REGULARITY OF TENSOR PRODUCTS OF k-ALGEBRAS
S. BOUCHIBA and S. KABBAJ∗
Abstract
This paper tackles a problem on the possible transfer of regularity to tensor products of algebras over a fieldk. The main result establishes necessary and sufficient conditions for a Noetherian tensor product of two extension fields ofkto inherit regularity in various settings of separability.
Thereby, we provide some applications as well as several original examples to illustrate or delimit the scope of the established results.
1. Introduction
All algebras considered are commutative with identity elements and, unless otherwise specified, are assumed to be non-trivial. All ring homomorphisms are unital. Throughout,k stands for a field. A Noetherian local ring(R,ᒊ) is regular if its Krull and embedding dimensions coincide; i.e., dim(R) = embdim(R), where embdim(R)denotes the dimension ofᒊ/ᒊ2as anR/ᒊ- vector space. Regular local rings were first introduced by Krull, and then be- came prominent once Zariski showed that, geometrically, a regular local ring corresponds to a smooth point on an algebraic variety. Later, Serre found a ho- mological characterization for a local ringRto be regular; that is,Rhas finite global dimension. Finite global dimension is preserved under localization, so that localizations of regular local rings at prime ideals are again regular. Geo- metrically, this corresponds to the intuition that if a surface contains a smooth curve, then the surface is smooth near the curve. Consequently, the definition of regularity got globalized as follows: A Noetherian ringR is regular if its localizations with respect to all prime ideals are regular. Using homological techniques, Auslander and Buchsbaum proved in 1950’s that every regular local ring is a UFD.
A Noetherian local ring(R,ᒊ)is a complete intersection if the completion ˆ
RofRwith respect to theᒊ-adic topology is the quotient ring of a regular local ring modulo an ideal generated by a regular sequence. The ringRis Gorenstein if its injective dimension (as anR-module) is finite; andRis Cohen-Macaulay
∗This work was supported by KFUPM under DSR Research Grant # FT100021.
Received 17 July 2012.
if grade and height coincide for every ideal ofR. These notions are globalized by carrying over to localizations with respect to the prime ideals. We have the following diagram of implications:
Regular ring
⇓
(Locally) Complete Intersection ring
⇓ Gorenstein ring
⇓
Cohen-Macaulay ring
⇓ Noetherian ring
In this paper we will tackle a problem, originally initiated by Grothendieck [10], on the possible transfer of regularity to tensor products of k-algebras.
Recently, it has been proved that a Noetherian tensor product ofk-algebras A⊗k B inherits from AandB the notions of locally complete intersection ring, Gorenstein ring, and Cohen-Macaulay ring [4], [11], [17], [19], [20]. In particular,K⊗kLis a locally complete intersection ring, for any two extension fieldsKandLofksuch thatK⊗kLis Noetherian [20, Proposition 5]. Notice at this point that tensor products of rings subject to the above concepts were recently used to broaden or delimit the context of validity of some homological conjectures; see for instance [12], [14].
As to regularity, the problem remains elusively open. Indeed, contrary to the above notions, a Noetherian tensor product of two extension fields ofkis not regular in general. In 1965, Grothendieck proved thatK⊗kLis a regular ring provided K or L is a finitely generated separable extension field of k [10, Lemma 6.7.4.1]. In 1969, Watanabe, Ishikawa, Tachibana, and Otsuka, showed that under a suitable condition tensor products of regular rings are complete intersections [23, Theorem 2, p. 417]. In 2003, Tousi and Yassemi proved that a Noetherian tensor product of twok-algebrasAandBis regular if and only if so areAandBin the special case wherekis perfect; i.e., every (algebraic) extension ofkis separable [20], [11].
Recall that regularity, though a topic of commutative Noetherian rings, proved to be well approached via homological methods. In fact, a characteriz- ation of regular homomorphismsR−→Sis given by the vanishing of the first André-Quillen homology functorD1(S/R,−). In the case of a homomorphism of fieldsk −→ K, the vanishing ofD1(K/k,−)totally characterizes separ- ability ofK over k. So that, under separability and Noetherianity, K ⊗k A inherits regularity via base change. Nevertheless, the case of tensor products
ofk-algebras involving purely inseparable extensions ofkremains unexplored.
