1![MAXIMAL MEASURE ALGEBRAS IN P(N)/Z0 ILIJAS FARAH Let Sκ be the forcing for adding κ side-by-side Sacks reals, with countable support. Let D be the family of all subsets of N that have density. Let Z0 be the ideal of se MAXIMAL MEASURE ALGEBRAS IN P(N)/Z0 ILIJAS FARAH Let Sκ be the forcing for adding κ side-by-side Sacks reals, with countable support. Let D be the family of all subsets of N that have density. Let Z0 be the ideal of se](https://www.pdfsearch.io/img/08723262764a30cdea2109da2b285a58.jpg) | Add to Reading ListSource URL: www.math.yorku.caLanguage: English - Date: 2003-06-13 22:10:12
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2![CONTRACTIBILITY OF THE MAXIMAL IDEAL SPACE OF ALGEBRAS OF MEASURES IN A HALF-SPACE AMOL SASANE Abstract. Let H[n] be the canonical half space in Rn , that is, H[n] = {(t1 , . . . , tn ) ∈ Rn \ {0} | ∀j, [tj 6= 0 and CONTRACTIBILITY OF THE MAXIMAL IDEAL SPACE OF ALGEBRAS OF MEASURES IN A HALF-SPACE AMOL SASANE Abstract. Let H[n] be the canonical half space in Rn , that is, H[n] = {(t1 , . . . , tn ) ∈ Rn \ {0} | ∀j, [tj 6= 0 and](https://www.pdfsearch.io/img/01a304a41cc272f46fa4d42ae71b6929.jpg) | Add to Reading ListSource URL: www.cdam.lse.ac.ukLanguage: English - Date: 2017-04-12 10:30:38
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3![CATEGORIES AND HOMOLOGICAL ALGEBRA Exercises for May 24 Exercise 1. Let R = k[x, y] with k a field. (a) Consider the maximal ideal m = (x, y) ⊂ R and the R-module M = R/m. Calculate ExtiR (M, M ) for all i. CATEGORIES AND HOMOLOGICAL ALGEBRA Exercises for May 24 Exercise 1. Let R = k[x, y] with k a field. (a) Consider the maximal ideal m = (x, y) ⊂ R and the R-module M = R/m. Calculate ExtiR (M, M ) for all i.](https://www.pdfsearch.io/img/59c2eba06c009284e2cbe77bb0aa1063.jpg) | Add to Reading ListSource URL: www.math.ru.nlLanguage: English - Date: 2018-05-22 09:41:41
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4![Finitely generated modules over a PID Lemma. Let R be a commutative ring with 1 6= 0. If Rm ∼ = Rn as R-modules then m = n. Proof. Let m ⊂ R be a maximal ideal, and let k = R/m. Then Rm ∼ = Rn implies that Finitely generated modules over a PID Lemma. Let R be a commutative ring with 1 6= 0. If Rm ∼ = Rn as R-modules then m = n. Proof. Let m ⊂ R be a maximal ideal, and let k = R/m. Then Rm ∼ = Rn implies that](https://www.pdfsearch.io/img/ef22d9af3efed2b49573ee8d11bd3842.jpg) | Add to Reading ListSource URL: www.math.ru.nlLanguage: English - Date: 2018-03-04 10:31:35
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5![MIMS Technical Report No) THE EQUALITY OF ELIAS-VALLA AND THE ASSOCOATED GRADED RINGS OF MAXIMAL IDEAL KAZUHO OZEKI MIMS Technical Report No) THE EQUALITY OF ELIAS-VALLA AND THE ASSOCOATED GRADED RINGS OF MAXIMAL IDEAL KAZUHO OZEKI](https://www.pdfsearch.io/img/7a0fc3cee7c54116a905aaffc79b74a1.jpg) | Add to Reading ListSource URL: www.mims.meiji.ac.jp- Date: 2015-04-16 22:07:26
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6![ON THE SPECTRUM OF RINGS OF FUNCTIONS SOPHIE FRISCH Abstract. Let D be a domain and M a maximal ideal of D. The ring of integer-valued polynomials on a subset E of D as well as more general rings of Q ON THE SPECTRUM OF RINGS OF FUNCTIONS SOPHIE FRISCH Abstract. Let D be a domain and M a maximal ideal of D. The ring of integer-valued polynomials on a subset E of D as well as more general rings of Q](https://www.pdfsearch.io/img/1e65e9a8db27fe7272c1fb65e4132364.jpg) | Add to Reading ListSource URL: blah.math.tu-graz.ac.atLanguage: English - Date: 2016-04-17 08:51:17
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7![Qualifying exam, Spring 2006, DayLet φ : A → B be a homomorphism of commutative rings, and let pB ⊂ B be a maximal ideal. Set A ⊃ pA := φ−1 (pB ). (a) Show that pA is prime but in general non maximal. (b Qualifying exam, Spring 2006, DayLet φ : A → B be a homomorphism of commutative rings, and let pB ⊂ B be a maximal ideal. Set A ⊃ pA := φ−1 (pB ). (a) Show that pA is prime but in general non maximal. (b](https://www.pdfsearch.io/img/7212b45419c1783f5f7716e76d27fb72.jpg) | Add to Reading ListSource URL: math.harvard.eduLanguage: English - Date: 2016-02-04 13:15:17
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8![MAT 534 — HOMEWORK 12 DUE ON FRIDAY 25 APRIL 1. (Ch. 14, #6) Find all maximal ideals in Z10 and Z12 . (Remember that an ideal is in MAT 534 — HOMEWORK 12 DUE ON FRIDAY 25 APRIL 1. (Ch. 14, #6) Find all maximal ideals in Z10 and Z12 . (Remember that an ideal is in](https://www.pdfsearch.io/img/bb41a7957d73a8de8fabfa3e1e79bc25.jpg) | Add to Reading ListSource URL: www.leuschke.org- Date: 2014-04-23 14:40:18
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9![Lemma. (1) If p is an ideal which is maximal with respect to the property that p is not finitely generated then p is prime. (2) If p is an ideal maximal with respect to the property that p is not principal, then p is pri Lemma. (1) If p is an ideal which is maximal with respect to the property that p is not finitely generated then p is prime. (2) If p is an ideal maximal with respect to the property that p is not principal, then p is pri](https://www.pdfsearch.io/img/8fea520ff22821dba508040fe9f5479b.jpg) | Add to Reading ListSource URL: www.math.hawaii.eduLanguage: English - Date: 2001-04-07 05:35:52
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10![Qualifying exam, Spring 2006, DayLet φ : A → B be a homomorphism of commutative rings, and let pB ⊂ B be a maximal ideal. Set A ⊃ pA := φ−1 (pB ). (a) Show that pA is prime but in general non maximal. (b Qualifying exam, Spring 2006, DayLet φ : A → B be a homomorphism of commutative rings, and let pB ⊂ B be a maximal ideal. Set A ⊃ pA := φ−1 (pB ). (a) Show that pA is prime but in general non maximal. (b](https://www.pdfsearch.io/img/11cb86b66870cc27e33e331039c9c61a.jpg) | Add to Reading ListSource URL: www.math.harvard.eduLanguage: English - Date: 2006-02-02 22:49:40
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