Homework problems from Chapters IVVI: answers and solutions


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1 Homework problems from Chapters IVVI: answers and solutions IV In this problem we have to describe the field F of quotients of the domain D. Note that by definition, F is the set of equivalence classes [(n 1 + m 1 i, n 2 + m 2 i)] of D D with equivalence relation if and only if (n 1 + m 1 i, n 2 + m 2 i) (k 1 + l 1 i, k 2 + l 2 i) (n 1 + m 1 i)(k 2 + l 2 i) = (n 2 + m 2 i)(k 1 + l 1 i). (1) We will show that F is isomorphic to the following field: Q[i] := {a + bi a, b Q}. The isomorphism ϕ : F Q[i] is given by the formula ϕ([(n 1 + m 1 i, n 2 + m 2 i)]) := n 1 + m 1 i n 2 + m 2 i or in short ϕ([z, w]) := z w for (z, w) D D. In what follows we show that ϕ is an isomorphism. The map ϕ is welldefined because of (1). The map ϕ is a homomorphism of rings because: ϕ([(z 1, w 1 )]+[(z 2, w 2 )]) = ϕ([z 1 w 2 +w 1 z 2, w 1 w 2 ]) = z 1w 2 + w 1 z 2 w 1 w 2 = z 1 w 1 + z 1 w 1 = ϕ([(z 1, w 1 )])+ϕ([(z 2, w 2 )]) Similarly ϕ([(z 1, w 1 )] [(z 2, w 2 )]) = ϕ([(z 1, w 1 )]) ϕ([(z 2, w 2 )]). The map ϕ is onto because for z w Q[i] we have ϕ([(z, w)]) = z w. The map ϕ is onetoone because if ϕ([(z, w)]) = 0 then z = 0, i.e. [(z, w)] = [(0, 1)]. IV Solved in class. IV The elements of Q(Z 4, {1, 3})are [(0, 1)] = [(0, 3)], [(1, 1)] = [(3, 3)], [(1, 3)], [(2, 1)], [(2, 3)], [(3, 1)], i.e. total 6 distinct elements. IV f(x)g(x) = x IV We have that 106 = , 99 = , 53 = , and thus φ 4 (3x x x 53 ) = (4 6 ) (4 6 ) (4 6 ) 8 = =...( mod 7), since 4 6 = 1( mod 7) by Fermat s Little Theorem.
2 IV Solved in class. IV (a) The units of D[x] are the constant polynomials f(x) = u, where u is a unit of D. (b) The units of Z[x] are the constant polynomials f(x) = ±1. (b) The units of Z 7 [x] are the constant polynomials f(x) = u, for u = 1, 2, 3, 4, 5, 6. IV (a) You must show that D is a homomorphism without using any properties of derivatives (done in class). (b) The kernel is the set of all constant functions. (c) The image F [x] is F [x] itself, i.e. D is onto (this needs a proof). IV The factorization is x = (x + 1)(x + 2)(x + 3)(x + 4) (you simply verify that any nonzero element of Z 4 is a root of x = 0). IV Verify that there f(x) has no rational root.the possible rational roots are ±1. ± 2, ±4, ±8. Note that one can not apply Eisenstein s Criterion for any p in this case. IV Eisenstein s Criterion does not apply for any p. IV Eisenstein s Criterion applies for p = 5. IV The polynomial has always a linear factor x + a, since a is a root (due to Fermat s Little Theorem). IV Done in class. V The sets are isomorphic as abelian groups (identical addition tables), but not as rings (as the multiplication tables are different). One has to write down the tables. V One must show that (a + b) p = a p + b p and (ab) p = a p b p. The first one is done in class using the binomial formula and the second one follows by the fact that ab = ba. V Let N(R) := {a R a n = 0 for some n Z + }. be the nilradical of R. One must show that:
3 (1) N(R) is an abelian subgroup of R, i.e. a + b N(R), 0 R N(R), a N(R), whenever a, b N(R); (2) If a N(R) and b R then ab N(R). For the addition part of (1) one uses the binomial formula to show that, if a n = 0 and b m = 0, then (a + b) n+m = 0 (done in class). Not that in order to prove ( a) n = 0 one must use the definition of a and not the formula ( a) n = ( 1) n a n as R might have not identity. The second axiom follows from the identity (ab) n = a n b n. V The reasoning is similar to the on in V V Take R = Q[x]. (a) One example is N := x 2, i.e. the principal ideal of R generated by the polynomial f(x) = x 2. Then x N, but x / N. (b) One example is N := x, i.e. the principal ideal of R generated by the polynomial g(x) = x. Then N = N. V The prime and maximal ideals of Z 12 (all primes are maximals in this case) are those that are isomorphic to Z 4 and Z 6, since M is prime iff Z 12 /M is an integral domain iff Z 12 /M = Z 2, Z 3. Thus the answer is: I 1 = 3 = 9 = Z 4 and I 2 = 2 = 10 = Z 6. V All nonzero proper ideals are both maximal and prime. V Solved in class. The shortest way to find the prime and maximal ideals is again to apply the theorems: M is a maximal (respectively, prime) ideal of a commutative ring R iff R/M is a field (resp., integral domain). So we must find all M for which Z 2 Z 4 /M is an integral domain. But if M is proper and nontrivial then Z 2 Z 4 /M, as an abelian group, is isomorphic to one of the following: Z 2, Z 4, Z 2 Z 2. In this list there is only one integral domain: Z 2, so we need to list all ideals M of Z 2 Z 4 which are isomorphic as abelian groups to Z 2 Z 2 or Z 4. These will be both the maximal and prime ideals. The answer is Z 2 {0, 2}, {0} Z 4. V Find all c for which x 2 + x + c is irreducible, i.e. does not have a root. The answer is c = 1, 2 V Same reasoning as in V V Solved in class. Relates to the Theorem on page 181. V Solved in class. Must show that if a principal ideal f(x) is prime then f(x) is irreducible (here we use theorems and 27.25). The proof is by contradiction.