The main goal of this paper is to handle such a case. Actually, our main result (Theorem 2.4) establishes necessary and sufficient conditions for a Noetherian tensor product of two extension fields ofkto inherit regularity; and hence gen- eralizes Grothendieck’s aforementioned result. As a prelude to this, we revisit the constructions of the formA⊗k BwhereAorBis geometrically regular (Lemma 2.1) and then offer a new direct proof (without use of André-Quillen homology). We close with a discussion of the correlation betweenA⊗kBand its fiber rings when subject to regularity. It turns out that, in caseA(orB) is assumed to be residually separable,A⊗kBis regular if and only if so areA andB(Theorem 2.11). This is a slight improvement of [20, Theorem 6(c)]. All along the paper, several original examples are provided to illustrate or delimit the scope of the established results.
2. Transfer of regularity to tensor products ofk-algebras
A transcendence baseBof an extension fieldKoverkis called a separating transcendence base ifKis separable algebraic overk(B); andKis said to be separable overkif every finitely generated intermediate field has a separating transcendence base overk. Finally, recall that a Noetherian ringAcontaining a fieldkis said to be geometrically regular overkifA⊗kFis a regular ring for every finite extensionF ofk; and a homomorphismϕ:A→Bof Noetherian rings is said to be regular ifϕis flat andB⊗AκA(p)is geometrically regular overκA(p)for eachp ∈ Spec(A), whereκA(p) denotes the residue field of Ap[16, §32, pp. 255–256].
In 1965, Grothendieck proved that ifKandLare two extension fields ofk such that eitherKorLis finitely generated overkand ifKis separable overk, thenK⊗kLis regular [10, Lemma 6.7.4.1]. More generally, ifKis a separable extension field ofkandAis a regular finitely generatedk-algebra, thenK⊗kA is regular; indeed, separability implies thatk → K is regular. Then a base change via the finite type homomorphismk →Ayields thatA →K⊗kA is regular since regularity of the fibers is preserved (as the residue fields ofA are finitely generated extensions ofk). By [16, Theorem 32.2(i)],K⊗kAis regular.
Now, let us substitute the assumption “K⊗kAis Noetherian” for “Ais a finitely generatedk-algebra.” In this case, regularity is transferred toK⊗kA through base change of regular homomorphisms via André-Quillen homology (which requires no finite type assumption). Indeed, by [13, (6.3)],Dn(K⊗ A/A,−)∼=Dn(K/k,−)for everyn∈Z(since we are in the trivial case where Torkn(K, A)=0 for everyn≥1). Then, by [13, Theorem 9.5],A→K⊗A is regular if and only if k → K is regular. So that, under separability and
Noetherianity,K⊗k Ais regular if and only if so isA. We were not able to locate any explicit reference for this result. Next we record this fact in a slightly more general form and also offer a new direct proof (without use of André- Quillen homology) via the prime ideal structure (cf. [3, Proposition 4.14]).
Lemma2.1.LetAandB be twok-algebras such thatAis geometrically regular. Then the following assertions are equivalent:
(i) A⊗kBis regular;
(ii) Bis regular andA⊗kBis Noetherian.
Proof. The implication (i)⇒(ii) is straightforward by [20, Corollary 4].
Next, we prove (ii)⇒(i) via two steps.
Step 1. Suppose thatB=Kis an extension field ofksuch thatA⊗k Kis Noetherian. Letdenote the set of all finitely generated extension fields ofk contained inKand let
D:=A⊗kK= lim→
E∈
D(E)
whereD(E):= A⊗kEfor eachE ∈ . Fix a prime idealP ofDand let PE :=P ∩D(E)for eachE∈.
Claim1.IfF ∈such thatPED= PFDfor eachE∈containingF, thenP =PFD.
