← Expository Notes

The Arithmetic Surface \(\operatorname{Spec}\mathbb{Z}[x]\)

Last edited August 2026

Below we will look at Mumford's drawing of \(\operatorname{Spec}\mathbb{Z}[x]\) as a surface fibered over \(\operatorname{Spec}\mathbb{Z}\). This note classifies the primes of \(\mathbb{Z}[x]\) with proof, identifies that fibration, and then reads the splitting, inertia, and ramification of primes in number fields off the resulting geometry.

In the figure below, vertical lines are the fibers \(\mathbb{A}^1_{\mathbb{F}_p}\), the dashed line at the right is the generic fiber \(\mathbb{A}^1_{\mathbb{Q}}\), and the horizontal line at the bottom is the base \(\operatorname{Spec}\mathbb{Z}\). On a fiber, small dots are the \(\mathbb{F}_p\)-rational points \((p, x-a)\), hollow dots are closed points whose residue field is larger than \(\mathbb{F}_p\), and the smudge at the top is the fiber's generic point.

A horizontal curve \(V(f)\) is drawn as \(\deg f\) strands, one for each sheet of the covering \(V(f) \to \operatorname{Spec}\mathbb{Z}\). They land on separate points where \(p\) splits, merge into a single hollow dot where \(p\) is inert, and touch where \(p\) ramifies, converging at the right to the one point of \(\mathbb{A}^1_{\mathbb{Q}}\) the curve came from. Every line, dot, strand, and smudge is a prime ideal of \(\mathbb{Z}[x]\); clicking one gives its generators, height, residue field, and local ring. The next section explains why a single prime is drawn as a whole line or curve rather than a dot.

Figure. \(\operatorname{Spec}\mathbb{Z}[x]\), after Mumford. Click any line, dot, strand, or smudge to see the prime ideal it denotes. A hollow dot is a closed point with residue field larger than \(\mathbb{F}_p\); a shaded halo is multiplicity, i.e. ramification. The primes are spaced evenly rather than to scale. The height of an \(\mathbb{F}_p\)-rational point \((p, x-a)\) is \(\bigl(a+\tfrac12\bigr)/p\), which is a real invariant of the point and not decoration: it is why \(V(2x-1)\) settles toward the middle of the plate, since its root mod \(p\) is \((p+1)/2\) and the curve is tracking the rational number \(\tfrac12\). Heights of higher-degree points are arbitrary, though consistent across curves, since such a point has no natural coordinate. All incidence data, such as which curve meets which point and with what \(e\), \(f\), and residue field, is computed rather than drawn by hand.

1Points and their closures

Fix a commutative ring \(R\). The points of \(\operatorname{Spec} R\) are the prime ideals of \(R\), and the closed sets are exactly the sets \[ V(I) \;=\; \{\,\mathfrak{p} \in \operatorname{Spec} R \;:\; I \subseteq \mathfrak{p}\,\} \] for ideals \(I \subseteq R\); for a subset \(Z \subseteq \operatorname{Spec} R\) write \(I(Z) = \bigcap_{\mathfrak{p} \in Z} \mathfrak{p}\). The residue field at \(\mathfrak{p}\) is \(\kappa(\mathfrak{p}) = \operatorname{Frac}(R/\mathfrak{p})\), and we write \(\mathbb{A}^1_R = \operatorname{Spec} R[x]\) for the affine line over \(R\). We say a nonempty closed set is irreducible if it is not the union of two strictly smaller closed subsets. Points of this space may not be closed, and the two results below say what a point morally is: it is a closed set, and exactly an irreducible one. That is why a single prime ideal appears in the drawing as a line or a curve rather than as a dot.

Lemma (closure of a point).
For every \(\mathfrak{p} \in \operatorname{Spec} R\) we have \(\overline{\{\mathfrak{p}\}} = V(\mathfrak{p})\). Hence \(\mathfrak{q} \in \overline{\{\mathfrak{p}\}}\) if and only if \(\mathfrak{p} \subseteq \mathfrak{q}\), and \(\{\mathfrak{p}\}\) is closed if and only if \(\mathfrak{p}\) is maximal.
Proof. \(V(\mathfrak{p})\) is closed and contains \(\mathfrak{p}\), because \(\mathfrak{p} \subseteq \mathfrak{p}\). It is the smallest such set: if \(V(I)\) is any closed set containing \(\mathfrak{p}\), then \(I \subseteq \mathfrak{p}\), so every \(\mathfrak{q} \in V(\mathfrak{p})\) satisfies \(I \subseteq \mathfrak{p} \subseteq \mathfrak{q}\) and therefore lies in \(V(I)\); that is, \(V(\mathfrak{p}) \subseteq V(I)\). Since the closure is the smallest closed set containing the point, \(\overline{\{\mathfrak{p}\}} = V(\mathfrak{p})\). The first consequence is just the definition of \(V(\mathfrak{p})\). For the second, \(V(\mathfrak{p}) = \{\mathfrak{p}\}\) says exactly that no prime strictly contains \(\mathfrak{p}\), and since every proper ideal lies in a maximal one, that is maximality.
Theorem.
The assignment \(\mathfrak{p} \mapsto V(\mathfrak{p})\) is a bijection \[ \{\text{primes of } R\} \;\xrightarrow{\ \sim\ }\; \{\text{irreducible closed subsets of } \operatorname{Spec} R\}, \] with inverse \(Z \mapsto I(Z)\). It reverses inclusions: \(\mathfrak{p} \subseteq \mathfrak{q}\) if and only if \(V(\mathfrak{q}) \subseteq V(\mathfrak{p})\). Moreover \(\mathfrak{p}\) is the unique point of \(V(\mathfrak{p})\) whose closure is all of \(V(\mathfrak{p})\), i.e. it is the generic point.
Proof.

