Algebraic Geometry
... In this course, we have attached an affine algebraic variety to any algebra finitely generated over a field k. For many reasons, for example, in order to be able to study the reduction of varieties to characteristic p ¤ 0, Grothendieck realized that it is important to attach a geometric object to ev ...
... In this course, we have attached an affine algebraic variety to any algebra finitely generated over a field k. For many reasons, for example, in order to be able to study the reduction of varieties to characteristic p ¤ 0, Grothendieck realized that it is important to attach a geometric object to ev ...
Math 113 Final Exam Solutions
... those of the form (0, n)H and (1, n)H, where n can be any integer. Suppose (r, n)H = (s, m)H where s, r = 0 or 1. Then we have (r − s, n − m) ∈ H. Note that |r − s| = 0 or 1. At the same time, since H = h(2, 4)i, we have 2|r − s, so r − s = 0 necessarily. But then n − m = 0 · 4 = 0, and so we have t ...
... those of the form (0, n)H and (1, n)H, where n can be any integer. Suppose (r, n)H = (s, m)H where s, r = 0 or 1. Then we have (r − s, n − m) ∈ H. Note that |r − s| = 0 or 1. At the same time, since H = h(2, 4)i, we have 2|r − s, so r − s = 0 necessarily. But then n − m = 0 · 4 = 0, and so we have t ...
Lecture 1 File
... every element of R is equal to 0. This ring with one element is called the trivial ring. This also shows that in a non-trivial ring 0 ∈ / R∗ . Definition 1.5 (of skew field and field): A non-trivial, ring with unity (R, +, ∗) is said to be a skew field or a division ring if R − {0} = R∗ , i.e. every ...
... every element of R is equal to 0. This ring with one element is called the trivial ring. This also shows that in a non-trivial ring 0 ∈ / R∗ . Definition 1.5 (of skew field and field): A non-trivial, ring with unity (R, +, ∗) is said to be a skew field or a division ring if R − {0} = R∗ , i.e. every ...
SOLUTIONS FOR THE TRAINING FINAL Remember : the final exam
... equations are the integers modulo 10 that are cogruent to 4̄ modulo 5, that is 4̄ and 9̄. 7.– Compute φ(18). Compute 51000 (mod 18). Solutions: φ(18) is the numebr of integer prime to 18 between 1 and 18. Those are 1, 5, 7, 11, 13, 17, so φ(18) = 6. By Euler’s theorem, since 5 is relatively prime to ...
... equations are the integers modulo 10 that are cogruent to 4̄ modulo 5, that is 4̄ and 9̄. 7.– Compute φ(18). Compute 51000 (mod 18). Solutions: φ(18) is the numebr of integer prime to 18 between 1 and 18. Those are 1, 5, 7, 11, 13, 17, so φ(18) = 6. By Euler’s theorem, since 5 is relatively prime to ...
MSM203a: Polynomials and rings Chapter 3: Integral domains and
... are polynomials q(X), r(X) and an element a 6= 0 of R such that ap(X) = q(X)s(X) + r(X) and the degree of r(X) is less than that of s(X). Integral domains have no zero divisors, and consequently allow one to solve equations of the form x(y − z) = 0 in the way one expects: either x = 0 or else y = z. ...
... are polynomials q(X), r(X) and an element a 6= 0 of R such that ap(X) = q(X)s(X) + r(X) and the degree of r(X) is less than that of s(X). Integral domains have no zero divisors, and consequently allow one to solve equations of the form x(y − z) = 0 in the way one expects: either x = 0 or else y = z. ...
Week 10 Let X be a G-set. For x 1, x2 ∈ X, let x 1 ∼ x2 if and only if
... Let X be a G-set. For x1, x2 ∈ X, let x1 ∼ x2 if and only if ∃ g ∈ G s.t. gx1 = x2. The ∼ is an equivalence relation. Each cell in X/∼ is an orbit and the orbit contains x is denoted by Gx. Theorem Let X be a G-set and x ∈ X. Then |Gx| = (G : Gx ) where Gx = {g ∈ G | gx = x}. If |G| is finite, then ...
