Handout - Algebra Review

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1 Algebraic Geometry Instructor: Mohamed Omar Handout - Algebra Review Sept 9 Math 176 Today will be a thorough review of the algebra prerequisites we will need throughout this course. Get through as much of this handout as you can. Some of the concepts may not be review but will be good to know, and we will revisit some of these concepts as they come up. We start with a few definitions that will be used throughout the course: A ring will always mean a commutative ring with a multiplicative identity. A ring homomorphism φ : R S (where R, S are rings) is a function that preserve addition and multiplication in the respective rings. That is φ(r + r ) = φ(r) + φ(r ) and φ(rr ) = φ(r)φ(r ) for all r, r R A domain (also called an integral domain) is a ring (with at least two elements) in which the cancellation law holds. That is for any x, y, z 0 in R, xy = xz y = z. A field is a domain in which every nonzero element is a unit. That is, every nonzero element has a multiplicative inverse. We will use Z, Q, R, C to denote the ring of integers, rational numbers, real numbers and complex numbers (under their usual operations), respectively. 1. Determine all integers n 2 for which Z/nZ, under its usual addition and multiplication (modular arithmetic), is a domain. 2. If φ : R S is a ring homomorphism, prove that ker(φ) is an ideal of R. (Recall ker(φ) = {r R φ(r) = 0}.) 1

2 Let R be a ring. The ring of polynomials in n variables of R is typically written as R[x 1,..., x n ]. When n = 2 or 3 we often write R[x, y] and R[x, y, z] respectively. The monomials in R[x 1,..., x n ] are the polynomials x i 1 1 x in n where i j is a nonnegative integer for all j. Every polynomial f R[x 1,..., x n ] can be written uniquely as a sum a (i) x (i) where x (i) is some monomial in x 1, x 2,..., x n and a (i) R. We say F is homogeneous or a form of degree d if all the coefficients a (i) are zero except for when x (i) is a monomial of degree d. Any polynomial can be written as a sum of homomgeneous polynomials f = f 0 + f f d where f i is a form of degree i and f d 0. The integer d is called the degree of f and is denoted deg(f). If R is a domain, deg(fg) = deg(f) + deg(g) for any f, g R[x 1,..., x n ]. 1. Is the product of two homogeneous polynomials in R[x 1, x 2,..., x n ] necessarily homogeneous? 2. If R is a domain, prove R[x 1,..., x n ] is a domain. 2

3 Recall that a nonzero element u in a ring R is a unit if there is some other nonzero element v R for which uv = vu = 1. An element a in a ring R is irreducible if it is not a unit and not zero, and for any factorization a = bc with b, c R, either b or c is a unit. A domain R is a unique factorization domain, written UFD, if every nonzero element in R can be factored uniquely into a product of irreducible elements (up to multiplication by units and the ordering of factors). For example, let R = Z. The units in R are ±1. Suppose a R. If a is prime and a = bc then one of b or c is ±1. Furthermore if a is compositive, then it can be written as a product bc where neither b nor c is ±1. Hence the irreducible elements in R are the primes. Now any integer can be written as a product of primes uniquely, up to multiplication by ±1 and reordering of the primes. If R is a UFD, then one can prove R[x] is also a UFD. Hence, for any field k, k[x 1,..., x n ] is a UFD (since we can repeatedly appeal to the fact that for any m, k[x 1, x 2,..., x m ] = (k[x 1, x 2,..., x m 1 ])[x m ]. 1. The ring k[x, y] is a UFD. Factor y 3 x 4 y in k[x, y] into irreducibles. Explain why your irreducible elements are indeed irreducible. 3

4 Rings have various types of ideals that we will come across. These include: principal ideals: I is principal if there is some r R for which I = (r). maximal ideals: I is maximal if the only ideals J in R for which I J R are J = I and J = r. prime ideals: I R is prime ideal of R if for any ab I, either a I or b I. As an example, consider the ring C[x]. Every ideal in this ring is a principal ideal. Indeed, if I C[x] is an ideal that is not the 0 ideal, then any polynomial p(x) with smallest degree in I must generate the ideal I (that is, I = (p(x)). To see why, suppose otherwise. Then I contains some polynomial a(x) that is not a multiple of p(x). By the Division Algorithm, there exists q(x), r(x) C[x] where 0 < deg(r(x)) < deg(p(x)) such that a(x) = b(x)p(x) + r(x). But then r(x) = a(x) b(x)p(x) I and has smaller degree than p(x), a contradiction. As another example, consider the ring Z. By the same argument as in C[x], every ideal in Z is a principal ideal. Suppose I is an ideal in Z and I = (m) for some m Z. How can we determine if m is a prime ideal? First, observe that if m = ±1 then I = R so we rule this out by definition. Suppose first that m is composite, say m = ab and neither a nor b is ±1. Then ab I, but neither a nor b is in I, so I is not a prime ideal. Now suppose m is prime. Then if ab I, we have m ab and since m is prime, m a or m b and hence a I or b I. 1. Determine all maximal ideals in the ring k[x] where k is a field. 2. Suppose I = (p(x)) for some p(x) k[x], p(x) 0. Prove I is a prime ideal if and only if p(x) is an irreducible polynomial. 4

5 Let I be an ideal of a ring R. The quotient ring R/I (pronounced R modulo I ) is the set of equivalence classes of elements in R under the equivalence relation a b if a b I. The equivalence class containing an element a is often written as a or a + I. We will use the notation a to be consistent with Fulton s book. The multiplication and addition in R/I are formed in a way as to make the reduction map π : R R/I a ring homomorphism. In other words, if r, s R/I then r s = rs, r + s = r + s For example, let k be a field, and I be the ideal I = (x 3 + x + 1) in k[x]. Then k[x]/i = {a + bx + cx 2 a, b, c k} (Why? Hint: divide any polynomial by x 3 + x + 1.) If p(x) = 1 + x + x 2 and q(x) = 1 x + x 2 in k[x]/i then p(x)q(x) = 1 + x + x 2 1 x + x 2 = x 4 + x = x(x 3 ) + x = x( x 1) + x = x If k is a field, briefly explain why k[x 1, x 2,..., x n ]/I is a k-vector space for any ideal I in k[x 1, x 2,..., x n ]. 2. Find a basis for the k-vector space k[x, y]/(xy 1). 3. Let R be a ring and I be an ideal of R. (a) Prove that I is maximal if and only if R/I is a field. (b) Prove that I is prime if and only if R/I is a domain. 5

Section III.6. Factorization in Polynomial Rings

III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)