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At a closed point of a variety over an algebraically closed field, the cotangent space is ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. The universal property of Kähler differentials identifies its dual vector space with $k$-derivations of the local ring into the residue field, hence with the Zariski tangent space. The Kähler differential sheaf has the same fibre because a derivation kills constants and products of two elements of the maximal ideal.

Let $W$ be an integral effective Cartier divisor in an integral variety $V$ over a perfect field, and suppose $W$ is not contained in the singular locus of $V$. On a dense open subset both varieties are smooth. The Zariski tangent space of $W$ there has codimension one in that of $V$, so the differential of the local defining equation is nonzero. Thus the conormal sheaf map is injective generically. Its kernel is a subsheaf of a line bundle on the integral variety $W$ and is therefore a torsion-free sheaf; generic vanishing implies zero everywhere. Right exactness is the Conormal exact sequence for Kähler differentials.

For an effective divisor on an algebraic curve $D$ on a smooth projective curve, the Zariski tangent space to its symmetric product of a curve is $H^0({\mathcal{O}}_D(D))$. The derivative of the Abelian sum map is the connecting homomorphism of $0\to {\mathcal{O}}_X\to {\mathcal{O}}_X(D)\to {\mathcal{O}}_D(D)\to 0$. Its dual under Serre duality is restriction of holomorphic differential forms to $D$, including all jets at repeated points. Hence its rank is $\mathrm{deg}D+1-h^0({\mathcal{O}}_X(D))$, and its image annihilates $H^0(K_X(-D))$.

For a finite generating space $V\subseteq H^0(X,\mathcal{L})$ on a projective scheme, its map to projective space is a closed immersion exactly when, after algebraic closure of the ground field, the displayed evaluation is surjective for every length-two closed subscheme $E$. Two distinct points test point separation; a double point tests a direction in the Zariski tangent space. For the complete section space, this characterizes a very ample line bundle.

For an affine plane algebraic curve $F(x,y)=0$, a point over an algebraic closure is nonsingular when the displayed condition holds. Its Zariski tangent space has dimension one, and a local coordinate can be chosen in the direction of a nonzero partial derivative. A smooth algebraic curve has all points nonsingular. Over a complete local field, the Hensel lemma lifts a nonsingular residue point by solving for a coordinate with unit partial derivative.

The infinitesimal base change action on quiver representations is ${\xi }_x(u{)}_{\rho }=u_{t(\rho )}{x}_{\rho }-x_{\rho }{u}_{s(\rho )}$. Its image is the Zariski tangent space to the orbit, because the stabilizer is a smooth open subset of the endomorphism ring. The extension complex of quiver representations identifies the quotient of the ambient tangent space by this image with ${\mathrm{Ext}}_Q^1(X,X)$. Consequently rigid quiver representations have open orbits.

Every projective line on the smooth algebraic surface through $P$ lies in the projective tangent plane $T_{P}S$, since its tangent direction is contained in the surface's Zariski tangent space. The plane section $S\cap T_{P}S$ is a nonzero plane cubic. It cannot be the entire plane, since a smooth cubic surface is irreducible and has no plane component.

Each distinct projective line through $P$ is therefore a distinct degree-one factor of the cubic restricted to $T_{P}S$. A degree-three polynomial has at most three such factors. Hence there are at most three lines through any point. This is the lines through a point of a smooth cubic surface bound; equality is allowed when the tangent section consists of three concurrent lines.

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.

Any isomorphism would permute irreducible components and preserve their intersections. In each configuration, the origin is the unique point lying on all three components, so an isomorphism must take origin to origin.

The coordinate ring of the spatial configuration isThe displayed ideal is indeed the intersection of the three coordinate-axis ideals. Its generators have no linear parts, so at the origin the residue classes of $x,y,z$ form a basis of $\mathfrak{m}/{\mathfrak{m}}^2$. Hence its algebraic cotangent space and Zariski tangent space have dimension three.

For the plane configuration, the coordinate ring is $B=k[s,t]/(st(s-t))$. The three linear factors are distinct in every characteristic, so this principal ideal is radical. It likewise has no linear part, but now $\mathfrak{n}/{\mathfrak{n}}^2$ has basis $s,t$ and dimension two. An isomorphism of local rings induces an isomorphism of these cotangent spaces, which is impossible:Thus the varieties are not isomorphic. This is embedding dimension distinguishes three-branch curves: matching incidence of branches does not determine their local algebra.

