Describe \(\Spec \RR[x]\). How does its topological space compare to the set \(\RR\)? To \(\CC\)?
Solution
Put \[ X=\operatorname{Spec}\mathbb R[x]. \]
1. The prime ideals of \(\mathbb R[x]\) are\[ (0) \]and the principal ideals\[ (p(x)) \]generated by irreducible polynomials \(p(x)\).
The polynomial ring \(\mathbb R[x]\) is a principal ideal domain. Every nonzero prime ideal in a PID is generated by an irreducible element. Since \(\mathbb R[x]\) is an integral domain, \((0)\) is prime as well.
2. Every irreducible polynomial in \(\mathbb R[x]\) has degree \(1\) or \(2\).The irreducibles are, up to multiplication by a nonzero scalar,\[ x-a, \qquad a\in\mathbb R, \]and\[ (x-a)^2+b^2, \qquad a\in\mathbb R, \quad b>0. \]
By the fundamental theorem of algebra, every nonconstant real polynomial factors over \(\mathbb C\) into linear factors. Nonreal roots occur in conjugate pairs.
If a real polynomial is irreducible and has a real root, it must be linear. If it has no real root, choose a nonreal root \[ a+ib, \qquad b\ne0. \] Its conjugate \(a-ib\) is also a root, so the real quadratic \[ (x-a-ib)(x-a+ib) =(x-a)^2+b^2 \] divides the polynomial. Irreducibility forces equality up to a scalar. Taking \(b>0\) makes the representation unique.
3. Thus the points of \(X\) are\[ \boxed{ \eta=(0), \quad x_a=(x-a)\ (a\in\mathbb R), \quad x_{a,b}=((x-a)^2+b^2)\ (a\in\mathbb R,b>0). } \]The point \(\eta\) is generic and all the other points are closed.
Step 1 and 2 list all prime ideals. The closure of \((0)\) is all of \(X\), since \[ V(0)=X. \] Every nonzero prime ideal in the PID \(\mathbb R[x]\) is maximal, so each corresponding point is closed.
4. The residue fields at the closed points are\[ \boxed{ \kappa(x_a)=\mathbb R, \qquad \kappa(x_{a,b})\cong\mathbb C. } \]
For a maximal ideal generated by an irreducible polynomial \(p\), \[ \kappa((p)) =\mathbb R[x]/(p). \] For \(p=x-a\) this quotient is \(\mathbb R\).
For \[ p=(x-a)^2+b^2, \qquad b>0, \] the class of \[ \frac{x-a}{b} \] squares to \(-1\), so the quotient is isomorphic to \(\mathbb C\).
5. Every proper closed subset of \(X\) is finite and consists only of closed points.
Every closed subset is \[ V(I)=V(f) \] for some \(f\in\mathbb R[x]\), because the ring is a PID.
If \(f=0\), then \[ V(f)=X. \] If \(f\ne0\), the generic prime \((0)\) does not contain \(f\). A closed point \((p)\) lies in \(V(f)\) exactly when \[ p\mid f. \] A nonzero polynomial has only finitely many irreducible factors. Hence \(V(f)\) is a finite set of closed points.
6. The real line \(\mathbb R\) identifies with the subset of real closed points\[ \{(x-a):a\in\mathbb R\}\subseteq X, \]and the induced Zariski topology on this subset is the cofinite topology.
The map \[ a\longmapsto(x-a) \] is a bijection from \(\mathbb R\) to the closed points whose residue field is \(\mathbb R\).
By 5, the intersection of a proper closed subset of \(X\) with these points is finite. Conversely, any finite set \[ \{a_1,\ldots,a_r\} \] is the zero set of \[ \prod_{i=1}^r(x-a_i). \] Thus the closed subsets of the induced topology are exactly the finite subsets and the whole set.
So this is the Zariski, not the Euclidean, topology on the real line.
7. Set-theoretically, the closed points of \(X\) are naturally\[ \boxed{ \mathbb C/\{z\sim\bar z\}. } \]
A real point \(a\in\mathbb R\) is fixed by conjugation and corresponds to the linear prime \[ (x-a). \] A nonreal conjugate pair \[ \{a+ib,a-ib\}, \qquad b>0, \] corresponds to its minimal polynomial over \(\mathbb R\), \[ (x-a)^2+b^2. \] Step 2 shows that every closed point occurs uniquely in one of these ways.
8. More canonically, the finite integral extension\[ \mathbb R[x]\hookrightarrow\mathbb C[x] \]induces a surjective closed map\[ \pi: \operatorname{Spec}\mathbb C[x] \longrightarrow \operatorname{Spec}\mathbb R[x] \]which sends\[ (x-z) \quad\text{and}\quad (x-\bar z) \]to the same point when \(z\notin\mathbb R\) and is one-to-one on the real points and on the generic point.
The extension \(\mathbb C[x]/\mathbb R[x]\) is finite, hence integral. Lying over makes the induced map on spectra surjective, and integral morphisms are closed.
For \(z\in\mathbb C\), the contraction of \[ (x-z)\subseteq\mathbb C[x] \] to \(\mathbb R[x]\) is the kernel of evaluation \[ \mathbb R[x]\to\mathbb C, \qquad f\mapsto f(z). \] If \(z=a\in\mathbb R\), this kernel is \((x-a)\). If \[ z=a+ib, \qquad b\ne0, \] the kernel is generated by the real minimal polynomial \[ (x-a)^2+b^2, \] which is also the contraction of \((x-\bar z)\).
The zero ideal contracts to the zero ideal, so the generic point maps to the generic point.
9. Thus one may picture \(\operatorname{Spec}\mathbb R[x]\) as\[ \boxed{ \text{one generic point together with the conjugation quotient of }\mathbb C \text{ as its closed points}, } \]with every proper closed subset finite.
This combines 3, 5, and <1>7–<1>8.
10. Q.E.D.
Steps <1>3–<1>9 give the requested description and comparison with \(\mathbb R\) and \(\mathbb C\).