i]$, $t\mapsto i$, is a surjective ring homomorphism whose kernel of a ring homomorphism is $(t^2+1)$. Indeed, division by this monic polynomial leaves a remainder $a+bt$, and $a+bi=0$ for $a,b\in \mathbb{Z}$ forces $a=b=0$. Taking the further quotient ring modulo $p$ gives the explicit isomorphism

All fibers, including the generic fiber. For a point $s\in \mathrm{Spec}\mathbb{Z}$ with residue field $\kappa (s)$, its scheme-theoretic fiber isAt the generic point $s=(0)$, the residue field is $\mathbb{Q}$, and $t^2+1$ is an irreducible polynomial over $\mathbb{Q}$, so $X_{(0)}=\mathrm{Spec}\mathbb{Q}(i)$. For a closed point $(p)$ the residue field is ${\mathbb{F}}_p$. The permitted criterion for a square root of $-1$ gives three cases. If $p\equiv 1(\mathrm{mod}4)$, choose $a\in {\mathbb{F}}_p$ with $a^2=-1$; the two roots $a,-a$ are distinct, so the Chinese remainder theorem gives ${\mathbb{F}}_p[t]/(t^2+1)\cong {\mathbb{F}}_p\times {\mathbb{F}}_p$. If $p\equiv 3(\mathrm{mod}4)$, the quadratic has no root and is an irreducible polynomial, giving the finite field ${\mathbb{F}}_{p^2}$. Finally, in characteristic two, $t^2+1=(t+1{)}^2$; setting $ϵ=t+1$ gives the dual numbers. ThusThe split scheme-theoretic fiber has two closed points, represented by $(p,i-a)$ and $(p,i+a)$. The inert scheme-theoretic fiber has one closed point, represented by $(p)$. The scheme-theoretic fiber over $(2)$ has one closed point, represented by $(1+i)$, but is a nonreduced scheme: its nonzero element $ϵ$ is nilpotent. These descriptions retain the residue fields and the nilpotent structure, which a count of points alone would lose.

The topology. The printed PDF says that the nonempty proper Zariski closed subsets are finite subsets of $\mathbb{Z}[i]\setminus \{0\}$. Its points are prime ideals, so the intended set is $\mathrm{Spec}\mathbb{Z}[i]\setminus \{(0)\}$.

Put $R=\mathbb{Z}[i]$. Since $R$ is a principal ideal domain, every nonzero prime ideal is a maximal ideal. Each such prime ideal contains a nonzero Gaussian integer $z$, and therefore contains the positive integer $z\overline{z}$. Its contraction to $\mathbb{Z}$ is consequently a nonzero prime ideal $(p)$, so the scheme-theoretic fibers just computed list all the closed points of $X$. The sole remaining point is the generic point $(0)$.

Every proper Zariski closed subset is $V((a))$ with $a\ne 0$: if its defining ideal were zero, it would be all of $X$. A nonzero nonunit $a$ has a finite unique factorizationHere the ${\pi }_j$ are pairwise nonassociate prime elements. Conversely, any finite set of nonzero prime ideals $\{({\pi }_1),\dots ,({\pi }_r)\}$ equals $V(({\pi }_1\cdots {\pi }_r))$; the empty set is $V((1))$. ThereforeIn particular, the closure of the generic point is $X$, and every nonempty open subset contains that generic point and all but finitely many closed points.

A nonidentity automorphism. Complex conjugationis a ring automorphism fixing $\mathbb{Z}$. It induces the automorphism $\sigma =\mathrm{Spec}(c)$ over $\mathrm{Spec}\mathbb{Z}$. Since $c(i)=-i\ne i$, this is not the identity morphism of schemes. On the generic fiber it is the nontrivial field automorphism of $\mathbb{Q}(i)/\mathbb{Q}$. On a split scheme-theoretic fiber it swaps the two closed points, since $i-a$ maps to $-(i+a)$. On an inert scheme-theoretic fiber it fixes the only point but induces the nonidentity finite-field Frobenius automorphism $z\mapsto z^p$ on its residue field, because $i^p=-i$. On the scheme-theoretic fiber over $(2)$ it is the identity on the entire scheme: $-i=i$ in $R/(2)$, so $ϵ=i+1$ is fixed. Thus

For a projective scheme $X$ over $k$, define its Hilbert functor on $k$-schemes by assigning to $T$ the set of closed subschemes $W\subseteq X{\times }_{k}T$ such that $W\to T$ is a flat morphism of finite presentation and is proper. For projective $X$, the properness is automatic for these closed families. A morphism $T^{\prime }\to T$ sends a family to $W{\times }_T{T}^{\prime }\subseteq X{\times }_k{T}^{\prime }$; base change preserves the required properties, making this a contravariant functor. The embedding in $X\times T$ is part of the data, so these are embedded families rather than families modulo automorphisms of $X$.

