9 Solutions for Section 2

Transcription

1 9 Solutions for Section 2 Exercise 2.1 Show that isomorphism is an equivalence relation on rings. (Of course, first you ll need to recall what is meant by an equivalence relation. Solution Most of this was done in class. After noting that, for any ring R, the identity function on R is an isomorphism from R to itself (hence reflexivity of isomorphism, we have to show that if R is isomorphic to S, say θ : R S is an isomorphism then there is an isomorphism from S to R (symmetry; also that if R is isomorphic to S, say θ does the job, and S is isomorphic to the ring T, say ψ : S T is an isomorphism, then R is isomorphic to T (transitvity. For the symmetry, show that the inverse, θ 1 : S R, of θ, defined by θ 1 (s = r iff s = θ(r is an isomorphism. For transitivity, show that the composition, ψθ : R T, defined by (ψθ(r = ψ(θ(r is an isomorphism. + ( 1 0 Exercise 2.5 Show that the ring of 2 2 matrices of the form a ( ( ( 0 1 i i b + c + d with a, b,c, d R (and i a square 1 i i 0 root of 1 is isomorphic to the ring of quaternions. Proof: Let R denote the ring of matrices ( of the above form. Define the map a + ci b + di θ : H R by θ(a + bi + cj + dk =. Showing that this b + di a ci is 1-1 and onto is just as in Example 2.4 (which was done in lectures. Clearly θ(1 is the identity matrix. Take a + bi + cj + dk, a + b i + c j + d k H. To check that θ preserves addition we have to show that θ ( (a + bi + cj + dk + (a + b i + c j + d k = θ(a + bi + cj + dk + θ(a + b i + c j + d k, which is, when you begin to write it down, obvious. To check multiplication we have to show that θ ( (a+bi+cj +dk (a +b i+ c j+d k = θ(a+bi+cj+dk θ(a +b i+c j+d k. The LHS equals θ ( (aa bb cc dd +(ab +a b+cd c di+(ac +a c+b d bd j+(ad +a d+bc b ck ( = (aa bb cc dd + (ac + a c + b d bd i (ab + a b + cd c d + (ad + a d + bc b ci (ab + a b + cd c d + (ad + a d + bc b ci (aa bb cc dd (ac + a c + b d bd i which, if you mul- ( ( a + ci b + di a + ci b + di The RHS equals b + di a ci b + di a ci tiply it out, gives exactly the matrix above. Exercise 2.7 Show that all rings with three elements are isomorphic. Solution Suppose that R is a ring with three elements. Two of these elements must be 0,1; denote the third by a. It will be enough to show that there is no choice in drawing up the addition and multiplication tables of R, for then, if S is another ring with three elements, 0,1, b say, then the map taking 0 to 0, 1 to 1 and a to b must preserve addition and multiplication (since there is no choice over these, so must be an isomorphism. Addition: what is 1 + 1? It can t be 1 because from = 1 we d get the contradiction 1 = 0. Could = 0? No, because by the kind of cancelling argument just used, a + 1 cannot equal a or 1 hence a + 1 = 0, but then we d have 1+1 = a+1 so a = 1 contradiction. So we must have 1+1 = a, and we ve also noted that a+1 = 0, that is = 0 hence the ring is of characteristic. 48

2 3 (actually you could also have got to this point by using Lagrange s Theorem. It follows that a + a = = 1. So now we have the complete addition table. Multiplication: since 0 times anything is 0 and 1 times anything is that thing, the only question to be answered is what is a a? But a = 2 and the ring has characteristic 3, so a a = 1. There was no choice anywhere, so we have proved that, up to isomorphism, there is just one ring with three elements (familiar as Z 3, the integers modulo 3. Example 2.8 There are rings with four elements which are not isomorphic. In fact, the tables in Examples 1.3 do all give different rings (indeed, there is even one more. Comment The fourth ring is Z 2 Z 2, the product of Z 2 with itself. You can see from the 1s on the diagonal of the multiplication table of this ring that it is not isomorphic to any of the other three. Exercise 2.10 Give an example to show that if θ : R S is a