\(V(\mathfrak{p})\) is irreducible. By the lemma it is \(\overline{\{\mathfrak{p}\}}\), and the closure of a point is always irreducible: it is nonempty, and if \(\overline{\{\mathfrak{p}\}} = Z_1 \cup Z_2\) with \(Z_i\) closed, then \(\mathfrak{p}\) lies in one of them, say \(Z_1\), and \(Z_1\) being closed forces \(\overline{\{\mathfrak{p}\}} \subseteq Z_1\), so \(Z_1\) is not strictly smaller.

Injectivity. If \(V(\mathfrak{p}) = V(\mathfrak{q})\), then \(\mathfrak{q} \in V(\mathfrak{q}) = V(\mathfrak{p})\) gives \(\mathfrak{p} \subseteq \mathfrak{q}\), and the same argument with the roles exchanged gives \(\mathfrak{q} \subseteq \mathfrak{p}\).

Surjectivity. Let \(Z\) be irreducible closed and set \(\mathfrak{p} = I(Z)\). First, \(\mathfrak{p}\) is prime. It is proper, since \(Z\) is nonempty and so \(\mathfrak{p}\) is contained in some prime. Suppose \(fg \in \mathfrak{p}\). For any prime \(\mathfrak{q}\) we have \(fg \in \mathfrak{q}\) if and only if \(f \in \mathfrak{q}\) or \(g \in \mathfrak{q}\), so \(V(fg) = V(f) \cup V(g)\), and therefore \[ Z \;=\; \big(Z \cap V(f)\big) \,\cup\, \big(Z \cap V(g)\big), \] a union of two closed subsets of \(Z\). Irreducibility forces one of them to be all of \(Z\), say \(Z \subseteq V(f)\), which says \(f \in \mathfrak{q}\) for every \(\mathfrak{q} \in Z\), i.e. \(f \in I(Z) = \mathfrak{p}\). Second, \(V(\mathfrak{p}) = Z\). Writing \(Z = V(J)\), every \(\mathfrak{q} \in Z\) contains \(J\), so \(J \subseteq I(Z) = \mathfrak{p}\) and hence \(V(\mathfrak{p}) \subseteq V(J) = Z\); conversely if \(\mathfrak{q} \in Z\) then \(I(Z) \subseteq \mathfrak{q}\) by definition of \(I(Z)\), so \(\mathfrak{q} \in V(\mathfrak{p})\).

Inclusions and the generic point. If \(\mathfrak{p} \subseteq \mathfrak{q}\) and \(\mathfrak{r} \in V(\mathfrak{q})\), then \(\mathfrak{p} \subseteq \mathfrak{q} \subseteq \mathfrak{r}\), so \(V(\mathfrak{q}) \subseteq V(\mathfrak{p})\); conversely \(V(\mathfrak{q}) \subseteq V(\mathfrak{p})\) applied to \(\mathfrak{q} \in V(\mathfrak{q})\) gives \(\mathfrak{p} \subseteq \mathfrak{q}\). Finally, if \(\mathfrak{q} \in V(\mathfrak{p})\) has \(\overline{\{\mathfrak{q}\}} = V(\mathfrak{p})\), then \(V(\mathfrak{q}) = V(\mathfrak{p})\), so \(\mathfrak{q} = \mathfrak{p}\) by injectivity.

Let’s see what our work tells us so far. In the picture above, the vertical line over \(2\) is the single prime ideal \((2) \subset \mathbb{Z}[x]\), drawn as its closure. A horizontal curve is a single prime ideal \((f)\). The haze covering the whole plate is the single prime \((0)\), whose closure is everything. Only the dots at the crossings are what a classical geometer would call points, and by the lemma those are exactly the maximal ideals.

Note: the smudges are drawn that way for a reason: a generic point is not located anywhere in particular, it is everywhere in its closure at once. Fuzz is the honest notation.

2Finding the primes of \(\mathbb{Z}[x]\)

\(\mathbb{Z}[x]\) is a Noetherian UFD but not a PID, e.g. the ideal \((2,x)\) needs two generators. That failure of principality is exactly what gives the space its second dimension. Now, the classification of primes needs one classical input, and that input is about content, so we define it first on \(\mathbb{Q}[x]\) as well as on \(\mathbb{Z}[x]\), since the proof moves between the two.

Definition.

For a nonzero \(f = a_nx^n + \cdots + a_0 \in \mathbb{Z}[x]\), the content is \[ c(f) \;=\; \gcd(a_0, \ldots, a_n) \;\in\; \mathbb{Z}_{>0}, \] and \(f\) is primitive if \(c(f) = 1\). Dividing through, \(f = c(f)\,f_0\) with \(f_0 \in \mathbb{Z}[x]\) primitive.