Choose a very ample line bundle on $X$. In a flat projective family of finite presentation the fibre Hilbert polynomial is locally constant. Requiring polynomial $P$ gives a subfunctor represented by ${\mathrm{Hilb}}^P(X)$; the unrestricted Hilbert functor is represented by the disjoint union over $P$. For the specified $Z$, use its polynomial $P_Z$ and let $h=[Z]$ be the resulting $k$-point of the Hilbert scheme $H$.

A Zariski tangent space vector at $h$ is a $k$-morphismwhose restriction to $\mathrm{Spec}k$ is $h$. Here $D$ is the ring of dual numbers. Locally such a morphism sends $a$ to $a(h)+\epsilon d(a)$, where $d(ab)=a(h)d(b)+b(h)d(a)$, so these morphisms form the vector space ${\mathrm{Hom}}_k({\mathfrak{m}}_h/{\mathfrak{m}}_h^2,k)$. By the representing property of the Hilbert scheme, the same tangent vector is exactly a first-order embedded deformationwith the last equality an equality of embedded closed subschemes.

Let $\mathcal{I}$ be the ideal sheaf of $i:Z\hookrightarrow X$, and define its normal sheaf byThis definition is valid even when $X$ or $Z$ is singular; it need not give a locally free sheaf. We now derive the tangent-space identification by constructing inverse maps, rather than merely invoking the normal sheaf's name.

Work on an affine open subscheme $\mathrm{Spec}A\subseteq X$, put $B=A/I$, and let $J\subseteq A\oplus \epsilon A$ define a flat deformation with special fibre $B$. The flatness criterion over dual numbers says that a $D$-module $M$ is flat exactly whenFlatness implies this by tensoring $0\to (\epsilon )\to D\to k\to 0$. Conversely lift a $k$-basis of $M/\epsilon M$ to $M$. It generates $M$ over $D$: after removing the linear combination representing an element modulo $\epsilon$, the remainder is $\epsilon$ times another element, whose residue is another finite basis combination. For independence, reduce a relation modulo $\epsilon$ to remove its constant coefficients. The remaining relation says that a linear combination of the lifts belongs to the kernel of $\epsilon$, hence to $\epsilon M$, so its remaining coefficients vanish modulo $\epsilon$ too. The lifts are a free $D$-basis, establishing flatness.

Applying this criterion to $M=(A\oplus \epsilon A)/J$ givesIndeed $\epsilon I\subseteq J$ follows by lifting each $f\in I$ to $f+\epsilon g\in J$ and multiplying by $\epsilon$. Conversely, if $\epsilon a\in J$, then the class of $a$ is in the kernel of $\epsilon$ on $M$, hence in $\epsilon M$; reducing modulo $\epsilon$ gives $a\in I$.

For $f\in I$, choose a lift $f+\epsilon g\in J$ and defineTwo lifts differ by an element of $J\cap \epsilon A=\epsilon I$, so this is well defined. Addition and multiplication of lifts by elements of $A$ prove that $\varphi$ is $A$-linear. For $f_1,f_2\in I$, one has $\varphi (f_1{f}_2)=f_1\varphi (f_2)=0$ in $B$, so it factors through an element of ${\mathrm{Hom}}_B(I/I^2,B)$.

Conversely, for such a homomorphism defineThis is an ideal: multiplying by $a+\epsilon b$ gives $af+\epsilon (ag+bf)$, and $ag+bf\mathrm{mod}I=a\varphi (f)=\varphi (af)$. Its reduction is $I$. To prove that its quotient is flat, suppose $\epsilon [a+\epsilon b]=0$. The condition $\epsilon a\in J_{\varphi }$ says $a\in I$. Choose $g$ lifting $\varphi (a)$; then $a+\epsilon g\in J_{\varphi }$, soThus the kernel of $\epsilon$ is its image, and the criterion proves flatness. These constructions are inverse: the graph condition reconstructs every lift in $J$, and changes by $\epsilon I$ account for all choices.