homomorphism and if r R is such that θ(r is invertible in S, it need not be the case that r is invertible in R. Solution For instance, take the canonical homomorphism Z Z 5 which takes a to [a] 5. Then 2 Z is not invertible but its image, [2] 5 is (its inverse is [3] 5. Exercise 2.12 Prove that there is no homomorphism from Z n to Z l if l is not an integer factor of n. [Hint, make use of the notion of characteristic] Proof: Suppose there is such a homomorphism θ. Since [1] n n = 0 we must have θ([1] n n = 0, that is, [1] l n = 0. But the characteristic of Z l is l so, see Lemma 1.5(2, it must be that l divides n, as required. Exercise 2.15 Show that if θ : R S and β : S T are homomorphisms such that the composition βθ : R T is an embedding then it need not be the case that β is an embedding. Solution There are many possible examples. For instance take θ : Q Q[X] to be the embedding which takes a rational r Q to that same rational, regarded as a constant polynomial. And take β : Q[X] Q to be evaluation (of a polynomial at, say 0. The composition simply returns r Q to r, so certainly is an embedding. But β is not: it takes any polynomial, such as X, with zero constant term to 0. Exercise 2.20 Show that the map from Z[ 2] to itself given by sending a+b 2 to a b 2 is an automorphism. Show that there are no more automorphisms of Z[ 2] apart from the identity map. Solution First we show that this map, θ say, is a homomorphism. Clearly θ(1 = 1. Also θ ( (a+b 2(c+d 2 = θ ( (ac+2bd+(ad+bc 2 = (ac+2bd (ad + bc 2. On the other hand θ(a + b 2θ(c + d 2 = (a b 2(c d 2 = (ac + 2bd (ad + bc 2. So θ preserves multiplication and it is even easier to show that it preserves addition, hence is a homomorphism. Certainly θ is onto: given a typical element a + b 2 of the ring, this is the image of a b 2 under θ. Also θ is 1-1: directly or since θ 2 is the identity map (exercise: make sure you see why this observation is enough. To see that there are no more automorphisms: suppose that β : Z[ 2] 49

3 Z[ 2] is an automorphism of Z[ 2]. Since β(1 = 1 it must also be that, for any positive integer n = (n 1s we have β(n = β( = β(1 + + β(1 = = n. Since β( n = β(n and since β(0 = 0 we deduce that β(n = n for every n Z. So β(a + b 2 = a + bβ( 2 and it remains to determine what β( 2 can be. We have β( 2 2 = β(( 2 2 = β(2 = 2 so β( 2 must be a square root of 2. Since Z[ 2] is contained in the field of complex numbers and since the only square roots of 2 in C are ± 2 it follows that β( 2 = 2 - in which case β is the identity map - or β( 2 = 2 - in which case β is the map θ above. Therefore there are just these two automorphisms of Z[ 2]. Exercise 2.24 Give an example of a ring R and a subset H of R which contains 0 and is closed under addition and multiplication but which is not an ideal of R. Solution For instance, take R = Z and H = N. Exercise 2.27 Let ( R = M 2 (Z be the ring of 2 2 matrices with integer 1 0 entries. Let a =. Compute the right ideal generated by a, the left ideal generated by a and the (two-sided ideal generated by a. ( ( ( 1 0 a b a b Solution = so the right ideal generated by c d ( ( 1 0 a b a = is the set of all matrices of the form. ( ( ( a b 1 0 a 0 Also = so the left ideal generated by a c d ( c 0 a 0 consists of all matrices of the form. c 0 The elements ( of the ( (two-sided ( ideal generated by a will be sums of terms a b 1 0 a b of the form c d c d. So it is necessary to do a bit of experimentation to see what can be got in this way. You might find: ( ( ( ( ( 1 0. So, since both and are in this ideal, so is ( 1 0. But any ideal which contains the identity element must equal the ( 1 0 whole ring. So we conclude that the two-sided ideal generated by is all of M 2 (Z. Exercise 2.29 Show that if R is any ring then R is a division ring iff the only right ideals of R are {0} and R (and equivalently for left ideals. It is the case (and showing this is a substantial exercise that the first Weyl algebra, defined at Example 1.31, is a (non-commutative ring which has no ideals except {0} and R yet which is not a division ring. So the analogue of 2.28 for non-commutative rings must refer to one-sided ideals: the condition on two-sided ideals is not enough to give a division ring. Proof: Suppose that R is a division ring and let I be any non-zero ideal of R. ( = 50