In fact, letF ∈such thatPED= PFDfor eachE ∈containingF. Letx ∈P. Then there existsE ∈such thatx∈D(E), and thusx ∈PED.
Whence,x ∈PF (E) =PFD, whereF (E)denotes the composite field ofF andEinK. It follows thatP =PFD, proving the claim.
Claim2.There existsE∈such thatP =PED.
Assume, by way of contradiction, that PEDP for anyE ∈ (notice that under this hypothesisK is necessarily infinitely generated over k; i.e., K /∈ ). Choose E1 ∈ . By Claim 1, there existsE2 ∈ containingE1 such thatPE1DPE2D. Iterating this process yields the following infinite chain of ideals inD
PE1DPE2D· · ·PEnD· · ·P
where theEj ∈. This leads to a contradiction sinceDis Noetherian. Hence there existsE∈such thatP =PED, as desired.
Claim3.P DP is generated by aDP-regular sequence.
Indeed, by Claim 2,P =PEDfor someE∈. Now, observe that DP :=(A⊗kK)P ∼=
D(E)PE⊗EK
P
and P DP ∼=
PED(E)PE⊗EK DP
withPED(E)PE being the maximal ideal ofD(E)PE. AsEis finitely gener- ated overk,D(E)is regular (recall that Ais geometrically regular). Hence D(E)PE is a regular local ring. By [15, Theorem 169],PED(E)PE is gener- ated by aD(E)PE-regular sequence x1, x2, . . . , xr. Further, it is easily seen thatx1⊗k 1, x2⊗E1, . . . , xr ⊗E1 is aD(E)PE ⊗E K-regular sequence of PED(E)PE ⊗E K. AsPED(E)PE⊗EKDP ∼= P DP, we get, by [15, The- orem 133], x1⊗1E1,x2⊗1E1, . . . ,xn⊗1E1 is aDP-regular sequence ofP DP. Now, sincePED(E)PE =(x1, x2, . . . , xn)D(E)PE, we get
P DP =
x1⊗E1
1 ,x2⊗E1
1 , . . . ,xn⊗E1 1
DP
establishing the claim.
It follows, by [15, Theorem 160], that DP is a regular local ring. Con- sequently,Dis a regular ring, as desired.
Step 2. Suppose thatBis a regulark-algebra such thatA⊗kBis Noetherian.
Letq ∈Spec(B). First, asA⊗kBis Noetherian,A⊗k kB(q), being a local- ization of a quotient ofA⊗kB, is Noetherian. Then, by Step 1,A⊗kkB(q) is regular for eachq∈Spec(B). Now, [20, Corollary 2] yields thatA⊗kBis regular, completing the proof of the theorem.
In particular, ifKis a separable extension field ofkandAis ak-algebra, thenK⊗kAis regular if and only ifAis regular andK⊗kAis Noetherian.
Example 2.12 shows that this result is not true, in general, if one substitutes pure inseparability for separability; and that, however, this latter condition is not necessary.
Recall that ifKandLare two extension fields ofksuch that one of them is finitely generated, thenK⊗kLis Noetherian [23]. The converse is not true in general; e.g.,
Q(x1, x2, . . .)⊗Q(√ 2,√
3, . . .)∼=Q(√ 2,√
3, . . .)(x1, x2, . . .) is a field, wherex1, x2, . . .are infinitely many indeterminates overQ. However, the converse holds in the caseK =L[9, Corollary 3.6] or [21, Theorem 11].
These facts combined with Lemma 2.1 yield the following remark, where the separability assumption is required only for regularity.
Remark2.2. LetKandLbe two extension fields ofkand assume thatKis separable overk. Then:KorLis finitely generated⇒K⊗kLis Noetherian⇔ K⊗kLis regular. The special case whereK=Lis handled by Corollary 2.6.
For an arbitraryk-algebraA(not necessarily a domain), the transcendence degree overkis given by (cf. [22, p. 392])
t.d.(A:k):=Sup{t.d.(A/p:k)|p∈Spec(A)}.