For a nonzero \(g \in \mathbb{Q}[x]\), choose \(d \in \mathbb{Z}_{>0}\) with \(dg \in \mathbb{Z}[x]\), e.g. the common denominator of the coefficients, and set \[ c(g) \;=\; \frac{c(dg)}{d} \;\in\; \mathbb{Q}_{>0}. \]

Lemma (Normalization).
The content of a rational polynomial is well defined, agrees with the integral definition on \(\mathbb{Z}[x]\), and satisfies \(c(\lambda g) = |\lambda|\,c(g)\) for \(\lambda \in \mathbb{Q}^{\times}\). Every nonzero \(g \in \mathbb{Q}[x]\) factors uniquely as \[ g \;=\; c(g)\, g_0, \qquad g_0 \in \mathbb{Z}[x] \text{ primitive}, \] and \(g \in \mathbb{Z}[x]\) if and only if \(c(g) \in \mathbb{Z}\).
Proof. For \(m \in \mathbb{Z}_{>0}\) and \(h \in \mathbb{Z}[x]\) we have \(c(mh) = m\,c(h)\), since the gcd of the coefficients scales. If \(d, d'\) both clear denominators of \(g\), then \(d'\,c(dg) = c(d'dg) = d\,c(d'g)\), so \(c(dg)/d = c(d'g)/d'\) and \(c(g)\) is well defined; taking \(d = 1\) recovers the integral definition. For \(\lambda = a/b\) with \(a,b \in \mathbb{Z}_{>0}\), choosing \(d\) for \(g\) makes \(bd\) clear denominators of \(\lambda g\), and \(c(\lambda g) = c(adg)/(bd) = a\,c(dg)/(bd) = \lambda\,c(g)\); the case \(\lambda < 0\) is the same since content ignores sign.

For the factorization, pick \(d\) with \(dg \in \mathbb{Z}[x]\) and write \(dg = c(dg)\,g_0\) with \(g_0\) primitive; dividing by \(d\) gives \(g = c(g)\,g_0\). It is unique because \(g_0 = g/c(g)\) is determined by \(g\): if \(g = \lambda h_0\) with \(\lambda \in \mathbb{Q}_{>0}\) and \(h_0\) primitive, then \(c(g) = \lambda\,c(h_0) = \lambda\). Finally, if \(c(g) \in \mathbb{Z}\) then \(g = c(g)g_0 \in \mathbb{Z}[x]\), and conversely \(c\) is integral on \(\mathbb{Z}[x]\) by definition.

Lemma (Gauss).
Content is multiplicative: \(c(fg) = c(f)\,c(g)\) for all nonzero \(f,g \in \mathbb{Q}[x]\). In particular a product of primitive polynomials in \(\mathbb{Z}[x]\) is primitive. Consequently, if \(f \in \mathbb{Z}[x]\) is primitive then \(f\mathbb{Q}[x] \cap \mathbb{Z}[x] = f\mathbb{Z}[x]\), and a primitive \(f\) of degree \(\ge 1\) is irreducible in \(\mathbb{Z}[x]\) if and only if it is irreducible in \(\mathbb{Q}[x]\).
Proof. Write \(f = c(f)f_0\) and \(g = c(g)g_0\) with \(f_0, g_0 \in \mathbb{Z}[x]\) primitive. By homogeneity of the content, \(c(fg) = c(f)c(g)\,c(f_0g_0)\), so everything reduces to showing \(f_0g_0\) is primitive. Let \(p\) be prime. Reduction mod \(p\) is a ring map \(\mathbb{Z}[x] \to \mathbb{F}_p[x]\), and \(\mathbb{F}_p[x]\) is a domain. Primitivity says that \(\bar f_0\) and \(\bar g_0\) are both nonzero, so \(\overline{f_0g_0} = \bar f_0\,\bar g_0\) is nonzero, i.e. \(p \nmid c(f_0g_0)\). As \(p\) was arbitrary, \(c(f_0g_0) = 1\).

Now let \(f\) be primitive and \(h = fq \in \mathbb{Z}[x]\) with \(q \in \mathbb{Q}[x]\) nonzero. Then \(c(h) = c(f)c(q) = c(q)\), and \(c(h) \in \mathbb{Z}\) because \(h \in \mathbb{Z}[x]\); so \(c(q) \in \mathbb{Z}\) and therefore \(q \in \mathbb{Z}[x]\) by normalization. That gives \(f\mathbb{Q}[x] \cap \mathbb{Z}[x] = f\mathbb{Z}[x]\).

For irreducibility, a factorization in \(\mathbb{Z}[x]\) into factors of degree \(\ge 1\) is already one in \(\mathbb{Q}[x]\). Conversely if \(f = gh\) in \(\mathbb{Q}[x]\) with \(\deg g, \deg h \ge 1\), normalize both: \(f = c(g)c(h)\,g_0h_0\), and taking contents gives \(1 = c(f) = c(g)c(h)\), so \(f = g_0h_0\) is a factorization in \(\mathbb{Z}[x]\) with both factors of degree \(\ge 1\).

Theorem (classification of primes).
Every prime ideal of \(\mathbb{Z}[x]\) is of exactly one of the following four types.
  1. \((0)\), of height \(0\);
  2. \((p)\), for \(p\) a rational prime, of height \(1\);
  3. \((f)\), for \(f \in \mathbb{Z}[x]\) primitive and irreducible of degree \(\ge 1\), of height \(1\);
  4. \((p, f)\), for \(p\) a rational prime and \(f \in \mathbb{Z}[x]\) a lift of an irreducible \(\bar f \in \mathbb{F}_p[x]\), of height \(2\) and maximal.
Proof. Let \(\mathfrak{p} \subset \mathbb{Z}[x]\) be prime. The contraction \(\mathfrak{p} \cap \mathbb{Z}\) is prime in \(\mathbb{Z}\), so it is \((0)\) or \((p)\). These two cases are the whole proof, and in each one we reduce to a principal ideal domain.

