# Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 2

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1 Solutions to odd-numbered exercises Peter J Cameron, Introduction to Algebra, Chapter 1 The answers are a No; b No; c Yes; d Yes; e No; f Yes; g Yes; h No; i Yes; j No a No: The inverse law for addition A3 fails There is no natural number b such that 1 + b = 0 In fact, if your convention is that zero is not a natural number, it doesn t satisfy A either b We adopted the convention that the zero polynomial doesn t have a degree, in which case this set is not a ring since A fails If, however, you decide that 0 has degree 1 or some such, then the set is still not a ring for n > 0: the closure law for multiplication M0 fails if n > 0 The polynomials x n and x n both belong to our set, but their product x n does not [If n = 0, then we have just the constant polynomials, in other words, the real numbers, which do indeed form a ring] c Yes: Rather than laboriously check all the axioms, let us take it for granted that real polynomials form a ring, and apply the subring test Certainly Z[x] is nonempty If f x and gx are polynomials in Z[x] ie with integer coefficients, then so are f x gx and f xgx So Z[x] passes the Second Subring Test Alternatively, using the theorem that the polynomials over a ring form a ring, it is clear that Z[x] is a ring d Yes: This set is a non-empty subset of Z[x], which we have just shown to be a ring; so we can apply the Subring Test again If f x and gx are polynomials with integer coefficients having constant term 0, then so are f x gx and f xgx e No: x and x 3 both belong to this set, but their product x 5 does not f Yes: Apply the Subring Test If f = g = 0, then f g = f g = 0 and f g = f g = 0 [We are using here the fact that, if we subtract or multiply polynomials and then make a substitution, we get the same answer as if we make the substitution and then subtract or multiply Why is this?] g Yes: If m and n are divisible by 3, then so are m n and mn h No: the matrices and are both non-singular they have 0 1 determinant 1, but their sum is, which is singular So A0 fails 0 0 i Yes: Apply the subring test since C is a ring j No: This set contains x + 1 and x 1 but not x 1, so M0 fails 1

2 3 In all cases we can apply the Subring Test since all are contained in M R a Not a ring For =, that is, the product of symmetric matrices need not be symmetric 1 0 b Not a ring For =, that is, the product of skew-symmetric matrices need not be skew-symmetric d e c This set is a ring For, if A = and B = are upper triangular, then 0 c 0 f a d b e ad ae + b f A B = and AB = are both upper triangular 0 c f 0 c f This ring is not commutative, as we saw in the solution to Exercise There is an identity, namely which is upper triangular It is not a division ring since the non-zero matrix has no inverse 0 0 d This set is also a ring The argument is similar to that in c This time, 0 a 0 b 0 0 = ; in other words, we have a zero ring all products are zero So the multiplication is commutative; there is no identity; it is not a division ring c d e Let A = and B = Then b a d c a c b d ac bd ad + bc A B = and AB = b d a c ad + bc ac bd Both are of the correct form to belong to the set R we are considering So it passes the subring test Calculation shows that the multiplication is commutative; the 1 0 identity belongs to the set; and, if a and b are not both zero, then the inverse of turns out with a little calculation to be b a a a +b b a +b b a +b a a +b [Multiply it out and see!] So R is a commutative division ring, that is, a field

