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In a vector space over a field $k$, an affine line is a translate of a one-dimensional vector subspace. Its direction is that vector subspace, so multiplying $v$ by a nonzero scalar preserves the direction. Over a finite field of size $q$ it contains exactly $q$ distinct points. This usage concerns affine subsets of vector spaces, rather than the affine line as an affine scheme.

The multiplicative group scheme represents units: its $R$-points are $R^{\times }$. Multiplication is specified by $t\mapsto t\otimes t$ on the coordinate ring. It is the principal open subset $D(t)$ of the affine line and the one-dimensional split algebraic torus.

// Target: ringed-space.bigb

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.

For fixed $u\ne 0$, the parallel affine lines $l_x=x+\mathrm{span}(u)$ are precisely the cosets of the indicated quotient vector space. Changing representatives adds a multiple of $u$, so $l_x+l_y=l_{x+y}$ and $a{l}_x=l_{ax}$ are well-defined. The zero is the whole line $l_0$, not a single point. If $(u,b_1,b_2)$ is a basis, the two line classes $l_{b_1},l_{b_2}$ form a basis of the quotient by spanning and independence modulo $\mathrm{span}(u)$.

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.

Use the usual convention that a variety is an irreducible variety. For a quasi-projective algebraic set $T$, its algebraic dimension is the supremum of the lengths $r$ of strict chainsof nonempty irreducible closed subsets of $T$. It is the maximum of the algebraic dimensions of its irreducible components. On an affine algebraic set, the correspondence between irreducible closed subsets and prime ideals reverses inclusion, so this is the Krull dimension of the coordinate ring. On a quasi-projective algebraic set, it is the supremum of the Krull dimensions of the coordinate rings of its affine open subsets. At a closed point $x$, the Krull dimension of the local ring measures chains through $x$.

Here is a closed-point dimension lemma for affine domains that avoids transcendence degree. Put $A=k[X]$. By Noether normalization, there is a integral extensionThe number of variables is $d$ because integral extensions preserve Krull dimension and $\dim B=d$. For any maximal ideal $\mathfrak{m}$ of $A$, its contraction $\mathfrak{n}$ to $B$ is a maximal ideal. Since $k$ is an algebraically closed field, $\mathfrak{n}=(z_1-c_1,\dots ,z_d-c_d)$ has height of a prime ideal $d$. The going-down theorem applies because $B$ is an integrally closed domain and $A$ is a integral domain. It lifts a length-$d$ chain below $\mathfrak{n}$ to one below $\mathfrak{m}$. HenceThe opposite inequality follows from $\dim A=d$.

For the nonempty open subset $U$, choose a nonempty principal open subset $D(g)\subset U$ and a closed point $x\in D(g)$. Every prime ideal below ${\mathfrak{m}}_x$ avoids $g$, so the preceding chain survives in the localization $A_g$. Consequently $\dim D(g)\ge d$. Conversely, any chain of irreducible closed subsets in $U$ gives a chain of the same length after taking closures in $X$: intersecting those closures with $U$ recovers the original subsets. Therefore

For the principal hypersurface dimension lemma, let $P$ be a minimal prime ideal over $(f)$. Since $f\ne 0$ in the integral domain $A$, $P\ne (0)$. The Krull principal ideal theorem gives $\mathrm{ht}P=1$. Choose a closed point $x$ on $V(P)$ lying on none of the other finitely many irreducible components of $V(f)$. Such a point exists because those other components cut out proper closed subsets of the irreducible variety $V(P)$, and closed points are dense. Set $R=A_{{\mathfrak{m}}_x}$. Then $\dim R=d$ andWrite $s=\dim R/fR$. Choose a system of parameters ${\bar{y}}_1,\dots ,{\bar{y}}_s$ in $R/fR$ and lift it to $R$. The ideal $(f,y_1,\dots ,y_s)$ has radical equal to the maximal ideal of $R$. The Krull height theorem yields $d\le s+1$. On the other hand, any chain of prime ideals containing $f$ can be extended strictly at the bottom by the zero prime ideal of the integral domain $R$, giving $s\le d-1$. Thus $s=d-1$. Since $R/PR$ is a localization of $A/P$, $\dim A/P\ge d-1$. Extending a chain in $A/P$ by $(0)⊊P$ also gives $\dim A/P\le d-1$. Hence every component has the required dimension:This argument uses Noether normalization, going-down theorem, and the Krull height theorem, never the dimension from the function field theorem. Irreducibility is essential: if the word variety were instead allowed to mean an arbitrary reducible affine algebraic set, neither assertion would hold without extra hypotheses. For example, a disjoint union of an affine plane and an affine line has an open component of smaller algebraic dimension; a function equal to $1$ on the plane and a coordinate on the line has a nonempty zero set of algebraic dimension $0$.

An irreducible scheme is a nonempty scheme whose underlying topological space cannot be expressed as the union of two proper closed subsets. Equivalently, any two nonempty open subsets meet. A reduced scheme is one whose local rings have no nonzero nilpotent elements. Equivalently, every affine open subscheme has a reduced ring of regular functions. We may define an integral scheme as a nonempty scheme for which the coordinate ring of every nonempty affine open subscheme is an integral domain. We shall show that this is equivalent to being reduced and irreducible. The nonempty convention matters: the empty scheme is reduced but is not irreducible or integral.