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.

In an affine chart, the Zariski tangent space at $P$ consists of vectors $v$ with $d{f}_P(v)=0$ for every polynomial vanishing on the variety. Intrinsically it is the dual of ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. For a finite set of generators in an ambient $N$-space, its dimension is $N-\mathrm{rank}J(P)$. The condition that this be at least $r$ is the vanishing of all $(N-r+1)$-minors, hence is closed. The affine-chart descriptions agree intrinsically; being closed is local on the variety, so the same conclusion holds projectively.

For the two given quadrics the gradient rows areTheir intersection has dimension two: the quadrics have no common factor, so form a codimension-two complete intersection. Its singular points are where the two rows are dependent. If the first row vanishes, one obtains $[1:0:0:0:0]$. Otherwise write a dependence with nonzero coefficient of the second row. Its first coordinate forces $X_1=0$. The defining equations then imply $X_3{X}_4=0$ and $X_3^2+X_4^2=0$, hence $X_3=X_4=0$. Conversely every point of the resulting line lies on the variety and has dependent rows. ThereforeOn this line at least one of $X_0,X_2$ is nonzero, so the Jacobian rank is exactly one. The projective ambient tangent dimension is four, giving $\dim T_{V,P}=3$ at every singular point. Away from the line the rank is two and the tangent dimension is two.

The ideal $(f)$ is prime, since an irreducible element of the polynomial ring is prime. By the Hilbert Nullstellensatz, it is the full vanishing ideal of $Z(f)$. Its Zariski tangent space at a point $a$ is therefore the kernel of the single linear form $d{f}_a$, and has dimension $n-1$ or $n$.

At least one partial derivative $f_j$ is nonzero. In characteristic zero this is immediate for a nonconstant polynomial. In characteristic $p$, if all partial derivatives vanished, all exponents would be multiples of $p$. Since an algebraically closed field is perfect, $f$ would be a $p$th power of a nonconstant polynomial, contradicting irreducibility.

The nonzero $f_j$ cannot vanish everywhere on $Z(f)$: the Nullstellensatz would give $f_j^r\in (f)$, hence $f\mid f_j$, contradicting $\mathrm{deg}{f}_j<\mathrm{deg}f$. Thus some point has $d{f}_a\ne 0$, and the minimum tangent-space dimension is

For an irreducible affine variety, its dimension is the minimum of its Zariski tangent space dimensions:The minimum is attained on the smooth locus. Equivalently $T_{x}X$ is the dual of ${\mathfrak{m}}_x/{\mathfrak{m}}_x^2$, or the simultaneous kernel of the differentials of the defining equations at $x$. If “variety” permits reducible sets, apply this definition to each irreducible component and take the maximum; taking the minimum over an arbitrary reducible set with components of unequal dimension would be incorrect.

The parametrization $t\mapsto (t^3,t^2)$ maps ${\mathbb{A}}^1$ to the cusp. Off the origin its inverse is $t=x/y$, since $(x/y{)}^2=y$ and $(x/y{)}^3=x$ on $x^2=y^3$ with $y\ne 0$. This is an isomorphism between dense open subsets, so the cusp is birational to ${\mathbb{A}}^1$.

At the origin the differential of the defining polynomial is zero, giving a two-dimensional Zariski tangent space. The tangent space of ${\mathbb{A}}^1$ at every point is one-dimensional. An isomorphism induces tangent-space isomorphisms, so the cusp is not isomorphic to ${\mathbb{A}}^1$. The rational inverse cannot extend regularly through the singular point.

We use algebraic dimension and tangent spaces, equivalently after extending scalars to an algebraic closure; these are not dimensions of the finite set of rational points when $k$ is finite. The condition rank at most $r$ is equivalent to the vanishing of all $(r+1)\times (r+1)$ matrix minors. These are polynomials in the $nm$ entries, so define an affine determinantal variety $X^{n,m,r}$. The same equations make it a Zariski closed subset of $X^{n,m,r+1}$.