Case \(\mathfrak{p} \cap \mathbb{Z} = (p)\). Then \(p \in \mathfrak{p}\), so \(\mathfrak{p}\) is the preimage of a prime of \(\mathbb{Z}[x]/(p) \cong \mathbb{F}_p[x]\). But \(\mathbb{F}_p[x]\) is a PID, so its primes are \((0)\) and \((\bar f)\) with \(\bar f\) irreducible. Pulling back gives \(\mathfrak{p} = (p)\) or \(\mathfrak{p} = (p, f)\) for any lift \(f\) of \(\bar f\). In the latter case \(\mathbb{Z}[x]/(p,f) \cong \mathbb{F}_p[x]/(\bar f)\) is a finite field, so \(\mathfrak{p}\) is maximal.

Case \(\mathfrak{p} \cap \mathbb{Z} = (0)\). Then \(\mathfrak{p}\) misses the multiplicative set \(S = \mathbb{Z}\setminus\{0\}\), so \(\mathfrak{p} \mapsto \mathfrak{p}\,\mathbb{Q}[x]\) is a bijection onto the primes of \(S^{-1}\mathbb{Z}[x] = \mathbb{Q}[x]\), with inverse \(\mathfrak{q} \mapsto \mathfrak{q}\cap\mathbb{Z}[x]\). Again \(\mathbb{Q}[x]\) is a PID. If \(\mathfrak{p}\,\mathbb{Q}[x] = (0)\) then \(\mathfrak{p} = (0)\). Otherwise \(\mathfrak{p}\,\mathbb{Q}[x] = (g)\) for some irreducible \(g \in \mathbb{Q}[x]\). By normalization \(g = c(g)\,f\) with \(f \in \mathbb{Z}[x]\) primitive, so \(f\mathbb{Q}[x] = g\mathbb{Q}[x]\) and \(f\) is irreducible in \(\mathbb{Z}[x]\) by Gauss. Then \[ \mathfrak{p} \;=\; f\mathbb{Q}[x] \cap \mathbb{Z}[x] \;=\; f\mathbb{Z}[x], \] the last equality being Gauss again.

The types are mutually exclusive by inspecting contractions to \(\mathbb{Z}\) and heights, and the height claims follow from the chains exhibited in the corollary that follows.

Corollary.
\(\dim \mathbb{Z}[x] = 2\). Every maximal chain of primes has length \(2\) and looks like \[ (0) \subsetneq (p) \subsetneq (p,f) \qquad\text{or}\qquad (0) \subsetneq (f) \subsetneq (p,f). \] The closed points are exactly the \((p,f)\), and every residue field at a closed point is finite, namely \(\kappa(p,f) \cong \mathbb{F}_{p^{\deg \bar f}}\).

The two shapes of maximal chain are the two ways to walk down to a closed point, and they are already the vertical/horizontal dichotomy. Notice also the last sentence: all closed points have finite residue field, and the residue characteristic varies from point to point. Unlike \(\mathbb{A}^2_k\), there is no field over which the whole picture lives.

This classification also settles the local structure at a closed point. Recall that a Noetherian local ring \((A,\mathfrak{m})\) of Krull dimension \(d\) is regular if \(\mathfrak{m}\) can be generated by \(d\) elements, and that such a generating set is called a regular system of parameters.

Corollary (local rings at closed points).
For every closed point \(\mathfrak{m} = (p,f)\) of \(\operatorname{Spec}\mathbb{Z}[x]\), the local ring \(\mathbb{Z}[x]_{\mathfrak{m}}\) is regular local of dimension \(2\), and \(p, f\) is a regular system of parameters.
Proof. \(\mathbb{Z}[x]\) is Noetherian, so \(\mathbb{Z}[x]_{\mathfrak{m}}\) is a Noetherian local ring, and its Krull dimension is the height of \(\mathfrak{m}\), which the classification gives as \(2\). Its maximal ideal is generated by the images of the two elements \(p\) and \(f\).

3The fibration over \(\operatorname{Spec}\mathbb{Z}\)

A ring map \(\varphi : R \to A\) induces \(\pi : \operatorname{Spec} A \to \operatorname{Spec} R\) by \(\mathfrak{q} \mapsto \varphi^{-1}(\mathfrak{q})\), which is continuous because \(\pi^{-1}(V(I)) = V(\varphi(I)A)\). The fiber of \(\pi\) over \(\mathfrak{p} \in \operatorname{Spec} R\) is the preimage \(\pi^{-1}(\mathfrak{p})\) with its subspace topology: the set of primes of \(A\) that contract to \(\mathfrak{p}\).

Here \(\mathbb{Z} \hookrightarrow \mathbb{Z}[x]\) gives \(\pi : \mathbb{A}^1_{\mathbb{Z}} \to \operatorname{Spec}\mathbb{Z}\) with \(\pi(\mathfrak{q}) = \mathfrak{q} \cap \mathbb{Z}\), which is exactly the case division the proof of the classification used. So the classification is the fiber decomposition!

Proposition (the fibers).

Let \(p\) be a rational prime. Then \(\pi^{-1}\big((p)\big) = V(p)\), and reduction \(\mathbb{Z}[x] \twoheadrightarrow \mathbb{F}_p[x]\) induces a homeomorphism \(\pi^{-1}\big((p)\big) \cong \mathbb{A}^1_{\mathbb{F}_p}\).