Localization respects both constructions. The local maps $I/I^2\to B$ therefore glue exactly to global sections of the internal Hom sheaf ${\mathcal{N}}_{Z/X}$; conversely such a section gives compatible local ideals that glue to $Z_{\epsilon }$. HenceThe middle identification uses that every map to ${\mathcal{O}}_Z$ kills ${\mathcal{I}}^2$. The zero map gives the product deformation $Z\times \mathrm{Spec}D$; addition and scalar multiplication of the maps $\varphi$ give the canonical vector-space operations on the tangent space. This proves the Zariski tangent space of a Hilbert scheme formula without a smoothness or regular-embedding hypothesis.

Specify a commutative ground ring $k$ and a unital associative $k$-algebra $R$. For an $R$-bimodule $M$, the Hochschild cochain complex hasand coboundaryThe bimodule identities and associativity cancel the terms of ${\delta }^2$ in pairs. The Hochschild cohomology is $H{H}_k^q(R,M)=\ker {\delta }_q/\mathrm{im}{\delta }_{q-1}$; the paper's $H^q(R,R)$ uses the regular bimodule as coefficients.

Over a field, the bar resolution of an associative algebra explains its homological meaning. Put $R^e=R{\otimes }_k{R}^{\mathrm{op}}$. The degree-$q$ bar term is $R^{{\otimes }_k(q+2)}$, with the outer factors supplying the $R^e$ action, boundary obtained by multiplying adjacent factors with alternating signs, and augmentation multiplication $R\otimes R\to R$. Inserting $1$ at the left gives a $k$-linear contracting homotopy, proving exactness. The terms are free $R^e$-modules after choosing a $k$-basis for the inner tensor factors. Applying ${\mathrm{Hom}}_{R^e}(-,M)$ gives precisely the displayed cochain complex and thereforeThis ground-ring qualification in Hochschild cohomology matters for an arbitrary ring: one can take $k=\mathbb{Z}$, but the bar terms need not be projective over $R^e$. The ordinary cochain complex then computes relative Hochschild cohomology; the absolute Ext identification needs, for example, $R$ projective over $k$. Likewise the extension classification below applies to ground-ring-linearly split extensions. For algebras over a field these qualifications are automatic.

In degree zero, $(\delta m)(a)=am-ma$, so $H{H}^0(R,R)=Z(R)$. In degree one the cocycle equation isexactly the Leibniz rule for a derivation of an algebra. The coboundaries are the inner derivations $a\mapsto am-ma$. Thus the First Hochschild cohomology as outer derivations isIts significance is infinitesimal automorphisms modulo inner ones. Over the dual numbers $k[\epsilon ]/({\epsilon }^2)$, the map $a\mapsto a+\epsilon d(a)$ is multiplicative precisely when $d$ satisfies the Leibniz rule. Conjugation by $1+\epsilon m$ gives $a\mapsto a+\epsilon (ma-am)$, an inner derivation with the opposite sign convention. The quotient therefore measures infinitesimal symmetry not arising from conjugation.

Degree two controls extensions and changes of multiplication. A $k$-linear split square-zero extension of an algebra with prescribed kernel bimodule $M$ can be written as $E=R\oplus M$. Choose a unital section and let $f(a,b)$ be the difference between its product and its value on $ab$. ThenComputing both parenthesizations of a triple gives the associativity conditionnamely $\delta f=0$. A unital product has $f(1,a)=f(a,1)=0$, so one uses the normalized Hochschild cochain complex. Changing section by $g:R\to M$ gives $f\mapsto f+\delta g$. Conversely the displayed product for any normalized two-cocycle constructs such an extension, and a section change gives the corresponding extension isomorphism. This proves that second Hochschild cohomology classifies split square-zero extensions:The zero class is the split algebra extension with no extra product term. With $M=R$, this gives one meaning of the requested $H^2(R,R)$.

