局部化多項式環中單項式理想的約化

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(1)國立臺灣師範大學數學系碩士班碩士論文. 指導教授：劉容真博士. Reductions of Monomial Ideals in k[x,y](x,y) and k[x,y,z](x,y,z). 研究生：葉柔吟. 中華民國 102 年 6 月.

(2) 感謝詞終於到了撰寫感謝詞的時候，表示我的研究所碩班生涯即將正式畫下句點。除了一年級的時候是全職學生之外，二、三年級的我都是一邊在高中職教書、一邊進修，因此我的學業和這本論文可以順利完成真的多虧了很多人的幫助，他們都是我的貴人，藉著這篇感謝詞，我想對這些貴人說出由衷的感謝，謝謝你們的各種幫忙，我才能夠完成。首先是我的指導教授劉容真老師。從大學的時候就喜歡上老師的課，因為老師總是不厭其煩把複雜的課程講得很清楚，上了研究所能夠當老師的學生更是最幸運的事。感謝老師這段日子以來體諒我身兼教職，對我很包容也很有耐心，在寫論文的這一年裡，從構思到撰寫、每次的修改終至逐步完成，每個階段都多虧了老師的用心指導，我真的很感恩我的指導教授是您。從老師身上所感受到對待學生的親切、耐心和關懷也將讓我帶到教職生涯中去幫助我的學生。感謝論文口試委員江謝宏任老師和洪有情老師，謝謝您們百忙之中抽空擔任我的口試委員，並且提出寶貴的意見讓我可以把論文修改得更加完善。感謝我服務過的大安高工和嘉義女中，謝謝學校給我工作同時進修的機會，讓我可以同時從事這兩件想做的事。而在寫論文的最後這一年裡，我要特別感謝嘉義女中學務處同事們的幫忙：謝謝我的職務代理人寇姐，每週一請公假北上 meeting 都是妳代為處理衛生組的各項事務，辛苦了；謝謝我的主管皮皮主任，行政業務上有賴妳罩我，從我到學務處以來受妳照顧和幫忙很多很多；謝謝鏗鏗、小鳳、毓萱、玉芬、清隆、珮瑜、沂蓁阿姨的幫忙，每當我不在的時候，有幫我代收回收金的、有幫我處理學生問題的等等，還有你們正面或是反面(激將法？)的鼓勵和加油打氣，因為有你們這些讓人安心和貼心的同事，我才能工作之餘撥出心力完成論文。感謝我的好姊妹陳姳伃、林方魚、謝依紋，除了一直以來的互相加油與支持外，妳們都在我北上進修的時候收留過我，讓我在台北有住的地方，尤其是阿亮陳姳伃，這一年來我大多都是去打擾妳的，後來還索性在妳家放了一些我的東西，哈，有妳這個好朋友真是我的福氣！謝謝和我同時寫論文枚娟，雖然我們在不同的學校進修，但有妳一起互相鼓勵並且有時還提供諮詢，讓我受益良多。最後要感謝我最親愛的家人─爸、媽、弟、妹，你們永遠是我最忠實可信賴的依靠，因為有你們做我的後盾，我才能一直去做自己想做的事，我愛你們！最後的最後，謝謝台灣師範大學這個培育我的母校，從大學到研究所，是我求學生涯裡待最久的地方，這次畢業真的要離開你了！希望往後教職之路上我可以繼續運用所學、做得越來越好，成為一個可以榮耀母校並回饋社會的好老師！葉柔吟謹誌 102 年 6 月.

(3) Abstract We consider monomial ideals in the two-dimensional localized polynomial ring k[x, y](x,y) where k is an infinite field. In C-Y. Jean Chan and Jung-Chen Liu’s paper, they determine a sufficient condition under which an ideal containing xa y b + xc y d is a reduction of an ideal containing xa y b and xc y d. In this thesis, we use another approach to prove the above result. Furthermore, we extend the sufficient condition to the three-dimensional localized polynomial ring k[x, y, z](x,y,z) where k is an infinite field. Keywords. localized polynomial ring, monomial ideal, reduction.. 1.

(4) Contents 1 Introduction. 1. 2 Preliminaries. 2. 3 Reductions in k[x, y](x,y). 5. 4 Reductions in k[x, y, z](x,y,z). 15. References. 24.

(5) 1. Introduction. While reading papers related to reductions of monomial ideals, it was brought to my attention that in V. C. Qui˜ nonez’s research report [Q1] and in C-Y. Jean Chan and Jung-Chen Liu’s paper [CL], they had achieved similar corollaries ([Q1, Corollary 3.6] and [CL, Corollary 3.7]) regarding minimal reductions of monomial ideals in two-dimensional power series rings and two-dimensional localized polynomial rings from two different theorems ([Q1, Theorem 3.3] and [CL, Theorem 3.3]). In particular, in [CL], Chan and Liu consider monomial ideals in the two-dimensional localized polynomial ring k[x, y](x,y) where k is an infinite field and determine a sufficient condition under which an ideal I is a reduction of the minimal monomial ideal containing I ([CL, Theorem 3.3]). For example, we can check straightforwardly from the graph whether an ideal containing xa y b + xc y d is a reduction of an ideal containing xa y b and xc y d. To me, this method is very direct and concrete. I wonder whether this sufficient condition can be extended to the three-dimensional localized polynomial ring k[x, y, z](x,y,z). In addition to inspiration from reading Qui˜ nonez’s research reports, the experience of teaching in a senior high school also brings me new ideas. In this thesis, we revisit the two-dimensional case in Section 3 and extend the results to the three-dimensional case in Section 4. More precisely, we use some property of vectors in the real plane, that I taught last semester, to prove Lemma 3.3 and Lemma 3.4, and use them to replace the key lemmas [CL, Lemma 3.5 and 3.6] for proving the main theorem ([CL, Theorem 3.3]) for the two-dimensional localized polynomial ring k[x, y](x,y) . We also give one example to demonstrate this algorithm. Moreover, since the property of vectors using in Section 3 still holds true in three-dimensional real space, we give a similar lemma (Lemma 4.2) and prove a theorem (Theorem 4.3) for the three-dimensional localized polynomial ring k[x, y, z](x,y,z). As in Section 3, we provide a few examples to illustrate this algorithm at the end of Section 4.. 1.