4 Choose a I with a 0. Since every non-zero element of R is invertible we have 1 = aa 1 I and hence I = R (for then, if r R we have r = 1r = a(a 1 r I. Conversely if there are no right ideals except {0} and R, take a R with a 0. Certainly the right ideal, ar, generated by a is non-zero (it contains a so, by assumption it must equal R. In particular 1 belongs to this ideal, so there is r R with ar = 1. Hence a has a right inverse. Of course the same applies to r (since certainly r 0, say rs = 1. Then we have a = a1 = ars = 1s = s. Thus r is also a left inverse for a and hence every non-zero element of R is right and left invertible, that is, R is a division ring. (For the comment, Exercise 2.12 on the More Examples sheet goes more than half-way towards suggesting a proof of the fact that the first Weyl algebra has no two-sided ideals apart from {0} and the whole ring. Exercise 2.30 Show that if A is a subset of a ring R then A is the smallest ideal of R which contains A. Proof: Any ideal containing A must be closed under right and left multiplication and under addition, hence every element of the form r 1 a 1 s r n a n s n with the a i A and the r i, s i R must be in every ideal containing A. On the other hand, it is pretty clear that the set of such elements is closed under right and left multiplication and under addition, hence forms an ideal. Thus the set of these elements is indeed the smallest ideal containing A. Example/Exercise 2.31 Let R = Z[X] and let I = 2, X be the ideal of Z[X] generated by {2, X}. Prove that I is not a principal ideal. [Hint: suppose, for a contradiction, that it is, choose a generator,...] Proof: Suppose, for a contradiction, that this ideal were principal, say is generated by the polynomial p. Since 2 p it must be that 2 = pq for some polynomial q. Comparing degrees it follows that p must have degree 0, that is, p must be a constant polynomial, that is, an integer, k say. But also X must be a multiple, X = kq of p = k. Comparing leading coefficients, we deduce that k divides 1, hence that k = ±1. Whichever, it follows that the ideal generated by p contains 1 (hence is the whole of Z[X]. But, we claim, the ideal generated by {2, X} does not contain 1 for, if it did, we would have 2q +Xq = 1 for some polynomials q, q. Substitute X = 0 on both sides to deduce 2q(2 = 1, hence that 1 is an integer multiple of 2, which it is not, the required contradiction. Exercise 2.32 The ideal 4,6 of Z generated by 4 and 6 together is principal - find a generator for it. Generalise this. Solution 2 is a generator: certainly 2 = 6 4 is in the ideal 4,6. Conversely, every element of that ideal must be divisible by 2 (since each of the generators is, that is, lies in the ideal generated by 2. More generally, a, b = gcd(a, b. For the gcd has the form ak+bl for some integers k,l, hence is in the ideal generated by a and b, thus gcd(a, b a, b and, conversely, each of a and b is a multiple of gcd(a, b, hence is in the ideal generated by gcd(a, b, hence a, b gcd(a, b and so these ideals are equal. Exercise 2.36 Let R = Z and let I = 3. What is I 2? What is I n? Let J = 12. What is IJ? Solution I 2 = 9 ; I n = 3 n ; IJ = 36. Exercise 2.37 Let I, J be ideals of a ring R. Show that IJ I J. Give an 51