For a commutative ring $A$, the spectrum of a ring is reduced exactly when $A$ is a reduced ring. One direction follows since localization preserves reducedness. Conversely, if $a$ is a nonzero nilpotent element, choose a prime ideal containing its proper annihilator. Then $a/1$ cannot vanish at that localization, contradicting reducedness of its local ring.

The spectrum of a ring is irreducible exactly when its nilradical $N=\sqrt{(0)}$ is a prime ideal. To see the essential implication directly, if $ab\in N$, then $D(a)\cap D(b)=D(ab)$ is empty. Irreducibility forces $D(a)$ or $D(b)$ to be empty, hence $a\in N$ or $b\in N$. Also $N$ is proper because the spectrum is nonempty. Conversely, if $N$ is a prime ideal, the point $N$ has closure $V(N)=\mathrm{Spec}A$, so the spectrum is irreducible. Combining the two criteria gives

Now suppose $X$ is reduced and irreducible. Every nonempty affine open subscheme is also reduced and irreducible: irreducibility passes to nonempty open subsets, because their nonempty open subsets are nonempty opens of $X$. The affine criterion makes each coordinate ring an integral domain, so $X$ is integral. Conversely, suppose all its nonempty affine coordinate rings are integral domains. Their localizations show that $X$ is reduced. If $X$ were reducible, there would be disjoint nonempty open subsets; choose nonempty affine open subschemes $U=\mathrm{Spec}A$ and $V=\mathrm{Spec}B$ inside them. Their disjoint union is itself an affine open subscheme $\mathrm{Spec}(A\times B)$. Since $A,B$ are nonzero, $(1,0)(0,1)=0$ contradicts the domain condition. Therefore

The generic point of an integral scheme $X$ is the unique point $\eta$ with $\overline{\{\eta \}}=X$. For existence, take a nonempty affine open subscheme $V=\mathrm{Spec}A$. Its point $(0)$ has closure containing $V$, which is dense in $X$, so its closure in $X$ is all of $X$. For uniqueness, both proposed generic points lie in every nonempty open subset, hence in $V$, where the only dense point is $(0)$. The function field isHere the stalk description shows that the field of fractions is independent of the choice of nonempty affine open subscheme.

Every nonempty open subscheme $U$ contains $\eta$, so taking a germ there defines $\Gamma (U,{\mathcal{O}}_X)\to K(X)$. If a section has zero germ, restrict it to any nonempty affine open subscheme $V\subseteq U$. Its image in $\mathrm{Frac}\Gamma (V,{\mathcal{O}}_X)$ is zero. Since this coordinate ring is an integral domain, the section is zero on $V$. Such affines cover $U$, and the sheaf gluing axiom makes the section zero on $U$. Thus the generic-point embedding of regular functions is

For a nonreduced reducible fibre between integral schemes, take both source and target to be the affine line over $\mathbb{C}$, and use the ring homomorphismBoth coordinate rings are integral domains, so both schemes are integral. At the target closed point $t=0$, the scheme-theoretic fibre has ringby the Chinese remainder theorem. Its underlying space has two distinct closed points, so it is reducible. The class of $x(x-1)$ is nonzero but has square zero, so it is not reduced. Even a morphism between integral schemes can have a fibre consisting of two nonreduced double points.

TakeThe global regular functions on projective space give $\Gamma (X,{\mathcal{O}}_X)={\mathbb{F}}_p$, on which the Absolute Frobenius morphism is the identity. On the standard affine line $\mathrm{Spec}{\mathbb{F}}_p[t]$, however, its map on functions is $t\mapsto t^p$, which is not surjective. Thus $F_X$ is not an isomorphism of schemes.

Choose the rational closed point $x$ given by $t=0$. The affine formula for a fibre product of schemes gives its scheme-theoretic fiber:This is the Fibre of absolute Frobenius over a rational point of the affine line: it is a one-point, length-$p$ nonreduced scheme, rather than a reduced point.

Suppose an affine line lies in $W$. Parametrize it asThe defining equation would give the polynomial identityThe only units of $k[t]$ are nonzero constants. Since a product is a unit only when every factor is a unit, all three linear polynomials must be constant, so $b_1=b_2=b_3=0$. The alleged line is a point, a contradiction. Therefore

A regular function on an open subset $U$ of a scheme $X$ is a section of ${\mathcal{O}}_X(U)$. On an affine variety it is locally a quotient of polynomial functions with a denominator that does not vanish.

On a variety, a regular function is a regular map to the affine line; the arbitrary-target regular-map concept is broader.

Now take one equation for two unknowns:The linear system $x_1+2x_2=3$ has an entire affine line of exact solutions. Since $A$ has full row rank, its Moore-Penrose inverse isThe kernel is $\mathrm{span}\{(-2,1{)}^T\}$, and all exact solutions areBecause $x_{\ast }$ is orthogonal to $(-2,1{)}^T$,The Euclidean norm is therefore smallest exactly at $t=0$. The zero residual alone cannot select a solution; the minimum-norm rule supplies the additional criterion.

This selection depends on the specified inner product. Rescaling one unknown changes what counts as a small Euclidean norm, so units and weighting matter in an application. A minimum-norm answer is mathematically canonical for the chosen geometry, rather than an automatic guarantee of physical plausibility.