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Write $x=3+18k$. The second modular congruence becomes $18k\equiv -1(\mathrm{mod}23)$. The Bezout identity above says that $9$ is the inverse of $18$ modulo $23$, so $k\equiv -9\equiv 14(\mathrm{mod}23)$. ThereforeThe checks are $255=18\cdot 14+3=23\cdot 11+2$. The Chinese remainder theorem explains uniqueness modulo $18\cdot 23=414$, since the moduli are coprime integers.

The idele group is the restricted productwith componentwise multiplication and the restricted product topology on the idele group relative to the unit groups at finite places. We need weak approximation for number fields only at the finitely many specified coordinates, not density in the full restricted-product topology.

Here is a proof including the Archimedean places. At each finite place in $S$, choose $a_v\in K$ sufficiently close to $x_v$, using the definition of the completion. Choose a nonzero $d\in {\mathcal{O}}_K$ clearing their denominators. Pick powers ${\mathfrak{p}}_v^{m_v}$ so that $|z/d{|}_v<\epsilon$ whenever $z\in {\mathfrak{p}}_v^{m_v}$. Put $I={\prod }_{v\in S,v\text{ finite}}{\mathfrak{p}}_v^{m_v}$. For each positive rational integer $M$ coprime to the rational primes below these places, the Chinese remainder theorem gives $b_M\in {\mathcal{O}}_K$ with $b_M\equiv Md{a}_v(\mathrm{mod}{\mathfrak{p}}_v^{m_v})$. All of $b_M+I$ satisfy the same congruences.

Under the Minkowski embedding of a number field, $I$ is a full lattice in ${\mathbb{R}}^{r_1}\times {\mathbb{C}}^{r_2}$. A fundamental parallelepiped is bounded, so every real vector is within a fixed distance of this lattice, uniformly over its translates. Choose a target vector whose coordinates at the specified Archimedean places are those of $Md{x}_v$, assigning arbitrary values at the other Archimedean places. There is $b\in b_M+I$ within that fixed distance of the target. Dividing by $Md$ makes all specified Archimedean errors tend to zero as $M\to \infty$ through the allowed integers. At finite places $M$ is a unit, so the earlier congruence bounds persist. By taking the initial approximations sufficiently close, the ultrametric inequality gives the requested errors there too. Thus some $\boxed{y=b/(Md)\in K}$ satisfies every required inequality. If there are no finite places, set $I={\mathcal{O}}_K$ and omit the congruences.

Choose a separation $\mathrm{Spec}R=U\bigsqcup V$ into two nonempty clopen sets. Both are closed in the Zariski topology, so write $U=V(I)$ and $V=V(J)$ for ideals $I,J$. Their disjointness gives $V(I+J)=\varnothing$. A proper ideal lies in a maximal ideal, so this forces $I+J=R$.

Their union is all of the spectrum of a commutative ring, and $V(I)\cup V(J)=V(IJ)$. Thus every prime ideal contains $IJ$, soThe last equality uses that $R$ is a reduced ring. For comaximal ideals, $I\cap J=IJ$: if $i+j=1$ and $r\in I\cap J$, then $r=ri+rj\in IJ$. Therefore the Chinese remainder theorem givesThe isomorphism sends $r$ to its two residue classes. Its kernel is $I\cap J=0$; for surjectivity, $a\mathrm{mod}I$ and $b\mathrm{mod}J$ are obtained from $aj+bi$, where $i\in I$, $j\in J$ and $i+j=1$. The two factors are nonzero because $U,V$ are nonempty, so $I,J$ are proper.

This is a disconnected reduced spectrum product decomposition. The elements $i,j$ also satisfy $ij=0$, $i^2=i$, $j^2=j$, exhibiting the ring product decomposition by an idempotent. No Noetherian hypothesis is required.

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.

Factor $X^3+X=X(X-i)(X+i)$. These factors generate pairwise comaximal ideals, so the Chinese remainder theorem givesThe three copies of $\mathbb{C}$ have different module structures: $X$ acts by $0,i,-i$, respectively. Thus $q(X)$ acts on $(a,b,c)$ as $(q(0)a,q(i)b,q(-i)c)$, making $\Phi$ a module homomorphism.

The inverse, obtained by Lagrange interpolation, isEvaluating at $0,i,-i$ recovers $(a,b,c)$. Conversely, a polynomial whose three evaluations vanish is divisible by $X(X-i)(X+i)$, so these maps are mutual inverses on the quotient module. Three copies with identical trivial $X$-action would not give the required isomorphism.

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.

Interpret the language of ordered rings in the standard natural numbers, with nonlogical symbols $0,1,+,\times ,<$. We give a single first-order formula $F(n,y)$ defining $y=n!$, rather than a separate expression with $n$ multiplication signs. The key is Gödel beta-function sequence coding. The remainder relationuses only the allowed symbols; all variables range over $\mathbb{N}$. With $c>0$, it assigns a unique remainder $r$ at each position $i$.

The factorial graph formula via remainder coding asserts that one code contains the initial value $1$, the final value $y$, and every recurrence step. Here is the fully expanded formula, with no remainder-function or factorial symbol:The first equation represents remainder $1$ at position zero: its omitted explicit remainder bound follows from $c>0$. The second line represents remainder $y$ at position $n$. The universally checked equations represent the adjacent remainders and impose multiplication by $i+1$. Uniqueness of remainders therefore forces them, successively, to be $0!,1!,\dots ,n!$, proving $F(n,y)\Rightarrow y=n!$ by mathematical induction. When $n=0$, the recurrence clause is empty and the two endpoint clauses force $y=1$.

Conversely, choose a positive $c$ larger than every value in the finite sequence $0!,1!,\dots ,n!$ and divisible by each positive integer at most $n$. The moduli $d_i=1+(i+1)c$, $0\le i\le n$, are pairwise coprime. Indeed, for $i<j$, a common divisor of $d_i,d_j$ divides $(j-i)c$ and is coprime to $c$, so divides $j-i$. Since $j-i$ divides $c$, the common divisor must be one. The Chinese remainder theorem gives a nonnegative $b$ with $b\equiv i!(\mathrm{mod}{d}_i)$ for every $i$. The chosen values are smaller than their moduli, so they are the actual remainders; the appropriate quotient witnesses then make $F(n,n!)$ true. ThusThe formula has fixed length, independent of $n$. For example, writing its displayed syntax in plain text with one-character variables, ordinary parentheses, and explicit logical connectives takes fewer than $1000$ characters and fewer than $500$ logical-symbol tokens. Its length is therefore $O(1)$ as a uniform definition of the graph. Numeral names for particular input values, if substituted for the variables, add their own encoding lengths. Over the ordered ring $\mathbb{Z}$, restrict every quantifier to nonnegative integers and require $n,y\ge 0$; this gives the same definition on its nonnegative part. It is the chosen arithmetic structure, not ring axioms alone, that makes the formula define factorial.