5 example to show that this inclusion may be proper. Solution The typical element of IJ is a sum of terms of the form ab with a I and b J. Note that any such element ab is both in I and in J, hence is in I J. Hence any sum of such elements is in I J. That is, IJ I J. Look to the previous exercise for an example: take I = 3 and J = 12, then IJ = 36, whereas I J = 12 (note that Exercise 2.38 Let R = Z. Show that = Z. Compute 2 5 and the product 2 5, writing each of these as a principal ideal, that is, in the form n for some integer n. Now replace 2 and 5 by arbitrary integers a and b. What are a + b, a b and a b? [Hint: unless you ve guessed the answers this might be difficult to answer, let alone prove, so try with some other pairs of integers in place of 2 and 5. In other words, explore until you see what s happening.] Compute ( 2 : 5 and ( 6 : 8. What about the general case ( a : b? Solution Note that 1 = = 2,5. Both 2 5 and 2 5 equal 10. More generally a, b = gcd(a, b, a b = hcf(a, b and a b = ab. These can be proved by, for example, considering the prime factorisations of a and b. Finally ( 2 : 5 = 2 and ( 6 : 8 = 3 and, in general, ( a : b = a gcd(a,b (since a. a gcd(a,b is the smallest integer which turns b into a multiple of Exercise 2.39 Find all ring homomorphisms from Z Z to itself. Solution Set e 1 = (1,0 and e 2 = (0,1. If θ is a homomorphism from Z Z to itself then, once we know the values of θ on these idempotents, we know θ (for a typical element of Z Z is (m, n = me 1 +ne 2 so θ(m, n = mθ(e 1 +nθ(e 2. Any homomorphism preserves idempotents and the only idempotents of Z Z are 0, 1 = (1,1, e 1 and e 2, so each of θ(e 1, θ(e 2 has to be one of these four. But since e 1 + e 2 = 1 we must have θ(e 1 + θ(e 2 = 1. So let s look at the possibilities. If θ(e 1 = e 1 then θ(e 2 = 1 e 1 = e 2 so this is the identity map. If θ(e 1 = e 2 then θ(e 2 = 1 e 2 = e 1 so this is the map which switches the two copies of Z. If θ(e 1 = 1 then θ(e 2 = 1 1 = 0 so this is the map which takes (m, n to (m, m and it s easy to check that this is indeed a homomorphism. If θ(e 1 = 0 then θ(e 2 = 1 0 = 1 so this is the homomorphism (symmetric to the previous one which takes (m, n to (n, n. { ( a 2b Exercise 2.40 Let R = Z[ 2] and let S = map θ : R S by θ(a + b ( a 2b 2 = b a : a, b Z }. Define the b a. Prove that θ is an isomorphism. Solution First we show that θ is a homomorphism. So choose any a+b 2, c+ d 2 R. Then θ(a + b 2 + θ(c + d ( ( a 2b c 2d 2 = + = b a d c ( a + c 2b + 2d and θ((a + b 2 + (c + d 2 = θ((a + c + (b + d 2 = b + d a + c 52

6 ( a + c 2(b + d. Thus θ(a+b 2+θ(c+d 2 = θ((a+b 2+(c+d 2, b + d a + c so θ does preserve addition. Similarly compute θ((a + b 2 (c + d 2 = θ((ac + 2bd + (ad + bc 2 = = θ(a + b 2 θ(c + d 2. Furthermore θ(1 = θ(1+0 b is, by definition of θ, the identity matrix. All this shows that θ is a ring homomorphism. Also θ is onto (=surjective since the matrix in S with parameters a, b is the image under θ of the element a + b 2 in R. And θ is 1-1 since if θ(a + b 2 = θ(c+d ( ( a 2b c 2d 2 then, by definition of θ, = so (by definition of b a d c what it means for matrices to be equal a = c and b = d, hence a+b 2 = c+d 2. (You might also be careful and note that θ is well-defined since each member of R can be written in only one way in the form a + b 2. Exercise 2.41 Prove that Z 6 Z 2 Z 3. Is Z 10 Z 2 Z 5? Is Z 8 Z 2 Z 4? Justify your answers. Solution To show that the rings are isomorphic we produce a map from one to the other and show that it is an isomorphism. So you have a choice, define a map from Z 6 to Z 2 Z 3 or define a map from Z 2 Z 3 to Z 6. Let s do the first (at the end I give a map in the other direction. Define the map θ from Z 6 to Z 2 Z 3 by θ(a = (a 2, a 3 for a Z 6 : here by a n I mean the reduction of a modulo n. (Thus θ(0 = (0,0, θ(1 = (1,1, θ(2 = (0, 2, θ(3 = (1, 0, θ(4 = (0, 1, θ(5 = (1, 2 is a complete definition of θ. Each of the reduction modulo n maps from Z 6 to Z n (n = 2,3 is a homomorphism (by the notes so it follows directly (I won t