GENERIC FIBER RINGS OF MIXED POLYNOMIAL/POWER SERIES RINGS

William Heinzer Purdue University

joint with Christel Rotthaus and Sylvia Wiegand

MOTIVATION

A question raised by Mel Hochster:

Question. Can one describe or somehow classify the local maps R ↪ S of complete local domains R and S such that

P ∈ Spec(S), P ≠ (0) ⟹ P ∩ R ≠ (0)?

Hochster remarks that if, for example, R is equal characteristic zero, such extensions arise as

R = K[[x_{1}, …, x_{m}]] ↪ T = L[[x_{1}, …, x_{m}, y_{1}, …, y_{n}]] → T/P = S,

where K is a subfield of L and P ∈ Spec(T) is maximal with respect to P ∩ R = {0}. Partial Answer. In Hochster’s set-up,

[L : K] < ∞ ⟹ dim S = 2 or dim S = m.

Main Results

Let K be a field, m and n positive integers, X = {x_{1}, …, x_{m}}, Y = {y_{1}, …, y_{n}} sets of independent variables over K. Let A be the localized polynomial ring K[X]_{(X)}.

Theorem 1. Every prime ideal P of Â = K[[X]] that is maximal with respect to P ∩ A = (0) has height m − 1.

Let B := K[[X]][Y]_{(X,Y)} and C := K[Y]_{(Y)}[[X]].

Theorem 2. Every prime ideal P of B̂ = Ĉ that is maximal with respect to either P ∩ B = (0) or P ∩ C = (0), has height m + n − 2.

Theorem 3. Every prime ideal P of K[[X, Y]] that is maximal with respect to P ∩ K[[X]] = (0) has height either n or m + n − 2.

Local Embeddings

There exist local embeddings:

A = K[X]_{(X)} ↪ Â:= K[[X]],  Â ↪ B̂ = Ĉ = K[[X, Y]]

B = K[[X]] [Y]_{(X,Y)} ↪ C = K[Y]_{(Y)}[[X]] ↪ B̂ = Ĉ = K[[X]] [[Y]].

Matsumura observes there exist P ∈ Spec Â with ht P = m − 1 and P ∩ A = (0), and also that there exist P ∈ Spec K[[X, Y]] with ht P = m + n − 2 such that

P ∩ C = (0) or P ∩ B = (0).

However he does not address the question of whether all primes maximal with these properties have the same height.

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.

Let (R, m) be a Noetherian local integral domain and let R̂ denote the m-adic completion of R.

The generic formal fiber ring of R is the localization (R \ (0))^{−1}R̂ of R̂.

The formal fibers of R are the fibers of the map Spec R̂ → Spec R. For P ∈ Spec R, the formal fiber over P is Spec( (R_{P} /P R_{P}) ⊗_{R} R̂).

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.

Variations on a theme of Weierstrass

The Weierstrass Preparation Theorem is our main technical tool. Let P be a prime ideal in the power series ring Â = K[[X]], where X = {x_{1}, …, x_{n}} is a set of n variables over the field K. Here A = K[X]_{(X)} is the localized polynomial ring. Notation. By a change of variables, we mean a finite sequence of ‘polynomial’ change of variables over K of the form

x_{1} ↦ x_{1} + x_{n}^{e_{1}}, … x_{n−1} ↦ x_{n−1} + x_{n}^{e_{n−1}}, x_{n} ↦ x_{n}, perhaps followed by

z_{1} ↦ z_{1}, z_{2} ↦ z_{2} + z_{1}^{f_{2}}, … z_{n} ↦ z_{n} + z_{1}^{f_{n}},

where e_{i}, f_{i} ∈ ℕ and z_{i} denotes the image of x_{i} under the first map.

The Weierstrass Preparation Theorem

Theorem (Weierstrass) Let (R, m) be a complete local ring, let f ∈ R[[x]] be a formal power series and let \overline{f} denote the image of f in (R/ m)[[x]]. Assume that \overline{f} ≠ 0 and that ord \overline{f} = s > 0. There exists a unique ordered pair (u, F) such that u is a unit in R[[x]] and F ∈ R[x] is a distinguished monic polynomial of degree s such that f = uF. Here F = x^{s} + a_{s−1}x^{s−1} + ⋯ + a_{0} ∈ R[x] is distinguished if a_{i} ∈ m for 0 ≤ i ≤ s − 1.

Corollary. The ideal fR[[x]] is extended from R[x] and R[[x]]/(f) is a free R-module of rank s. Every g ∈ R[[x]] is of the form g = qf + r, where q ∈ R[[x]] and r ∈ R[x] is a polynomial with deg r ≤ s − 1.