Further, ifAandBare twok-algebras such thatA⊗kB is Noetherian, then necessarily A and B are Noetherian rings and either t.d.(A : k) < ∞ or t.d.(B :k) <∞(cf. [4, p. 69]). Also, for any two extension fieldsKandLof k, [18, Theorem 3.1] asserts that
dim(K⊗kL)=min{t.d.(K:k),t.d.(L:k)}.
These facts allow one to give illustrative examples of regular tensor products (of fields) of arbitrary dimension.
Example 2.3. Let x1, x2, . . . be infinitely many indeterminates over k.
Then, for any positive integern,k(x1, . . . , xn)⊗k(x1, x2, . . .)is ann-dimen- sional regular ring.
Note thatk(x1, . . . , xn)andk(x1, x2, . . .)are (non-algebraic) separable ex- tensions ofk by Mac Lane’s Criterion. For the algebraic separable case, see Example 2.9.
LetKandLbe two extension fields ofk. Assume thatKis purely insepar- able overkand letLbe an algebraic closure ofL. Then there exists a unique k-homomorphismu:K→L[5, Proposition 3, p. V.25], and the isomorphic imageu(K)is obviously purely inseparable overk. In this vein, we can always viewKandLas subfields of a common fieldL. Recall Mac Lane’s notion of linear disjointness; namely,KandLare linearly disjoint overkif every subset ofKwhich is linearly independent overkis also linearly independent overL;
equivalently, ifK⊗kLis a domain.
In the sequel, given an extension fieldKofk,Ks andKi will denote the (not necessarily algebraic) separable closure and (algebraic) purely inseparable closure of k in K, respectively. Notice that K is an extension field of the composite fieldKsKi and the equalityKsKi = Kholds, for instance, when Kis separable, purely inseparable, or normal overk.
The next main result of this paper handles the tensor products of two ex- tensions fields, which will be used to generate new and original examples of regular tensor products of extension fields. It is worthwhile noting that this result falls beyond the scope of André-Quillen homology (since purely insep- arable field extensions are not geometrically regular).
Theorem2.4.LetKandLbe two extension fields ofksuch thatK⊗kLis Noetherian. Assume thatK = KsKi and letKi = k(S)for some generating subsetSofKi. Then the following assertions are equivalent:
(i) K⊗kLis regular;
(ii) Ki⊗kLis a domain;
(iii) Ki⊗kLis a field;
(iv) [k(S):k]=[L(S):L]for each finite subsetSofS;
(v) Ki∩L(S)=k(S)for each finite subsetSofS.
Proof. Letp:=char(k). The theorem easily holds whenp=0 (in which casek is perfect). Next, assumep ≥ 1. SinceKs is a separable extension of k,Ks ⊗kKi is reduced [24, Chap. III, §15, Theorem 39]. Further, sinceKi is algebraic overk,Ks⊗kKi is zero-dimensional [18, Theorem 3.1] and hence a von Neumann regular ring [15, Ex. 22, p. 64]. By [21, Proposition 2(c)], Ks ⊗k Ki has one unique minimal prime ideal. It follows that Ks ⊗k Ki is local and therefore a field. Now, consider the surjective ring homomorphism ϕ : Ks ⊗kKi →Ks(Ki), given on generators ofKs ⊗kKi bya⊗b →ab (asKsandKi may be contained in a common field). Soϕis an isomorphism;
that is,Ks⊗kKi ∼=KsKi =K. By Lemma 2.1,K⊗kL∼=Ks⊗k(Ki ⊗kL) is regular if and only ifKi⊗kLis regular. Hence, for the rest of the proof, we may suppose thatKis a purely inseparable algebraic extension field ofk(i.e., K=Ki) with char(k)=p=0. Same arguments as above yieldK⊗kLis a zero-dimensional local ring and, therefore, (i)⇒(ii)⇒(iii)⇒(i). Moreover, the assumption “K⊗kLis a domain” is equivalent to saying that “KandLare linearly disjoint overk,” as mentioned above. So that we get (ii)⇔(iv) by [5, Proposition 5(a), p. V.13] and (ii)⇒(v) by [5, p. V.13] and via the isomorphism K⊗kL∼=K⊗k(S)(k(S)⊗kL)for each finite subsetSofS.