(6) 2. Preliminaries. Let R be a commutative ring with identity and let J ⊆ I be ideals of R. 1. I is integral over J, if for every element u in I, there exist elements ai in J i such that un + a1 un−1 + a2 un−2 + . . . + an−1 u + an = 0. 2. J is a reduction of I, if I m+1 = JI m for some positive integer m. When R is Noetherian, these two definitions are equivalent, i.e, I is integral over J if and only if J is a reduction of I. Let R be a Noetherian local ring with maximal ideal m. Given an ideal I = (a1 , a2 , . . . , as ) in R, there are two rings that are used frequently in the study of reduction, namely the Rees algebra R[It] of I and the f iber cone R[It]/mR[It] of I. Note that R[It] = R ⊕ I ⊕ I 2 ⊕ I 3 ⊕ . . . = R ⊕ It ⊕ I 2 t2 ⊕ I 3 t3 ⊕ . . . = R[a1 t, a2 t, a3 t, . . . , as t] = {f (a1 t, a2 t, . . . , as t) | f (x1 , x2 , . . . , xs ) ∈ R[x1 , x2 , . . . , xs ]} ⊆ R[t]. Let J be a subideal of I, i.e., J ⊆ I. Then R[Jt] = R ⊕ Jt ⊕ J 2 t2 ⊕ . . . ⊆ R[It], i.e., R[Jt] is a subring of R[It]. Following directly from the definition, we have that for every element u in I, un + b1 un−1 + b2 un−2 + . . . + bn−1 u + bn = 0 with bi ∈ J i if and only if (ut)n + (b1 t)(ut)n−1 + . . . + (bn−1 tn−1 )(ut) + bn tn = 0, with bi ti ∈ J i ti ⊆ R[Jt]. From this observation, we see that I is integral over J if and only if R[It] is integral over R[Jt] as rings. In fact, a similar result holds for their fiber cone. R[Jt] R[It] Consider the homomorphism ϕ : −→ induced by the inclusion mR[Jt] mR[It] 2.

(7) map R[Jt] ֒→ R[It]. We say R[It]/mR[It] is integral over R[Jt]/mR[Jt] provided that R[It]/mR[It] is integral over Im ϕ. Therefore, I is integral over J if and only if R[It]/mR[It] is integral over R[Jt]/mR[Jt]. Let R = k[x, y](x,y) be the polynomial ring k[x, y] localized at the maximal ideal (x, y) where k is an infinite field. Since k is a field, k is a Noetherian ring. This implies that k[x, y] is Noetherian by Hilbert’s Basis Theorem. Furthermore, a localization of a Noetherian ring is again Noetherian. Hence k[x, y](x,y) is a Noetherian local ring. Similarly, k[x, y, z](x,y,z) is also a Noetherian local ring. Note that by multiplying a suitable unit to a generator, we see that every ideal in R is generated by polynomials in k[x, y]. For every element f in k[x, y], f can P be written as a linear combination of monomials, i.e., f = ηij xi y j with ηij ∈ k where we assume no repeated like terms in the expression. We use the following notation for the collection of the finitely many monomials occurring in f X Γ(f ) = {xi y j | f = ηij xi y j and ηij 6= 0}.. Let I be an ideal of R and suppose it is generated by f1 , . . . , fm ∈ k[x, y]. If I ′ is a monomial ideal containing I, then it is clear that I ′ contains Γ(f1 ) ∪ . . . ∪ Γ(fm ). Hence the smallest monomial ideal containing I is generated by Γ(f1 ) ∪ . . . ∪. Γ(fm ). We denote this monomial ideal by I ∗ . Note that since I ∗ is the smallest monomial ideal containing I, it does not depend on the choice of generators of I. A similar argument can be applied to ideals I in any localized polynomial rings k[x1 , x2 , . . . , xn ](x1 ,x2 ,...,xn ) and define the ideal I ∗ . In particular, we discuss the case of n = 3 in Section 3, namely ideals in k[x, y, z](x,y,z) . In the end of this section, we include two exercises and one property which are applied in later sections. Let R be a commutative ring with identity and let I be an ideal of R. Consider the homomorphism of polynomial rings ϕ :. R[x]. −→. (R/I)[x]. a0 + a1 x + . . . + an xn 7−→ a0 + a1 x + . . . + an xn where ai = ai + I ∈ R/I. Then ϕ is a ring epimorphism and ker ϕ = IR[x]. Thus, R[x]/IR[x] ∼ = (R/I)[x]. 3.

(8) We further assume that R is local with maximal ideal m and let ϕ : A → B be an isomorphism of extension rings of R. Consider the epimorphism ϕ. π. ψ:A− →B− → B/mB where π is the canonical epimorphism. Since ker π = mB, ker ψ = ϕ−1 (mB) = mA. Then A/mA ∼ = B/mB. Lastly, we present a property of vectors that will be applied in later proofs. Suppose O, P, Q are three non-collinear points and suppose C is a point such that −→ −→ −→ OC = αOP + β OQ with α, β ∈ R. Then P, Q, C are collinear if and only if α + β = 1. On the other hand, if C is not on the line through P and Q, then α + β 6= 1. More precisely, α + β < 1 if and only if C and O are on the same side ←→ of P Q; equivalently, α + β > 1 if and only if C and O are on different two sides ←→ of P Q. Furthermore, if C lies inside the two-dimensional region bounded by the −→ −→ rays OP and OQ, then α and β are both positive. Therefore, it can be concluded from the above properties that α + β = 1 with 0 ≤ α, β ≤ 1 if and only if C is on the line segment P Q.. 4.

