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For a scheme over $\mathbb{R}$, its base change of a morphism of schemes to $\mathbb{C}$ has scheme-theoretic fibre over a point $p$ equal to $\mathrm{Spec}(\kappa (p){\otimes }_{\mathbb{R}}\mathbb{C})$. This is the spectrum of $\kappa (p)[s]/(s^2+1)$. It has two points if $-1$ is a square in the residue field, and otherwise one point with a quadratic field extension. In both cases it is reduced, because the quadratic polynomial has no repeated root in characteristic zero. The distinction applies to generic points as well as closed points.

For an injective ring homomorphism between nonzero integral domains $A\to B$, the scheme-theoretic fibre over the generic point of $\mathrm{Spec}A$ is $\mathrm{Spec}(S^{-1}B)$ with $S=A\setminus \{0\}$. Localization at these nonzero elements is a nonzero integral domain, so the generic fibre is a nonempty integral scheme. A surjective morphism between these affine schemes necessarily gives an injective ring map: a prime above $(0)$ contains its kernel.

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.

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 $y$ on the target and $x$ on the source, so the map of coordinate rings sends $y$ to $x^4$. A closed point of the real affine line corresponds to a maximal ideal generated either by $y-a$, $a\in \mathbb{R}$, or by a monic irreducible quadratic $q(y)$. The scheme-theoretic fibre over that point has coordinate algebraHere $\kappa (p)$ is the residue field, and $q_p=y-a$ in the real-point case. The fibre is a scheme, so its nilpotents must be retained.

For $a>0$, put $r=a^{1/4}$. The factors $x-r$, $x+r$, $x^2+r^2$ are pairwise coprime, with the last one irreducible over $\mathbb{R}$. The Chinese remainder theorem identifies $B_p$ with $\mathbb{R}\times \mathbb{R}\times \mathbb{C}$, giving three irreducible components.

For $a<0$, put $r=(-a{)}^{1/4}$. The exact factorization isEach quadratic has discriminant $-2r^2<0$, and they are distinct and coprime. Thus $B_p\cong \mathbb{C}\times \mathbb{C}$, giving two irreducible components.

For $a=0$, $B_p=\mathbb{R}[x]/(x^4)$. It has one prime ideal, generated by the class of $x$, so one irreducible component. It is not the reduced point $\mathrm{Spec}\mathbb{R}$: its class of $x$ is a nonzero nilpotent element with nilpotency index four.

Finally write the irreducible quadratic as $q_p(y)=(y-z)(y-\overline{z})$, with $z\notin \mathbb{R}$. Over $\mathbb{C}$, $q_p(x^4)$ has four roots above $z$ and four above $\overline{z}$. They are all simple, because $z\ne 0$, and the two sets are disjoint. None is real. Conjugation pairs them into four distinct irreducible real quadratics. Hence $B_p\cong {\mathbb{C}}^4$ and this fibre has four irreducible components.

The real fourth-power fibre classification is:

All fibres in each row are isomorphic as affine schemes over $\mathbb{R}$. The rows are distinct: their numbers of irreducible components already distinguish them, and the zero fibre is additionally nonreduced. ThereforeThese are ordinary scheme components, not geometric components counted after complexifying the base. Counting only real roots would miss both the complex-residue factors and the nonreal closed points.

A ring homomorphism is understood to preserve $1$. Start with $\phi :A\to B$. Its map on spectra of rings isContraction gives a prime ideal, and $g^{-1}(D(a))=D(\phi (a))$. Thus $g$ is continuous for the Zariski topology. On each principal open subscheme, the structure sheaf map is the ring homomorphismThese maps commute with restrictions, so they define a sheaf morphism ${\mathcal{O}}_X\to g_{\ast }{\mathcal{O}}_Y$. At $\mathfrak{q}$, with $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, its stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. The inverse image of the maximal ideal $\mathfrak{q}{B}_{\mathfrak{q}}$ is $\mathfrak{p}{A}_{\mathfrak{p}}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, let $(g,g^{\#})$ be a morphism of locally ringed spaces. Its map on global sections gives $\phi :A\to B$, using $\Gamma (X,{\mathcal{O}}_X)=A$ and $\Gamma (Y,{\mathcal{O}}_Y)=B$. Fix $\mathfrak{q}\in Y$ and put $\mathfrak{p}=g(\mathfrak{q})$. Compatibility with the stalk maps and the local homomorphism property giveConsequently $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, so the underlying map is forced. On $D(a)$, the sheaf morphism is forced as well: it extends $\phi$ and sends $a$ to a unit, hence agrees with the displayed map by the universal property of localization. The principal open subschemes form a basis, so the entire sheaf morphism is determined. Taking global sections of the construction recovers $\phi$. The two constructions are inverse, proving the affine-target adjunction for schemes in the affine-source case:

To describe the real affine plane scheme points, write $R=\mathbb{R}[t_1,t_2]$. It is a unique factorization domain of Krull dimension two. Its points are exactly the following prime ideals:

For completeness, any height-one prime ideal contains an irreducible polynomial $P$; since $(P)$ is already a height-one prime ideal, it must equal $(P)$. Every remaining nonzero prime ideal has height two and is maximal, by Krull dimension. For a nonreal pair, evaluation generates all of $\mathbb{C}$ over $\mathbb{R}$, so its kernel is maximal. Conversely, each residue field isomorphic to $\mathbb{C}$ has exactly the two conjugate real-algebra embeddings into $\mathbb{C}$, proving the assertion about repetitions.

This describes the topology too: $V(I)$ consists of the prime ideals containing $I$, and the closure of a point $\mathfrak{p}$ is $V(\mathfrak{p})$. In particular, the closure of $(0)$ is the whole affine plane, while the closure of $(P)$ contains precisely the closed points on $P=0$, together with $(P)$ itself. The spectrum is much larger than the set ${\mathbb{R}}^2$. For example, $(t_1^2+1)$ is a height-one point although its curve has no real points. Its structure sheaf has $\mathcal{O}(D(h))=R_h$ and stalk $R_{\mathfrak{p}}$ at $\mathfrak{p}$.

The induced morphism of schemesis contraction of prime ideals, with the structure sheaf maps given above. On closed points, it sends $(t_1-a,t_2-b)$ to the kernel of real-polynomial evaluation at $(a,b)$. A real closed point has one complex point above it; a nonreal closed point has two, interchanged by complex conjugation. The source generic point maps to the target generic point. The source height-one points are generated by irreducible complex polynomials $Q$. Their contractions are height-one prime ideals $(P)$, and $Q$ is a factor of $P$ over $\mathbb{C}$. An irreducible real $P$ either stays irreducible over $\mathbb{C}$ or splits into two distinct conjugate irreducible factors. Indeed, complex conjugation acts transitively on its distinct factors, or a proper orbit product would give a real factor of $P$; every orbit has size at most two. Repeated factors are excluded by separability in characteristic zero. Thus one or two height-one points lie above $(P)$.

The complexification fibres of a real scheme give a uniform description of all scheme-theoretic fibres, including the nonclosed points, is especially useful. The extension of coordinate rings isa free $R$-module with basis $1,s$. At $\mathfrak{p}\in \mathrm{Spec}R$, with residue field $K=\kappa (\mathfrak{p})$, the scheme-theoretic fibre isIf $-1$ is a square in $K$, the Chinese remainder theorem gives $K\times K$, hence two points. Otherwise it is a quadratic field extension, hence one point. The polynomial has no repeated root in characteristic zero, so all these scheme-theoretic fibres are reduced schemes. This also proves surjectivity. Conjugation acts on each two-point fibre by exchanging its points and fixes each one-point fibre. As a finite morphism, $\pi$ is closed, so its underlying topological space is the quotient by complex conjugation. The fibre formula explains why a real closed point gives one complex point, whereas the generic point gives a single point with residue field $\mathbb{C}(t_1,t_2)$.

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.

For a point $y\in Y$ with residue field $\kappa (y)$, the scheme-theoretic fibre is the base change of a morphism of schemesThe map $\mathrm{Spec}\kappa (y)\to Y$ is the canonical point map. This fibre product of schemes retains the structure sheaf, including any nilpotents, rather than just the set $f^{-1}(y)$. For an affine scheme map induced by $A\to B$, at $\mathfrak{p}\in \mathrm{Spec}A$ it is

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 hyperplane sections through $P$ are spanned by the three linear forms $x_1,x_2,x_0-x_3$. Their common zero on $X$ is precisely $P$, so away from $P$ the linear system of divisors is base-point-free and defines the projection from a point on a smooth quadricIn particular, the chart-ratio construction in part (a) makes this a morphism of schemes on $X\setminus \{P\}$, not merely a rational map there.