(v)⇒(iii) Letx ∈Sand letpm =[k(x):k] withman integer≥0. Then a := xpm ∈ k. We wish to show that k(x)⊗k Lis a field. We may assume x /∈k. By (v),xpr ∈K∩L=k for each positive integerr < m. Therefore, x ∈ L\L, where Ldenotes an algebraic closure ofL, forcing(Xpm −a) (=(Xpr−xpr)pm−r for each positive integerr < m) to be irreducible inL[X].
It follows that
k(x)⊗kL∼=k[X]/(Xpm−a)⊗kL∼=L[X]/(Xpm −a)∼=L[x]=L(x) whereXdenotes an indeterminate overL. Sok(x)⊗kLis a field. Next, let x1, . . . , xn ∈S. We have
k(x1, . . . , xn)⊗kL∼=k(x1, . . . , xn)⊗k(x1,...,xn−1)(k(x1, . . . , xn−1)⊗kL).
By induction onn,k(x1, . . . , xn−1)⊗kL∼=L(x1, . . . , xn−1)is a field and, by (v), we get
k(x1, . . . , xn)∩L(x1, . . . , xn−1)⊆K∩L(x1, . . . , xn−1)=k(x1, . . . , xn−1) so that
k(x1, . . . , xn)∩L(x1, . . . , xn−1)=k(x1, . . . , xn−1).
Hence, the first step yields
k(x1, . . . , xn)⊗kL∼=k(x1, . . . , xn−1)(xn)⊗k(x1,...,xn−1)L(x1, . . . , xn−1) is a field. Letdenote the set of all finite subsetSofSand observe that
K⊗kL= lim→
S∈
k(S)⊗kL.
Thus,k(S)⊗kLis a field, for eachS∈, and so is their direct limitK⊗kL, establishing (iii) and completing the proof of the theorem.
One can use Theorem 2.4(v) to build new examples of regular tensor products of fields, as illustrated by the next example.
Example 2.5. Letp be a prime element of Zand let y1, y2, . . . , ym, x1, x2, . . . , xn, . . .be indeterminates overZ/pZ. Let
k:=(Z/pZ)(y1p, y2p2, . . . , ympm, x1p, x2p2, . . . , xnpn, . . .), K:=k(x1, x2, . . . , xn, . . .),
L:=k(y1, y2, . . . , ym).
ThenK⊗kLis a regular ring.
Indeed, notice thatK and Lare purely inseparable extension fields of k with [L:k]<∞. Also, we have
K=(Z/pZ)(y1p, y2p2, . . . , ympm, x1, x2, . . . , xn, . . .), L=(Z/pZ)(y1, y2, . . . , ym, x1p, x2p2, . . . , xnpn, . . .).
Next, letxi1, xi2, . . . , xir be a finite subset of{x1, x2, . . . , xn, . . .}. Then K∩L(xi1, xi2, . . . , xir)
=(Z/pZ)(xi1, xi2, . . . , xir, x1p, x2p2, . . . , xnpn, . . . , y1p, y2p2, . . . , ympm)
=k(xi1, xi2, . . . , xir).
Hence, by Theorem 2.4(v),K⊗kLis regular, as desired.
As a consequence of Theorem 2.4(v), under the Noetherianity assumption, separability rises as a necessary (and sufficient) condition for regularity in the special case whereK = Las shown in the next corollary. It also refines [6, Exercice 28, Chap. 8, p. 98] and links regularity ofK⊗kKto the projectivity ofKas aK⊗kK-module whenKis a finitely generated extension field ofk [8, Theorem 7.10].
Corollary2.6.LetKbe an extension field ofk. The following assertions are equivalent:
(i) K⊗kKis regular;
(ii) K⊗kKis Noetherian andKis separable overk;
(iii) Kis a finitely generated separable extension field ofk;
(iv) K⊗kLis regular for each extension fieldLofk;
(v) Kis a finitely generated extension field ofkand a projectiveK⊗kK- module.