3 Remark The ring R in part e is isomorphic to the field of complex numbers Check that the rules for addition and multiplication for complex numbers a + bi and for matrices work in exactly the same way b a 5 a We can argue informally: m x is the sum of m terms equal to x So, if we add m x to n x, we add m xs to n xs, giving m+n altogether; and if we add up n x m times, the effect is to add mn xs So the results hold More formally, we can use induction We can define n x by the rules: 1 x = x; for n 1, n + 1 x = n x + x Now let us prove the first identity by induction on n Starting the induction for n = 1: the left-hand side is m + 1 x and the right is m x + x, which are equal according to our definition The inductive step Suppose that m + n x = m x + n x Then m + n + 1 x = m + n x + x by definition = m x + n x + x by the induction hypothesis = m x + n x + x by the associative law = m x + n + 1 x by definition So the result holds with n + 1 replacing n, and is true for all n by induction The proof by induction of the second equation is for you to try! b We have n x = x + + x n terms = x by the distributive law = n 1x = 0x by assumption = 0 7 a Since x is the unique additive inverse of x, it is enough to show that 1x is also an inverse of x, that is, that x + 1x = 0 This holds because x + 1x = 1x + 1x = 1 + 1x = 0x = 0 b Again, it suffices to show that y x is an inverse of x + y: x + y + y x = x + y y x = x + 0 x = x x = 0 c Suppose that all the axioms hold except possibly the commutative law for addition Check that the properties of inverses, and in particular the results of a and b above, both hold There is it more to be done here: for example, in a, as well as 3

4 showing that x+ 1x = 0, we have also to show that 1x+x = 0; but the argument is quite similar Now we have x y = 1x + y = x + y = y x So addition of x and y is commutative, for any x and y Since any element has an inverse, this actually shows that addition of arbitrary elements is commutative 9 To show that R S is a ring, it is necessary to check the ring axioms Everything is very straightforward, since if we evaluate anything in R S, we just get the corresponding expressions in the two coordinates For a simple case, consider A4: r 1,s 1 + r,s = r 1 + r,s 1 + s = r + r 1,s + s 1 = r,s + r 1,s 1 One point should be noted When we write r 1 + r,s 1 + s or r 1 r,s 1 s, the addition and multiplication in the first coordinate are those of the ring R, while those in the second coordinate are those of S So, for example, the zero element of the ring R S is 0 R,0 S, where 0 R is the zero of R and 0 S is the zero of S If you just write 0,0, you must make clear that 0 means two different things in the two positions The proof of the commutative law for R S, assuming the commutative law for R and for S, is much like the proofs of the other axioms To prove the converse the only if part, argue by contradiction If r 1 r r r 1, then r 1,0r,0 r,0r 1,0 So, if R is not commutative, then R S is not commutative Similarly for S So, if R S is commutative, then both R and S are commutative The argument for the identity is similar If 1 R and 1 S are identities in R and S respectively, then 1 R,1 S is the identity of R S Conversely, if u,v is an identity of R S, then u and v are identities in R and S respectively The answer to the last part is: R S is a field if and only if one of R and S consists of just one element namely, 0, and the other is a field For the forward implication, argue by contradiction Suppose that both R and S have more than one element Let r and s be non-zero elements of R and S respectively Then r,0 S and 0 R,s are non-zero elements of R S; but their product is zero, so R S has divisors of zero, and cannot be a field If, say, R is zero, then R S is isomorphic to S by means of the mapping θ defined by 0 R,sθ = s; so R S is a field if and only if S is a field 11 This exercise requires the verification of a whole list of axioms The displayed identity is easily checked: a 1 + bi + cj + dka 1 bi cj dk = a + b + c + d 1 + ab ba + cd dci + ac ca bd + dbj + ad da + bc cbk Now, letting N = a + b + c + d, we have a 1 + bi + cj + dka/n 1 b/ni c/nj d/nk = 1 if N 0; so non-zero elements have multiplicative inverses 4