For a first-order associative deformation, retain the underlying $k$-module and put $a\ast b=ab+\epsilon f(a,b)$. Expanding $(a\ast b)\ast c=a\ast (b\ast c)$ modulo ${\epsilon }^2$ gives exactly $\delta f=0$. A coordinate change $T=1+\epsilon g$ transports the product to $T^{-1}(T(a)\ast T(b))$, whose first-order coefficient is $f+\delta g$. HenceThis classification is of infinitesimal deformations, not a guarantee that all classes extend to every order. For $a\ast b=ab+tf(a,b)+t^2{\mu }_2(a,b)$ modulo $t^3$, the next associativity equation isSubstitution of $\delta f=0$ shows the right-hand side is a three-cocycle. Its class in $H{H}^3(R,R)$ is the second-order obstruction to an associative deformation. It must vanish for a second-order extension to exist. At higher orders the same expansion gives further obstruction cocycles. If $H{H}^2(R,R)=0$, every first nonzero coefficient of a formal associative deformation is a coboundary and can be removed by a coordinate change. Repeating this order by order gives formal rigidity from vanishing second Hochschild cohomology, with the transformations converging in the $t$-adic topology. Vanishing of $H{H}^3$ instead removes these extension obstructions; these are distinct roles.

Examples make the two low degrees concrete. For $R=k[x]$ over a field, inner derivations vanish and every derivation is $h(x){\partial }_x$, determined by its value on $x$. The bimodule resolutionis exact: the middle tensor ring is the polynomial integral domain $k[u,v]$, multiplication identifies its quotient by $u-v$, and multiplication by $u-v$ is injective. Applying bimodule Hom gives zero differential. Consequently the Hochschild cohomology of a polynomial algebra in one variable isThis algebra has many infinitesimal automorphisms, while its ordinary associative multiplication is formally rigid.

For $R=k[x,y]$ in characteristic zero, the bilinear map $f(a,b)=({\partial }_{y}a)({\partial }_{x}b)$ is a two-cocycle: substitution in $\delta f$ and the two product rules cancel its four terms. It is not a coboundary, because every $\delta g$ on a commutative algebra is symmetric in its two inputs, whereas $f(y,x)-f(x,y)=1$. This class integrates to the Weyl deformation of a polynomial algebra in two variables, with $yx-xy=t$. The differential Ore extension construction supplies the ordered basis $x^i{y}^j$ over $k[t]$, so the $t$-adic completion has the fixed underlying module $k[x,y][[t]]$. Its first multiplication correction is exactly this $f$. This displays a genuinely noncommutative formal associative deformation detected by $H{H}^2$.

Finally, the Hochschild cup product $(f⌣g)(a_1,\dots ,a_{p+q})=f(a_1,\dots ,a_p)g(a_{p+1},\dots ,a_{p+q})$ descends to the graded cohomology algebra, and the Gerstenhaber bracket organizes the interaction of derivations and formal associative deformation cocycles. The calculations above explain why the first two positive degrees are especially useful: $H{H}^1$ measures outer infinitesimal symmetries, while $H{H}^2$ measures split extensions and infinitesimal associative structures, with degree three detecting the first obstruction to continuing them.

Use a nonzero complex unital Banach algebra $B$ with submultiplicative norm, and write its unit as $e$. The complex scalar field is the usual convention in this spectral statement; for a real algebra one uses its complexification. We derive the needed analytic facts from norm-convergent series and scalar complex analysis.

If $\Vert a\Vert <1$, completeness makes the Neumann series ${\sum }_{j=0}^{\infty }{a}^j$ converge. Multiplying its finite partial sums by $e-a$ gives $e-a^{m+1}$, which tends to $e$. Hence its sum is the inverse of $e-a$. If $b$ is invertible, then $b+h=b(e+b^{-1}h)$ is invertible whenever $\Vert b^{-1}h\Vert <1$. This proves that the invertible elements form an open set and that inversion is locally norm-continuous.

Define the Banach algebra spectrum and the resolvent of an element byFor $|\lambda |>\Vert x\Vert$, the Neumann series givesThus $\sigma (x)$ is bounded by $\Vert x\Vert$ and is closed, because invertibility is open. Also $R(\lambda )\to 0$ in norm at infinity; for example its norm is at most $\Vert e\Vert /(|\lambda |-\Vert x\Vert )$, since $\Vert e\Vert \ge 1$ in a nonzero submultiplicative unital algebra. The standard normalization $\Vert e\Vert =1$ is not needed.

For points where the Banach algebra resolvent exists $\lambda ,\mu$, multiply out the inverses to get the resolvent identityFor ${\lambda }_0$ outside the Banach algebra spectrum, put $R_0=R({\lambda }_0)$. Locally,This series and its differentiated series converge uniformly on every smaller disk. It therefore proves $R$ is a Banach-space-valued holomorphic function, establishing complex differentiability in the Banach-space norm and gives $R^{\prime }=-R^2$, without assuming a Banach-algebra-valued analytic theorem. In particular every continuous complex linear functional $\varphi$ makes $\varphi (R(\lambda ))$ a scalar holomorphic function.