Over the generic point, the inclusion \(\mathbb{Z}[x] \hookrightarrow \mathbb{Q}[x]\) induces a homeomorphism \(\pi^{-1}\big((0)\big) \cong \mathbb{A}^1_{\mathbb{Q}}\).

Proof. For the first, \(\mathfrak{q} \cap \mathbb{Z} = (p)\) if and only if \(p \in \mathfrak{q}\): one direction is immediate, and conversely \(\mathfrak{q} \cap \mathbb{Z}\) is a prime of \(\mathbb{Z}\) containing \(p\) and not containing \(1\), hence equals the maximal ideal \((p)\). So \(\pi^{-1}((p)) = V(p)\). Pulling back along the quotient map \(\mathbb{Z}[x] \to \mathbb{Z}[x]/(p) \cong \mathbb{F}_p[x]\) identifies the primes of \(\mathbb{F}_p[x]\), bijectively and inclusion-preservingly, with the primes of \(\mathbb{Z}[x]\) containing \(p\), and it matches \(V(\bar J)\) with \(V(J + (p))\), so it is a homeomorphism.

For the second, \(\mathfrak{q} \cap \mathbb{Z} = (0)\) says exactly that \(\mathfrak{q}\) is disjoint from \(S = \mathbb{Z} \setminus \{0\}\). Primes of \(S^{-1}\mathbb{Z}[x] = \mathbb{Q}[x]\) correspond to primes of \(\mathbb{Z}[x]\) disjoint from \(S\), again compatibly with closed sets.

Reading this against the classification of primes: the fiber over \((p)\) consists of its generic point \((p)\) together with the closed points \((p,f)\), and the fiber over \((0)\) consists of \((0)\) together with the primes \((f)\). Both fibers are affine lines, one over \(\mathbb{F}_p\) and one over \(\mathbb{Q}\), so \(\mathbb{A}^1_{\mathbb{Z}}\) is a family of lines parametrized by \(\operatorname{Spec}\mathbb{Z}\). That is all "surface" means here: a one-dimensional base with one-dimensional fibers. (The general statement, which we do not need, is that \(\pi^{-1}(\mathfrak{p}) \cong \operatorname{Spec}\big(A \otimes_R \kappa(\mathfrak{p})\big)\); the two computations above are the cases \(\mathbb{Z}[x] \otimes \mathbb{F}_p = \mathbb{F}_p[x]\) and \(\mathbb{Z}[x] \otimes \mathbb{Q} = \mathbb{Q}[x]\).)

Vertical and horizontal

Let \(Z = V(\mathfrak{p})\) be irreducible closed of dimension \(1\), so \(\mathfrak{p}\) has height \(1\) and the classification leaves two possibilities. Recall that \(\pi|_Z\) is said to be dominant if its image is dense in \(\operatorname{Spec}\mathbb{Z}\).

Exactly which characteristics is worth pinning down explicitly, because it is where the one non-monic curve in the figure misbehaves.

Lemma (which fibers a horizontal curve meets).
Let \(f \in \mathbb{Z}[x]\) be primitive irreducible of degree \(\ge 1\). Then \(V(f)\) meets the fiber over \(p\) if and only if the reduction \(\bar f \in \mathbb{F}_p[x]\) is nonconstant. If \(f\) is monic this holds for every \(p\), so \(\pi|_{V(f)}\) is surjective.
Proof. \(V(f)\) meets the fiber over \(p\) exactly when some prime contains both \(p\) and \(f\), i.e. when the ideal \((p,f)\) is proper, i.e. when \(\bar f\) is not a unit in \(\mathbb{F}_p[x]\). Since \(f\) is primitive, \(\bar f\) is nonzero, so it is a unit precisely when it is a nonzero constant. If \(f\) is monic then \(\bar f\) is monic of the same degree \(\ge 1\).

So \(V(2x-1)\) misses the fiber over \(2\), where its reduction is the constant \(1\); every monic curve meets every fiber.

Horizontal curves are orders

An order in a number field \(K\) of degree \(n\) is a subring \(\mathcal{O} \subseteq K\) that is finitely generated as a \(\mathbb{Z}\)-module and spans \(K\) over \(\mathbb{Q}\); equivalently, a subring free of rank \(n\) as a \(\mathbb{Z}\)-module.

Proposition (horizontal curves are orders).
Let \(f\) be primitive irreducible of degree \(n \ge 1\), let \(K = \mathbb{Q}[x]/(f)\), and let \(\alpha\) be the image of \(x\) in \(\mathbb{Z}[x]/(f)\). Then \(V(f) = \operatorname{Spec}\mathbb{Z}[\alpha]\), the residue field at the generic point \((f)\) is \(\kappa\big((f)\big) \cong K\), and if \(f\) is monic then \(\mathbb{Z}[\alpha]\) is an order in \(K\).
Proof. \(V(f)\) is the set of primes containing \(f\), which correspond to primes of \(\mathbb{Z}[x]/(f) = \mathbb{Z}[\alpha]\). Since \((f)\) is prime this ring is a domain, and \(\kappa((f)) = \operatorname{Frac}(\mathbb{Z}[\alpha])\) by definition. That fraction field is \(K\): it sits inside \(K = \mathbb{Q}[x]/(f)\), and every element of \(K\) has the form \(\tfrac1m h(\alpha)\) with \(h \in \mathbb{Z}[x]\) and \(m \in \mathbb{Z}\), hence is a ratio of elements of \(\mathbb{Z}[\alpha]\). If \(f\) is monic, division with remainder by \(f\) is available in \(\mathbb{Z}[x]\), so every class mod \(f\) has a unique representative of degree \(< n\); thus \(\mathbb{Z}[\alpha]\) is free with basis \(1, \alpha, \ldots, \alpha^{n-1}\), of rank \(n = [K:\mathbb{Q}]\), and it spans \(K\).

