M373K, Homework #8 April 3, 2013 From the book, please work out: x3.2 #1 (a + b)(c + d) = ac + ad + bc + bd. The point to this problem is that multiplication is non-commutative. So bc does not necessarily equal cb, for example. x3.2 #6 Let D be an integral domain with finite characteristic m. Recall that this means ma = 0 for every a 2 D (and m is a positive integer). Wlog we can assume that m is the smallest positive integer for which this holds. Suppose j; k are positive integers and jk = m. Then for any a; b 2 D,(ja)(kb) = jkab = 0 (by the previous problem). However, D is an integral domain, so it has no zero divisors. So either ja = 0 or kb = 0. In particular, by setting a = b we see that either ja = 0 for every a 2 R or ka = 0 for every a 2 R. So if m is not prime then we can choose j; k so that both j and k are smaller than m. But this contradicts our choice of m. So m must be prime. x3.2 #10 Let D be an integral domain and a; b; c 2 D with ab = ac and a 6= 0. Then ab − ac = 0 so a(b − c) = 0. Because a 6= 0 and D has no zero divisors, this implies b − c = 0 which implies b = c. On the other hand, suppose that D is a commutative ring which satisfies the cancellation property: for every a; b; c 2 D with a 6= 0 and ab = ac, we must have b = c. We need to show that D is an integral domain; i.e. that D has no zero divisors. So suppose u; v 2 D are nonzero and uv = 0. Then uv = u0. But u is nonzero so by the cancellation property, v = 0. This contradiction shows that D has no zero divisors so it must be an integral domain. x3.4 #1 U must contain 1 · r = r for every r 2 R. So U = R. x3.4 #2 If I is an ideal and a 2 I is nonzero then for any element r 2 F , ra−1a 2 I. But ra−1a = r. So I = R. x3.4 #4b Consider the ring R of all 2 × 2 real matrices. Let 0 1 A = 0 0 Show that fAB : B 2 Rg 6= fBA : B 2 Rg. (This is easy - AB is a matrix in which the bottom row is zero and BA is a matrix in which the second column is zero). So fAB : B 2 Rg is not a 2-sided ideal. 1 1. If α1; : : : ; αn are complex numbers then Z[α1; : : : ; αn] standsp for the smallest subring of C containing Z andp the numbers α1; : : : ; αn. For example Z[ 2] consists of all numbers of the form a + b 2 where a; b are integers. Q[α1; : : : ; αn] is defined similarly. p (a) Give an explicit description of Z[21=3] (similar to the one given for Z[ 2] above). Solution. Z[21=3] = fa + b21=3 + c22=3 : a; b; c 2 Zg. (To verify this, you have to show that this set is closed under multiplication - the rest is obvious). p p p p (b) Show that [ 2; 3] = [ 2 + 3]. Q p Qp p p p p p pSolution. Because 2 + 3 2 Q[ 2; 3], it follows that Q[ p2; p3] ⊃ Q[p2 + p3]. So let's show the other direction. It suffices to show that 2; 3 2 Q[ 2 + 3]. First note that p p p p p ( 2 + 3)2 = 5 + 2 6 2 [ 2 + 3]: p p p Q Therefore, 6 2 Q[ 2 + 3]. Next note that p p p p p p p 6( 2 + 3) = 2 3 + 3 2 2 Q[ 2 + 3]: Therefore p p p p p p p 2 3 + 3 2 − 2( 2 + 3) = 2 2 [ 2 + 3]: p p p Q Now that we know 2 2 Q[ 2 + 3] it follows that p p p p p p ( 2 + 3) − 2 = 3 2 Q[ 2 + 3]: 2. A element r 2 R is called a unit if it has a multiplicative inverse. For example, f1; −1g are the units of Z. Every nonzero element of Q is a unit of Q. The set of units always forms a multiplicative group. (a) Find the group of units Z[i]. Solution. The group is f1; −1; i; −ig. To see this, suppose that a + bi 2 Z[i] is a unit. Then its inverse is in Z[i]. However Z[i] does not contain any element with absolute value < 1. Since j(a + bi)−1j = ja + bij−1 we must have that ja + bij ≤ 1 (and therefore ja + bij = 1). In other words, a2 + b2 = 1. Since a and b are integers, this implies either a = 0; b = ±1 or a = ±1 and b = 0. (b) Find the group of units of Z=10Z. Solution. f1; 3; 7; 9g. These are the numbers relatively prime to 10 (mod 10). (c) Find the group of units of R[x]. Recall that R[x] is the ring of polynomials with real coefficients. n Solution. let p(x) = a0 + a1x + ··· + anx be a polynomial. You can see that 1 p(x) is not a polynomial unless the degree of p is 0. One way to see this is to 1 realize that p(r) is not defined if r is a root of p and positive degree polynomials always have roots (even though they may all be complex). So if p is a unit then p(x) = a0 for some nonzero a0. In other words, the group of units of R[x] is exactly the nonzero real numbers. 2.