write out the details but you can that this map θ is a homomorphism. It s clear from the above explicit list that θ is a bijection, hence an isomorphism. However, that last bit is not a good proof because something general is going in and the fact that we can see it s a bijection directly (because the numbers involved are small hides this. So let s see what the general point is, in the next paragraph. Suppose we start with Z m and we have two factors, l,n of m. Then we can define a map θ : Z m Z l Z n by θ(a = (a l, a n. Because both l and n divide m, this map is well-defined and it s a homomorphism for the same general reason as in the previous paragraph. Let s see if it s injective. So suppose θ(a = θ(b, that is, (a l, a n = (b l, b n, that is, a b mod l and a b mod n (here I m reading a and b as being in Z. Now, it s a fact from basic number theory (not so difficult to prove that what follows from this is that a b mod lcm{l,n}. So in order to conclude that a = b, and hence that θ is injective, we need to have that m is the least common multiple of l and n. What about surjectivity? In order that θ be surjective, for each pair (c, d Z l Z n there should be some integer a (modulo m such that a c mod n and a d mod n. A result from number theory, called the Chinese Remainder Theorem says that this is possible iff l and n are coprime (that is, have no common factor greater than 1. So the conclusion, in the general case, is that the natural homomorphism from Z m to Z l Z n is an isomorphism if l and n are coprime and ln = m (note that, since they are coprime, ln is the lowest common multiple of l and n. [In the original case a map η from Z 2 Z 3 to Z 6 which does the job is given by η(b, c = 3b + 2c mod6. You can check that this is a homomorphism and is a bijection (alternatively you can check that it is the inverse of θ.] By the general considerations a couple of paragraphs above, we do have 53

7 Z 10 Z 2 Z 5 but we don t have Z 8 Z 2 Z 4 (because 2 and 4 aren t relatively prime. Well, for the latter, that s not really a proof because, although the natural map is not an isomorphism, there could be some other map which is. But note that the characteristic of Z 8 is 8, whereas the characteristic of Z 2 Z 4 is 4, so they can t be isomorphic rings. Exercise 2.42 Prove that if θ : R S is a surjective homomorphism of rings and R is commutative then S must be commutative. Give an example to show that the conclusion may be false if θ is not surjective. Solution Let s, s S (we must show that ss = s s. Since θ is surjective there are r, r R such that θ(r = s and θ(r = s. Since R is commutative rr = r r, so θ(rr = θ(r r. Since θ is a homomorphism, the latter equation gives θ(rθ(r = θ(r θ(r, that is, ss = s s, as required. For an example, choose your favourite non-commutative ring for S and take R = Z and let θ be the map which sends n Z to n S (meaning n 1 S s added together. (For a specific example take S = M 2 (Z so then θ is a map n ( n 0 0 n. Exercise 2.43 Let R be the ring of all infinitely differentiable functions from the unit interval [0, 1] to R (with pointwise addition and multiplication as the operations. Let D : R R be the map which takes a function f to its derivative f. Is D a ring homomorphism? Solution No, although it does preserve addition (D(f + g = D(f + D(g it fails to preserve multiplication (D(fg equals D(fg + fd(g, not D(fD(g (and also does not take 1 to 1: the identity element of R is the constant function which maps everything to 1 R and this has derivative the zero function. Exercise 2.44 Suppose that R is a commutative ring of characteristic 3. Prove that the map θ : R R defined by θ(r = r 3 is a homomorphism. Suppose also that R has no non-zero nilpotent elements: prove that θ is injective. Solution Check: θ preserves 1 since 1 3 = 1; θ preserves multiplication since θ(rθ(s = r 3 s 3 = (rs 3 (since