Let \(k = \FF_p\) be the finite field with \(p\) elements. Describe \(\Spec k[x]\). What are the residue fields of its points? How many points are there with a given residue field?
Solution
Put \[ X=\operatorname{Spec}\mathbb F_p[x]. \]
1. The points of \(X\) are the generic point\[ \eta=(0) \]and one closed point\[ x_f=(f) \]for every monic irreducible polynomial\[ f\in\mathbb F_p[x]. \]
The ring \(\mathbb F_p[x]\) is a principal ideal domain. Its nonzero prime ideals are therefore generated by irreducible polynomials. Multiplying a generator by a nonzero scalar does not change the ideal, so every nonzero prime has a unique monic irreducible generator. Since the ring is a domain, \((0)\) is prime as well.
2. The generic point has residue field\[ \boxed{\kappa(\eta)=\mathbb F_p(x).} \]
For an affine scheme, \[ \kappa((0)) =\operatorname{Frac}(\mathbb F_p[x]/(0)) =\mathbb F_p(x). \]
3. If \(f\) is irreducible of degree \(d\), then\[ \boxed{ \kappa(x_f) =\mathbb F_p[x]/(f) \cong\mathbb F_{p^d}. } \]
The quotient by the maximal ideal \((f)\) is a field. As an \(\mathbb F_p\)-vector space it has basis \[ 1,x,\ldots,x^{d-1}, \] so it has \[ p^d \] elements. Up to isomorphism there is exactly one finite field with \(p^d\) elements, denoted \(\mathbb F_{p^d}\).
4. Thus the closed points with residue field isomorphic to \(\mathbb F_{p^d}\) are exactly the points defined by monic irreducible polynomials of degree \(d\).
Step 3 shows that every degree-\(d\) irreducible gives residue field \(\mathbb F_{p^d}\).
Conversely, if \[ \kappa(x_f)\cong\mathbb F_{p^d}, \] then \[ |\kappa(x_f)|=p^d. \] But 3 gives \[ |\kappa(x_f)|=p^{\deg f}, \] so \[ \deg f=d. \]
5. Let \(N_d\) denote the number of monic irreducible polynomials of degree \(d\) over \(\mathbb F_p\). Then\[ \boxed{ p^n=\sum_{d\mid n}dN_d } \]for every \(n\ge1\).
The roots of \[ T^{p^n}-T \] in an algebraic closure of \(\mathbb F_p\) are exactly the elements of the finite field \(\mathbb F_{p^n}\). Its derivative is \[ p^nT^{p^n-1}-1=-1, \] so the polynomial has no repeated roots.
A monic irreducible polynomial \(f\) over \(\mathbb F_p\) divides \[ T^{p^n}-T \] exactly when all its roots lie in \(\mathbb F_{p^n}\). If \(\deg f=d\), its roots lie in \(\mathbb F_{p^d}\), and \[ \mathbb F_{p^d}\subseteq\mathbb F_{p^n} \iff d\mid n. \] Hence \[ T^{p^n}-T = \prod_{d\mid n} \prod_{\substack{f\text{ monic irreducible}\\\deg f=d}} f(T). \] Taking degrees gives \[ p^n =\sum_{d\mid n}dN_d. \]
6. Möbius inversion gives\[ \boxed{ N_d =\frac1d \sum_{e\mid d}\mu(e)p^{d/e} =\frac1d \sum_{e\mid d}\mu(d/e)p^e. } \]
Define \[ a_d=dN_d. \] Step 5 says \[ p^n=\sum_{d\mid n}a_d. \] The Möbius inversion formula therefore gives \[ a_d =\sum_{e\mid d}\mu(e)p^{d/e}. \] Dividing by \(d\) yields the first formula. Replacing the divisor \(e\) by \(d/e\) yields the second.
7. Consequently the number of points of \(X\) with residue field isomorphic to \(\mathbb F_{p^d}\) is\[ \boxed{ \frac1d\sum_{e\mid d}\mu(e)p^{d/e}. } \]There is also exactly one point with residue field \(\mathbb F_p(x)\), namely the generic point.
6. The second is 2 together with uniqueness of the generic prime \((0)\).
8. Topologically, every proper closed subset of \(X\) is a finite set of closed points, while the generic point is dense.
Every closed subset is \[ V(g) \] for some \(g\in\mathbb F_p[x]\). If \(g=0\), this is all of \(X\). If \(g\ne0\), then \((0)\) does not contain \(g\), and a closed point \((f)\) belongs to \(V(g)\) exactly when the irreducible polynomial \(f\) divides \(g\). A nonzero polynomial has finitely many irreducible factors, so the closed set is finite.
Also \[ V(0)=X, \] so the generic point \((0)\) is dense.
9. Q.E.D.
Steps <1>1–<1>3 describe the points and residue fields, and <1>4–<1>7 give the requested count for each residue field.