M373K, Homework #8

April 3, 2013

From the book, please work out: §3.2 #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. §3.2 #6 Let D be an integral with finite characteristic m. Recall that this means ma = 0 for every a ∈ 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 ∈ D,(ja)(kb) = jkab = 0 (by the previous problem). However, D is an , 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 ∈ R or ka = 0 for every a ∈ 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. §3.2 #10 Let D be an integral domain and a, b, c ∈ 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 which satisfies the cancellation property: for every a, b, c ∈ 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 ∈ 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. §3.4 #1 U must contain 1 · r = r for every r ∈ R. So U = R. §3.4 #2 If I is an ideal and a ∈ I is nonzero then for any element r ∈ F , ra−1a ∈ I. But ra−1a = r. So I = R. §3.4 #4b Consider the ring R of all 2 × 2 real matrices. Let

 0 1  A = 0 0

Show that {AB : B ∈ R} 6= {BA : B ∈ R}. (This is easy - AB is a in which the bottom row is zero and BA is a matrix in which the second column is zero). So {AB : B ∈ R} is not a 2-sided ideal.

1 1. If α1, . . . , αn are complex numbers then Z[α1, . . . , αn] stands√ for the smallest subring of C containing Z and√ 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. √ (a) Give an explicit description of Z[21/3] (similar to the one given for Z[ 2] above). Solution. Z[21/3] = {a + b21/3 + c22/3 : a, b, c ∈ Z}. (To verify this, you have to show that this set is closed under multiplication - the rest is obvious). √ √ √ √ (b) Show that [ 2, 3] = [ 2 + 3]. Q √ Q√ √ √ √ √ √ Solution√ . Because 2 + 3 ∈ Q[ 2, 3], it follows that Q[ √2, √3] ⊃ Q[√2 + √3]. So let’s show the other direction. It suffices to show that 2, 3 ∈ Q[ 2 + 3]. First note that √ √ √ √ √ ( 2 + 3)2 = 5 + 2 6 ∈ [ 2 + 3]. √ √ √ Q Therefore, 6 ∈ Q[ 2 + 3]. Next note that √ √ √ √ √ √ √ 6( 2 + 3) = 2 3 + 3 2 ∈ Q[ 2 + 3]. Therefore √ √ √ √ √ √ √ 2 3 + 3 2 − 2( 2 + 3) = 2 ∈ [ 2 + 3]. √ √ √ Q Now that we know 2 ∈ Q[ 2 + 3] it follows that √ √ √ √ √ √ ( 2 + 3) − 2 = 3 ∈ Q[ 2 + 3]. 2. A element r ∈ R is called a if it has a multiplicative inverse. For example, {1, −1} are the units of Z. Every nonzero element of Q is a unit of Q. The set of units always forms a multiplicative .

(a) Find the group of units Z[i]. Solution. The group is {1, −1, i, −i}. To see this, suppose that a + bi ∈ 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 |(a + bi)−1| = |a + bi|−1 we must have that |a + bi| ≤ 1 (and therefore |a + bi| = 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. {1, 3, 7, 9}. 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.

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