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A Note on the Tensor Products of Algebras over a Field Mem. Fac. Educ., Kagawa Univ. II, 63（2013）, 1－ 4
A Note on the Tensor Products of Algebras over a Field
Kazunori F
UJITAINTRODUCTION. All rings considered in this paper are assumed to be commutative with identity, and all ring homomorphisms are unital. Throughout, k stands for a field. A ring R is said to be a Hilbert ring if it satisfies one of the following equivalent conditions. （1） Every prime ideal of R is an intersection of maximal ideals.
（2） If M is a maximal ideal in R［X］, then M ∩ R is a maximal ideal.
Let K and L be extension fields of k. If tr.degkK ≧ tr.degkL＝ n ＜ ∞, then K ⊗k L is a
Hilbert ring in which every maximal ideal has height n（［16］, Theorem 5）. This result is generalized in ［6］, ［11］, ［12］ and ［15］. In the article ［5］, the following question was given.
QUESTION: Let A and B be two Hilbert k-algebras such that A ⊗k B is noetherian. Is A ⊗k B
a Hilbert ring ? Is it equicodimensional ?
In this paper, we show some conditions of Hilbert k-algebras A and B when A ⊗k B is
a Hilbert ring. Any unreferenced material is standard, as in ［4］ and ［10］.
LEMMA 1. If R is a Hilbert ring, then R［X1, ･･･, Xn］is a Hilbert ring for each positive
integer n
PROOF. See［3, （31.8）］
PROPOSITION 1. Let A and B be two k-algebras. If A is finitely generated over k, and B is a Hilbert ring, then A ⊗k B is a Hilbert ring.
PROOF. Let A＝k［a1, ･･･, an］, and φ : k［X1, ･･･, Xn］ A be a k-algebra homomorphism
such that φ（Xi）＝ai. Since B is a Hilbert ring, k［X1, ･･･, Xn］⊗k B ≅ B［X1, ･･･, Xn］ is a
Hilbert ring by Lemma 1. Therefore its homomorphic image A ⊗k B is a Hilbert ring.
LEMMA 2. Let K be a field of uncountable cardinality, A be a countabuly generated K-algebra. Then A is a Hilbert ring.
－2－ Kazunori FUJITA
Theorem in［9, p.407］. ［{Xn｜n ∈N }］ is integral over B＝K［{Xn｜n ∈N}］, so B is
a Hilbert ring. A is a homomorphic image of B, hence A is a Hilbert ring.
PROPOSITION 2. Let k be a field of countable cardinality, K be a field of uncountable cardinality containing k. If A is a countably generated k-algebra, and if B is a countably generated K-algebra. Then A ⊗k B is a Hilbert ring.
PROOF. A ⊗k B is a countablly generated K-algebra. Therefore A ⊗k B is a Hilbert ring by
Lemma 2.
EXAMPLE 1. Q（X1, X2, ･･･, Xn）⊗QR［Y1, Y2, Y3, ･･･］ is a Hilbert ring by Proposition 2.
EXAMPLE 2. Q［X1, X2, X3, ･･･］⊗QR ≅ R［X1, X2, X3, ･･･］ is a Hilbert ring by Lemma 2, however Q［X1, X2, X3,･･･］ is not a Hilbert ring.
DEFINITIONS. （［17］, p.394） Let A be an integral domain over k. A is an AF-domain （altitude formula） if
ht（P）+ tr.deg（A／P）＝tr.degk （A）k
for each prime ideal P of A.
DEFINITIONS. （［17］, p.395） Let R be a k-algebra, p a prime ideal of R and 0 ≦ d ≦ s be integers. Set
Δ（s, d, p）：＝ht（p［X1, ･･･, Xs］）+ min（s, d + tr.deg（R／p））k
D（s, d, R）：＝max{Δ（s, d, p）｜p ∈ Spec（R）}
LEMMA 3 （［17］, Theorem 3.7）. Let A be an AF-domain with t＝tr.deg（A）and d ＝ dim（A）. k
Let R be any k-algebra. Then
dim（A ⊗kR）＝D（t, d, R）
＝max{ht（pR［X1, ･･･, Xt］）+min（t, d+tr.deg（R／p）｜p∈ Spec（R）}k
LEMMA 4 （［17］, Corollary 4.2）. Let A is an integral domain containing k. If A is noetherian, or is a Prüfer domain（e.g., a valuation domain）, or AF-domain, then
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A Note on the Tensor Products of Algebras over a Field
Let R be a ring, and let U be the multiplicatively closed subset of monic polynomials of R［X］. Then the ring U－1R［X］ is denoted by R〈X〉. Some basic properties of rings of this type are stydied in ［1］, ［14］. If R is a noetherian ring, then R〈X〉 is a Hilbert ring （［2］, ［13］）.
EXAMPLE 3. Let （V, M） be a discrete rank-one valuation domain containing k. Then （1） （V〈X〉）［Y］ is a Hilbert domain such that every maximal ideal is of height 2. （2） k（Y）⊗k V〈X〉 is a Hilbert domain.
PROOF. （1） V〈X〉 is a 1-dimensional noetherian Hilbert domain, so that （V〈X〉）［Y］ is a Hilbert domain and for every maximal ideal M of （V〈X〉）［Y］, ht（M）＝2.
（2） Lemma 3 implies that dim（k（Y）⊗kV〈X〉＝2. Let S＝k［Y］－{0}. Then it can be
shown that S－1（V〈X〉）［Y］／P is not semi-local for each height one prime ideal P of S－1 （V〈X〉）［Y］. Thus k（Y）⊗k V〈X〉 is a Hilbert domain.
LEMMA 5 （［4］, p.106）. Let R be a noetherian ring, and let s be a non-nilpotent element of rad（R）. Then Rs is a Hilbert ring.
EXAMPLE 4. Let （V, M） be a discrete rank-one valuation domain containing k such that tr.degkV＝n ＜∞. Then dim（V〈X〉⊗kV〈 X 〉）＝n + 2 by Lemma 4. Is V〈X〉⊗k V〈 X 〉
is a Hilbert ring ? Let B＝（1+YV［Y］）－1V［Y］, where Y＝1／X. Then V〈X〉＝B［1／Y］, B＝V［Y］（M, Y）, so that Y is contained in rad（B）by Proposition 1.4 in［8］. If B ⊗k B is
noetherian, and if Y ⊗ 1, 1 ⊗ Y ∈ rad（B ⊗k B）, then V〈X〉⊗kV〈X〉 is a Hilbert ring by
Lemma 5.
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