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MT310.01: Introduction to Abstract Algebra Final Examination Answers
- (10 points) The following sets are all commutative rings: Z , Q , R , C , Z / 2 Z , Z / 3 Z , Z / 4 Z , Z / 5 Z , and Z / 6 Z. Which of these rings are integral domains? Which are fields? You do not need to justify your answers to this question.
Answer : The integral domains are Z , Q , R , C , Z / 2 Z , Z / 3 Z , and Z / 5 Z. The fields are Q , R , C , Z / 2 Z , Z / 3 Z , and Z / 5 Z.
- (10 points) ( a ) State Eisenstein’s Criterion. ( b ) Let n be a positive integer, and let k be an integer which is bigger than 1. Prove that √ k 25 n + 15 is irrational.
Answer : ( a ) If f (x) = a n x n^ + · · · + a 1 x + a 0 is a polynomial in Z [x], and (1) p - a n. (2) p|a n − 1 ,.. ., p|a 1 , p|a 0. (3) p^2 - a 0. Then f (x) is irreducible in Q [x]. ( b ) Consider the polynomial x k^ − (25n + 15), with k and n as above. Eisenstein’s Criterion says that this polynomial is irreducible in Q [x], because (1) 5 - 1. (2) 5 | 25 n + 15. (3) 25 - 25 n + 15. Because the polynomial is irreducible in Q [x], we know that it cannot have a factor of the form x − pq. Therefore, the roots of the polynomial are irrational, so k
25 n + 15 is irrational.
- (10 points) Suppose that F is a field, and f (x) is an element of F [x]. Prove or give a counterexample to each of the following statements: ( a ) If f (x) is irreducible, then f (x^2 ) is irreducible. ( b ) If f (x^2 ) is irreducible, then f (x) is irreducible. A counterexample means finding a specific field F and specific polynomial f (x) that makes the statement false.
Answer : ( a ) This statement is false. One simple counterexample is given by f (x) = x − 1 , which is irreducible, while f (x^2 ) = x^2 − 1 = (x + 1)(x − 1). ( b ) This statement is true. Suppose that f (x) is reducible, so that f (x) = g(x)h(x), with deg (g(x)) ≥ 1 and deg (h(x)) ≥ 1. We then have f (x^2 ) = g(x^2 )h(x^2 ), with deg (g(x^2 )) ≥ 2 and deg(h(x^2 )) ≥ 2. This shows that f (x^2 ) is reducible.
- (10 points) Suppose that φ : G 1 → G 2 is a homomorphism of groups. Prove or give a counterexample: ( a ) If H 1 / G 1 , then φ(H 1 ) / G 2 ( b ) If H 2 / G 2 , then φ−^1 (H 2 ) / G 1. As usual, φ(H 1 ) is defined with the equation φ(H 1 ) = {φ(h 1 ) : h 1 ∈ H 1 } and φ−^1 (H 2 ) is defined with the equation φ−^1 (H 2 ) = {g 1 ∈ G 1 : φ(g 1 ) ∈ H 2 }.
Finding a counterexample means finding specific groups G 1 and G 2 , a specific homomorphism φ, and a specific normal subgroup which makes the statement false. Be sure not to assume in your proofs that φ is either injective or surjective.
Answer : ( a ) This statement is false. The easiest way to find a counterexample is to let G 1 be an abelian group, so that every subgroup is normal, and to let G 2 be S 3 , because we know from homework problems that {e, (12)} is a subgroup of S 3 which is not normal. One possibility is to let G 1 = Z , and define φ(n) = (12) n. In other words,
φ(n) =
(12) n is odd e n is even
Now, let H 1 = 3 Z , the subgroup of all multiples of 3. We know that 3 Z / Z , because Z is abelian, and φ(H 1 ) = {e, (12)}, which is not a normal subgroup of S 3. ( b ) This statement is true. Let h 1 ∈ φ−^1 (H 2 ), and g 1 ∈ G 1. We need to show that g 1 h 1 g 1 − 1 ∈ φ−^1 (H 2 ). Because h 1 ∈ φ−^1 (H 2 ), we have φ(h 1 ) ∈ H 2 , and therefore if g 2 is any element of G 2 , we know that g 2 φ(h 1 )g 2 − 1 ∈ H 2. In particular, if g 2 = φ(g 1 ), we have φ(g 1 h 1 g− 1 1 ) = φ(g 1 )φ(h 1 )φ(g 1 )−^1 ∈ H 2. This shows that g 1 h 1 g− 1 1 ∈ H 2.
