MIXED POLYNOMIAL/POWER SERIES RINGS AND RELATIONS AMONG THEIR SPECTRA

William Heinzer Purdue University

joint with Christel Rotthaus and Sylvia Wiegand

OVERVIEW 1

At first glance the rings

B := k[[y]] [x] and C := k[x] [[y]]

look similar. One has

B = k[[y]] [x] ↪ k[x] [[y]] = C,

but this is a strict inclusion. For example, 1 − xy is a nonunit of B, and

\frac{1}{1 − xy} = ∑_{i=0}^{∞} x^{n}y^{n} ∈ C,

so 1 − xy is a unit of C. Indeed, the rings B and C are not isomorphic: the intersection of the maximal ideals of B is (0), while y is in every maximal ideal of C.

OVERVIEW 2

Consider the mixed polynomial/power series rings

A := k[x, y] ↪ B := k[[y]] [x] ↪ C := k[x] [[y]] ↪ D := k[[x, y]], where k is a field. The inclusion maps here are all flat homomorphisms. The prime ideal structure of these rings is well understood. The above inclusions induce maps

Spec A ← Spec B ← Spec C ← Spec D. We are interested in describing these Spec maps.

OVERVIEW 3

Consider

k[x] [[y]] = C ↪ C[1/x] ↪ k[x, 1/x] [[y]] := E,

At first glance, it appears that E is a localization of C, but it is not. There are elements in E that are not in the fraction field of C. However, E is obtained from C by the localization C[1/x] followed by the (y)- adic completion of C[1/x]. Thus E is flat over C. The map C ↪ E induces Spec C ← Spec E, and again we are interested in describing this Spec map.

OVERVIEW 4

Also consider

C ↪ C_{1} := k[x] [[\frac{y}{x}]] ↪ ⋯ ↪ C_{n} := k[x] [[\frac{y}{x^{n}}]] ↪ ⋯ ↪ E.

The maps C ↪ C_{n} and C_{i} ↪ C_{n} for i < n are not flat, but C_{n} ↪ E = k[x, 1/x] [[y]] is the localization C_{n}[1/x] followed by the (y)-adic completion of C_{n}[1/x]. Thus C_{n} ↪ E is flat. These inclusion maps induce maps

Spec C ← Spec C_{1} ← ⋯ ← Spec C_{n} ← ⋯ ← Spec E. We are interested in describing these Spec maps.

GENERIC FIBER RINGS

Let R ↪ S be an injective homomorphism of commutative rings with R an integral domain. The generic fiber ring of the map R ↪ S is the localization (R \ {0})^{−1}S of S. With A := k[x, y] ↪ B := k[[y]] [x] ↪ C := k[x] [[y]] ↪ D := k[[x, y]], the generic fiber ring of A ↪ R is one-dim. for R ∈ {B, C, D}, while the generic fiber ring of R ↪ S is zero-dim for R ⊆ S in {B, C, D}.

TRIVIAL GENERIC FIBER EXTENSIONS

Let R be a subring of an integral domain S. Definition. R ↪ S is a trivial generic fiber extension or a TGF extension if

(0) ≠ P ∈ Spec S ⟹ P ∩ R ≠ (0).

One obtains a TGF extension S of R by considering R ↪ T → T/P := S,

where T is an extension ring of R and P ∈ Spec T is maximal with

respect to P ∩ R = (0).

Thus the generic fiber ring of R ↪ T is relevant to constructing TGF extensions S of R.

A TGF EXTENSION

Let x and y be indeterminates over a field k. Then

R := k[[x, y]] ↪ S := k[[x]] [[\frac{y}{x}]] is TGF.

Proof. It suffices to show P ∩ R ≠ (0) for each P ∈ Spec S with ht P = 1. This is clear if x ∈ P, while if x ∉ P, then k[[x]] ∩ P = (0), so k[[x]] ↪ R/(P ∩ R) ↪ S/P. Now S/P is one-dim local with residue field k. Hence by Cohen’s Theorem 8, S/P is finite over k[[x]]. Thus dim R/(P ∩ R) = 1, so P ∩ R ≠ (0).

Cohen’s Theorem 8

Theorem (Classical) Let I be an ideal of a ring R and let M be an R-module. Assume that R is complete in the I-adic topology and ⋂_{n=1}^{∞} I^{n}M = (0). If M/I is generated over R/I by elements \overline{w}_{1}, …, \overline{w}_{s} and w_{i} is a preimage in M of \overline{w}_{i} for 1 ≤ i ≤ s, then M is generated over R by w_{1}, …, w_{s}.

This is useful for proving that with

B := k[[y]] [x] ↪ C := k[x] [[y]] ↪ D := k[[x, y]],

then R ↪ S is TGF for R ⊆ S in {B, C, D}.

TGF EXTENSIONS

PROP. 1. Let R ↪ S and S ↪ T be injective maps, where R, S and T are integral domains.

(1) If R ↪ S and S ↪ T are TGF extensions, then so is R ↪ T. Equivalently if R ↪ T is not TGF, then at least one of the extensions R ↪ S or S ↪ T is not TGF.

(2) If R ↪ T is TGF, then S ↪ T is TGF.

(3) If the map Spec T → Spec S is surjective, then R ↪ T is TGF implies R ↪ S is TGF.

