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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 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.

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.

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.

A point $P$ belongs to the smooth locus of a variety if its local ring ${\mathcal{O}}_{X,P}$ is a regular local ring, meaningEquivalently, the Zariski tangent space has the expected local dimension. Over the algebraically closed field, this agrees with smoothness for a reduced variety. In affine coordinates, the Jacobian criterion expresses it by the rank of the matrix of first derivatives of defining equations.

For an affine variety $X\subseteq {\mathbb{A}}_k^n$ with vanishing ideal $I(X)$, its Zariski tangent space at $p$ isThe dimension from minimum tangent dimension is

Now let $X=Z(f)$ with $f$ irreducible. The Strong Hilbert Nullstellensatz gives $I(X)=(f)$, soIts dimension is at least $n-1$, and it is exactly $n-1$ wherever $\nabla f(p)\ne 0$.

It remains to find such a point in positive characteristic $q$. If every partial derivative of $f$ were the zero polynomial, every exponent occurring in $f$ would be divisible by $q$. Since the algebraically closed field $k$ is perfect, all coefficients have $q$th roots and $f$ would be a $q$th power, contradicting irreducibility. Thus some partial derivative $f_{x_i}$ is nonzero. If the gradient nevertheless vanished at every point of $Z(f)$, the Nullstellensatz would imply $f_{x_i}\in (f)$. This is impossible because $0\le \mathrm{deg}{f}_{x_i}<\mathrm{deg}f$. The smooth point on an irreducible hypersurface in positive characteristic therefore exists, and

Let $(a,b,c)$ be a tangent vector in the $x,y,z$ directions. At a point $(0,y,z)\in C$, the two isolated components are absent in a Zariski neighbourhood, so the Zariski tangent space is that of $C$. Differentiating $x$ and $y^2-z^3$ givesFor $(y,z)\ne (0,0)$ this is a line. At the cusp,which is two-dimensional.

Each of $P$ and $Q$ is a reduced isolated component with local ring $k$, soThis last conclusion must use all functions in $I(V)$, as required by the definition of the Zariski tangent space. The Jacobian of only the three displayed defining polynomials has an extra spurious tangent direction at $Q$, because those polynomials do not generate the radical ideal there; for example, $x(x^2-1)\in I(V)$ removes that direction.