Embeddings of Power Series Rings

Theorem (Classical) There exists a K-algebra embedding of the formal power series ring R := K[[x_{1}, …, x_{n}]] into the formal power series ring in two variables K[[y, z]]. Proof. Let f_{1}, …, f_{n} be power series in K[[y]] that are algebraically independent over K. Define φ : R → K[[y, z]] by φ(x_{i}) = f_{i}z, for 1 ≤ i ≤ n. Suppose g ∈ R is such that φ(g) = 0. Write g = ⨁_{k=0}^{∞} g_{k}, where g_{k} is a form in K[x_{1}, …, x_{n}] of degree k. Then φ(g_{k}) = z^{k}g_{k}(f_{1}, …, f_{n}) with g_{k}(f_{1}, …, f_{n}) ∈ K[[y]]. Hence

φ(g) = ⨁_{k=0}^{∞} g_{k}(f_{1}, …, f_{n})z^{k} = 0

implies g_{k}(f_{1}, …, f_{n}) = 0 for each k ∈ ℕ. Since f_{1}, …, f_{n} are algebraically independent over K, each g_{k} = 0, so g = 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}.

Corollary. Assume φ : (R, m) → (S, n) is a local homomorphism. If R is m-adically complete, m S is n-primary and S/ n is finite over R/ m, then S is a finitely generated R-module. If φ : R → K[[y, z]] = S is an embedding of the power series ring in n > 2 variables, then m S is not n-primary.

The Generic Formal Fiber Rings of B and C

Since B := K[[X]][Y]_{(X,Y)} ↪ K[Y]_{(Y)}[[X]] := C, if P ∈ Spec K[[X, Y]] and P ∩ C = (0), then P ∩ B = (0). We know P ∈ Spec K[[X, Y]] maximal with respect to P ∩ B = (0) or maximal with respect to P ∩ C = (0) ⟹ ht P = n + m − 2. Hence if P ∈ Spec K[[X, Y]] is maximal with P ∩ C = (0), then P is also maximal with respect to P ∩ B = (0). However, if n ≥ 2, the generic fiber of B ↪ C is nonzero. This means ∃ P ∈ Spec K[[X, Y]] maximal in the generic formal fiber of B but P not in the generic formal fiber of C.

The Generic Fiber of B ↪ C

Lemma. B = K[[x]][y_{1}, y_{2}] ↪ K[y_{1}, y_{2}][[x]] = C is not TGF. Proof. There exists σ ∈ K[y_{1}][[x]] that is transcendental over K[[x]][y_{1}]. Let q = (y_{2} − σx)K[y_{1}, y_{2}][[x]] and define π : K[y_{1}, y_{2}][[x]] → K[y_{1}, y_{2}][[x]]/ q ≅ K[y_{1}][[x]]. Thus π(y_{2}) = σx. If h ∈ q ∩(K[[x]][y_{1}, y_{2}]), then ∃ s, t ∈ ℕ so that h = ∑_{i=0}^{s} ∑_{j=0}^{t}(∑_{k∈ℕ} a_{ijk}x^{k})y_{1}^{i}y_{2}^{j}, where a_{ijk} ∈ K. Hence 0 = π(h) = ∑_{i=0}^{s} ∑_{j=0}^{t}(∑_{k∈ℕ} a_{ijk}x^{k})y_{1}^{i}(σx)^{j}. Since σ is transcendental over K[[x]][y_{1}], each a_{ijk} = 0. Therefore q ∩(K[[x]][y_{1}, y_{2}]) = (0), and B ↪ C is not TGF.

Lemma. Let R := K[[X]] and let P ∈ Spec R with x_{1} ∉ P and ht P = r, where 1 ≤ r ≤ n − 1. There exists a change of variables x_{1} ↦ z_{1} := x_{1} (x_{1} is fixed), x_{2} ↦ z_{2}, …, x_{n} ↦ z_{n} and a regular sequence f_{1}, …, f_{r} ∈ P so that, upon setting Z_{1} = {z_{1}, …, z_{n−r}}, Z_{2} = {z_{n−r+1}, …, z_{n}} and Z = Z_{1} ∪ Z_{2},

1. (f_{1}, …, f_{r}) ⊂ K[[Z_{1}]][Z_{2}] and P is a minimal prime of (f_{1}, …, f_{r})R.

2. The (Z_{2})-adic completion of K[[Z_{1}]] [Z_{2}]_{(Z)} is identical to the (f_{1}, …, f_{r})-adic completion and is R = K[[X]] = K[[Z]].