R is commutative = θ(rs; θ preserves addition since θ(r + s = (r + s 3 = r 3 + 3r 2 s + 3rs 2 + s 3 (again using that R is commutative, = r 3 + s 3 (since 3 = 0 in R, = θ(r + θ(s. An element r R is in the kernel of θ, i.e. θ(r = 0, iff r 3 = 0 so, if there are no non-zero nilpotent elements, then this implies r = 0, hence ker(θ = 0, hence θ is injective. Exercise 2.45 Show that if I is a right ideal of a ring R and if there is an element a I with a right inverse then I = R. Solution Say ar = 1. Since a I and r R we have, by definition of right ideal, that 1 I. But then, if s R we have s = 1 s I, so every element of R is in I, as required. Exercise 2.46 Let R be a commutative ring and set N(R = {r R : n, r n = 0} to be the set of all nilpotent elements of R. Prove that N(R is an ideal of R. Compute N(R for: (i R = Z 12 ; (ii R = Z 24 ; 54

8 { ( a b (iii R = : a, b,c Q }. 0 c Solution First, 0 clearly is in N(R. Next, suppose r N(R, say r n = 0, and let s R. Then, since R is commutative, (rs n = r n s n = 0, so rs N(R. Finally, if r, s N(R, say r n = 0 and s m = 0 then consider (r + s n+m. If this is expanded using the binomial theorem (valid since R is commutative then each term has the form a i r i s n+m i and so either i n or n + m i m: either way the term equals 0. So since (r + s n+m = 0 and hence r + s N(R. Therefore N(R is indeed an ideal of R. (i You can just take the elements in turn and see whether some power is 0 (they soon cycle so this doesn t take long: obviously 0 N(R; clearly neither 1 nor 1 = 11 is in N(R; the powers of 2 go 2,4,8,4,... so neither it nor 2 = 10 has 0 as a power, et cetera. You will find that 6 is nilpotent but that s it, so N(R = {0,6}. (ii A similar check will give N(R = {0,6,12,18}. Alternatively, if you think about what s happening, you might realise that the condition for a Z n to be nilpotent is that every prime factor of n is also a factor of a (then some high enough power of a will contain enough powers of each prime to have n as a factor, that is to be 0 modulo n and, on the other hand, if some prime is missing then no power of a will ever be divisible by n. ( a b (iii Computing successive powers of a generic matrix you will see 0 c (and can prove by induction that the diagonal entries are powers of a and c so the only way these can ever be 0 is if both a and c are 0. But ( any matrix with 0 b a = 0 = c has square 0 so is nilpotent. Therefore N(R = { : b Q}. Exercise 2.47 Prove that every non-zero ideal of the ring Z[i] contains a nonzero integer. Solution Let a + bi I be a non-zero element of an ideal I. Since a bi R and I is an ideal (a + bi(a bi = a 2 + b 2 I, so a 2 + b 2 is a non-zero integer in I. Exercise 2.48 Determine whether or not each of the following statements about ideals in the polynomial ring Q[X, Y ] is true: (i XY + Y 2, X 2 + XY + Y 2, X 2 + Y 2 = X 2, XY, Y 2 ; (ii X 2 + XY, XY Y 2, X 2 + Y 2 = X 2, XY, Y 2. Solution (i Since each of XY + Y 2, X 2 + XY + Y 2 and X 2 + Y 2 is a linear combination of X 2, XY and Y 2 certainly XY +Y 2, X 2 +XY +Y 2, X 2 +Y 2 X 2, XY, Y 2. To show the converse we have to produce each of X 2, XY and Y 2 as a linear combination of XY + Y 2, X 2 + XY + Y 2 and X 2 + Y 2. We have: XY = (X 2 + XY + Y 2 (X 2 + Y 2 ; Y 2 = (XY + Y 2 XY = (XY + Y 2 ((X 2 + XY + Y 2 (X 2 + Y 2 ; X 2 = (X 2 +Y 2 Y 2 = (X 2 +Y 2 ((XY +Y 2 ((X 2 +XY +Y 2 (X 2 +Y 2 ; X 2 = (X 2 + Y 2 Y 2, which you can simplify if you want, but already the question is answered: XY + Y 2, X 2 + XY + Y 2, X 2 + Y 2 = X 2, XY, Y 2. (ii As in the first part, we have X 2 +XY, XY Y 2, X 2 +Y 2 X 2, XY, Y 2. Notice that X 2 + Y 2 = (X 2 + XY (XY Y 2 so the generator X 2 + Y 2 is not actually needed and hence X 2 + XY, XY Y 2, X 2 + Y 2 = X 2 + XY, XY Y 2. 55