Proof. (i)⇒(ii) Assume thatK⊗kKis regular. ThenK⊗kKis Noeth- erian, so thatKis finitely generated overk. We claim thatK⊗EKis regular for any extension fieldEofkcontained inK. In effect, letEbe a field extension ofkcontained inK. Then
K⊗kK ∼=K⊗E(E⊗kK)
∼=K⊗E(K⊗kE)
∼=(K⊗EK)⊗kE (cf. [1, Ex. 2.15, p. 27]).
It follows, by [16, Theorem 23.7] and by localization, thatK⊗EKis regular, establishing the claim. Now, letB be a finite transcendence basis ofK over k and letE be the algebraic separable closure ofk(B) in K. Then, via the above claim, K ⊗E K is regular and K is purely inseparable over E. By Theorem 2.4(v),K=E. It follows thatKis separable overk, as desired.
(ii)⇒(iii) is handled by [9, Corollary 3.6] or [21, Theorem 11] as mentioned above, (iii)⇒(iv) follows from [10, Lemma 6.7.4.1], (iv)⇒(i) is trivial, and (iii)⇔(v) is a particular case of [8, Theorem 7.10], completing the proof of the corollary.
One can use Theorem 2.4(v) or Corollary 2.6 to build (zero-dimensional Noetherian local) tensor products of fields that are locally complete intersection but not regular, as shown below.
Example2.7. Let kK ⊆ Lbe extension fields such that Kis purely inseparable overkandK⊗kLis Noetherian. ThenK⊗kLis a locally complete
intersection ring [20, Proposition 5(a)] which is not regular by Theorem 2.4(v) (or Corollary 2.6). For instance, for any primep, one may simply take
k:=(Z/pZ)(xp) and K=L:=(Z/pZ)(x) wherexis an indeterminate overZ/pZ.
The next result handles the (algebraic) separable case featuring a slight generalization of [21, Proposition 8]. Recall, for convenience, that ifK is a separable extension ofk, then K⊗k Lis always reduced for any extension fieldLofk[24, Chap. III, §15, Theorem 39].
Corollary2.8.LetKandLbe two extension fields ofksuch thatK⊗kLis Noetherian. Assume thatKis algebraic overk. Then the following assertions are equivalent:
(i) K⊗kLis(von Neumann)regular;
(ii) K⊗kLis reduced;
(iii) K⊗kLis a finite product of fields.
If, in addition, K is separable and L is Galois over k such that K, L are contained in an algebraic closure ofk, then the above are equivalent to:
(iv) n:=[K∩L:k]<∞.
Moreover,K⊗kLis isomorphic to the product ofncopies of the fieldK(L).
Proof. By [18, Theorem 3.1], dim(K⊗kL)=0. Recall at this point that a zero-dimensional Noetherian ring is regular if and only if it is von Neumann regular. So a combination of [15, Theorem 164], [15, Ex. 22, p. 64], and [21, Lemma 0] yields (i)⇔(ii)⇔(iii). The last two statements are handled by [21, Proposition 8].
Next, we give an illustrative example for this corollary.
Example2.9. Let(pj)j≥1denote the sequence of all prime numbers. Let X:=
i, e2iπ3
∪√pj |j odd
and Y := {i} ∪√pj |jeven . Clearly,Q(X)(resp.,Q(Y )) is an infinite algebraic separable non-normal (resp., Galois) extension field ofQand hence by Corollary 2.8
Q(X)⊗Q(Y )∼=Q
i, e2iπ3 ,√ 2,√
3, . . .
×Q
i, e2iπ3 ,√ 2,√
3, . . . is a non-trivial zero-dimensional regular ring.