(9) 3. Reductions in k[x, y](x,y). In this section, we let R = k[x, y](x,y) , where k is an infinite field, and let I be an ideal of R generated by f1 , . . . , fm ∈ k[x, y]. We revisit [CL, Theorem 3.3] which states a sufficient condition, in terms of monomials in Γ(f1 ) ∪ . . . ∪ Γ(fm ), for I to be a reduction of I ∗ . We will give a slightly different proof for this theorem, which allows us to generalize this theorem to the three-dimensional localized polynomial ring k[x, y, z](x,y,z) . In order to approach this theorem, we include several lemmas first. In particular, we use Lemma 3.3 and Lemma 3.4, which we prove by applying a property of vectors in the real plane, to replace [CL, Lemma 3.5 and 3.6]. Theorem 3.1. ([CL, Theorem 3.3]) Let R = k[x, y](x,y) and |k| = ∞. Let I be an ideal of R generated by f1 , . . . , fm ∈ k[x, y]. Assume that the following is true: for all i = 1, 2, . . . , m and for any two distinct monomials xa y b and xc y d in Γ(fi ) with c < a and b < d, there exists xr y s ∈ Γ(fj ) for some j such that the point (r, s) lies on the left hand side of the line through (a, b) and (c, d). Then I is a reduction of I*. Prior to proving this theorem, we discuss several supporting lemmas. For completeness, we state the following lemma ([CL, Lemma 3.4]) and provide the proof. Lemma 3.2. ([CL, Lemma 3.4]) Let k[u1 , u2 , . . . , un ] be a k-algebra and consider its k-subalgebra k[η1 u1 + η2 u2 + . . . + ηn un ] for nonzero η1 , . . . , ηn in k. For all i 6= j, suppose there are positive integers αij , βij such that ui αij uj βij = 0. Then k[u1 , u2 , . . . , un ] is integral over k[η1 u1 + η2 u2 + . . . + ηn un ]. Proof. First, we prove the case where η1 = η2 = . . . = ηn = 1 by induction on n. For n = 1, it’s trivial that k[u1 ] is integral over k[u1 ]. For the case n = 2, we consider the k-algebra k[u1 , u2 ] and its k-subalgebra k[u1 + u2 ]. By assumption, there are positive integers α1 , α2 such that uα1 1 uα2 2 = 0. This implies uα1 1 u2α2 +1 = 0.. 5.

(10) So we may assume that α2 is odd. Then we compute as the following: u1α1 +α2 = uα1 1 uα2 2 + u1α1 +α2 = uα1 1 (uα2 2 + uα1 2 ) = uα1 1 {[(u1 + u2 ) − u1 ]α2 + uα1 2 } = uα1 1 [(u1 + u2 )α2 − (α1 2 )(u1 + u2 )α2 −1 u1 + . . . + (αα22 −1 )(u1 + u2 )u1α2 −1 − uα1 2 + uα1 2 ]. We conclude u1α1 +α2 −(αα22 −1 )(u1 +u2 )u1α1 +α2 −1 +. . .+(α1 2 )(u1 +u2 )α2 −1 u1α1 +1 −(u1 +u2 )α2 uα1 1 = 0. Thus u1 is integral over k[u1 + u2 ]. Then k[u1 + u2 ][u1 ] is a finitely generated k[u1 + u2 ]-module and so k[u1 + u2 ][u1 ] = k[u1 + u2 , u1 ] = k[u1 , u2 ] is integral over k[u1 + u2 ]. Assume n ≥ 3 and suppose the assertion holds for all k-algebras with n − 1 or less generators. For the k-algebra k[u1 , u2, . . . , un ] with n generators, we choose α = max{αij , βij }, then we have uαi uαj = 0 for all i 6= j. Consider the i,j. k-algebras k[u1 + . . . + un−1 + un ] ⊆ k[u1 + . . . + un−1 , un ] ⊆ k[u1 , . . . , un−1, un ]. It’s enough to show that (1)k[u1, . . . , un−1 , un ] is integral over k[u1 + . . . + un−1 , un ], and (2)k[u1 + . . . + un−1 , un ] is integral over k[u1 + . . . + un−1 + un ]. For (1), since uαi uαj = 0 for all i 6= j, by the induction hypothesis, we have that k[u1 , u2 , . . . , un−1] is integral over k[u1 +u2 +. . .+un−1 ] and so k[u1 , u2 , . . . , un−1 , un ] is integral over k[u1 + u2 + . . . + un−1 , un ]. For (2), consider the element (u1 + u2 + . . . + un−1 )(n−1)α uαn . Note that after expanding (u1 + u2 + . . . + un−1)(n−1)α , all terms are of the form α. n−1 uα1 1 uα2 2 . . . un−1 with α1 + α2 + . . . + αn−1 = (n − 1)α. If αi < α for all i, then. α1 + α2 + . . . + αn−1 < α(n − 1), which can never hold. Hence at least one of the αi is larger than or equal to α. Let αk ≥ α, then α. α. n−1 n−1 (uα1 1 uα2 2 . . . un−1 )uαn = (uα1 1 . . . un−1 )uαk k uαn = 0.. 6.

(11) Therefore, (u1 + u2 + . . . + un−1)(n−1)α uαn = 0. Thus, by the case of n = 2, k[u1 + . . . + un−1 , un ] is integral over k[u1 + . . . + un−1 + un ]. So it follows from (1) and (2) that k[u1 , . . . , un ] is integral over k[u1 + . . . + un ]. At last, we consider the α. β. general case. Note that ui ij uj ij = 0 implies (ηi ui )αij (ηj uj )βij = 0. By replacing uℓ by ηℓ uℓ for all ℓ = 1, 2, . . . , n, we have k[η1 u1 , η2 u2 , . . . , ηn un ] is integral over k[η1 u1 + η2 u2 + . . . + ηn un ]. Since η1 , η2 , . . . , ηn are all units in k, k[u1 , u2, . . . , un ] = k[η1 u1 , η2 u2 , . . . , ηn un ]. The proof is complete. Next, we use a new approach to prove the following two technical lemmas. Lemma 3.3. Let (a, b), (c, d), (e, f ) ∈ Z2≥0 with a > c and b < d. If the point (e, f ) lies within the triangular region with vertices (a, b), (c, d), (0, 0), including all boundaries except the line segment connecting (a, b) and (c, d), then there exist nonnegative integers α, β and positive integers γ, δ such that (xa y b)α (xc y d)β = xγ y δ (xe y f )α+β . Simply speaking, the above monomial equality holds if (e, f ) is in the following shaded region:. (c, d) b. b. b. b. (a, b). b. (0, 0) Figure 3.3. Proof. Suppose that (e, f ) is in the assumed region. Then there exist s, t ∈ R with 0 ≤ s, t < 1 and s + t < 1 such that (e, f ) = s(a, b) + t(c, d). So we have e = sa + tc, f = sb + td. 7.