3. If P_{1} := P ∩ K[[Z_{1}]] [Z_{2}]_{(Z)}, then P_{1}R = P, that is, P is extended from K[[Z_{1}]] [Z_{2}]_{(Z)}.

4. The extension: K[[Z_{1}]] ↪ K[[Z_{1}]] [Z_{2}]_{(Z)}/P_{1} ≅ K[[Z]]/P is finite (and integral).

R = K[[X]] = K[[Z_{1}, Z_{2}]] (X)R  D = K[[Z_{1}]] [Z_{2}]_{(Z)} P = P_{1}R  P_{1} = P ∩ D

(f_{1}, …, f_{r}) ⊂ P_{1} and K[[Z_{1}]] ↪ D/P_{1} ≅ K[[Z]]/P is finite.

Theorem. Let y and X = {x_{1}, …, x_{m}} be variables over the field K. Assume V is a DVR with K[y] ⊆ V ⊆ K[[y]] such that V has completion K[[y]]. Also assume the field K((y)) = K[[y]] [1/y] has uncountable transcendence degree over the fraction field Q(V) of V. Set R_{0} := V [[X]] and R = R̂_{0} = K[[y, X]]. Let P be a prime ideal of R such that:

P ⊆ (X)R (so y ∉ P) and dim(R/P) > 2. Then there exists a prime ideal Q of R such that

1. P ⊂ Q ⊂ XR,

2. dim(R/Q) = 2, and

3. P ∩ R_{0} = Q ∩ R_{0}. In particular, P ∩ K[[X]] = Q ∩ K[[X]].

R = K[[y, Z]] D := K[[y, Z_{1}]] [Z_{2}]_{(Z)}  Q(K[[y]]) = K[[y]] [1/y] = K((y)) K[[y]]  E := Q(T) = L(γ_{2}, …, γ_{n−r}) T := L(γ_{2}, …, γ_{n−r}) ∩ K[[y]] S := Q(V (Δ)) ∩ K[[y]] L := Q(S) = Q(V (Δ)) K[y] ⊆ V  Q(V)

16

Prime Ideals With The Same Contraction

Corollary. Let K be a field, y and X = {x_{1}, …, x_{m}} variables over K and let R = K[[y, X]]. Assume P ∈ Spec R is such that:

(i) P ⊆ (x_{1}, …, x_{m})R and (ii) dim(R/P) > 2. Then there exists a prime ideal Q of R such that

1. P ⊂ Q ⊂ XR,

2. dim(R/Q) = 2, and

3. P ∩ K[y]_{(y)}[[X]] = Q ∩ K[y]_{(y)}[[X]]. In particular, P ∩ K[[x_{1}, …, x_{m}]] = Q ∩ K[[x_{1}, …, x_{m}]].

\begin{matrix}R = K[[y, X]] & (X)R & R_{0} = K[y]_{(y)} [[X]] & P ⊂ Q & P ∩ R_{0} = Q ∩ R_{0} \\ dim R/Q = 2 and P ∩ K[[X]] = Q ∩ K[[X]] \end{matrix}

18

Generic Fibers of Power Series Extensions

Theorem. Let y and X = {x_{1}, …, x_{m}} be variables over the field K. Consider the extension Â = K[[X]] ↪ K[[X]][[y]] = B̂. Let P be a prime ideal of B̂ such that P ∩ Â = (0). Then

1. If P ⊈ XB̂, then dim B̂/P = m and P is maximal with respect to P ∩ Â = (0).

2. If P ⊆ XB̂, then there exists Q with P ⊆ Q, dim B̂/Q = 2, and Q is maximal with respect to Q ∩ Â = (0).

If m > 2, then for each prime Q maximal with respect to Q ∩ Â = (0), we have either

(i) dim B̂/Q = m and Â ↪ B̂/Q is finite, or (ii) dim B̂/Q = 2 and Q ⊂ XB̂.

Generic Fibers of Power Series Extensions 2

Theorem. Let K be a field, m and n positive integers, X = {x_{1}, …, x_{m}} and Y = {y_{1}, …, y_{n}} sets of independent variables over K. Consider the extension

Â = K[[X]] ↪ K[[X]][[Y]] = B̂.

Let Q be a prime ideal of B̂ maximal with respect to Q ∩ Â = (0).

1. If m = 1, then dim B̂/Q = 1 and Â ↪ B̂/Q is finite.

2. If m ≥ 2, there are two possibilities:

(i) Â ↪ B̂/Q is finite, in which case dim B̂/Q = dim Â = m, or (ii) dim B̂/Q = 2.

Mixed Polynomial/Power Series Rings

Consider

A := K[x, y] ↪ B := K[[x]][y] ↪ C := K[y][[x]]
↪ D := K[y][[x/y]] ↪ E := K[y, 1/y][[x]]

where K is a field and x and y are indeterminates over K.