The Banach algebra spectrum is nonempty. Otherwise $R$ would be defined and analytic everywhere. Its local norm continuity and decay at infinity make it globally bounded. For every continuous linear functional $\varphi$, the scalar entire function $\varphi (R(\lambda ))$ is bounded and tends to zero at infinity, so it is identically zero: the scalar Cauchy estimate for its derivative on circles of radius $r$ is $M/r$, which tends to zero, and the limiting value then fixes its constant as zero. Such functionals separate points by the Hahn-Banach theorem. Explicitly a bounded real functional can be extended from the real span of any nonzero element using its norm, and complexifying it by $\varphi (v)=\ell (v)-i\ell (iv)$ gives a bounded complex linear functional nonzero on that element. Therefore $R$ itself would be zero, contradicting $(\lambda e-x)R=e$. We have proved the Banach algebra spectrum is a nonempty compact subset of the complex plane.

We next establish the spectral radius formula. The power norms are submultiplicative. If some $x^m=0$, their root norms are eventually zero. Otherwise, for any fixed $m$, write $n=qm+r$ with $0\le r<m$ to obtainTaking roots gives ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$. Taking the infimum in $m$, and using that every root norm is at least that infimum, proves existence ofFor $|\lambda |>{\rho }_0$, the root test makes $\sum x^j/{\lambda }^{j+1}$ converge, even when $|\lambda |\le \Vert x\Vert$. The same telescoping multiplication proves it is a two-sided inverse. HenceFor the reverse bound fix $R>r_{\sigma }$. If $R_1>\max (R,\Vert x\Vert )$, integrate the uniformly convergent exterior series term by term on $|\lambda |=R_1$, obtainingThe same integral equals the one on $|\lambda |=R$: applying an arbitrary continuous linear functional reduces contour deformation through the Banach algebra spectrum-free annulus to the scalar Cauchy theorem, and separation of points restores equality in $B$. This proves the resolvent Cauchy coefficient formula rather than assuming vector-valued analytic calculus. Using the norm bound for the allowed vector integral givesTaking $n$th roots yields ${\rho }_0\le R$, and letting $R\downarrow r_{\sigma }$ finishes the proof:

For the first example take the commutative algebra of dual numbers $B=\mathbb{C}[\epsilon ]/({\epsilon }^2)$ with norm $\Vert a+b\epsilon \Vert =|a|+|b|$. Completeness follows from finite dimension, and the product estimate makes the norm submultiplicative. For $x=\epsilon$, $x\ne 0$ but $x^2=0$, so $\rho (x)=0$. Directly, $\lambda e-\epsilon$ has inverse ${\lambda }^{-1}e+{\lambda }^{-2}\epsilon$ for $\lambda \ne 0$, and its Banach algebra spectrum is $\{0\}$. For the second example take $B=\mathbb{C}$ with its absolute-value norm and $x=1$. Its Banach algebra spectrum is $\{1\}$, giving $\rho (x)=\Vert x\Vert =1$.

Since $M$ is a finitely generated module, choose a surjection from a finite free module, $F_0\twoheadrightarrow M$. A finite free module over a left Noetherian ring is a Noetherian module, so the kernel $K_1$ is finitely generated. Choose a finite free module $F_1$ surjecting onto $K_1$, then compose its map with $K_1\hookrightarrow F_0$. Its kernel is again finitely generated. Repeating constructs an exact sequencewith all $F_i$ finite free modules, hence finitely generated projective modules. Thus the requested projective resolution exists, without asserting that it has finite length.

Assume now that $M\ne 0$ and ${\mathrm{pd}}_{R}M=n<\infty$. We may arrange a length-$n$ projective resolution with all terms finite projective. Here is why the preceding construction gives one. Set $K_0=M$, with short exact sequences $0\to K_{j+1}\to F_j\to K_j\to 0$. Applying the Ext functor, whose positive degrees vanish on a projective first argument, gives by repeated dimension shiftingfor every left module $N$. The final equality follows from the existence of any length-$n$ projective resolution. Vanishing of these degree-one groups means that every short exact sequence ending in $K_n$ splits: its extension class is zero. In particular a free surjection onto $K_n$ splits, so $K_n$ is a projective module. It is finitely generated by the Noetherian kernel construction. Use $P_n=K_n$ and $P_j=F_j$ for $j<n$ to obtainFor a left finite projective module $P$, put $P^{\ast }={\mathrm{Hom}}_R(P,R)$, a right module under $(fr)(p)=f(p)r$. It is a right finite projective module, and its natural evaluation mapis an isomorphism. These claims hold for a finite free module by coordinates, and pass to direct summands, proving them for every finite projective module without any right-Noetherian hypothesis.