Next, we move to the general case, where we discuss the correlation between A⊗kB and its fiber rings when subject to regularity. LetAandBbe twok- algebras. By identifyingAandBwith their canonical images inA⊗kB, one
can viewA⊗kB as a free (hence faithfully flat) extension ofAandB. This very fact lies behind the known transfers of regularity betweenA⊗k B and its fiber rings over the prime ideals ofAorB. The next result collects these transfer results along with a slight generalization of [20, Theorem 6(c)]. We also provide an example, via Theorem 2.4, for the non-reversibility in general of the implications involved. For this purpose, we first make the following definition.
Definition2.10. Ak-algebraRis said to be residually separable, ifκR(P ) is separable overkfor eachP ∈ Spec(R), whereκR(P )denotes the residue field ofRP.
It is easily seen that a fieldk is perfect if and only if everyk-algebra is residually separable. More examples of residually separable k-algebras are readily available through localizations of polynomial rings or pullback con- structions [2], [7]. For instance, letx be an indeterminate overk andK ⊆ L two separable extension fields ofk. Let
R:=L[x](x) and S:=K+xL[x](x). Note that the extensions
k ⊆K⊆L⊆L(x)=qf(R)=qf(S)
are separable by Mac Lane’s Criterion and transitivity of separability. So that RandSare residually separablek-algebras.
Theorem2.11. LetAandBbe twok-algebras such thatA⊗kB is No- etherian. Consider the following assertions:
(i) A,B, andκA(P )⊗kκB(Q)are regular∀(P , Q)∈Spec(A)×Spec(B);
(ii) BandA⊗kκB(Q)are regular∀Q∈Spec(B);
(iii) AandκA(P )⊗kBare regular∀P ∈Spec(A);
(iv) A⊗kBis regular;
(v) AandBare regular.
Then(i)⇒(ii) (resp.,(iii))⇒(iv)⇒(v). IfA(orB)is residually separable, then all assertions are equivalent.
Proof. The first statement is a combination of Corollary 2 and Corollary 4 as well as the proof of Theorem 6 in [20].
Next, suppose thatAorBis residually separable. ThenκA(P )⊗kκB(Q) is always regular by Lemma 2.1 for anyP ∈Spec(A)andQ∈Spec(B); and, hence, so areκA(P )⊗kBandA⊗kκB(Q). Moreover, recall that Noetherianity
carries over toκA(P )⊗kκB(Q) via localization of the general fact that ifI andJ are proper ideals ofAandB, respectively, then
(A⊗kB)/(I ⊗kB+A⊗kJ )∼=(A/I )⊗k(B/J ).
Thus, the five assertions in the theorem collapse to: “A⊗kBis regular if and only ifAandBare regular.”
The above implications are not reversible in general, as shown by the next example. This example shows also that the separable assumption in Lemma 2.1 is sufficient but not necessary and it does not hold, in general, for purely inseparable extensions.
Example 2.12. Let K be a purely inseparable extension field of k with char(k) = p = 0 and let u ∈ K with pe := [k(u) : k] for some e ≥ 2.
Thena :=upe ∈k. Letxbe an indeterminate overk,r∈ {1,· · ·, e−1}, and A:=k[x](xpe−r−a). Then:
(i) Ais local regular with maximal idealᒊ:=(xpe−r −a)A.
(ii) k(u)⊗kAis regular.
(iii) k(u)⊗kA/ᒊis not regular.
Indeed, notice that(xpe−r −a)is a prime ideal ofk[x] and, hence,ᒊis the maximal ideal of A, since A/ᒊ ∼= k[x]/(xpe−r −a) ∼= k(upr). Moreover, k(u)⊗kA∼=S−1k(u)[x] is a regular ring, whereS:=k[x]\(xpe−r−a). This proves (i) and (ii). However,k(u)⊗k(A/ᒊ)∼=k(u)⊗kk(upr)is not regular, by Theorem 2.4(v), sincek=k(u)∩k(upr)=k(upr), proving (iii).
The assumption “A(orB) is residually separable” in Theorem 2.11 is not necessary, as shown by the following example.