(12) The fact a, b, c, d, e, f ∈ Z≥0 implies s, t ∈ Q. Let ℓ = ad − bc > 0, and set α = ℓs, β = ℓt. Then we obtain that α and β are nonnegative integers and that e(α + β) = eℓ(s + t) < eℓ = (sa + tc)ℓ = aα + cβ, f (α + β) = f ℓ(s + t) < f ℓ = (sb + td)ℓ = bα + dβ. By setting γ = (aα + cβ) − e(α + β) and δ = (bα + dβ) − f (α + β), we have (xa y b )α (xc y d )β = xγ y δ (xe y f )α+β in which α, β are nonnegative integers and γ, δ are positive integers as desired. Lemma 3.4. Let (a, b), (c, d), (e, f ) ∈ Z2≥0 with a > c and b < d. If the point (e, f ) lies within the triangular region bounded by the x-axis, the line through (a, b) and (c, d), and the line through (a, b) and (0, 0), including the x-axis but excluding the other two boundaries, i.e., the shaded region in Figure 3.4.1,. bb. (c, d). (c, d). bb. bb. (a, b) bb. bb. bb. bb. (a, b). bb. (0, 0). (0, 0). Figure 3.4.2.. Figure 3.4.1.. then there exist positive integers α, β, γ, δ such that (xa y b)α+β = xγ y δ (xe y f )α (xc y d)β . Symmetrically, if the point (e, f ) is in the shaded region in Figure 3.4.2, then there exist positive integers α, β, γ, δ such that (xc y d)α+β = xγ y δ (xa y b)α (xe y f )β . Proof. Note that if (e, f ) is in the shaded region in Figure 3.4.1, then (a, b) is on the right hand side of the line through (c, d) and (e, f ), and lies inside the 8.

(13) region bounded by the ray from (0, 0) toward (c, d) and the ray from (0, 0) toward (e, f ). This implies that there exist s, t ∈ R with s, t > 0 and s + t > 1 such that (a, b) = s(e, f ) + t(c, d). So we have a = se + tc, b = sf + td. The fact a, b, c, d, e, f ∈ Z≥0 implies s, t ∈ Q. Let ℓ = |ed − cf |, and set α = ℓs, β = ℓt. Then we obtain that α, β are positive integers and that a(α + β) = aℓ(s + t) > aℓ = (se + tc)ℓ = eα + cβ, b(α + β) = bℓ(s + t) > bℓ = (sf + td)ℓ = f α + dβ. By setting γ = a(α + β) − (eα + cβ) and δ = b(α + β) − (f α + dβ), we have (xa y b )α+β = xγ y δ (xe y f )α (xc y d )β in which α, β, γ, δ are all positive integers as desired. Symmetrically, if (e, f ) is in the shaded region in Figure 3.4.2, (c, d) is on the right hand side of the line through (e, f ) and (a, b), and lies inside the region bounded by the ray from (0, 0) toward (e, f ) and the ray from (0, 0) toward (a, b). This implies that there exist s, t ∈ R with s, t > 0 and s + t > 1 such that (c, d) = s(a, b) + t(e, f ). So we have c = sa + te, d = sb + tf. The fact a, b, c, d, e, f ∈ Z≥0 implies s, t ∈ Q. Let ℓ = |af − eb|, and set α = ℓs, β = ℓt. Then we obtain that α, β are positive integers and that c(α + β) = cℓ(s + t) > cℓ = (sa + te)ℓ = aα + eβ, d(α + β) = dℓ(s + t) > dℓ = (sb + tf )ℓ = bα + f β. By setting γ = c(α + β) − (aα + eβ) and δ = d(α + β) − (bα + f β), we have (xc y d )α+β = xγ y δ (xa y b )α (xe y f )β in which α, β, γ, δ are all positive integers as desired. 9.

(14) Now, we are ready to prove the main theorem of this section, Theorem 3.1. Proof. We express the generators of I as the following: f1 =. n1 X. η1j xa1j y b1j , f2 =. j=1. n2 X. η2j xa2j y b2j , . . . , fm =. nm X. ηmj xamj y bmj. j=1. j=1. with ηij 6= 0 in k. Then I ∗ is the ideal generated by xaij y bij for all i = 1, . . . , m and j = 1, . . . , ni . Let m = (x, y)R be the maximal ideal of R. In order to show that I is a reduction of I ∗ , it suffices to show that the fiber cone R[I ∗ t]/mR[I ∗ t] of I ∗ is integralover thefiber cone R[It]/mR[It] of I, that is, R[I ∗ t]/mR[I ∗ t] is integral R[It] R[It] R[I ∗ t] over Φ where Φ : −→ is the natural homomorphism mR[It] mR[It] mR[I ∗ t] induced by the inclusion map R[It] ֒→ R[I ∗ t]. Consider the polynomial ring R[Uij ] = R[Uij | i = 1, . . . , m, j = 1, . . . , ni ] and the ring epimorphism ϕ : R[Uij ] −→ Uij. R[I ∗ t].. 7−→ xaij y bij t. Then we have R[I ∗ t] ∼ = R[Uij ]/ ker ϕ with xaij y bij t 7→ Uij + ker ϕ and R[I ∗ t] ∼ R[Uij ]/ ker ϕ ∼ R[Uij ] = = ∗ mR[I t] m(R[Uij ]/ ker ϕ) (mR[Uij ] + ker ϕ) with xaij y bij t+mR[I ∗ t] 7→ Uij +ker ϕ +m(R[Uij ]/ ker ϕ) 7→ Uij +(mR[Uij ]+ker ϕ). Since. R[Uij ]/mR[Uij ] R[Uij ] ∼ = (mR[Uij ] + ker ϕ) (mR[Uij ] + ker ϕ)/mR[Uij ] with Uij + (mR[Uij ] + ker ϕ) 7→ Uij + mR[Uij ] + (mR[Uij ] + ker ϕ)/mR[Uij ] ,. we consider the epimorphism ψ : R[Uij ] −→ (R/m)[Uij ]. Since ker ψ = mR[Uij ], ψ induces the isomorphism ψ : (R[Uij ]/mR[Uij ]) −→ (R/m)[Uij ]. Then we have R[Uij ]/mR[Uij ] (R/m)[Uij ] ∼ . = (mR[Uij ] + ker ϕ)/mR[Uij ] ψ (mR[Uij ] + ker ϕ)/mR[Uij ] 10.