For $1\le r<\min (n,m)$, every line $t{E}_{ij}$ lies in $X$ because its matrices have rank at most one. Its tangent direction $E_{ij}$ therefore belongs to the Zariski tangent space at zero. The matrix units span all $nm$ directions, so $T_{0}X=k^{nm}$. Meanwhile a rank-$r$ chart with an invertible $r\times r$ block has the formThe free blocks have $r^2+r(m-r)+(n-r)r=r(n+m-r)$ entries. Rank-$r$ matrices are dense in $X$: the factorization $UV$ parametrizes all rank-at-most-$r$ matrices, and the full-rank factor pairs form a dense open subset. This also proves irreducibility, so the local dimension at zero equals the dimension of $X$. Since $r(n+m-r)<nm$, zero is singular when $1\le r<\min (n,m)$.

There is a necessary qualification to the printed claim: for $r=0$, $X=\{0\}$ is a smooth point, with zero-dimensional tangent space. Thus the claim is false if zero is allowed as a rank bound.

Finally, for $n=5,m=2,r=1$, the formula gives $\boxed{\dim X^{5,2,1}=6}$. Concretely, on a chart where an entry of the first column is nonzero, choose that column freely and make the second an arbitrary scalar multiple of it, giving five plus one parameters.

A blowup of an algebraic variety supplies a reduced special fibre with all three requested defects. Let $B={\mathbb{A}}^3$ have coordinates $(x,y,z)$, let $p=(0,1,0)$, and let $\pi :X={\mathrm{Bl}}_{p}B\to B$. Take $Y={\mathbb{A}}^1$ and the regular mapFor an explicit construction, $X$ is the closed subvariety of ${\mathbb{A}}^3\times {\mathbb{P}}^2$, with projective coordinates $[u:v:w]$, defined byThese equations say that $(x,y-1,z)$ is proportional to $(u,v,w)$. The three standard projective charts are affine spaces ${\mathbb{A}}^3$, so $X$ is a smooth variety which is an irreducible variety embedded in a quasi-projective algebraic set. Its algebraic exceptional divisor is $E={\pi }^{-1}(p)\cong {\mathbb{P}}^2$.

If $t\ne 0$, the surface $xy=t$ misses $p$, and the blowup of an algebraic variety is an isomorphism over it. HenceIt is a smooth variety, an affine variety, an irreducible variety, and has algebraic dimension $2$.

For $t=0$, the fibre of a morphism has three irreducible components: the strict transform of an algebraic subvariety $\widetilde{V(x)}$, the unchanged plane $V(y)$, and the algebraic exceptional divisor $E\cong {\mathbb{P}}^2$. The centre $p$ lies on $V(x)$ but not on $V(y)$; this is why all three components have multiplicity one. To verify reducedness and the crossing directly near $E$, use the chart $v\ne 0$, puttingHere $x=ab$, $y=1+b$, $z=bc$, and $f=ab(1+b)$. Near $b=0$, the factor $1+b$ is a unit, so the fibre of a morphism is $ab=0$, a reduced union of two crossing planes. The $u$-chart gives $f=x(1+xv)$, whose second factor is a unit along $E$; the $w$-chart gives $f=zu(1+zv)$ and the same reduced crossing. Away from $E$, reducedness follows from $xy=0$ in $B$. Thus the scheme-theoretic fibre of a morphism is itself reduced.

It has singular points of an algebraic variety already along $V(x,y)$, which misses $p$: locally its equation is $xy=0$, with Zariski tangent space of dimension of a vector space $3$ but local dimension of an algebraic variety $2$. Finally, $E$ is a closed subvariety of $X_0$. A closed subvariety of an affine variety is an affine variety, whereas ${\mathbb{P}}^2$ is not: all its global regular functions are constant, which cannot be the coordinate ring of a positive-dimensional affine variety. Consequently $X_0$ is singular, reducible, and nonaffine, as required. Choosing the centre off the intersection of the original planes avoids introducing a nonreduced special fibre of a morphism.

A unipotent algebraic group admits a faithful linear representation in which every group element is a unipotent matrix. For $E={\mathrm{End}}_Q(X)$, invertibility is the nonvanishing condition ${\prod }_i\det h_i\ne 0$. Therefore ${\mathrm{Aut}}_Q(X)=E^{\times }$ is a nonempty Zariski-open subset of the vector space $E$.