Monicity is not just decoration here. For \(f = 2x-1\) we get \(\mathbb{Z}[x]/(2x-1) \cong \mathbb{Z}[\tfrac12]\), which is not finitely generated as a \(\mathbb{Z}\)-module, since no finite set of elements has denominators bounded enough to contain every \(1/2^k\); so it is not an order in \(\mathbb{Q}\).

Proposition (spreading out).
Taking closures is a bijection \[ \{\text{closed points of } \mathbb{A}^1_{\mathbb{Q}}\} \;\xrightarrow{\ \sim\ }\; \{\text{horizontal curves in } \mathbb{A}^1_{\mathbb{Z}}\}, \qquad \mathfrak{q} \longmapsto \overline{\{\mathfrak{q}\}}, \] whose inverse sends a horizontal curve to its generic point, equivalently to its intersection with the generic fiber.
Proof. By the proposition on fibers, the primes of \(\mathbb{Q}[x]\) correspond to the primes \(\mathfrak{q}\) of \(\mathbb{Z}[x]\) with \(\mathfrak{q} \cap \mathbb{Z} = (0)\), which by our previous classification result are \((0)\) and the \((f)\). Discarding the zero ideal on both sides leaves the closed points of \(\mathbb{A}^1_{\mathbb{Q}}\) matched with the primes \((f)\), whose closures are by definition the horizontal curves. The dictionary of §1 says a nonempty irreducible closed set has a unique generic point, which gives the inverse; and that generic point is the unique element of \(V(f)\) contracting to \((0)\), i.e. \(V(f) \cap \pi^{-1}((0))\).

So a closed point of the generic fiber and a horizontal curve are the same datum: a number field \(K\), presented as a point of a \(\mathbb{Q}\)-line, spreads out into a curve over \(\mathbb{Z}\). In the figure each horizontal curve converges on the right to the single dot it came from.

4Where the curves meet the fibers

Fix \(f\) monic irreducible of degree \(n\) and a prime \(p\), and factor the reduction into distinct monic irreducibles in \(\mathbb{F}_p[x]\), \[ \bar f \;=\; \prod_{i=1}^{r} \bar g_i^{\,e_i}, \qquad f_i := \deg \bar g_i . \] Call \(e_i\) the ramification index and \(f_i\) the residue degree at \(\bar g_i\). Following the usual terminology, \(p\) is ramified for \(f\) if some \(e_i > 1\); inert if \(r = 1\) with \(e_1 = 1\) and \(f_1 = n\); and split completely if \(r = n\), forcing every \(e_i = f_i = 1\).

Proposition (intersections).
\(V(f) \cap V(p) = V(p,f)\), and this is the finite set of closed points \((p, g_i)\) for \(i = 1, \ldots, r\), with residue fields \(\kappa\big((p,g_i)\big) \cong \mathbb{F}_{p^{f_i}}\).
Proof. A prime contains both \(p\) and \(f\) if and only if it contains the ideal they generate, so \(V(f) \cap V(p) = V\big((p) + (f)\big) = V(p,f)\); these correspond to the primes of \(\mathbb{Z}[x]/(p,f) \cong \mathbb{F}_p[x]/(\bar f)\). The Chinese remainder theorem splits that ring as \(\prod_i \mathbb{F}_p[x]/(\bar g_i^{\,e_i})\), and each factor is local with maximal ideal \((\bar g_i)\), so the primes are the \((\bar g_i)\) and the residue field at \((p,g_i)\) is \(\mathbb{F}_p[x]/(\bar g_i)\), a field with \(p^{f_i}\) elements.

We can define the intersection number of the curve with the fiber to be the \(\mathbb{F}_p\)-dimension of the intersection, \[ V(f) \cdot V(p) \;:=\; \dim_{\mathbb{F}_p} \mathbb{Z}[x]/(p,f), \] which counts the intersection points with multiplicity: the point \((p,g_i)\) contributes \(e_i f_i\), its ramification index times the degree of its residue field.

Proposition (constant intersection number).
Let \(f\) be monic of degree \(n\). Then \(\mathbb{Z}[x]/(f)\) is free of rank \(n\) over \(\mathbb{Z}\), and for every prime \(p\), \[ V(f)\cdot V(p) \;=\; \sum_{i=1}^{r} e_i f_i \;=\; n . \]
Proof. Division with remainder by the monic \(f\) gives every class in \(\mathbb{Z}[x]/(f)\) a unique representative of degree \(< n\), so \(1, x, \ldots, x^{n-1}\) is a \(\mathbb{Z}\)-basis. Reducing mod \(p\), \(\mathbb{Z}[x]/(p,f) \cong \big(\mathbb{Z}[x]/(f)\big) \otimes_{\mathbb{Z}} \mathbb{F}_p\) has \(\mathbb{F}_p\)-dimension \(n\). For the middle equality, the decomposition in the previous proof gives \(\dim_{\mathbb{F}_p} \mathbb{F}_p[x]/(\bar f) = \sum_i \dim_{\mathbb{F}_p} \mathbb{F}_p[x]/(\bar g_i^{\,e_i}) = \sum_i e_i f_i\), since \(\deg \bar g_i^{\,e_i} = e_i f_i\).