(15) Since ψ (mR[Uij ] + ker ϕ)/mR[Uij ] = ψ(mR[Uij ] + ker ϕ) = ψ(ker ϕ), (R/m)[Uij ] (R/m)[Uij ] = . ψ(ker ϕ) ψ (mR[Uij ] + ker ϕ)/mR[Uij ]. Let uij = Uij + ψ(ker ϕ) in (R/m)[Uij ]/ψ(ker ϕ) and identify R/m with k. Then (R/m)[Uij ] = k[uij ]. So we have ψ(ker ϕ) R[Uij ]/mR[Uij ] ∼ = k[uij ] (mR[Uij ] + ker ϕ)/mR[Uij ] with Uij + mR[Uij ] + (mR[Uij ] + ker ϕ)/mR[Uij ] 7→ uij . Therefore, we have R[I ∗ t] ∼ an isomorphism = k[uij ] with xaij y bij t + mR[I ∗ t] 7→ uij . Hence with the mR[I ∗ t] homomorphism Ψ:. R[It] mR[It] fi t + mR[It]. R[I ∗ t] mR[I ∗ t]. Φ. −−−−→ 7−→. Φ =. ni X. ∼ =. ηij xaij y bij t + mR[It] j=1 ! ni X ηij xaij y bij t + mR[I ∗ t]. k[uij ]. j=1. =. ni X. aij bij. ∗. ηij x y t + mR[I t]. j=1. where R[It] = R[f1 t, f2 t, . . . , fm t], Ψ. . R[It] mR[It]. . =k. n1 X j=1. . 7−→. η1j u1j , . . . ,. ni X. ηij uij. j=1. nm X j=1. ηmj umj .. Thus, showing that R[I ∗ t]/mR[I ∗ t] is integral over R[It]/mR[It] is equivalent to n1 nm X X ηmj umj . By Lemma 3.2, showing that k[uij ] is integral over k η1j u1j , . . . , j=1. j=1. for each i = 1, . . . , m, it’s enough to prove that for all ℓ 6= j, uαijℓ uβiℓℓ = 0 for some. positive integers αℓ and βℓ . Note that Uij (resp. Uiℓ ) corresponds to xaij y bij t (resp. xaiℓ y biℓ t) in the epimorphism ϕ. Without loss of generality, we assume aij ≥ aiℓ . If aij = aiℓ and bij < biℓ (resp. bij > biℓ ), then xaiℓ y biℓ t = y biℓ −bij (xaij y bij t) (resp. xaij y bij t = y bij −biℓ (xaiℓ y biℓ t)). This shows Uiℓ − y biℓ −bij Uij ∈ ker ϕ (resp. Uij − y bij −biℓ Uiℓ ∈ ker ϕ) and so Uiℓ ∈ mR[Uij ] + ker ϕ (resp. Uij ∈ mR[Uij ] + R[Uij ] ∼ ker ϕ). With the isomorphism = k[uij ], we obtain uiℓ = 0 (mR[Uij ] + ker ϕ) 11.

(16) (resp. uij = 0) and so uij uiℓ = 0. Similarly for aij > aiℓ and bij ≥ biℓ , we have xaij y bij t = xaij −aiℓ y bij −biℓ (xaiℓ y biℓ t). This shows Uij − xaij −aiℓ y bij −biℓ Uiℓ ∈ ker ϕ and so Uij ∈ mR[Uij ] + ker ϕ. That is uij = 0 and so uij uiℓ = 0 is true. The last case is that aij > aiℓ and bij < biℓ . By the assumption of the theorem, there exists xahs y bhs ∈ Γ(fh ) such that (ahs , bhs ) lies on the left hand side of the line through (aij , bij ) and (aiℓ , biℓ ) as shown in the following shaded region:. bb. (aiℓ , biℓ ) bb. (aij , bij ) bb. bb. bb. (0, 0). Figure 3.1.1.. We divide Figure 3.1.1. into the following three parts:. (aiℓ , biℓ ) bb. (aij , bij ) bb. bb. (0, 0). Figure 3.1.2.. bb. (aiℓ , biℓ ) bb. (aiℓ , biℓ ) bb. (aij , bij ) bb. bb. bb. bb. (aij , bij ). bb. (0, 0). (0, 0). Figure 3.1.3.. Figure 3.1.4.. 12.

(17) If (ahs , bhs ) lies in the shaded region in Figure 3.1.2., then, by Lemma 3.3, there exist nonnegative integers α, β and positive integers γ, δ such that (xaij y bij t)α (xaiℓ y biℓ t)β = xγ y δ (xahs y bhs t)α+β . α+β ∈ ker ϕ. That is Uijα Uiℓβ ∈ mR[Uij ] + ker ϕ and Thus, we have Uijα Uiℓβ − xγ y δ Uhs. so uαij uβiℓ = 0. If (ahs , bhs ) lies in the shaded region in Figure 3.1.3., then, by Lemma 3.4, there exist positive integers α, β, γ, δ such that (xaij y bij t)α+β = xγ y δ (xahs y bhs t)α (xaiℓ y biℓ t)β , and this implies uα+β = 0 as above. Similarly, if (ahs , bhs ) lies in the shaded region ij in Figure 3.1.4., we may apply Lemma 3.4 and get uα+β = 0 for some positive iniℓ tegers α, β. These prove that for each i = 1, . . . , m, k[ui1 , . . . , uini ] is integral over ni n1 nm X X X k ηij uij and so k[uij ] is integral over k ηmj umj . Hence I η1j u1j , . . . , j=1. j=1. j=1. is a reduction of I ∗ and the proof is completed.. We end this section with an example which illustrates the algorithm provided in Theorem 3.1. Example 3.5. Consider the ideal I = hx12 + x10 y 2 + x3 y 4 + xy 8 , x10 y + x5 y 3 + y 9, x6 y 2 + x2 y 8i in k[x, y](x,y) with |k| = ∞. Note that the monomials occurring in the first generator of I are x12 , x10 y 2 , x3 y 4, xy 8 . From Figure 3.5.1, we see that for every line through two of the above 4 monomials, either x6 y 2 or y 9 is on the left of the line. Similarly, from Figure 3.5.2, we see that for each of the three lines determined by the three monomials occurring in the second generator of I, either x6 y 2 or x3 y 4 is on the left of the line. From Figure 3.5.3, x3 y 4 is on the left of the line through the two monomials occurring in the third generator of I. Hence, by Theorem 3.1, I ∗ = hx12 , x10 y, x6y 2 , x5 y 3 , x3 y 4, xy 8 , y 9i is integral over I.. 13.