4 V The second statement follows by using a proof by contradiction. Namely, one assumes that f and g are irreducible and shows that N = F [x]. The proof relies on the polynomial version of Euclidean Algorithm whose general statement and proof can be found on p age 404. VI f(x) = (x 2 5) 2 24 = x 4 10x VI f(x) = (x 1) VI f(x) = (x 2 1) 3 2. VI irr(α, Q) = (x 1) = x 2 2x + 9 and deg(α, Q) = 2. One shows that x 2 2x + 9 is irreducible by verifying that it does not have a rational root. VI irr(α, R) = (x 1) = x 2 2x + 2 and deg(α, R) = 2. VI irr(α, R) = x π and deg(α, R) = 1. VI π 2 is transcendental over Q. Proof by contradiction: must show that if π 2 is algebraic, then π is also algebraic over Q. VI irr(α, Q(π)) = x π 2 and deg(α, Q(π)) = 1. In particular, F (α) = F. VI Solved in class. irr(α, Q(π 3 )) = x 3 π 6 and deg(α, Q(π 3 )) = 3. Must show that x 3 π 6 is irreducible in Q(π 3 )[x] (done in class). VI Solved in class. VI Since α is algebraic over F (β) one has that a 0 + a 1 α a n α n = 0 for some a i F (β). Every element u of F (β) has the form u = f(β) g(β), for some polynomials f, g F [x]. Thus there are polynomials p i, q i, i = 0,..., n for which a i = p i(β) q. But then i(β) p0(β) q 0 (β) + p 1(β) q 1 (β) α p n(β) q n (β) αn = 0 Multiplying by the least common denominator q 0 (β)...q n (β) we find that P (α, β) = 0 for some polynomial P (x, y) of two variables in F [x, y]. Then if Q(x) := P (α, x) F [α][x] F (α)[x] we have that Q(β) = 0 and we complete the proof. VI (a) The shortest way to solve this problem is first to find all irreducible (and reducible) monic polynomials in Z 3 of degree 2. We have that all reducible ones are x 2, x(x+1), x(x+2), (x+1) 2, (x+1)(x+2), (x+2) 2. Thus we
5 have 3 irreducible ones. Then we easily count all reducible monic polynomials of degree 3: we have 9 that are product of irreducibles of degree one and degree two, and 10 more that are product of three (not necessarily different!) of degree one. So we have total 19 reducible monic polynomials and thus 8 irreducible ones. VI [Q( 2, 3, 18) : Q] = 4, and a basis is {1, 2, 3, 6}. Here we prove Q( 2, 3, 18) = Q( 2, 3) and prove that f(x) = (x 2 5) 2 24 = x 4 10x is irreducible. (Must show more work!) VI [Q( 3 2, 3) : Q] = 6, and basis is {1, 3, 3 2, 3 2 3, 3 4, 3 4 3}. (Must show more work!) VI [Q( 2 + 3) : Q( 3)] = 2, and basis is {1, 2}. Here we use Q( 2 + 3) = Q( 2, 3). VI They are the same fields. (Must show more work!) VI Use the hint I provided in class. VI Solved in class. VI In class we showed Q( 2+ 3) = Q( 2, 3). Use similar reasoning. VI Solved during the review session. We look at [F (α) : F (α 2 )]. This degree is either 1 or 2 since f(α 2 ) = 0 for f(x) = x α 2 F (α 2 )[x]. We show that it can not be 2 using the assumption in the problem. Then we use the fact that [F (α) : F ] = [F (α) : F (α 2 )][F (α 2 ) : F ].
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