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Given $f:X\to Y$ and $g:Y^{\prime }\to Y$, the base change of $f$ along $g$ is the projectionProperties such as being proper and flat are stable under arbitrary base change.

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.

Use the toric scheme over a base ring construction over $\mathbb{Z}$. Take the cocharacter lattice of an algebraic torus $N={\mathbb{Z}}^2$ and dual character lattice of an algebraic torus $M=\mathrm{Hom}(N,\mathbb{Z})$. For every $i\in \mathbb{Z}$, setThese cones in toric geometry, together with their rays and the zero cone, form an infinite fan in toric geometry. Adjacent two-dimensional cones meet in their common ray; all other pairs meet only at zero. Thus the face and intersection axioms hold even though there are infinitely many charts.

The consecutive vectors $v_i,v_{i+1}$ are a basis of $N$, since their determinant is $-1$. The dual cone has integral generatorswith $⟨a_i,v_i⟩=0$, $⟨a_i,v_{i+1}⟩=1$, $⟨b_i,v_i⟩=1$, $⟨b_i,v_{i+1}⟩=0$. Its integral points form the free commutative monoid $\mathbb{N}{a}_i+\mathbb{N}{b}_i$. Hence the associated semigroup algebra chart is simplyOn the dense algebraic torus, write $t={\chi }^{(1,0)}$ and $q={\chi }^{(0,1)}$. The chart coordinates areAlthough these expressions use inverses on the torus, $x_i,y_i,q$ are regular on the indicated charts.

The common-ray chart of $U_i$ and $U_{i+1}$ is $D(y_i)\subset U_i$, identified with $D(x_{i+1})\subset U_{i+1}$ byThe inverse is $y_i=x_{i+1}^{-1}$, $x_i=y_{i+1}{x}_{i+1}^2$. Nonadjacent charts meet in the common dense torus, and all transition maps agree there with the same characters of an algebraic torus $t,q$, so their cocycle identities follow. By gluing of schemes along open subschemes, these charts construct a scheme $X$.

The lattice projection $(a,b)\mapsto b$ carries every ${\sigma }_i$ into the nonnegative ray of the fan of the affine line. It therefore defines a toric morphismExplicitly its pullback on each chart sends $q$ to $x_i{y}_i$. The adjacent-chart formula preserves that product, so the local morphisms glue.

Compute its scheme-theoretic fibre over $q=0$:This is the union of the two coordinate affine lines, each with multiplicity one. Let $D_i$ denote the toric divisor corresponding to ${\rho }_i$. Its part in $U_i$ is $y_i=0$, with coordinate $x_i$; its part in $U_{i-1}$ is $x_{i-1}=0$, with coordinate $y_{i-1}$. On the overlap these coordinates satisfy $x_i=y_{i-1}^{-1}$. Consequently these two affine lines glue toThe neighbouring components $D_i,D_{i+1}$ meet at the section $p_i=(x_i,y_i)=(0,0)$, isomorphic to $\mathrm{Spec}\mathbb{Z}$. No other components meet. Every special-fibre chart has exactly these two branches and no embedded or extra components. ThusThe union is an infinite chain, not a finite-type curve disguised as one: its infinitely many components account for the infinitely many toric charts.

Over the principal open subset $D(q)$, both $x_i$ and $y_i$ are units because their product is $q$. Every chart then becomes the same torus chart:The transition maps identify the coordinate $t$ exactly, soThis is the generic region described in the question; over the actual generic point of $\mathrm{Spec}\mathbb{Z}[q]$ its fibre is ${\mathbb{G}}_{m,\mathbb{Q}(q)}$.

The construction is also flat. Indeed, each node chart has a free $\mathbb{Z}[q]$-module basisEvery monomial $x_i^u{y}_i^v$ reduces uniquely by taking out $q^{\min (u,v)}$, and the resulting monomials are independent. Thus the chart morphisms are flat morphisms. This verifies directly that the semistable infinite chain from a toric fan is a degeneration of the multiplicative group scheme rather than an unrelated special fibre.

A morphism of schemes $f:X\to Y$ is a flat morphism when each local ring ${\mathcal{O}}_{X,x}$ is a flat module over ${\mathcal{O}}_{Y,f(x)}$: tensoring with it preserves exact sequences. An open immersion induces isomorphisms on the local rings at its points, so it is flat.

For a closed immersion, the printed equivalence is valid with the additional hypothesis of finite presentation, for example over a locally Noetherian target, and with nonempty source. Here is the proof with that hypothesis. On an affine chart the map is $A\to A/I$, with $I$ finitely generated. If $\mathfrak{p}\supseteq I$, the flat cyclic module $A_{\mathfrak{p}}/I_{\mathfrak{p}}$ is free over the local ring $A_{\mathfrak{p}}$, and its residue-field dimension is one. Therefore its quotient map from $A_{\mathfrak{p}}$ is an isomorphism and $I_{\mathfrak{p}}=0$. Finite generation gives a neighbourhood of $\mathfrak{p}$ on which $I$ vanishes. The closed image $V(I)$ is consequently open. A nonempty clopen image in a connected target is the whole target; then $I$ vanishes at every prime and is zero. The immersion is an isomorphism. The converse is immediate. This proves the flat closed immersion of finite presentation assertion.

