= Solution
A <ring homomorphism> is understood to preserve $1$. Start with $\varphi:A\to B$. Its map on <spectra of rings> is
$$
g:\operatorname{Spec}B\longrightarrow\operatorname{Spec}A,
\qquad \mathfrak q\longmapsto\varphi^{-1}(\mathfrak q).
$$
Contraction gives a <prime ideal>, and $g^{-1}(D(a))=D(\varphi(a))$. Thus $g$ is <continuous> for the <Zariski topology>. On each <principal open subscheme>, the <structure sheaf> map is the <ring homomorphism>
$$
A_a\longrightarrow B_{\varphi(a)},\qquad
\frac{b}{a^m}\longmapsto\frac{\varphi(b)}{\varphi(a)^m}.
$$
These maps commute with restrictions, so they define a <sheaf morphism> $\mathcal O_X\to g_*\mathcal O_Y$. At $\mathfrak q$, with $\mathfrak p=\varphi^{-1}(\mathfrak q)$, its <stalk> map is $A_{\mathfrak p}\to B_{\mathfrak q}$. The inverse image of the <maximal ideal> $\mathfrak qB_{\mathfrak q}$ is $\mathfrak pA_{\mathfrak p}$, 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 $\varphi:A\to B$, 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 give
$$
a\in\mathfrak p
\iff a/1\in\mathfrak pA_{\mathfrak p}
\iff g^\#_{\mathfrak q}(a/1)\in\mathfrak qB_{\mathfrak q}
\iff\varphi(a)\in\mathfrak q.
$$
Consequently $\mathfrak p=\varphi^{-1}(\mathfrak q)$, so the underlying map is forced. On $D(a)$, the <sheaf morphism> is forced as well: it extends $\varphi$ 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 $\varphi$. \b[The two constructions are inverse], proving the <affine-target adjunction for schemes> in the affine-source case:
$$
\boxed{\operatorname{Hom}_{\mathrm{LRS}}(\operatorname{Spec}B,\operatorname{Spec}A)
\cong\operatorname{Hom}_{\mathrm{Ring}}(A,B).}
$$
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>:
* $(0)$, the <generic point> of the whole <affine plane>.
* $(P)$ for each nonconstant <irreducible polynomial> $P\in R$, taken up to multiplication by a nonzero real constant. These are the height-one points, each the <generic point> of the <integral scheme> $V(P)$.
* The <maximal ideals>, or <closed points>. By the <Zariski lemma>, their <residue fields> are finite algebraic extensions of $\mathbb R$. Because $\mathbb R$ is a <real closed field>, those fields are $\mathbb R$ or $\mathbb C$. The first type is $(t_1-a,t_2-b)$ with $(a,b)\in\mathbb R^2$. The second type is the kernel of evaluation $R\to\mathbb C$ at a nonreal pair $(a,b)\in\mathbb C^2\setminus\mathbb R^2$; the pairs $(a,b)$ and $(\bar a,\bar b)$ give the same <maximal ideal>, and these are the only repetitions.
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. \b[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 schemes>
$$
\pi:\operatorname{Spec}\mathbb C[t_1,t_2]\longrightarrow\operatorname{Spec}R
$$
is 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> is
$$
\mathbb C[t_1,t_2]\cong R[s]/(s^2+1),
$$
a free $R$-<module> with basis $1,s$. At $\mathfrak p\in\operatorname{Spec}R$, with <residue field> $K=\kappa(\mathfrak p)$, the <scheme-theoretic fibre> is
$$
\boxed{\pi^{-1}(\mathfrak p)_{\mathrm{sch}}
=\operatorname{Spec}\bigl(K[s]/(s^2+1)\bigr).}
$$
If $-1$ 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> $\mathbb C(t_1,t_2)$.
Back to article page