Example2.13. Letk,K, andLbe defined as in Example 2.5 andx, ytwo indeterminates overk. Let
A:=K[x](x)=K+ᒊA with ᒊA:=xA B:=L[y](y)=L+ᒊB with ᒊB :=yB
ThenA and B are regular local k-algebras which are not residually separ- able over k (since K and L are purely inseparable over k as seen in Ex- ample 2.5). Moreover,A⊗kBis Noetherian (in fact, regular via localization) and(A/ᒊA)⊗k(B/ᒊB)∼=K⊗kLis a regular ring. Consequently,AandB satisfy all assertions of Theorem 2.11, as desired.
The next example illustrates the slight improvement (of [20, Theorem 6(c)]) featured in the last statement of Theorem 2.11. Namely, we provide original
examples wherekis an arbitrary field,A, Bare regulark-algebras withA⊗kB Noetherian andAis residually separable overk.
Example2.14. Letkbe an arbitrary field,Kany separable extension field ofk, andx, y, t three indeterminates overk. Consider theK-algebra homo- morphism
ϕ:K[x, y]→K[[t]]
defined by ϕ(x) = t and ϕ(y) = s :=
n≥1tn!. Sinces is known to be transcendental overK(t),ϕis injective. This induces the following embedding of fields
ϕ:K(x, y)→K((t)).
It is easy to check that A := ϕ−1(K[[t]]) is a discrete rank-one valuation overring ofK[x, y] and thatA=K+ᒊwithᒊ=xA. Then,Ais a residually separable regular ring. Now, letB be any regular ring such that A⊗k B is Noetherian. For instance, one may chooseBto be any finitely generated regular k-algebra or any (purely inseparable) finitely generated extension field ofk.
By Theorem 2.11,A⊗kBis a regular ring.
It is worthwhile noticing that, in most examples, the non-regularity was ensured by the negation of “Ki ∩L = k.” One might wonder if this weak property may generate the condition (v) of Theorem 2.4; namely, letK be a finite dimensional purely inseparable extension field of k and let Lbe an extension field ofk. Do we have:K∩L=k ⇔K⊗kLregular? The answer is negative as shown by the next example.
Example2.15. Letx, y, zbe three indeterminates overZ/2Z. Let k :=(Z/2Z)(x4, y4),
K :=k(x2, y2)=(Z/2Z)(x2, y2),
L:=k(x2(y2+z), z)=(Z/2Z)(x4, x2(y2+z), z).
ThenK∩L=kandK⊗kLis not a regular ring.
Indeed, clearly, K is a purely inseparable extension field ofk. Further, note that {1, x2} is a basis of K over k(y2) and, as (x2(y2+ z))2 ∈ k(z), {1, x2(y2+ z)} is a basis of L over k(z). Let f ∈ K∩L. So there exist g0, g1∈k(y2)andf0, f1∈k(z)such that
f =g0+g1x2
=f0+f1x2(y2+z).
As (x2)2 ∈ k(y2, z) and x2 ∈ k(y2, z) = (Z/2Z)(x4, y2, z), then {1, x2} is, as well, a basis ofk(x2, y2, z)overk(y2, z). It follows thatf0 = g0 and
f1(y2+z)=g1. Hence,f0∈k(z)∩k(y2)=k. Moreover, observe that{1, y2} is a basis ofk(y2, z)overk(z)and ofk(y2)overk. Hence, asg1=f1z+f1y2 andg1 ∈ k(y2), we getf1z ∈ k, so thatf1 = 0. Consequently, f ∈ k and thereforeK∩L=k, as claimed.
Now, L(x2) = k(x2, y2, z) = K(z). Hence K∩L(x2) = K = k(x2).
Then, by Theorem 2.4(v),K⊗kLis not regular, as desired.
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DEPARTMENT OF MATHEMATICS UNIVERSITY OF MEKNES MEKNES 50000 MOROCCO
E-mail:bouchiba@fs-umi.ac.ma
DEPARTMENT OF MATHEMATICS AND STATISTICS KFUPM, DHAHRAN 31261
KSA
E-mail:kabbaj@kfupm.edu.sa