(18) y9 b. xy 8 bb. x3 y 4 bb. b. x6 y 2. bb. x10 y 2 bb. x12. bb. Figure 3.5.1.. y9 bb. x3 y 4 b bb. x5 y 3 b. x5 y 3 b. x6 y 2. x2 y 8. bb. bb. x10 y. x6 y 2. Figure 3.5.3.. Figure 3.5.2.. 14.

(19) 4. Reductions in k[x, y, z](x,y,z). In V. C. Qui˜ nonez’s research report [Q1], although she proves a more general theorem on reductions of ideals ([Q1, Theorem 3.3]), it is not as concrete as Theorem 3.1 ([CL, Theorem 3.3]). Hence, we are interested in extending Theorem 3.1 to the three-dimensional case. In this section, we consider R = k[x, y, z](x,y,z) where k is an infinite field and let I be an ideal of R generated by f1 , . . . , fm ∈ k[x, y, z]. Similarly to the proofs of Lemma 3.3 and Lemma 3.4, we apply the property of vectors in the three-dimensional real space to obtain Lemma 4.2. With the help of this new lemma, we are able to extend Theorem 3.1 to Theorem 4.3, which gives a sufficient condition for I to be a reduction of I ∗ in the ring k[x, y, z](x,y,z) . Notation 4.1. Let m = xi y j z ℓ be a monomial in k[x, y, z]. We use the following notation for the three-dimensional region corresponding to the ideal hmi subtracting the point m: M(m) = {(a, b, c) ∈ R3≥0 | a ≥ i, b ≥ j, c ≥ ℓ} \ {(i, j, ℓ)}. Except Lemma 3.3 and Lemma 3.4, the remaining part of the proof for Theorem 3.1 is still applicable in the three-dimensional localized polynomial ring k[x, y, z](x,y,z) . Therefore, we need some supporting lemma which is available in the three-dimensional case to take the place of Lemma 3.3 and Lemma 3.4 for the two-dimensional case. With such a lemma, the following Lemma 4.2, we can extend Theorem 3.1 to the three-dimensional case. Thus, it is a crucial part of this section. Lemma 4.2. Let (a, b, c), (d, e, f ), (g, h, i) ∈ Z3≥0 . Suppose that the line through (a, b, c) and (d, e, f ) intersects M(xg y h z i ). Divide the line into three parts as the following:. I. II. b. b. (a, b, c). (d, e, f ). b bb. b. 15. III.

(20) Case 1. If part I intersects M(xg y h z i ), then there exist nonnegative integers α, β, γ, δ, λ such that γ, δ, λ are not all zero and that (xa y bz c )α (xd y e z f )β = xγ y δ z λ (xg y h z i )α+β . Case 2. If part II intersects M(xg y h z i ), then there exist nonnegative integers α, β, γ, δ, λ such that γ, δ, λ are not all zero and that (xa y bz c )α+β = xγ y δ z λ (xg y h z i )α (xd y e z f )β . Case 3. If part III intersects M(xg y h z i ), then there exist nonnegative integers α, β, γ, δ, λ such that γ, δ, λ are not all zero and that (xd y e z f )α+β = xγ y δ z λ (xa y b z c )α (xg y h z i )β . Proof. Since the line through (a, b, c) and (d, e, f ) intersects M(xg y h z i ), there exist s, t ∈ R with s + t = 1 such that s(a, b, c) + t(d, e, f ) ∈ M(xg y h z i ). So we have sa + td ≥ g, sb + te ≥ h, sc + tf ≥ i, and at lease one of the above inequalities is strict since M(xg y h z i ) does not include the point (g, h, i). Note that (s, t) ∈ R2 is a solution of the system   ax + dy ≥ g,      bx + ey ≥ h,   cx + f y ≥ i,     x + y = 1.. Hence we know that the line x + y = 1 passes through the region determined by the system     ax + dy ≥ g, bx + ey ≥ h,.   . cx + f y ≥ i. 16.

(21) If x + y = 1 intersects the region at only one point (s, t), then (s, t) must be the intersection of two of the lines ax + dy = g, bx + ey = h, and cx + f y = i. Since a, b, c, d, e, f, g, h, i ∈ Z≥0 , s, t ∈ Q. If x + y = 1 intersects the region at more than one point, then the intersection must contain a line segment and so we can find a point on this line segment with rational coordinates. In either case, we may assume s, t ∈ Q. Let ℓ be a positive integer such that sℓ, tℓ ∈ Z, then we have (sa + td)ℓ ≥ gℓ = gℓ(s + t), (sb + te)ℓ ≥ hℓ = hℓ(s + t), (sc + tf )ℓ ≥ iℓ = iℓ(s + t), and at lease one of the above inequalities is strict. Case 1: Since part I intersects M(xg y h z i ), 0 ≤ s, t ≤ 1. Then we can take α = ℓs and β = ℓt to obtain that aα + dβ = (sa + td)ℓ ≥ gℓ(s + t) = g(α + β), bα + eβ = (sb + te)ℓ ≥ hℓ(s + t) = h(α + β), cα + f β = (sc + tf )ℓ ≥ iℓ(s + t) = i(α + β), where at lease one of the inequalities is strict. By setting γ = (aα + dβ) − g(α + β), δ = (bα + eβ) − h(α + β), λ = (cα + f β) − i(α + β), we have (xa y bz c )α (xd y e z f )β = xγ y δ z λ (xg y h z i )α+β in which α, β, γ, δ, λ are nonnegative integers and γ, δ, λ are not all zero. Case 2: Since part II intersects M(xg y h z i ), s > 0 and t < 0. Then we can take α = ℓ = ℓ(s + t) and β = −ℓt to obtain that a(α + β) = aℓs ≥ gℓ − tdℓ = gα + dβ, b(α + β) = bℓs ≥ hℓ − teℓ = hα + eβ, c(α + β) = cℓs ≥ iℓ − tf ℓ = iα + f β, 17.

