Homework for UTK M351 Algebra I Fall 2013, Jochen Denzler, MWF 10:10 11:00 Each part separately graded on a [0/1/2] scale. Problem 1: Recalling the field axioms from class, prove for any field F (i.e., using only the field axioms): (a) There is a unique element 0, such that 0+x = x for all x F. (b) For each x F there is a unique element y such that x+y = 0. (c) For each x F, it holds 0 x = 0. Problem 2: Find exactly which field axioms are violated in (R 2 2,+, ), the set of real 2 2 matrices together with matrix addition and multiplication. Problem 3: (a) Generalize Euclid s lemma in Z, namely prove: If gcd(a,b) = 1 and a bx, then a x. (b) Prove: If a c and b c and gcd(a,b) = 1, then ab c. Show by counterexample that the hypothesis gcd(a,b) = 1 cannot be omitted. Problem 4: Let m Z. Show that the relation congruent modulo m as defined by is an equivalence relation on Z. a b (mod m) : m b a Problem 5: Prove: If a b (mod m) and c d (mod m), then ac bd (mod m). Problem 6: (a) Take the following set of six rational functions {I,F 0,F 1,F,M,W} given by I(z) = z, F 0 (z) = z/(z 1), F 1 (z) = 1/z, F (z) = 1 z, M(z) = (z 1)/z, W(z) = 1/(1 z). Don t try to understand the rationale behind the names. Use composition as operation. This is a group. Construct the group table. (b) Is the set of all rational functions, together with, a group? Give a reason for your answer. (A rational function is a function that can be written in the form f(z) = p(z)/q(z), where p and q are polynomials.) (c) Show that the set of those rational functions of the form f(z) = az+b cz+d with a,b,c,d real and ad bc 0 forms a group, again under the composition as operation. Problem 7: The set O 2 (R) R 2 consists of pairs (A,a) where A is an orthogonal 2 2 matrix and a is a 2-vector. We define the multiplication by (A,a) (B,b) = (AB,Ab + a). Check that this is a group. It is called the isometry group of the plane, because the motivation behind the definition of is that (A,a) represents the mapping x Ax+a,R 2 R 2, namely: Showthatiff(x) = Ax+aandg(x) = Bx+b, then(f g)(x) = Cx+c, where(c,c) = (A,a) (B,b). Problem 8: Given a group (G, ) and an element a G, we define C(a) := {x G ax = xa}. Show that C(a) is a subgroup of G. Problem 9: Given an abelian group (G, ) show that the set of squares H := {a a a G} is a subgroup.
Problem 10: In a ring, we denote by ( x) the unique additive inverse of the element x. Prove that ( x)y = (xy) = x( y) and that ( x)( y) = xy. You may only use ring axioms, including the uniqueness of 0 and of additive inverses, and 0x = 0, in this proof. Problem 11: Given a set S, define the power set P(S) as the set that consists of all subsets of S. For A,B P(S) (i.e., for A,B S), we define A+B := (A\B) (B \A) A B := A B (These notations + and are only for the purpose of the present problem and wouldn t be understood automatically outside this context.) Show that (P(S),+, ) is a commutative ring with one. Which sets are the 0 and the 1 in this ring? You may skip the trivial proofs of (C+), (C ) and (Ass ), but do show the proof of (Ass+), and the remaining axioms. Problem 12: Suppose (R,+, ) is a ring. Consider the set R R together with the following definition of + and : (a,b)+(c,d) := (a+c,b+d) and (a,b) (c,d) = (ac bd,ad+bc). Here, the minus sign stands for adding the additive inverse, as usual. (a) Show that R R, together with these operations is again a ring. We will call this ring R[i]. (b) Show that R[i] has (1,0) an identity element if R has 1 as identity element. Also show that R[i] is commutative, if R is. Problem 13: Show that R := {0,2,4,6,8} is a subring of Z 10 with addition and multiplication modulo 10. The ring R does contain a multiplicative identity. Which is it?