Without finite presentation, connectedness alone does not suffice. Take $A=C([0,1],\mathbb{R})$ and let $\mathfrak{m}=\{a:a(0)=0\}$. Let $I$ consist of functions vanishing on some neighbourhood of zero. The map $A\to A_{\mathfrak{m}}$ has kernel $I$: multiplying such a function by a continuous cutoff equal to one at zero kills it, whereas $ga=0$ with $g(0)\ne 0$ forces $a$ to vanish near zero. This map is surjective, because the reciprocal of any denominator nonzero at zero can be extended continuously from a small neighbourhood of zero to the whole interval. ThusThe map $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$ is both a closed immersion and a flat localization. Its source is connected because its ring is local. Its target is connected because continuous idempotent functions on the connected interval can only be the constants zero and one. It is not an isomorphism: $I$ contains nonzero functions supported away from zero. This supplies a concrete counterexample to the unrestricted assertion. The kernel is a pure ideal; this is the general mechanism behind such examples, as explained in the Stacks Project's treatment of pure ideals.

For the finite-flat fibre assertion, first note the local algebra fact that a finitely generated flat module $M$ over a local ring $(A,\mathfrak{m})$ is free. Choose generators $m_1,\dots ,m_r$ whose residues form a basis of $M/\mathfrak{m}M$, using Nakayama's lemma. If a row $a=(a_1,\dots ,a_r)$ gives a relation $\sum a_i{m}_i=0$, the equational criterion for flatness supplies elements $n_j$ and a matrix $B=(b_{ij})$ such that $m_i={\sum }_j{b}_{ij}{n}_j$ and $aB=0$. Write each $n_j$ as a linear combination of the $m_i$, with coefficient matrix $C$. Then $m=BCm$. Independence modulo $\mathfrak{m}$ makes $BC$ congruent to the identity. This square matrix is invertible over the local ring, and $aBC=0$ implies $a=0$. Thus the chosen generators are a basis. This also justifies the cyclic-module step above; see the finite-flat local lemma.

Now let $\mathcal{A}=f_{\ast }{\mathcal{O}}_X$ for the finite flat morphism. On each affine chart this is the finite flat algebra defining its inverse image. Hence ${\mathcal{A}}_y$ is free over ${\mathcal{O}}_{Y,y}$, of some finite rank $r_y$. Let $\eta$ be the generic point of the irreducible target. Localizing that free module at $\eta$ givesEvery point specializes from the same $\eta$, so every $r_y$ equals the fixed rank at $\eta$. This remains valid if the irreducible target is nonreduced: one compares free ranks over its nonzero generic local ring, not necessarily over a field. The fibre is affine, andConsequentlyThis is finite flat rank over an irreducible scheme. No unjustified replacement of finite flat by finite locally free on an arbitrary base is needed.

For an algebraic curve of finite type over a field, its normalization of an integral scheme is finite and birational. If flat, the preceding rank is one, since the generic fibre has the same function field. Locally its normalization algebra $B$ is therefore a free $A$-module of rank one. The element $1_B$ generates: its image cannot vanish in the one-dimensional fibre algebra, and Nakayama's lemma applies. Thus $A\to B$ is an isomorphism. Conversely the identity normalization is flat.

The parenthetical definition in the PDF does not explicitly impose finite type. One can prove the conclusion even in that broader setting without assuming a finite normalization. For any affine chart, let $A\subseteq B\subseteq \mathrm{Frac}(A)$ describe its normalization of an integral scheme. Integrality makes the map on spectra surjective; if $B$ is flat it is faithfully flat. Faithful flatness gives contraction of ideals, $vB\cap A=vA$. For $b=u/v\in B$, with $u,v\in A$ and $v\ne 0$, one has $u=vb\in vB\cap A=vA$. Thus $u=va$ for some $a\in A$, so $b=a$. Therefore $B=A$ on every chart. We have proved the flat normalization is an isomorphism result and, in particular,

A morphism of schemes $f:X\to Y$ is a flat morphism when every local-ring map ${\mathcal{O}}_{Y,f(x)}\to {\mathcal{O}}_{X,x}$ makes ${\mathcal{O}}_{X,x}$ a flat module.

In (i), the coordinate map is $k[t]\to k[y]$, $t\mapsto y^2$, andIt is therefore a free module of rank two and the morphism is flat, including in characteristic two.

In (ii), $k[x]$ is finite over the cusp ring $R=k[x^2,x^3]$ and has generic rank one. Were it flat, finite flatness over the local ring at the cusp would make it free of rank one. Its fiber there is insteadwhich has dimension two, so this morphism is not flat.

In (iii), the base coordinate $t$ acts as $x$, and the nonzero element $y$ satisfies $ty=xy=0$. Thus the coordinate ring has torsion as a $k[t]$-module. Since $k[t]$ is a principal ideal domain and a module over it is flat exactly when it is torsion-free, this morphism is not flat. Consequently

An ideal $I$ is pure when the quotient is a flat module. Equivalently every $a\in I$ satisfies $a=ab$ for some $b\in I$. The latter condition makes $I$ vanish after localizing at primes containing it, while at primes not containing it the quotient is zero. Pure ideals need not be finitely generated. They describe flat morphisms that are closed immersions on affine charts.