5 13 We use the First Isomorphism Theorem We define a function θ from R[x] to R/I[x] by the rule that a n x n θ = a n x n, where a = I +a R/I; that is, θ replaces each coefficient of a polynomial by its image under the canonical homomorphism from R to R/I Now θ is a homomorphism: for example, if f = a n x n and g = b n x n, then f gθ = n k a k b n k x n = n k a k b n k x n = f θgθ, with a similar but easier calculation for addition The kernel of θ consists of all polynomials a n x n R[x] for which a n = 0 that is, a n I for all n; this is just I[x] So I[x] is an ideal of R[x] and R[x]/I[x] = R/I[x] 15 a Recall that mz is the set of all multiples of m If mz contains nz then, in particular, n mz, so n is a multiple of m, or m divides n Conversely, if m divides n, say n = mk, then nx = mkx for all x; so every element of nz is in mz, or mz contains nz b The Second Isomorphism Theorem says that there is ijection between ideals of Z/60Z and ideals of Z containing 60Z Since Z is a PID, every ideal has the form mz By a, mz contains 60Z if and only if m divides 60 So there are 1 ideals of Z/60Z, corresponding to the twelve divisors of 60, viz 1,,3,4,5,6,10,1,15,0,30,60 Again it follows from the Second Isomorphism Theorem that maximal ideals correspond We proved in lectures that an ideal of a PID is maximal if and only if its generator is irreducible So there are three maximal ideals of Z/60Z, corresponding to the prime divisors,3,5 c By exactly the same argument, the number of ideals of Z/nZ is the number of divisors of n, and the number of maximal ideals is the number of prime divisors Any divisor of n = p a 1 1 pa r r has the form p b 1 1 pb r r, where b i lies between 0 and a i inclusive So there are a i +1 choices of b i for each i These choices are independent, so we multiply them together to get the number of divisors, which is a a r + 1 For n = 60 = , this formula gives = 1, in agreement with b above The number of prime divisors is clearly r 17 This question really asks us to prove that the formulae for addition and multiplication of polynomials work also when we think of a polynomial as a function on the ring R, so that, letting f u denote the result of substituting u for x, we have f + gu = f u + gu and f gu = f ugu Both follow easily from the axioms but note that the second equation does require M4, the commutative law for multiplication! 19 The homomorphism is given by mnz + xθ = nz + x 5

6 It is not clear that it is well defined independent of the choice of coset representative To show this, suppose that mnz + x = mnz + y Then x y is divisible by mn, and so certainly by n; thus nz + x = nz + y, as required Checking that θ is a homomorphism is straightforward It is clearly onto 1 We showed in Exercise 3c that R is a ring We are going to prove the whole thing in one blow, using the First Isomorphism Theorem You can prove parts a and b directly without too much difficulty, but a direct proof of c is harder A useful tip is that, if you are ever asked to prove that R/I = S, find a homomorphism θ : R S whose kernel is I and whose image is S This is usually much easier than fiddling round with cosets; the only problem is in finding the homomorphism Define θ : R R by θ = 0 c a 0 0 c This appears to be the only reasonable definition Now d e a + d b + e a + d 0 + θ = θ =, 0 c + f 0 c + f d e a 0 d 0 a + d 0 θ + θ = + = 0 c + f Similarly a b d e θ = d e θ θ = ad ae + b f 0 c f a 0 d 0 ad 0 θ = 0 c f = ad 0 0 c f, So θ is a homomorphism The image of θ clearly is S, the set of all diagonal matrices Its kernel is { } a 0 : = O = I 0 c 0 c Thus, by the three parts of the First Isomorphism Theorem, we conclude that S is a subring of R; that I is an ideal of R; and that R/I = S 3 Using x to denote the coset 1Z + x, the units of Z/1Z are 1,5,7,11 the cosets whose representatives are coprime to 1, and so the associate classes are {0},{1,5,7,11},{,10},{3,9},{4,8},{6} 5 If R is an integral domain, then deg f g = deg f + degg for any two nonzero polynomials f and g in R[x] For if f and g have leading terms a m x m and b n x n respectively, with a m,b n 0, then f g has leading term a m b n x m+n, and a m b n 0 since R is an integral domain Thus, a polynomial of degree greater than 0 can never be a unit, since multiplying it by any non-zero polynomial increases the degree A polynomial of degree zero is a constant, and is a unit in R[x] if and only if it is a unit in R 6