So a degree-\(n\) horizontal curve crosses every vertical fiber in a scheme of dimension exactly \(n\), without exception. What changes from prime to prime is only how that \(n\) is distributed: spread across \(n\) separate points, concentrated at one point with a large residue field, or piled up at one point with multiplicity. Those three are split, inert, and ramified.

Two classical inputs

To keep going, we will need to blackbox some classical results. To this end, let \(K\) be a number field of degree \(n\) and \(\mathcal{O}_K\) its ring of integers, the integral closure of \(\mathbb{Z}\) in \(K\); it is an order, and it contains every order, so any \(\mathbb{Z}[\alpha] \subseteq \mathcal{O}_K\) has finite index. The next theorem is the one result in this note we do not prove.

Theorem (Dedekind's factorization criterion).
Let \(f\) be monic irreducible with root \(\alpha\), let \(K = \mathbb{Q}(\alpha)\), and suppose \(p \nmid [\mathcal{O}_K : \mathbb{Z}[\alpha]]\). With \(\bar f = \prod_i \bar g_i^{\,e_i}\) as above, \[ p\,\mathcal{O}_K \;=\; \prod_{i=1}^{r} \mathfrak{P}_i^{\,e_i}, \qquad \mathfrak{P}_i = \big(p,\; g_i(\alpha)\big), \] with the \(\mathfrak{P}_i\) distinct primes of \(\mathcal{O}_K\) and \([\mathcal{O}_K/\mathfrak{P}_i : \mathbb{F}_p] = f_i\).

So under that hypothesis the intersection of the horizontal curve with the fiber over \(p\) is the factorization of \(p\) in \(\mathcal{O}_K\), and the \(e_i, f_i\) defined above by polynomial factorization agree with the usual ramification indices and residue degrees. The index hypothesis is the price of drawing \(\operatorname{Spec}\mathbb{Z}[\alpha]\) rather than \(\operatorname{Spec}\mathcal{O}_K\), and the figure flags the curves where it can fail.

The second input is the discriminant. For monic \(f\) of degree \(n\) with roots \(\alpha_1, \ldots, \alpha_n\) in an algebraic closure, \(\operatorname{disc}(f) = \prod_{i<j}(\alpha_i - \alpha_j)^2\). It is a universal polynomial with integer coefficients in the coefficients of \(f\), so \(\operatorname{disc}(f) \in \mathbb{Z}\) and reduction commutes with it: \(\operatorname{disc}(\bar f) = \overline{\operatorname{disc}(f)}\).

Proposition (ramification is a discriminant condition).
For \(f\) monic in \(\mathbb{Z}[x]\) and \(p\) prime, \(p\) is ramified for \(f\) if and only if \(p \mid \operatorname{disc}(f)\).
Proof. A monic polynomial over a field has vanishing discriminant if and only if it has a repeated root in an algebraic closure. Over \(\mathbb{F}_p\) that is the same as having a repeated irreducible factor: irreducible polynomials over a finite field are separable, so distinct irreducible factors have no common root, and a factor appearing to a power \(e_i > 1\) forces a repeated root. Applying this to \(\bar f\) and using \(\operatorname{disc}(\bar f) = \overline{\operatorname{disc}(f)}\), some \(e_i > 1\) if and only if \(\operatorname{disc}(f) \equiv 0 \pmod p\).

Since the discriminant of an irreducible polynomial is nonzero, only finitely many fibers are ramified for a given horizontal curve: the ones dividing the discriminant. Those are the primes where the strands in the figure touch instead of crossing.

5Reading the picture

Three primes for \(x^2+1\)

Let’s think about our above work in the context of Mumford’s drawing. Go turn on the curve \(x^2+1\) in the figure. It is \(V(x^2+1) = \operatorname{Spec}\mathbb{Z}[i]\), monic of degree \(2\), so two strands, and \(\operatorname{disc}(x^2+1) = -4\). Three things happen.

\(p = 5\), split. \(x^2+1 = (x-2)(x+2)\) in \(\mathbb{F}_5[x]\). The two strands land on the distinct rational points \((5,x-2)\) and \((5,x+2)\), each with \(e = f = 1\) and residue field \(\mathbb{F}_5\). Correspondingly \(5 = (2+i)(2-i)\) splits in \(\mathbb{Z}[i]\).
\(p = 3\), inert. \(x^2+1\) has no root mod \(3\), so it is irreducible in \(\mathbb{F}_3[x]\). The two strands merge into the single closed point \((3, x^2+1)\), with \(e = 1\) and \(f = 2\), residue field \(\mathbb{F}_9\): one point, but twice as big. The intersection number is still \(ef = 2\).
\(p = 2\), ramified. \(x^2 + 1 = (x+1)^2\) in \(\mathbb{F}_2[x]\), so \(e = 2\) and \(f = 1\) at the single point \((2, x+1)\), and the strands touch there rather than crossing. In \(\mathbb{Z}[i]\) this is \((1+i)^2 = 2i\), so \((2) = (1+i)^2\) as ideals. And indeed \(2 \mid \operatorname{disc} = -4\), as the discriminant criterion requires.