(22) and at lease one of the above inequalities is strict. By setting γ = a(α + β) − (gα + dβ), δ = b(α + β) − (hα + eβ), λ = c(α + β) − (iα + f β), we have (xa y bz c )α+β = xγ y δ z λ (xg y h z i )α (xd y e z f )β in which α, β, γ, δ, λ are nonnegative integers and γ, δ, λ are not all zero. Case 3: Since part III intersects M(xg y h z i ), s < 0 and t > 0. Then we can take α = −ℓs and β = ℓ = ℓ(s + t) to obtain that d(α + β) = dℓt ≥ gℓ − saℓ = aα + gβ, e(α + β) = eℓt ≥ hℓ − sbℓ = bα + hβ, f (α + β) = f ℓt ≥ iℓ − scℓ = cα + iβ, and at lease one of the above inequalities is strict. By setting γ = d(α + β) − (aα + gβ), δ = e(α + β) − (bα + hβ), λ = f (α + β) − (cα + iβ), we have (xd y e z f )α+β = xγ y δ z λ (xa y bz c )α (xg y h z i )β in which α, β, γ, δ, λ are nonnegative integers and γ, δ, λ are not all zero. Now, we can extend the main theorem in Section 3 to the three-dimensional case. In the following Theorem 4.3, we give a sufficient condition for an ideal I to be a reduction of I ∗ in the localized polynomial ring k[x, y, z](x,y,z) . Theorem 4.3. Let R = k[x, y, z](x,y,z) and |k| = ∞. Let I be an ideal of R generated by f1 , . . . , fm ∈ k[x, y, z]. Assume that the following is true: for all i = 1, 2, . . . , m and for any two distinct monomials xa y b z c and xd y e z f in Γ(fi ), there exists xr y s z t ∈ Γ(fj ) for some j such that the line through (a, b, c) and (d, e, f ) intersects M(xr y s z t ). Then I is a reduction of I*. 18.

(23) Proof. We express the generators of I as the following: n1 n2 nm X X X a1j b1j c1j a2j b2j c2j ηmj xamj y bmj z cmj f1 = η1j x y z , f2 = η2j x y z , . . . , fm = j=1. j=1. j=1. with ηij 6= 0 in k. Then I ∗ is the ideal generated by xaij y bij z cij for all i = 1, . . . , m and j = 1, . . . , ni . Similar to the proof of Theorem 3.1, we consider the polynomial ring R[Uij ] = R[Uij | i = 1, . . . , m, j = 1, . . . , ni ] and the ring epimorphism ϕ : R[Uij ] −→ Uij. R[I ∗ t].. 7−→ xaij y bij z cij t. R[I ∗ t] ∼ R[Uij ] Then we have R[I ∗ t] ∼ R[U ]/ ker ϕ and , in which = = ij mR[I ∗ t] (mR[Uij ] + ker ϕ) m = (x, y, z)R is the maximal ideal of R. Let uij denote the homomorphic image of Uij in R[Uij ]/(mR[Uij ] + ker ϕ). In order to show that I is a reduction of I ∗ , it suffices to show that R[I ∗ t]/mR[I ∗ t] is integral over R[It]/mR[It]. This is equivalent to showing that the k-algebra n1 nm X X ηmj umj . Thus, by Lemma 3.2, for k[uij ] is integral over k η1j u1j , . . . , j=1. j=1. each i = 1, . . . , m, it’s enough to prove that for all ℓ 6= j, uαijℓ uβiℓℓ = 0 for some. positive integers αℓ and βℓ . Note that Uij (resp. Uiℓ ) corresponds to xaij y bij z cij t (resp. xaiℓ y biℓ z ciℓ t) in the epimorphism ϕ. By the assumption of the theorem, there exists xahs y bhs z chs ∈ Γ(fh ) such that the line through (aij , bij , cij ) and (aiℓ , biℓ , ciℓ ) intersects M(xahs y bhs z chs ). We divide the line into the following three parts: I. II. b. b. (aij , bij , cij ). (aiℓ , biℓ , ciℓ ). bb bb. III. If part I intersects M(xahs y bhs z chs ), then, by Lemma 4.2, there exist nonnegative integers α, β, γ, δ, λ with γ, δ, λ not all zero such that (xaij y bij z cij t)α (xaiℓ y biℓ z ciℓ t)β = xγ y δ z λ (xahs y bhs z chs t)α+β . 19.

(24) α+β Thus, we have Uijα Uiℓβ − xγ y δ z λ Uhs ∈ ker ϕ. That gives Uijα Uiℓβ ∈ mR[Uij ] + ker ϕ. and uαij uβiℓ = 0. If part II intersects M(xahs y bhs z chs ), then, by Lemma 4.2, there exist nonnegative integers α, β, γ, δ, λ with γ, δ, λ not all zero such that (xaij y bij z cij t)α+β = xγ y δ z λ (xahs y bhs z chs t)α (xaiℓ y biℓ z ciℓ t)β and this implies uα+β = 0 as in the proof of Theorem 3.1. Similarly, if part III inij tersects M(xahs y bhs z chs ), we may apply Lemma 4.2 and get uα+β = 0 for some noniℓ negative integers α, β. These prove that for each i = 1, . . . , m, k[ui1 , . . . , uini ] is inni n1 nm X X X ηij uij and so k[uij ] is integral over k tegral over k ηmj umj η1j u1j , . . . , j=1. j=1. j=1. for all i, j. Hence I is a reduction of I ∗ . The proof is completed.. In the end of this section, we include three examples to illustrate the above algorithm. The first two examples Example 4.4 and Example 4.5 are adopted from [Q1]. Instead of using [Q1, Theorem 3.3], we apply Theorem 4.3 to these two examples. Example 4.4. Consider the ideal I = hx3 yz 2 +xy 3 z, x2 y 2z+y 4 z 2 i in k[x, y, z](x,y,z) with |k| = ∞. Note that the monomials occurring in the first generator of I are x3 yz 2 and xy 3 z. A point on the line through these two monomials can be written as s(3, 1, 2) + t(1, 3, 1) with s, t ∈ R and s + t = 1. Take s =. 1 2. and t = 12 , then the. point (2, 2, 23 ) ∈ M(x2 y 2 z). In other words, the line through (3, 1, 2) and (1, 3, 1) intersects M(x2 y 2z) as indicated in the following Figure 4.4.1.. x3 yz 2. xy 3 z b. b. bb. x2 y 2 z Figure 4.4.1.. Similarly, a point on the line through x2 y 2 z and y 4 z 2 , which occur in the second generator of I, can be written as s(2, 2, 1) + t(0, 4, 2) for some s, t ∈ R with 20.

