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Math 403 Assignment 1. Due Jan. 2013. Chapter 11. 1. (1.2) Show that, for n 6= 0, cos(2π/n) is algebraic over Q. Solution. Since e2πi/n and e−2πi/n are each algebraic (each satisfies the equation xn −1 = 0), half their sum is algebraic too, i.e., cos(2π/n) = (1/2)(e2πi/n + e−2πi/n ) is algebraic. √ √ 2. (1.3) Let R = Q[ 2, 3] ⊂ R. √ √ √ √ (a). Is Q[ 2, 3] = Q[ 2 + 3]? √ √ √ √ Solution.√ Yes.√To begin, we have the fact that is a field. Hence, ( √2 + √3)−1 √ Q[√ 2 + √ 3] √ lies√in Q[√ 2 + 3]. Next, that ( 2 + 3)( 2 − 3) = −1, and so, 2 − 3 = √we observe √ −( 2 + 3)−1 lies in Q[ 2 + 3]. √ √ √ √ √ √ √ √ 3] lie in Q[ 2 + 3], so do their sum 2 2 and difference 2 3. Since 2 + √ 3√and Q[ √2 + √ Hence, Q[ 2, 3] = Q[ 2 + 3]. √ √ √ √ (b). Is Z[ 2, 3] = Z[ 2 + 3]? √ √ √ √ √ √ √ √ Solution. Q[ 2 + 3] is a Q-vector space with basis {1, 2 + 3, ( 2 + 3)2 , ( 2 + 3)3 }; hence, each of the elements of the ring has a unique expression as a linear combination of the basis. √ 2 √ √ √ √ √ = 5√+ 2 6, we see that 6 = −5/2 + (1/2)( 2 + 3)2 , which is not Since ( 2 + 3)√ an element of√Z[ 2 + 3] since the√coefficients are not all√integers (and expression is √ √ √ the √ unique). But 6 is an element of Z[ 2, 3]. Therefore, Z[ 2, 3] 6= Z[ 2 + 3]. 3. (1.6) Is {a/b ∈ Q|(b, 3) = 1} a subring of Q? Solution. Yes, a/b + c/d = (ad + bc)/bd and (a/b)(c/d) = ac/bd, where (bd, 3) = 1 when (b, 3) = 1 and (d, 3) = 1 since 3 is prime. 4. (1.8) Determine the units in the following rings: Z/(12)Z, Z/8Z, and Z/nZ. Solution. In a ring R, if a and b are nonzero elements of R such that ab = 0, then a and b are nonunits. To see that, if a had a multiplicative inverse a−1 , then b = a−1 (ab) = a−1 0 = 0, 1 contrary to the hypothesis that b 6= 0. In Z/nZ, if (m, n) 6= 1, then m + nZ is a nonunit since (m + nZ)(n/(m, n) + nZ) = 0 + nZ. m + nZ is a unit if (m, n) = 1, since by arithmetic, if (m, n) = 1, there are integers a and b such that am + bn = 1. Hence, 1 + nZ = am + bn + nZ = am + nZ = (a + nZ)(m + nZ. Units of Z/(12)Z: 1 + (12)Z, 5 + (12)Z, 7 + (12)Z, 11 + (12)Z. Units of Z/8Z: 1 + 8Z, 3 + 8Z, 5 + 8Z, 7 + 8Z. Units of Z/nZ: {m + nZ|0 ≤ m < n, (m, n) = 1}. 5. (1.9) Let R, +, · satisfy all the ring axioms except possibly the commutative rule for +. Use the distributative rule to prove that + is in fact commutative. (Thus, we want a distributative rule, we must base a ring on an abelian group R, +.) Solution. We want to show that a + b = b + a, i.e., that a + b − (b + a) = 0. First, we show that −c = (−1)c for each element c ∈ R. In fact, c + (−1)c = 1c + (−1)c, since 1c = c. Hence, c + (−1)c = 1c + (−1)c = (1 + (−1))c = 0c = 0, using the distributative law. Hence,−c = (−1)c. Therefore, −(b + a) = (−1)(b + a) = (−1)b + (−1)a = −b − a, using the distributative law. Therefore, we have the comlutation a + b − (b + a) = a + b − b − a = 0, as we wished to show. 6. (2.1) For which positive integers n does x2 + x + 1 divide x4 + 3x3 + x2 + 7x + 5 in the polynomial ring Z/nZ[x]? Solution. For the division to take place, we need to solve the equation (x2 + x + 1)(x2 + ax + b) = x4 + 3x3 + x2 + 7x + 5, for a and b in Z/nZ. Multiplying out the left-hand side, we get x4 + (a + 1)x3 + (b + 1 + a)x2 + (a + b)x + b = x4 + 3x3 + x2 + 7x + 5, 2 in Z/nZ[x]. Comparing coefficients, we have a + 1 = 3, b + 1 + a = 1, a + b = 7, and b = 5 in Z/nZ. By those equations, we see that a = 2 and b = 5; hence, by the equation b+ 1 + a = 1, we see that 8 = 1, i.e., 7 = 0 in Z/nZ, i.e., n = 7. 7. (2.2) Let F be a field and let F [[t]] be the ring {a0 + a1 t + a2 t2 + · · · aj tj + · · · = j Σ∞ j=0 aj t |aj ∈ F ; j = 0, 1, 2, ...} where addition is gotten from the addition of polynomials and multiplication is gotten from the distributative rule and the multiplication of polynomials. Find the units in this ring. j Solution. For Σ∞ j=0 aj t to be a unit, we need to solve the equation j ∞ j (Σ∞ j=0 aj t )(Σj=0 bj t ) = 1, for the coefficientsbj . The product equals Σj=0 (aj b0 + aj−1 b1 + · · · a1 bj−1 + a0 bj )tj . For that product to equal 1, we need a0 b0 = 1 and aj b0 + aj−1 b1 + · · · a1 bj−1 + a0 bj = 0. So long as a0 6= 0, we can solve successively for b0 , b1 ,...,bj ,.... Therefore, the units are the elements j Σ∞ j=0 aj t with a0 6= 0. 3