Take a Krull-Schmidt decomposition $X\cong {⨁}_{a=1}^r{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposable modules. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local endomorphism ring. Its residue division algebra is $k$: over an algebraically closed field, every element of a finite-dimensional division algebra has an eigenvalue and hence must be scalar. The semisimple quotient of a module endomorphism algebra consequently gives, for the Jacobson radical $J=J(E)$,is surjective with kernel $1+J$. The nilpotence of $J$ makes $(1+j{)}^{-1}=1-j+j^2-\cdots$ finite and each $1+j$ unipotent. The kernel is closed and normal. Acting on the multiplicity spaces embeds the product of general linear groups back into $E^{\times }$ and splits this quotient. This proves the Levi decomposition of a quiver automorphism groupSince $1+J$ is the unipotent radical, a nonzero $X$ is indecomposable exactly when ${\mathrm{Aut}}_Q(X)/R_u({\mathrm{Aut}}_Q(X))\cong {\mathbb{G}}_m$: the product has a single factor of size one.

For the base change action on quiver representations, the orbit map is $g\mapsto (g_j{x}_{\rho }{g}_i^{-1}{)}_{\rho :i\to j}$. Substituting $g_i=I+ϵu_i$, with ${ϵ}^2=0$, shows its differential isIts kernel is ${\mathrm{End}}_Q(X)$. The stabilizer is smooth because it is open in that vector space. Hence the differential has rank $\dim \mathrm{GL}(\mathbf{n})-\dim {\mathrm{Aut}}_Q(X)=\dim {\mathcal{O}}_X$, and its image is the Zariski tangent space $T_x{\mathcal{O}}_X$. The normal space to a quiver orbit is thereforeThe ambient quiver representation space is an irreducible affine space, and orbits are locally closed. An orbit is open exactly when its dimension equals that ambient dimension, equivalently when ${\mathrm{Ext}}_Q^1(X,X)=0$. This proves that rigid quiver representations have open orbits. Such an orbit is dense and unique, since two nonempty open subsets of an irreducible space intersect.

On an affine chart $\mathrm{Spec}A\subseteq V$, the module of Kähler differentials ${\Omega }_{A/k}$ is generated by symbols $da$ subject to $k$-linearity and $d(ab)=adb+bda$. It represents $k$-derivations. These modules commute with localization, so their associated quasi-coherent sheaves glue to the Kähler differential sheaf ${\Omega }_{V/k}^1$. It is coherent: if $A=k[x_1,\dots ,x_N]/I$ and $I=(f_1,\dots ,f_s)$, then it has the finite presentation of a moduleFor a closed point $P$, put $R={\mathcal{O}}_{V,P}$ with maximal ideal $\mathfrak{m}$ and residue field $k$. The Zariski tangent space is ${\mathrm{Hom}}_k(\mathfrak{m}/{\mathfrak{m}}^2,k)$, equivalently ${\mathrm{Der}}_k(R,k)$. A derivation kills constants and ${\mathfrak{m}}^2$, so it factors through $a\mapsto a-a(P)$; conversely every linear functional on $\mathfrak{m}/{\mathfrak{m}}^2$ defines such a derivation by the product rule. The universal property of Kähler differentials therefore identifies this space withIn particular, ${\Omega }_{V,P}^1/{\mathfrak{m}}_P{\Omega }_{V,P}^1\cong {\mathfrak{m}}_P/{\mathfrak{m}}_P^2$, the algebraic cotangent space.

A point is a smooth point of a variety when its local ring is a regular local ring; over this algebraically closed field this says ${\dim }_k{T}_{V,P}=n=\dim V$. Tensor the finite presentation of a module above with $k(P)$. The tangent space is the kernel of the Jacobian matrix $J(P)$, hence has dimension $N-\mathrm{rank}J(P)$. This proves the Jacobian criterionFor a smooth irreducible variety $V$, choose at each point an invertible $(N-n)$-minor and shrink the affine chart so that it remains invertible. Its relations eliminate $N-n$ of the generators of ${\Omega }_{A/k}$, yielding a surjection $A^n\to {\Omega }_{A/k}$. At the generic point, its target has dimension $n$: over a perfect ground field, a separating transcendence basis of the function field has differentials forming a basis. Equivalently this follows from the assumed density of the smooth locus. The kernel therefore becomes zero over the fraction field of the integral domain $A$. As a submodule of $A^n$, it is a torsion-free module, so it is already zero. Thus these maps give local isomorphisms with ${\mathcal{O}}_V^n$, proving local freeness of differentials on a smooth variety with rank $n$.