9 Seeing that, you might suspect that this ideal is really smaller than X 2, XY, Y 2 : how could these three independent homogeneous quadratic polynomials be generated using just two homogeneous quadratic polynomials? To prove that they can t we can take the idea of independence seriously. Consider the Q-vectorspace consisting of all homogeneous quadratic polynomials in X and Y : {rx 2 + sxy + ty 2 : r, s,t Q}. Observe that this is indeed a vector space and {X 2, XY,Y 2 } is one basis (in part (i we showed that {XY + Y 2, X 2 + XY + Y 2, X 2 + Y 2 } is another basis. So this vector space is 3-dimensional. In particular it is impossible to generate all of X 2, XY and Y 2 as linear combinations of the form u(x 2 + XY + v(xy Y 2 with u, v Q. We re not quite finished yet, however, since the question is whether we can generate all of X 2, XY and Y 2 as linear combinations of the form p (X 2 + XY + q (XY Y 2 with p, q polynomials in Q[X]. But note that the degree 2 homogeneous part of any polynomial of the form p (X 2 + XY + q (XY Y 2 is just the constant term of the polynomial p times X 2 + XY plus the constant term of q times XY Y 2, and that is what we ve covered with the linear algebra. So the conclusion is that it is impossible to obtain all of X 2, XY and Y 2 in the form p (X 2 + XY + q (XY Y 2, hence X 2 + XY, XY Y 2, X 2 + Y 2 X 2, XY, Y 2. Exercise 2.49 Suppose that I, J are ideals of the commutative ring R such that I + J = R. Prove that IJ = I J. Solution Since I and J are (two-sided ideals IJ is a subset of both I and J hence IJ I J. We have to show the converse, so let r I J. Since this converse is not true in general we must use the additional information that I +J = R, in particular that there is a I and b J such that a+b = 1. Then r = r1 = ra + rb. Now, ra JI and rb IJ. Since R is commutative JI = IJ so ra IJ, hence r IJ + IJ = IJ (this equality directly from the definition of IJ, as required. Exercise 2.50 Let R = R X,Y : Y X = XY + 1 be the first Weyl algebra (as defined in the notes. Use the defining relation Y X = XY + 1 to write each of the following elements of R as a sum of monomials of the form X i Y j (i, j 0: Y X 2 ; Y 2 X 2 ; Y X 3 ; Y X 2 X 2 Y ; Y X 3 X 3 Y. Prove that, for every n N, the ideal generated by X n is equal to R. Solution Y X 2 = (Y XX = (XY + 1X = XY X + X = X(XY X = X 2 Y + 2X; Y 2 X 2 = Y (Y X 2 = Y (X 2 Y + 2X = (XY + 1XY + 2(XY + 1 = XY XY + 3XY + 2 = (X(XY + 1Y + 3XY + 2 = X 2 Y 2 + 4XY + 2; Y X 3 = (Y X 2 X = (X 2 Y + 2XX = X 2 (XY X 2 = X 3 Y + 3X 2 ; Y X 2 X 2 Y = X 2 Y + 2X X 2 Y = 2X; Y X 3 X 3 Y = X 3 Y + 3X 2 X 3 Y = 3X 2. The second part depends on your noting that the last two equations can be seen as Y X n X n Y = nx n 1 for n = 2,3. This is generally true and is proved by induction (the induction step is: assuming Y X n X n Y = nx n 1, we have Y X n+1 X n+1 Y = (Y X n X n Y X + X n Y X X n+1 Y = ((n 1X n X + X n (XY + 1 X n+1 Y (using the induction assumption = (n 1X n+1 + X n+1 Y + X n X n+1 Y = nx n+1. Next, note that Y X n X n Y is in the ideal generated by X n, that is, nx n 1 is in the ideal X n. Since 1 n R this means that X n 1 X n. Therefore X n 2 X n 1 also is true and so 56

10 X n 2 X n. And so on (i.e. by induction, to conclude that 1 = X 0 X n. Exercise 2.51 Let R = R[X, Y ] and let S R 2 be any subset of the real plane. Set I(S = {p R : (r, s S, p(r, s = 0}. Prove that I(S is an ideal of R. [Note that this is the kernel of the two-dimensional version of the evaluation map for polynomials which is defined in the printed notes.] Solution Suppose that p, q I(S. Let (r, s S. Then (p+q(r, s = p(r, s+ q(r, s = 0. So p + q I(S. Let f R. Then (pf(r, s = p(r, sf(r, s = 0 f(r, s = 0 so pf(= fp I(S. Also the zero polynomial is in I(S. Hence I(S is an ideal of R. Exercise 2.52 Prove that if θ : R S is a surjective homomorphism of rings and I is a right ideal of R then θ(i is a right ideal of S. Give an example to show that the conclusion may be false if θ is not surjective. Solution