Stepping back from the three primes to the whole curve: the split primes are exactly those with \(p \equiv 1 \pmod 4\) and the inert ones exactly those with \(p \equiv 3 \pmod 4\). Fermat's sum of two-squares theorem and the splitting of primes in the Gaussian integers become one statement about how a single conic in \(\operatorname{Spec}\mathbb{Z}[x]\) crosses the vertical lines.

Sections

Turn on \(x\). Since \(\mathbb{Z}[x]/(x - c) \cong \mathbb{Z}\) for any \(c \in \mathbb{Z}\), the curve \(V(x-c)\) is a copy of \(\operatorname{Spec}\mathbb{Z}\) and \(\pi\) restricts to an isomorphism on it: it is a section of \(\pi\). Degree \(1\) means the intersection number with every fiber is \(1\), so it meets each fiber in exactly one rational point, namely \((p,\, x - \bar c)\), always with \(e = f = 1\). A section is never ramified, and consistently \(\operatorname{disc}(x-c) = 1\), divisible by no prime.

Where two curves meet

Distinct horizontal curves can cross each other. With both \(x^2+1\) and \(x^2-5\) switched on, the figure shows the two curves sharing a point over \(2\) and another over \(3\). This is not an accident of the drawing.

Proposition (horizontal curves meet in closed points only).
Let \(f, g \in \mathbb{Z}[x]\) be primitive irreducible of degree \(\ge 1\) generating distinct ideals. Then \(V(f) \cap V(g)\) is a finite set of closed points, and it is empty for all but finitely many fibers.
Proof. By Gauss, \(f\) and \(g\) are non-associate irreducibles in \(\mathbb{Q}[x]\), hence coprime there, so \(uf + vg = 1\) for some \(u,v \in \mathbb{Q}[x]\). Clearing denominators gives a nonzero \(m \in \mathbb{Z}\) with \(m \in (f,g)\). Any prime \(\mathfrak{p} \supseteq (f,g)\) therefore contains \(m\), so \(\mathfrak{p} \cap \mathbb{Z}\) is nonzero, say \(\mathfrak{p} \cap \mathbb{Z} = (p)\) with \(p \mid m\). Also \(\mathfrak{p}\) is not \((p)\) itself: otherwise \(f \in (p)\), meaning \(p\) divides every coefficient of \(f\), contradicting primitivity. By the classification \(\mathfrak{p}\) is then a closed point \((p,h)\). Finitely many \(p\) divide \(m\), and each fiber contains only finitely many points of \(V(f)\).

For \(f = x^2+1\) and \(g = x^2-5\) this is concrete: \(f - g = 6\), so \((f,g) = (6,\, x^2+1)\), and a prime containing it contains \(2\) or \(3\). Modulo \(2\) both reduce to \((x+1)^2\), and modulo \(3\) both reduce to the irreducible \(x^2+1\). So \[ V(x^2+1) \,\cap\, V(x^2-5) \;=\; \big\{\,(2,\,x+1),\; (3,\,x^2+1)\,\big\}, \] exactly the two shared dots in the figure. Two number fields that are different over \(\mathbb{Q}\) can still collide modulo small primes, and that collision is a literal crossing of curves on the surface.

The curve that gets away

Turn on \(2x-1\). It is primitive and irreducible, so \((2x-1)\) is an honest height-one prime, but it is not monic: \(\mathbb{Z}[x]/(2x-1) \cong \mathbb{Z}[1/2]\). Over \(p = 2\) its reduction is the constant \(1\), a unit, so by the lemma on which fibers a curve meets, \(V(2x-1)\) misses the fiber over \(2\) entirely, and the strand runs off the top of the box. The map to \(\operatorname{Spec}\mathbb{Z}\) is neither finite nor surjective, and the constant intersection number fails precisely because monicity fails. This is the arithmetic version of an asymptote.

When the picture lies

Turn on \(x^2-5\), with \(\operatorname{disc} = 20\), and look at \(p = 2\): the reduction is \((x+1)^2\), so the drawing shows a tangency and reports ramification. But \(2\) is in fact inert in \(\mathcal{O}_K = \mathbb{Z}\!\left[\tfrac{1+\sqrt5}{2}\right]\), where \(K = \mathbb{Q}(\sqrt5)\). There is no contradiction, because \([\mathcal{O}_K : \mathbb{Z}[\sqrt5]] = 2\), so the hypothesis of Dedekind's criterion fails at \(p = 2\) and the factorization of \(\bar f\) is not entitled to describe the factorization of \(2\mathcal{O}_K\). What the tangency records instead is that \(\operatorname{Spec}\mathbb{Z}[\sqrt5]\) is singular at \((2, x+1)\): the curve drawn is an order, not the maximal one. The figure flags the curves where this can happen.

What the drawing leaves out

We have at least three omissions. Only six fibers are drawn, and only finitely many points on each, whereas every fiber \(\mathbb{A}^1_{\mathbb{F}_p}\) has one closed point for each monic irreducible in \(\mathbb{F}_p[x]\) and so is infinite. The vertical position of a closed point of degree \(> 1\) carries no information at all, since such a point has no coordinate to be placed by, unlike \((p, x-a)\), which sits at height \(\bigl(a+\tfrac12\bigr)/p\). And the drawing records the topological space together with which prime each mark denotes; the structure sheaf, which is what makes \(\operatorname{Spec}\mathbb{Z}[x]\) a scheme rather than a space, is not visible anywhere in it. Multiplicity is the one place where non-reduced structure surfaces, and even there it is annotated rather than seen.

6References