For an affine chart $\mathrm{Spec}A\subseteq V$, write $W\cap \mathrm{Spec}A=\mathrm{Spec}(A/I)$ and $\mathcal{M}=\tilde{M}$. The restriction of a module sheaf to a closed subvariety is $\widetilde{M{\otimes }_{A}A/I}=\widetilde{M/IM}$. The Conormal exact sequence for Kähler differentials isIt follows from the generators and relations: passing to $A/I$ imposes precisely the additional relations $df=0$ for $f\in I$. The first map is well defined because $d(I^2)$ lies in $I{\Omega }_{A/k}$. Glue these exact module sequences, using exactness of localization, to obtainIn the final assertion, interpret locally principal subvariety as a proper local hypersurface. Its ideal on each chart is $(f)$ with $f\ne 0$. Since $V$ is an irreducible variety and reduced, $f$ is a non-zero-divisor, and $A/(f)\to (f)/(f^2)$, $\bar{a}\mapsto af$, is an isomorphism. These local rank-one descriptions make the conormal sheaf ${\mathcal{I}}_W/{\mathcal{I}}_W^2$ invertible.

Because $W$ is not contained in the singular locus of $V$, there is a dense open subset of $W$ where both $V$ and $W$ are smooth varieties. At a closed point there, $\dim W=\dim V-1$ by the Krull principal ideal theorem. The tangent description $T_{W,P}=\ker (df:T_{V,P}\to k)$ then forces $df\ne 0$. Hence the conormal map is injective at the generic point of $W$. Its kernel is a subsheaf of a line bundle on the integral variety $W$, so it is a torsion-free sheaf; a torsion-free sheaf with zero generic fibre is zero. This proves conormal injectivity for a generically smooth Cartier divisor. If zero equations were allowed in the phrase locally principal, $W=V$ would be a counterexample to invertibility; the proper-hypersurface convention is essential.

A line bundle $\mathcal{L}$ is a globally generated line bundle when the evaluation mapis surjective. Equivalently, its global sections span each fibre of the line bundle; locally at every point, some section is a generator.

Choose a finite generating family $s_0,\dots ,s_n$. On the open set where $s_i$ generates $\mathcal{L}$, the ratios $s_j/s_i$ are regular functions, giving a morphism of schemes into the standard chart $D_{+}(x_i)$ of projective space. The ratios obey the usual transition rules on overlaps, so these chart maps glue toThe tuple is computed using any local trivialization of $\mathcal{L}$; changing that trivialization multiplies all entries by the same invertible function. The construction has ${\phi }^{\ast }\mathcal{O}(1)\simeq \mathcal{L}$ and pulls back the coordinate sections to the chosen $s_i$.

There is a finiteness qualification for an arbitrary $X$: global generation alone need not provide such a finite family. It does if $X$ is quasi-compact, since the open sets on which individual sections generate have a finite subcover. Without that hypothesis, take $X={\coprod }_{m\ge 1}{\mathbb{P}}_k^m$ and let $\mathcal{L}$ restrict to $\mathcal{O}(1)$ on each component. This line bundle is globally generated, but any $N$ global sections have a common zero on a component ${\mathbb{P}}^m$ with $m\ge N$. Thus this is a globally generated line bundle without finite generators. This is finite global generation on a quasi-compact scheme. The finite-family construction is automatic in the projective case asked next.

For projective nonsingular $X$, put $V=⟨s_0,\dots ,s_n⟩$. The length-two criterion for a very ample linear system says that $\phi$ is a closed immersion precisely when, after extending to an algebraic closure, $V$ separates distinct points and tangent directions. Equivalently, the evaluationis surjective for every length-two geometric closed subscheme $E\subseteq X$. Two distinct points give point separation; a nonreduced length-two subscheme supported at one point gives separation of a direction in the Zariski tangent space. For the complete space $V=H^0(X,\mathcal{L})$, this says exactly that $\mathcal{L}$ is a very ample line bundle. For a chosen smaller $V$, it is the chosen linear system of divisors which must be a very ample linear system; mere global generation is insufficient.

For $X\subset {\mathbb{A}}^n$ with vanishing ideal $I(X)$, the Zariski tangent space at $P$ isIt is the kernel of the Jacobian matrix of generators for the ideal, or intrinsically the dual of ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. A point is a singular point of an algebraic variety when this tangent dimension exceeds the local dimension of the variety; otherwise it is nonsingular. For an irreducible affine variety the local dimension at a closed point equals its dimension.