Suppose $n\ge 1$ and ${\mathrm{Ext}}_R^n(M,R)=0$. Computing the Ext functor from this resolution shows that its last dual mapis surjective. Since $P_n^{\ast }$ is projective as a right module, this map has a right-module section $h:P_n^{\ast }\to P_{n-1}^{\ast }$, with $d_n^{\ast }h=1$. Dualizing gives $h^{\ast }{d}_n^{\ast \ast }=1$. Under the natural double-dual identifications, $d_n^{\ast \ast }$ is $d_n$, so this gives a left-module retraction of $d_n$.

Therefore $P_{n-1}=d_n(P_n)\oplus Q$ for a finite projective module $Q$. The next differential has kernel $d_n(P_n)$, so restricting it to $Q$ gives an exact resolution with $Q$ replacing $P_{n-1}$ and the initial $P_n$ removed. This has length $n-1$, contradicting the definition of projective dimension. For $n=0$, the nonzero $M$ is a finite projective module and a summand of some $R^d$. A nonzero vector in that summand has a nonzero coordinate, so one coordinate projection restricts to a nonzero map $M\to R$. Thus the degree-zero Hom functor is nonzero as well. We have provedThis is top Ext detects finite projective dimension over a left Noetherian ring. The nonzero-module convention matters when $n=0$: if one assigns projective dimension zero to the zero module, then ${\mathrm{Hom}}_R(0,R)=0$ is an exception. With the convention ${\mathrm{pd}}_{R}0=-\infty$, it is outside the asserted finite nonnegative case.

For an example of infinite projective dimension, take the dual numbers $R=k[\epsilon ]/({\epsilon }^2)$ over a field $k$, and $M=k=R/(\epsilon )$. The periodic resolution over dual numbers isIt is exact because both the kernel and image of multiplication by $\epsilon$ are the ideal $(\epsilon )$, and that is also the kernel of $R\to k$. Applying ${\mathrm{Hom}}_R(-,k)$ gives a copy of $k$ in every degree and zero differentials, since $\epsilon$ acts as zero on $k$. ConsequentlyA finite projective resolution would make these groups zero above its length, so

For a real Lie group $G$, its Lie algebra is the real tangent space $T_{e}G$ at the identity. Each tangent vector extends uniquely to a left-invariant vector field, and the commutator of those fields defines the Lie bracket. For a matrix group this bracket is $[X,Y]=XY-YX$.

For a complex affine algebraic group, the Lie algebra of an affine algebraic group is the complex Zariski tangent space $T_{e}G={\mathrm{Hom}}_{\mathbb{C}}({\mathfrak{m}}_e/{\mathfrak{m}}_e^2,\mathbb{C})$. Its bracket is the commutator of left-invariant derivations. In a faithful matrix realization, one can compute the tangent space using dual numbers: $X$ belongs to it exactly when $I+\epsilon X$ satisfies the group equations modulo ${\epsilon }^2$. This is a complex Lie algebra, whereas compact groups such as $U_n$ and $S{U}_n$ have real Lie algebras.

Differentiate $g^{\ast }g=I$ at the identity to get $X^{\ast }+X=0$. Conversely $e^{tX}$ is unitary for every skew-Hermitian $X$. Differentiating the determinant gives $\det (I+tX)=1+t\mathrm{tr}X+O(t^2)$, while $\det e^{tX}=e^{t\mathrm{tr}X}$ proves sufficiency of the trace condition. Finally, $G{L}_n(\mathbb{C})$ is an open subset of all complex matrices, so its tangent space has no linear constraint. ThereforeTheir dimensions are respectively $n^2,n^2,n^2-1,n^2-1$, with real dimensions for the first and third and complex dimensions for the second and fourth. In every case the Lie bracket is the matrix commutator.

An irreducible complex affine algebraic group is connected. It is a reductive algebraic group when its unipotent radical $R_u(G)$, the largest connected normal unipotent subgroup, is trivial. This is an algebraic condition; in characteristic zero it also implies complete reducibility of finite-dimensional rational representations, as proved in Question 2.