Problem 14: Suppose that in a ring R, the following crossover cancellation axiom holds: If ab = ca and a 0, then b = c. Show that then R is commutative. Problem 15: Give an example of a ring R and elements a,b R such that ab = 0, but ba 0. Problem 16: In the ring R := Z Z Z (the set Z 3, with componentwise defined addition and multiplication), decide whether S := {(a,b,c) Z 3 a+b = c} is a subring. If yes, prove it; if no, specify which property/ies fail. Problem 17: A ring R is called a Boolean ring, if a 2 = a for all a R. (a) Show that in such a ring, a = a for all a. (b) Show that a Boolean ring is commutative. Hints: Consider (a+a) 2 and (a+b) 2. Problem 18: (a) Let R be a ring with unity 1, and consider the ring R[i]. Show that the element i := (0,1) R[i] satisfies i 2 = 1. (b) Furthermore suppose R is a field in which the equation x 2 = 1 does not have a solution. Show that R[i] also is a field. (But now in this field, the equation x 2 = 1 does have a solution.) (c) Now assume instead that R is a field in which the equation x 2 = 1 already has a solution (call it j). Show that now in R[i], we have (j i)(j +i) = 0, so R[i] is not a field, because it has zero divisors.
Problem 19: (Construction of Q from Z). Consider Z as an integral domain (a commutative ring with 1 where 1 0, and without zero divisors). You may do the following tasks for Z specifically, but be advised that any integral domain would do. We denote Z := Z\{0}. (a) On the set Z Z, define the relation by Show that is an equivalence relation. (a,b) (c,d) : ad = bc. (b) Continuing with (a), denote by (a,b) the equivalence class of (a,b) and define QZ as the set of all equivalence classes. (Later we ll denote the equaivalence class (a,b) as a b and the set QZ as Q.) On QZ define + and as follows: For q,r QZ, choose representatives (a,b) q and (c,d) r and define qr to be the equivalence class of (ac,bd) and q +r as the equivalence class of (ad +bc,bd). Show that the operations are well-defined, i.e., they do not depend on the choice of representatives. In other words, show: If (a,b) (a,b ) and (c,d) (c,d ), then (ad+bc,bd) (a d +b c,b d ) and (ac,bd) (a c,b d ). Also show for any x Z that (ax,bx) (a,b). (c) Show that QZ with the operations + and defined above is a commutative ring with 1. In particular, identify the 0 and 1 element in this ring, as well as the additive inverse of (a,b). Also show that every (a,b) other than the zero element has a multiplicative inverse, i.e., that QZ is a field. Problem 20: Given aset S andasubseta S, we definethefunction χ A : S Z 2 by χ A (x) = 0 if x / A and χ A (x) = 1 if x A. Give an example. χ A is called the characteristic function, or, indicator function of A. Recall the ring P(S) with intersection as multiplication and symmetric difference as addition. Also recall the ring F(S R) consisting of all functions from a set S to a given ring R, with pointwise defined addition and multiplication, i.e., (f +g)(x) := f(x)+g(x) and (f g)(x) := f(x)g(x) for all x S. Show that χ : P(S) F(S Z 2 ),A χ A is a ring isomorphism. Problem 21: (a) Show that every nonzero elemet of Z n is either a unit (i.e., invertible), or a zero divisor. How do you distinguish for a given k Z n which of the two cases applies? (b) Find a nonzero element in an appropriate commutative ring with 1, that is neither invertible nor a zero-divisor. Problem 22: Give an example of a commutative ring without zero-divisors that is not an integral domain. (If all else fails, do what you should have done in the first place, namely to recall the precise definition...)