Since 0 I we have 0 S = θ(0 R θ(i. Also if s, s θ(i, say s = θ(r and s = θ(s then s + s = θ(r + θ(r = θ(r + r θ(i since r + r I. Let s θ(i, say s = θ(r with r I. Also let t S. Since θ is surjective there is u R with θ(u = t. Then st = θ(rθ(u = θ(ru (since θ is a homomorphism. Since r I and I is a right ideal of R, we have ru I, hence θ(ru θ(i. But θ(ru = θ(rθ(u = st, so st I. Thus θ(i contains the zero element of S, is closed under addition and is closed under multiplication by all elements of S, so is a right ideal of S, as required. The part of the argument that seems to need θ to be surjective is where we took t S and could say that there is u R with θ(r = u. That observation can help in trying to find an example (to show failure without the surjectivity assumption. Here s one: take R = Z, I = R, S = M 2 ((Z and θ : R S the n 0 map which takes an integer n to the diagonal matrix. The image 0 n of R consists of all diagonal matrices with equal diagonal entries but this is not a right ideal: even though it contains the zero matrix and is closed under addition, it is not closed under multiplication by arbitrary matrices, for instance take t = ( Exercise 2.53 Let a be an element of a ring R. Show that the right annihilator of a, rann(a = {r R : ar = 0} is a right ideal of R. If also b R show that rann(a + b rann(a rann(b. Show that ( the inclusion can be proper. 2 3 Compute rann(a where R = M 2 (Z 6 and a =. Show that rann(a need not be a left ideal of R. Solution Certainly 0 rann(a. If r, s rann(a then ar = 0 = as so a(r+s = 0 hence r + s rann(a. Finally, if r rann(a, so ar = 0 and if t R then a(rt = (art = 0t = 0 so rt rann(a. Thus rann(a is a right ideal of R. Next, suppose r rann(a rann(b, so ar = 0 = br. Then (a + br = 0 so r rann(a + b, showing that rann(a rann(b rann(a + b. To see that the inclusion can be proper take, say b = a: certainly rann( a = rann(a so the intersection is just rann(a, but rann(a + b = rann(0 = R which can be larger than rann(a (e.g. take R = Z, a = 1. Now we compute the right annihilator of ( 2 3 ( 2 3. We have ( a b c d = 57

11 ( 2a 2b + 3d so this is zero iff 2a = 0 and 2b = 3d (note 3 = 3 in Z 6. So a = 0 or 3 and there is no restriction on c. The values of (b, d satisfying 2b = 3d are (0,0, (0,2, (0,4, (3,0, (3,2, (3,4. Thus rann ( ( 2 3 is ( a b the set of matrices where c can take any value, a is either 0 or 3 and c d the pair (b, d takes one of the six values above. Finally, to see that ( rann(a need not be( a left ideal, ( take (for( instance, r s t u R = M 2 (Z and let a =. We have = ( { ( t u r s r s so rann(a iff t = 0 = u, that is, rann(a = : r, s Z }. t u ( But this set is not a left ideal since, for instance is in the set but ( ( ( = is not. 1 1 Exercise 2.54 Let p be a prime integer and set Z (p = { m n Q : p n}. Prove that Z (p is a subring of Z. Show that the only proper non-trivial ideals of Z (p are those generated by powers of p. Solution Certainly Z (p contains 1 = 1/1. If r = m n and s = k l are in Z (p then (assuming that these fractions are written in simplest form, hence with numerator and denominator having no non-trivial factor p n and p l. Now r s = m n k l = ml nk nl and, since p is prime, p nl, so r s Z (p. Moreover, rs = mk nl and p nl so rs Z (p. Thus Z (p is a subring of Q. To find the ideals, take any non-zero r Z (p. Write r = p u m n where p is not a factor of m or n. Note that n m Z (p, so the ideal generated by r contains the ideal generated by p u = r n m, indeed the ideal generated by r is equal to that generated by p u since each of these is an invertible element times the other. It follows that the cyclic ideals of Z (p are Z (p > p > p 2 > >...{0}. It follows that every ideal is cyclic (since, given two elements, one is already in the ideal generated by the other. Therefore there are no more ideals other than those just listed. 58