We are interested in the prime spectra of these rings and the maps on the spectra determined by the inclusion maps on the rings. For example, do there exist nonzero primes of one of the larger rings that intersect a smaller ring in zero (i.e. non-TGF extensions).

Motivation 2

From the introduction to a paper by Alonzo-Tarrio, Jeremias-Lopez and Lipman: If a map between noetherian formal schemes can be factored as a closed immersion followed by an open one, can it also be factored as an open immersion followed by a closed one?

Brian Conrad observed that a counterexample can be constructed for every triple (R, x, p), where

1. R is an adic domain, that is, R is a Noetherian domain that is separated and complete with respect to the powers of a proper ideal I.

2. x ∈ R is nonzero, the completion of R[1/x] with respect to the powers of IR[1/x], denoted S := R_{{x}}, is an integral domain.

3. p is a nonzero prime ideal of S that intersects R in (0).

Motivation 2 continued

If (R, x, p) is such a triple with S = R_{{x}}, then the composition R → S → S/p determines a map on formal spectra Spf(S/p) → Spf(S) → Spf(R) that is a closed immersion followed by an open one. For a surjection such as S → S/p of adic rings gives rise to a closed immersion Spf(S/p) → Spf(S), while a localization like that of R with respect to x followed by the completion of R[1/x] with respect to the powers of IR[1/x] to obtain S gives rise to an open immersion Spf(S) → Spf(R). However, the map Spf(S/p) → Spf(R) cannot be factored as an open immersion followed by a closed one. For a closed immersion into Spf(R) corresponds to a surjective map of adic rings R → R/J, where J is an ideal of R.

Motivation 2 continued

If the immersion Spf(S/p) → Spf(R) factored as an open immersion followed by a closed one, we would have R-algebra homomorphisms from R → R/J → S/p, where Spf(S/p) → Spf(R/J) is an open immersion. Since p ∩ R = (0), we must have J = (0). This implies Spf(S/p) → Spf(R) is an open immersion, that is, the composite map Spf(S/p) → Spf(S) → Spf(R), is an open immersion. But also Spf(S) → Spf(R) is an open immersion. It follows that Spf(S/p) → Spf(S) is both open and closed. Since S is an integral domain this implies Spf(S/p) ≅ Spf(S). This is a contradiction since p is nonzero. One of our motivations is to describe such triples (R, x, p).

An example of (R, x, p).

An example described in [AJL]: Let w, x, y, z be indeterminates over a field K and consider

R := K[w, x, z][[y]] ↪ S := R_{{x}} = K[w, x, 1/x, z][[y]].

Let P = (w, z)R and T := R_{P} ⊂ S_{P S} := G. Then T ⊂ G are 2-dim regular local domains, and the residue field of G (i.e. the fraction field of K[x, 1/x][[y]]) is transcendental over that of T (i.e. the fraction field of K[x][[y]]). It follows by [HR, p. 364, Theorem 1.12] that there exist infinitely many height-one prime ideals of G in the generic fiber over T. Any one of these prime ideals of G contracts in S to a prime p as above.

Is There A Simpler Example?

In some correspondence to Lipman, Conrad asked:

Question. Is there a nonzero prime ideal of E := K[x, 1/x][[y]] that intersects C = K[x][[y]] in zero?

If there were such a prime ideal, then

C := K[x][[y]] ↪ E := K[x, 1/x][[y]]

would be a simpler counterexample to the assertion that a closed immersion followed by an open one also has a factorization as an open immersion followed by a closed one. We show there is no such prime ideal.

A two-dimensional TGF extension.

Let K be a field and let x and y be indeterminates over K.

Remark. C := K[x][[y]] ↪ E := K[x, 1/x][[y]] is a TGF extension.

It is true that Q(E) has infinite transcendence degree over Q(C). Notice that dim C = dim E = 2 and y is in every maximal ideal of C or E. Let P be a nonzero prime of E. If y ∈ P, then P ∩ C ≠ (0). Assume y ∉ P and P ≠ (0). Then K[[y]] ↪ C/(P ∩ C) ↪ E/P and E/P is finite over K[[y]] and hence also over C/(P ∩ C), so P ∩ C ≠ (0).

Question. Is K[x, z][[y]] ↪ K[x, 1/x, z][[y]] a TGF extension?

An Example of a TGF extension.

Example. The extension R := K[[x, y]] ↪ S := K[[x]][[\frac{y}{x}]] is a TGF extension of the type described in Hochster’s question.

To see this extension is TGF, 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 and S/P is finite over K[[x]]. Therefore dim R/(P ∩ R) = 1, so P ∩ R ≠ (0).

Remark. The extension K[[x, y]] ↪ K[[x, \frac{y}{x}]] is, up to

isomorphism, the same as the extension K[[x, xy]] ↪ K[[x, y]].