A NON-TGF EXTENSION

PROP. 2. R = k[[x]][y, z] ↪ k[y, z][[x]] = S is not TGF. Proof. There exists σ ∈ k[y][[x]] that is transcendental over k[[x]][y]. Let q = (z − σx)k[y, z][[x]]. Define π : k[y, z][[x]] → k[y, z][[x]]/ q ≅ k[y][[x]]. Thus π(z) = σx. If h ∈ q ∩(k[[x]][y, z]), then ∃ s, t ∈ ℕ so that h = ∑_{i=0}^{s} ∑_{j=0}^{t}(∑_{ℓ∈ℕ} a_{ijℓ}x^{ℓ})y^{i}z^{j}, where a_{ijℓ} ∈ k. Hence 0 = π(h) = ∑_{i=0}^{s} ∑_{j=0}^{t}(∑_{ℓ∈ℕ} a_{ijℓ}x^{ℓ})y^{i}(σx)^{j}. Since σ is transcendental over k[[x]][y], each a_{ijℓ} = 0. Therefore q ∩(k[[x]][y, z]) = (0), and R ↪ S is not TGF.

POWER SERIES RINGS

Lemma. Let R[[y]] denote the power series ring in the variable y over the commutative ring R. Then

(1) Each maximal ideal of R[[y]] has the form (m, y)R[[y]], where m is a maximal ideal of R. Thus y is in every maximal ideal of R[[y]].

(2) If R is Noetherian with dim R[[y]] = n and x_{1}, …, x_{m} are indeterminates over R[[y]], then y is in every maximal ideal of height n + m of the polynomial ring R[[y]] [x_{1}, …, x_{m}].

Lemma. Let R be an n-dim. Noetherian domain, let y be an indeterminate over R, and let q be a prime ideal of height n in the power series ring R[[y]]. If y ∉ q, then q is contained in a unique maximal ideal of R[[y]]. Proof. Let S := R[[y]]/ q. The assertion is clear if q is maximal. Otherwise, dim S = 1. Moreover, S is complete in its yS-adic topology and every maximal ideal of S is a minimal prime of the principal ideal yS. Hence S is a complete semilocal ring. Since S is also an integral domain, it is local by [Mat., Theorem 8.15]. Thus q is contained in a unique maximal ideal of R[[y]].

The picture of Spec k[[y]] [x] is shown below:

β  (y)  β  β  ⋯ •  •  •  ⋯ (#{ bullets} = α) (0)

In the diagram, β is the cardinality of k[[y]], and α is the cardinality of the set of maximal ideals of k[x]; the boxed β means there are cardinality β height-one primes in that position with

respect to the partial ordering.

The picture of Spec k[x] [[y]] is shown below:

(y)  β  β  ⋯ •  •  •  ⋯ (#{ bullets} = α) (0)

Here α is the cardinality of the set of maximal ideals of k[x], and β is the uncountable cardinal equal to the cardinality of k[[y]].

The picture of Spec R[[y]] for a one-dim Noetherian domain R:

(y)  κ_{i}  κ_{j}  ⋯ •  •  •  ⋯ (#{ bullets} = α) (0)

Spec R[[y]]

Here κ_{i} and κ_{j} are uncountable cardinals.

ISOMORPHIC SPECTRA

REMARK. Let F be a field that is algebraic over a finite field. Roger Wiegand proved that as partially ordered sets or topological spaces

Spec ℚ[x, y] ≅ S̸pec F [x, y] ≅ Spec ℤ[y].

The spectra of power series extensions in y behave differently: We have

Spec ℤ[[y]] ≅ Spec ℚ[x] [[y]] ≅ Spec F [x] [[y]].

Higher dimensional mixed power series/polynomial rings

We display several extensions involving three variables:

k[x, y, z] ↪^{α} k[[z]] [x, y] ↪^{β} k[x] [[z]] [y] ↪^{γ} k[x, y] [[z]] ↪^{δ} k[x] [[y, z]],

k[[z]] [x, y] ↪^{ϵ} k[[y, z]] [x] ↪^{ζ} k[x] [[y, z]] ↪^{η} k[[x, y, z]],

We have been able to show most of these extensions are not TGF.

PROP. 3. k[[z]] [x, y] ↪^{β} k[x] [[z]] [y] is not TGF. Proof. Fix σ ∈ k[x] [[z]] that is transcendental over k[[z]] [x]. Define π : k[x] [[z]] [y] → k[x] [[z]] to be the identity map on k[x] [[z]] and π(y) = σz. Let q = ker π. Then y − σz ∈ q. If h ∈ q ∩(k[[z]] [x, y]), then

h = ∑_{j=0}^{s} ∑_{i=0}^{t}(∑_{ℓ∈ℕ} a_{ijℓ}z^{ℓ})x^{i}y^{j}, for some s, t ∈ ℕ and a_{ijℓ} ∈ k, and so

0 = π(h) = ∑_{j=0}^{s} ∑_{i=0}^{t}(∑_{ℓ∈ℕ} a_{ijℓ}z^{ℓ})x^{i}(σz)^{j} = ∑_{j=0}^{s} ∑_{i=0}^{t}(∑_{ℓ∈ℕ} a_{ijℓ}z^{ℓ+j})x^{i}σ^{j}.

Since σ is trans. over k[[z]] [x], x and σ are alg. indep. over k((z)). Thus each a_{ijℓ} = 0. Therefore q ∩(k[[z]][x, y]) = (0), and the embedding β is not TGF.

QUESTION. Is k[x, y] [[z]] ↪^{θ} k[x, y, 1/x] [[z]] TGF? REMARK. For k a field and x, y, u and z indeterminates over k, the extension k[x, y, u] [[z]] ↪ k[x, y, u, 1/x,] [[z]] is not TGF.