Maximal ideal

Results: 30



#Item
1MAXIMAL 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

Add to Reading List

Source URL: www.math.yorku.ca

Language: English - Date: 2003-06-13 22:10:12
    2CONTRACTIBILITY 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

    Add to Reading List

    Source URL: www.cdam.lse.ac.uk

    Language: English - Date: 2017-04-12 10:30:38
      3CATEGORIES 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.

      Add to Reading List

      Source URL: www.math.ru.nl

      Language: English - Date: 2018-05-22 09:41:41
        4Finitely 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

        Add to Reading List

        Source URL: www.math.ru.nl

        Language: English - Date: 2018-03-04 10:31:35
          5MIMS 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

          Add to Reading List

          Source URL: www.mims.meiji.ac.jp

          - Date: 2015-04-16 22:07:26
            6ON 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

            Add to Reading List

            Source URL: blah.math.tu-graz.ac.at

            Language: English - Date: 2016-04-17 08:51:17
              7Qualifying 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

              Add to Reading List

              Source URL: math.harvard.edu

              Language: English - Date: 2016-02-04 13:15:17
                8MAT 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

                Add to Reading List

                Source URL: www.leuschke.org

                - Date: 2014-04-23 14:40:18
                  9Lemma. (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

                  Add to Reading List

                  Source URL: www.math.hawaii.edu

                  Language: English - Date: 2001-04-07 05:35:52
                    10Qualifying 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

                    Add to Reading List

                    Source URL: www.math.harvard.edu

                    Language: English - Date: 2006-02-02 22:49:40