Write a target point as $[u:v:w]$. The corresponding line through $P$ consists of pointsSubstituting in the quadric equation givesRemoving $P$ means $\mu \ne 0$, so set $\mu =1$. Over the residue field $K$ of the target point, its scheme-theoretic fibre is consequentlyIf $u\ne v$, this is one reduced point, with $\lambda =-uw/(u-v)$. If $u=v$ and $uw\ne 0$, it is empty: the corresponding line meets the quadric only at the removed point, with intersection multiplicity two. If $u=v$ and $uw=0$, the line lies entirely on the quadric, and the scheme-theoretic fibre is ${\mathbb{A}}_K^1$, the line with $P$ removed.

The last case occurs at exactly $[0:0:1]$ and $[1:1:0]$. Their lines are respectivelythe two rulings of a smooth quadric surface through $P$. Thus the image consists of the complement of the line $u=v$, together with those two points. The line $u=v$ parametrizes directions in the tangent plane $x_1=x_2$ at $P$. The calculation describes the scheme-theoretic fibres over arbitrary target points and works in every characteristic.

Work first over an affine neighbourhood $\mathrm{Spec}A$ of $y=\mathfrak{p}$, with an affine chart $\mathrm{Spec}B$ in its inverse image. The scheme-theoretic fibre on this chart has ringwhere $\kappa (\mathfrak{p})$ is the residue field. The prime ideal correspondence for localization and the correspondence for a quotient identify its points with primes $\mathfrak{q}\subset B$ containing $\mathfrak{p}B$ and disjoint from $S$. These two conditions are exactly $\mathfrak{q}\cap A=\mathfrak{p}$.

Under this bijection, each principal open $D(b\otimes 1)$ corresponds to $D(b)\cap f^{-1}\{y\}$. These are bases for the two topologies: every element of the fibre ring is represented by $b/s$, with the image of $s$ invertible. Thus the bijection is a homeomorphism. The identifications agree on overlapping charts and glue, provingThe scheme-theoretic fibre may nevertheless carry nilpotents, so this topological statement does not identify its structure sheaf with a naive restriction of ${\mathcal{O}}_X$.

The rings $A,B$ are nonzero integral domains. Let $\eta$ be the generic point of $Y$, corresponding to $(0)\subseteq A$. Surjectivity provides a prime ideal $\mathfrak{q}\subseteq B$ with ${\alpha }^{-1}(\mathfrak{q})=0$. Since $\ker \alpha \subseteq {\alpha }^{-1}(\mathfrak{q})$, the ring homomorphism $\alpha$ is injective.

Put $K=\mathrm{Frac}A$ and $S=A\setminus \{0\}$. The scheme-theoretic fibre at $\eta$ isEvery element of $\alpha (S)$ is nonzero, so the localization $S^{-1}B$ is a nonzero integral domain. Its zero ideal is prime, ensuring that its spectrum is nonempty; an affine spectrum of an integral domain is an integral scheme. This proves the integrality of the generic fibre of an affine dominant morphism. In fact the proof only needs injectivity of $\alpha$, rather than surjectivity at every point.

Use the ring homomorphism $k[t]\to k[x]$ given by $t\mapsto x^2$. The source scheme $X=\mathrm{Spec}k[x]$ is an integral scheme. At the rational point $y=(t)$, its residue field is $k$, and its scheme-theoretic fibre isThe class of $x$ is nonzero but has square zero, so this is not a reduced scheme. This example works over every field. An integral total space therefore need not have reduced fibres.

The scheme-theoretic fibres over positive real points, negative real points, zero, and nonreal closed points have real coordinate algebras ${\mathbb{R}}^2\times \mathbb{C}$, ${\mathbb{C}}^2$, $\mathbb{R}[x]/(x^4)$, and ${\mathbb{C}}^4$, respectively. Their ordinary irreducible component counts are $3,2,1,4$, so there are four real-scheme isomorphism classes. The last case pulls back an irreducible quadratic to a squarefree degree-eight polynomial, with four conjugate pairs of nonreal roots. The zero fibre retains its nilpotents and is not a reduced point.