(25) s + t = 1. By direct calculating, we see that the line intersects M(xy 3 z) at the point. 1 (2, 2, 1) 2 ∗. + 12 (0, 4, 2), which we indicate in Figure 4.4.2. Hence, by. Theorem 4.3, I = hx3 yz 2 , xy 3 z, x2 y 2z, y 4 z 2 i is integral over I.. b. bb. y4 z2. xy 3 z. b. x2 y 2 z Figure 4.4.2.. Example 4.5. Consider the ideal I = hx4 +y 2 z, y 4 +z 2 x, z 4 +x2 yi in k[x, y, z](x,y,z) with |k| = ∞. Note that the monomials occurring in the first generator of I are x4 , y 2z. A point on the line through the above two monomials can be written as s(4, 0, 0) + t(0, 2, 1) for some s, t ∈ R with s + t = 1. Take s =. 1 2. and t = 21 , then. the point (2, 1, 12 ) is contained in M(x2 y). In other words, the line through x4 and y 2 z intersects M(x2 y) as in the following Figure 4.5.1.. b. y2z. Figure 4.5.1. bb. x4. x2 y. b. Similarly, a point on the line through y 4 and z 2 x, which occur in the second generator of I, can be written as s(0, 4, 0) + t(1, 0, 2) where s, t ∈ R with s + t = 1. By calculating directly, we get that this line intersects M(y 2 z) at 12 (0, 4, 0) + 21 (1, 0, 2) as in Figure 4.5.2. At last, we apply the same approach to the line through 21.

(26) the two monomials occurring in the third generator of I and get that it intersects M(z 2 x) at 21 (0, 0, 4) + 21 (2, 1, 0) as in Figure 4.5.3. Hence, by Theorem 4.3, I ∗ = hx4 , y 2 z, y 4 , z 2 x, z 4 , x2 yi is integral over I.. z4 b. z2 x. z2 x b. bb bb. y2z. b. y4. b. x2 y Figure 4.5.3.. Figure 4.5.2.. At last, we give a slightly more complicated example. In particular, one of the two generators of the ideal I contains three monomials in this example. Example 4.6. Let I = hxy 2 z 3 + x6 z 7 , y 4 z 4 + x2 y 3z 2 + x3 yz 5 i ⊆ k[x, y, z](x,y,z) with |k| = ∞. Note that the monomials occurring in the first generator of I are xy 2 z 3 and x6 z 7 . A point on the line through these two monomials can be written as s(1, 2, 3) + t(6, 0, 7) with s, t ∈ R and s + t = 1. Take s =. 1 2. and t = 12 , then the. point ( 27 , 1, 5) ∈ M(x3 yz 5 ). In other words, the line through (1, 2, 3) and (6, 0, 7) intersects M(x3 yz 5 ) as indicated in the following Figure 4.6.1.. x3 yz 5 bb. x6 z 7 b. b. xy 2 z 3. Figure 4.6.1.. 22.

(27) Similarly, y 4 z 4 , x2 y 3z 2 , and x3 yz 5 are the monomials that occur in the second generator of I. We indicate these points and the connecting line segments, and M(xy 2 z 3 ) in the following Figure 4.6.2.. y4 z4 xy 2 z 3 x3 yz 5 b. b. bb b. x2 y 3 z 2 Figure 4.6.2.. By direct calculation, we know that for every line through two of the above three monomials intersects M(xy 2 z 3 ). More precisely, ⋆ the point (1, 72 , 3) = 12 (0, 4, 4) + 21 (2, 3, 2) lies in both M(xy 2 z 3 ) and the line through y 4 z 4 and x2 y 3 z 2 ; ⋆ the point ( 37 , 37 , 3) = 23 (2, 3, 2) + 31 (3, 1, 5) lies in both M(xy 2 z 3 ) and the line through x2 y 3 z 2 and x3 yz 5 ; ⋆ the point (1, 3, 13 ) = 32 (0, 4, 4) + 31 (3, 1, 5) lies in both M(xy 2 z 3 ) and the line 3 through y 4 z 4 and x3 yz 5 . Hence, by Theorem 4.3, I ∗ = hxy 2 z 3 , x6 z 7 , y 4z 4 , x2 y 3 z 2 , x3 yz 5 i is integral over I. In [CL, Corollary 3.7], which is an immediate application of [CL, Theorem 3.3], they find a minimal reduction of a given monomial ideal I in k[x, y](x,y) where k is an infinite field. In fact, the original aims of our study are not only extending [CL, Theorem 3.3] to the three-dimensional case but also finding a minimal reduction of a given monomial ideal I in k[x, y, z](x,y,z). But we are not able to do so yet. This issue can be discussed in a further research.. 23.

(28) References [AM]. M. F. Atyiah, and I. G. Macdonald, Introduction to Commutative Algebra, Addison-Wesley, 1969.. [CLO] D. Cox, J. Little, and D. O’Shea, Ideals, Varieties, and Algorithms, Springer-Verge, Heidelberg, Berlin, 1992. [SH]. I. Swanson and C. Huneke, Integral Closure of Ideals, Rings, and Modules, Cambridge University Press, Cambridge, 2006.. [CL]. C-Y. J. Chan, and J.-C. Liu, A Note on Reductions of Monomial Ideals in k[x, y](x,y) , Contemporary Mathematics of AMS, PASI proceedings, 555(2011), 13–34.. [Q1]. V. C. Qui˜ nonez, Minimal reductions of monomial ideals, Research Reports in Mathematics, Number 10 (2004), Department of Mathematics, Stockholm University, available at http://www.math.su.se/reports/2004/10/.. [Q2]. V. C. Qui˜ nonez,. Integrally Closed Monomial Ideals and Powers. of Ideals, Research Reports in Mathematics, Department of Mathematics,. Stockholm University,. http://www.math.su.se/reports/2002/7/.. 24. Number 7 (2002), available at.

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