... Let X be a G-set. For x1, x2 ∈ X, let x1 ∼ x2 if and only if ∃ g ∈ G s.t. gx1 = x2. The ∼ is an equivalence relation. Each cell in X/∼ is an orbit and the orbit contains x is denoted by Gx. Theorem Let X be a G-set and x ∈ X. Then |Gx| = (G : Gx ) where Gx = {g ∈ G | gx = x}. If |G| is finite, then ...
selected solutions to Homework 11
... / S. Thus, S is not closed under addition and is not a subring. (b) This statement is true. Proof. Let R be a ring with unity, 1. Let a, b ∈ S. This means that there exist c, d ∈ R such that ac = 1 and bd = 1. Then (ab)(dc) = a(bd)c = a(1)(c) = (ac) = 1 Thus, (ab) ∈ S and S is closed under multiplic ...
... / S. Thus, S is not closed under addition and is not a subring. (b) This statement is true. Proof. Let R be a ring with unity, 1. Let a, b ∈ S. This means that there exist c, d ∈ R such that ac = 1 and bd = 1. Then (ab)(dc) = a(bd)c = a(1)(c) = (ac) = 1 Thus, (ab) ∈ S and S is closed under multiplic ...
Section IV.19. Integral Domains
... Note. In classical algebra, we solve polynomial equations by: setting equal to 0, factoring, and setting factors equal to 0. For example, x2 − 5x + 6 = 0 factors as (x−3)(x−2) = 0 which implies that x = 2 and x = 3 are (real) solutions. However, there are more solutions in different settings. Exampl ...
... Note. In classical algebra, we solve polynomial equations by: setting equal to 0, factoring, and setting factors equal to 0. For example, x2 − 5x + 6 = 0 factors as (x−3)(x−2) = 0 which implies that x = 2 and x = 3 are (real) solutions. However, there are more solutions in different settings. Exampl ...
SOLUTIONS TO HOMEWORK 9 1. Find a monic polynomial f(x) with
... common multiple of the denominators, we get a polynomial g(x) with integer coefficients and which is satisfied by β. Let d denote the new leading coefficient. Then dβ is an algebraic integer, as can be seen by multiplying g(x) by dn−1 and substituting β for x. So d is the product of the least common ...
... common multiple of the denominators, we get a polynomial g(x) with integer coefficients and which is satisfied by β. Let d denote the new leading coefficient. Then dβ is an algebraic integer, as can be seen by multiplying g(x) by dn−1 and substituting β for x. So d is the product of the least common ...
LECTURES MATH370-08C 1. Groups 1.1. Abstract groups versus
... Many fields consists of ordinary real or complex numbers. Then they necessarily contain the field Q of rational numbers. Examples of rings: Z, nZ, Z/(nZ), Z[ 12 ], Q, R, C, H. From these examples Q, R, C and Z/(pZ) for p prime are fields; H is a skew-field. The notion of a subring of a ring R is def ...
... Many fields consists of ordinary real or complex numbers. Then they necessarily contain the field Q of rational numbers. Examples of rings: Z, nZ, Z/(nZ), Z[ 12 ], Q, R, C, H. From these examples Q, R, C and Z/(pZ) for p prime are fields; H is a skew-field. The notion of a subring of a ring R is def ...
Chapter 1 (as PDF)
... We have a ring version of the First Isomorphism Theorem: Theorem 2.13 (First Isomorphism Theorem for Rings) If φ is a ring homomorphism from a ring R onto a ring S then the factor ring R/kerφ and the ring S are isomorphic by the map r + kerφ 7→ φ(r). We can use mappings to transfer a structure from ...
... We have a ring version of the First Isomorphism Theorem: Theorem 2.13 (First Isomorphism Theorem for Rings) If φ is a ring homomorphism from a ring R onto a ring S then the factor ring R/kerφ and the ring S are isomorphic by the map r + kerφ 7→ φ(r). We can use mappings to transfer a structure from ...
math.uni-bielefeld.de
... Let F be an arbitrary field of characteristic 6= 2, φ a non-degenerate (2n+1)-dimensional quadratic form over F (with n ≥ 1), X the orthogonal grassmanian of n-dimensional totally isotropic subspaces of φ. The variety X is projective, smooth, and geometrically connected; dim X = n(n + 1)/2. We write ...
... Let F be an arbitrary field of characteristic 6= 2, φ a non-degenerate (2n+1)-dimensional quadratic form over F (with n ≥ 1), X the orthogonal grassmanian of n-dimensional totally isotropic subspaces of φ. The variety X is projective, smooth, and geometrically connected; dim X = n(n + 1)/2. We write ...