7 Remark If R = Z 8, we have 1 + x1 x + 4x = 1, so that 1 + x is a unit So the condition that R is an integral domain is necessary for the proof 7 We have 1 + x1 x + x + 1 n 1 x n 1 = 1 + x x + x n 1 x n 1 + x n = n 1 x n = 1 9 a If a = x + yi and b = s +ti 0, then a x + yis ti xs + yt ys xt = b s +t = s + +t s +t i = u + vi, as claimed Now let m and n be the integers nearest to u and v respectively Then u m 1 and v n 1, so u + vi m + ni = 1/, as claimed This means that a = bq+r, where q = m+ni and r = bu m+v ni; we have r b / < b, and the Euclidean property is verified b The point of this proof is that there is an element of R whose distance from any given complex number is strictly less than 1, in fact at most 1/ This can be seen geometrically by noticing that the points of R are the vertices of the square lattice in the complex plane, and any point is at distance at most 1/ from some corner of the square containing it Now the Eisenstein integers are the points of the unit triangular lattice in the plane, and any point is at distance less than 1 in fact, at most 1/ 3 from some corner of the triangle containing it The rest of the proof proceeds as before 31 Since 9 = 3, we have to start with a field F with 3 elements which we take to be the integers mod 3, say {0,1,}, and an irreducible polynomial of degree over F which you can find by trial and error: there are three irreducible polynomials, one of which is x + 1, but any one would do [How to check? If a quadratic polynomial is reducible, it must be a product of two factors of degree 1, and hence it must have a root in F So we can check that x + 1 is irreducible by noting that = 1 0, = 0, + 1 = 0] Now let α be a root of the polynomial x + 1 = 0 Then the elements of K = F[x]/x +1 have the form c 0 +c 1 α, where c 0,c 1 F: there are 3 = 9 such elements We add and multiply them in the usual way, using the fact that α = 1 to ensure that no power of α higher than the first occurs 7

8 33 Each coset has a unique representative of degree less than n, of the form a 0 + a 1 α + + a n 1 α n 1, where α = f + x Each of the n coefficients a 0,,a n 1 can be chosen to be any of the q elements of F So there are q n cosets 35 a This is true by definition: R is an integral domain if and only if the product of non-zero elements cannot be zero b The proof is almost identical to that for the field of fractions of an integral domain The ring axioms are easily checked; the embedding of R is by the map a [a,1]; and [a,b] = ab 1, since [b,1][1,b] = [1,1] 37 a If I is an ideal with a I, then na I for any element n Z; and any element of the form sa, at, or s i at i for s,t,s i,t i R, belongs to I; hence any sum of such elements also belongs to I So a I To finish the argument we have to show that the set of such elements is an ideal in which case it is clearly the smallest Closure under subtraction follows from n 1 n a = n 1 a n a, s 1 s a = s 1 a s a, at 1 t = at 1 at Closure under multiplication on the right by an element r R follows from nar = anr, atr = atr, and s i at i r = s i at i r Closure under left multiplication is similar b If R has an identity then we can write na as n1a, and sa = sa1, at = 1at So every term in the sum is of the form s i at i c If a is central then we can replace each term s i at i by as i t i, and ny an1; and then as i t i = a s i t i d In this case, a is the set of elements of the form na + ar for n Z, a R The proof is over to you 39 a 1 e = 1 e + e = 1 e + e = 1 e, so 1 e is an idempotent b These elements are clearly idempotents To show that 1, 0 is central, observe that 1,0r,s = r,0 = r,s1,0 c Note that er and 1 er are ideals of R Define ijection θ from R to er 1 er by the rule rθ = er,1 er It is straightforward to show that θ is a homomorphism If r Kerθ, then er = 1 er = 0; adding, we see that r = 0 So θ is injective Now take es er and 1 et 1 er Let r = es + 1 et Then er = e s + e1 et = es, 1 er = 1 ees + 1 e t = 1 et, so rθ = es,1 et Thus θ is onto, and is an isomorphism 41 Suppose that r n = 1 Then 1 + r1 r + r + 1 n 1 r n 1 = n 1 r n = 1, so 1 + r is a unit we have found its inverse 8

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