- (10 points) Suppose that φ : R 1 → R 2 is a homomorphism of rings. Prove or give a counterexample: ( a ) If I 1 is an ideal of R 1 , then φ(I 1 ) is an ideal in R 2 ( b ) If I 2 is an ideal of R 2 , then φ−^1 (I 2 ) is an ideal in R 1. Finding a counterexample means finding specific rings R 1 and R 2 , a specific homomorphism φ, and a specific ideal which makes the statement false. Be sure not to assume in your proofs that φ is either injective or surjective.
Answer : ( a ) This statement is false. A simple counterexample is to take R 1 = Z , and R 2 = Q , with the homomorphism φ(n) = n. Let I 1 = 2 Z , the ideal of even integers. Then φ(I 1 ) is also the set of even integers, but inside the ring Q , the set of even integers is not an ideal. For example, 2 is an even integer, and 12 ∈ Q , and their product is not an even integer. ( b ) This statement is true. Suppose that i, j ∈ φ−^1 (I 2 ). Then we know that φ(i), φ(j) ∈ I 2 , and because I 2 is an ideal, we know that φ(i) + φ(j) ∈ I 2. That shows that φ(i + j) ∈ I 2 , or i + j ∈ φ−^1 (I 2 ). Now let r ∈ R 1. We need to show that ir and ri ∈ φ−^1 (I 2 ). We know that φ(i) ∈ I 2 , and therefore r 2 φ(i) and φ(i)r 2 ∈ I 2 for any r 2 ∈ R 2 , because I 2 is an ideal. In particular, we let r 2 = φ(r), and then φ(r)φ(i) and φ(i)φ(r) ∈ I 2. That shows that φ(ri) and φ(ir) ∈ I 2 , or ri, ir ∈ φ−^1 (I 2 ).
- (5 points) Suppose that R is a commutative ring, and the only ideals of R are { 0 } and R. Prove that R is a field.
Answer : We need to show that any non-zero element of R has an inverse in R. Let r be such an element. The ideal (r) cannot be the zero ideal, because r 6 = 0. Therefore (r) = R. Because 1 ∈ R, we know that there is some element s ∈ R so that rs = 1. This shows that r has an inverse.
that o(ab) = 6. Let c = ab, and we know that G is a cyclic group generated by c, so G = {e, c, c^2 , c^3 , c^4 , c^5 }. Now the function φ : G → Z / 6 Z given by φ(c n ) = [n] 6 is an isomorphism.
- (10 points) Suppose that G is a nonabelian group with 6 elements. Show that G is isomorphic to S 3.
Answer : By Cauchy’s Theorem, G contains an element a with o(a) = 2 and an element b with o(b) = 3. We can list the 6 elements of G as {e, b, b^2 , a, ab, ab^2 }. We need to compute ba. We cannot have ba = e, or else b = a−^1. We cannot have ba = b, or else a = e. We cannot have ba = b^2 , or else a = b. We cannot have ba = a, or else b = e. If ba = ab, then a and b commute, o(ab) = 6, G is cyclic, and therefore G is abelian. The only possibility is that ba = ab^2. We can now compute for example that b^2 a = b(ba) = b(ab^2 ) = (ba)b^2 = ab^4 = ab. Similar computations allow us to write out the complete group operation table for G: · e b b^2 a ab ab^2 e e b b^2 a ab ab^2 b b b^2 e ab^2 a ab b^2 b^2 e b ab ab^2 a a a ab ab^2 e b b^2 ab ab ab^2 a b^2 e b ab^2 ab^2 a ab b b^2 e Here is the operation table for S 3 : · e (123) (132) (12) (23) (13) e e (123) (132) (12) (23) (13) (123) (123) (132) e (13) (12) (23) (132) (132) e (123) (23) (13) (12) (12) (12) (23) (13) e (123) (132) (23) (23) (13) (12) (132) e (123) (13) (13) (12) (23) (123) (132) e Now inspection shows that these are the same group operation tables with the mapping φ(e) = e, φ(b) = (123), φ(b^2 ) = (132), φ(a) = (12), φ(ab) = (23), and φ(ab^2 ) = (13). Grade Number of people 70 1 52 1 51 1 48 1 38 1 37 1 36 1 24 1
Mean: 44. Standard deviation: 13.