Problem 23: Let R 1,R 2 be rings with 1 and f : R 1 R 2 be a ring isomorphism. Show that f(1) = 1. (Your proof will likely show that the same is true for f a ring homomorphism that is onto.) Comment: (read & understand, no hwk in here) SupposeS 1 and S 2 are sets, and S 1 S 2. Take the rings R 1 = P(S 1 ) and R 2 = P(S 2 ) with symmetric difference and intersection as ring operations. The mapping j : R 1 R 2, A A is an injective ring homomorphism that is not onto unless S 1 = S 2. In this example j(1) 1. Problem 24: Show that R = {0,2,4,6,8} with addition and multiplication modulo 10 is a field. (May use results from Pblm 13). I also claim that R is (ring-)isomorphic to Z 5. Give a specific mapping φ : R Z 5 (by value table) that is a ring isomorphism. Hint: Use Pblm 23 to find φ. Find φ(1), then φ(1+1), and φ(1+1+1) etc., if φ is a ring homormorphism. Then show that the φ so constructed is indeed a ring isomorphism. Problem 25: Suppose a and b belong to an integral domain. Show: If a 5 = b 5 and a 3 = b 3, then a = b. Now assume m,n are positive integers that are relatively prime, i.e., gcd(m,n) = 1. Show: If a m = b m and a n = b n then a = b. (Hint: How would you do it if m = n+1; then remember the gcd can be expressed as a linear combination). Problem 26: How many solutions does the equation x 2 5x+6 = 0 have in each of the following rings?: (a) Z 7 (b) Z 8 (c) Z 12 (d) Z 14 Problem 27: (a) Let n = 7 and define {[ ] } a b R := a,b Z n b a Show that R is a field. How many elements does it have? (b) Let n = 5 and take the same definition for R. Show that this R has zero divisors. (c) Let n be a positive integer, n 2. Show that the mapping φ : a + bi isomorphism from Z n [i] R. [ a b b a ] is a ring
Problem 28: Show that the rings Z 3 [i] and Z 9 are not isomorphic. Give two different proofs. Hint: Assume φ : Z 3 [i] Z 9 were an isomorphism. (1) What would φ(0), φ(1), φ(1 + 1), etc. have to be; derive a contradiction. (2) What equation would φ(i) have to satisfy in Z 9? Derive a contradiction. Problem 29: We have seen for p prime and R a ring of characteristic p, that x x p is a ring homomorphism R R. Let s try this for non-primes: Decide whether x x 4, Z 4 Z 4 is a ring homomorphism or not. Problem 30: Let R be a ring with 1 and R[X] the polynomial ring over R. Let z R. We define the evaluation of a polynomial p in z, written as ev z (p) as the result of plugging z into p : In other words, if p = (a 0,a 1,...,a n ), more conveniently written as p = a 0 +a 1 X +...+a n X n R[X], then ev z (p) := p(z) := a 0 +a 1 z +...+a n z n. (a) Now assumethat R is also commutative. Show that for each z R, the evaluation ev z : R[X] R, p ev z (p) = p(z) is a ring homomorphism. (b) If R is not commutative, e.g., R = M 2 (Q), find elements a,b,z R with p = ax and q = b such that ev z (p)ev z (q) ev z (pq). Hint: Write out diligently, what this last inequality means in terms of a,b,z, and then you ll probably find that this problem isn t as daunting as it looks. A corollary of part (a) is: If R is a commutative ring with 1, R[X] the polynomial ring over R and F(R R) the ring of functions from R to R with pointwise addition and multiplication, then cvf : R[X] F(R R), which assigns to a polynomial p the function z p(z), is a ring homomorphism. I have chosen the weird name cvf for this ring homomorphism to remind of what it does: convert to function. I am not asking you to write this out in formal detail, but want you to be aware of this conversion homomorphism. Problem 31: Give an example of polynomials p,q Z 4 [X] for which deg(pq) < deg(p)+deg(q). Also find a nonconstant polynomial in Z 4 [X] that has a multiplicative inverse in this ring. Problem 32: (a) Find q,r Q[X] such that X 4 + X 2 = q (X 2 + 3X + 1) + r with degr < deg(x 2 +3X +1). (b) Same question in Z 5 [X].