Examples are algebraic tori, $G{L}_n(\mathbb{C})$ and $S{L}_n(\mathbb{C})$, and products of these groups. A diagonal matrix that is unipotent must be the identity, so a torus has no nontrivial unipotent subgroup. For the two matrix groups, let $N$ be a normal unipotent subgroup. The Kolchin theorem supplies a nonzero fixed vector in their standard representation. The subspace fixed by $N$ is invariant under the whole group: for $g\in G$, $n(gv)=g(g^{-1}ng)v=gv$. The standard representation is irreducible, so every vector is fixed by $N$; its faithfulness makes $N$ trivial. A normal unipotent subgroup of a product projects to such a subgroup of each factor, proving the product assertion. By contrast the additive group and a nontrivial upper-unitriangular group are not reductive: their unipotent radicals are the whole group.

Compute each scheme-theoretic fibre by tensoring with the residue field of the chosen base point.

For the first map, put $L=\kappa (\mathfrak{p})$ for a point $\mathfrak{p}$ of $\mathrm{Spec}k[T]$ and let $t$ be the image of $T$ in $L$. ThenIf the characteristic of a field is not two and $t\ne 0$, the two factors $U-t$ and $U+t$ are coprime, so the Chinese remainder theorem gives $L[U]/(U^2-t^2)\simeq L\times L$: the scheme-theoretic fibre is two distinct $L$-points. If $t=0$, its ring is $L[U]/(U^2)$, a dual number ring, so it is a nonreduced double point. In characteristic two, $U^2-t^2=(U-t{)}^2$ at every point, giving a nonreduced double point in every scheme-theoretic fibre. This covers the generic point, where $L=k(T)$, as well as closed points defined by irreducible polynomials.

For the arithmetic map, the generic scheme-theoretic fibre isOver a closed point $(p)$ it is $\mathrm{Spec}{\mathbb{F}}_p[T]/(T^2+1)$. At $p=2$ this is $\mathrm{Spec}{\mathbb{F}}_2[ϵ]/({ϵ}^2)$, since $T^2+1=(T+1{)}^2$. For odd $p$, the finite field multiplicative group is cyclic, and $-1$ is a square exactly when $p\equiv 1(\mathrm{mod}4)$. ThusIn the second case there is one degree-two closed point over ${\mathbb{F}}_p$, which becomes two points after extending the residue field to an algebraic closure. The case $p=2$ remains nonreduced after such extension.

For $\mathrm{Spec}\mathbb{C}\to \mathrm{Spec}\mathbb{Z}$, the unique source point maps to the generic point $(0)$. Since all nonzero integers are invertible in $\mathbb{C}$,Indeed $\mathbb{C}{\otimes }_{\mathbb{Z}}\mathbb{Q}=\mathbb{C}$, whereas $\mathbb{C}{\otimes }_{\mathbb{Z}}{\mathbb{F}}_p=\mathbb{C}/p\mathbb{C}=0$. The distinction between a reduced split scheme-theoretic fibre and a nonreduced double point is essential in the first two examples.

The algebra must be finite-dimensional, and there is no universal bound on its dimension. Let $e=\epsilon m$ and write its coevaluation morphism as a finite sumFor any $v\in A$, one snake identity givesThus the finitely many $b_i$ span $A$, making it a finite-dimensional vector space. The other snake identity shows that $e$ is nondegenerate in the other variable as well, so the maps to the dual space determined by this pairing are isomorphisms. In the standard algebraic terminology this is a Frobenius algebra.

Every positive finite dimension of a vector space occurs: take $A=k^d$ with coordinatewise multiplication and $\epsilon (a_1,\dots ,a_d)={\sum }_i{a}_i$. The coordinate idempotents $e_i$ give $e(e_i,e_j)={\delta }_{ij}$ and $n(1)={\sum }_i{e}_i\otimes e_i$. If zero unital algebras are allowed, the zero object gives dimension zero too.

The pairing need not make $A$ a semisimple algebra. For example, the dual numbers $k[t]/(t^2)$ with $\epsilon (a+bt)=b$ have pairing matrixand coevaluation morphism $1\mapsto 1\otimes t+t\otimes 1$. This pairing is nondegenerate over every field, despite the nilpotent ideal $(t)$. Finite-dimensionality is the dimension conclusion; separability is not part of the given definition.

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.