Exercises on chapter 3

Transcription

1 Exercises on chapter 3 1. If M is a left R-module and X M, the annihilator of X, Ann R (X), denotes the set of all elements of R that act as zero on all elements of X, i.e. Ann R (X) = {a R ax = 0 for all x X}. (i) Check that Ann R (X) is always a left ideal of R, and that if in fact X is a submodule of M then Ann R (X) is a two-sided ideal of R. (ii) If M is a cyclic R-module, say M = Rx for x M, then M = R/Ann R (x) as R-modules. (iii) Prove that every non-zero cyclic module M = Rx has a maximal submodule (a proper submodule that is maximal amongst all proper submodules of M). (iv) Prove that every finitely generated left R-module M has a maximal submodule. (Hint: Let x 1,..., x n be a minimal spanning set for M and let K = Rx Rx n, noting that x 1 / K.) (v) Prove that an R module M is irreducible if and only if M = R/I for a maximal left ideal I of R. (i) You can check this. (ii) Define a map R M, r rx. It is an R-module homomorphism with kernel Ann R(x). Hence by the first isomorphism theorem you get induced an isomorphism of left R-modules R/Ann R(x) = M. (iii) By (ii), this follows from the fact that every left ideal of R can be embedded into a maximal left ideal. We proved almost the same thing in class for two-sided ideals using Zorn s lemma last term... the same argument works for this. (iv) As in the hint, we just need to show that M/K has a maximal submodule. Then by the lattice isomorphism theorem, its pre-image in M is a maximal submodule of M. But that follows by (iii) since M/K is cyclic, generated by the image of x 1. (v) Note R/I is irreducible if and only if I is a maximal left ideal of R, by the lattice isomorphism theorem. This proves one implication easily. Conversely, suppose that M is irreducible. Take 0 x M. Then Rx is a non-zero R-submodule of M, hence Rx = M by irreducibility. Hence M = R/I for some left ideal of R, necessarily maximal as M is irreducible. 2. Let I be a two-sided ideal of a ring R and let M be a left R-module. Explain a natural way to make M/IM into a left R/I-module. Define an action of r + I R/I on M/IM by (r + I)(m + IM) = rm + IM. We need to check this is well-defined. Say r + I = r + I and m + IM = m + IM, i.e. (r r ) I and (m m ) IM. Then (r + I)(m + IM) = r m + IM. So we need to see that r m rm IM. But r m rm = (r r)m + r(m m) and the right hand side belongs to IM. It is clear that this makes M/IM into a left R/I-module. 15

2 16 3. The goal of this question is to prove directly that Z 60 = Z 4 Z 3 Z 5 = Z 4 Z 3 Z 5. Of course we know this already by the Chinese remainder theorem since products and coproducts are the same thing in the category of abelian groups... (i) Find submodules M 4, M 3 and M 5 of Z 60 isomorphic to Z 4, Z 3 and Z 5 respectively with Z 60 = M 4 M 3 M 5. (ii) For each i = 4, 3, 5, write down the maps p i : Z 60 Z i so that Z 60 together with these maps is the direct product of Z 4, Z 3 and Z 5. (i) Let M 4 = Z15, M 3 = Z20 and M 5 = Z12. Note M 4 = Z4, M 3 = Z3 and M 5 = Z5. Note M 3+M 4 = Z5 (because 5 is the GCD of 15 and 20), hence since 5 and 12 are coprime M 5 (M 3 + M 4) = 0. Similarly M 3 (M 5 + M 4) = M 4 (M 3 + M 5) = 0. Hence the sum M 4 + M 3 + M 5 is direct. It is all of Z 60 because GCD(15, 20, 12) = 1 so you can write 1 = a15 + b20 + c12 for some a, b, c Z. (ii) Define p i by n n + Zi, i.e. reduce modulo i, for each i = 4, 3, 5. Since Z 60 = Z/Z60 it follows by the second isomorphism theorem that for i 60, Z 60/iZ 60 = Z/Zi. So this really does the job. To see that it is the direct product you need to check the universal property. Take maps q i from an abelian group A to Z 4, Z 3 and Z 5. We need to construct a map from A to Z Well, a A maps to the unique element of Z 60 that is congruent to q i(a) modulo i for each i = 4, 3, Let {M i } i I be a family of left R-modules and M be another left R-module. (i) Prove that Hom R ( i I M i, M) = i I Hom R (M i, M). (ii) Prove that (iii) Is it true that Hom R (M, i I How about if M is finitely generated? M i ) = Hom R (M, M i ). i I Hom R (M, M i ) = Hom R (M, M i )? i I (i) Suppose we are given an element of Q i I HomR(Mi, M). This means we are given maps fi : M i M for each i. Define a map f : L i I Mi M by f(p i mi) := P i fi(mi). Note this depends Q on all but finitely many m i s being zero to make sense... In this way we get a map from i I HomR(Mi, M) to HomR(L i I Mi, M). It is easy to see that it is actually a homomorphism of abelian groups (i.e. better than just a map of sets). Conversely given f Hom L R( i I Mi, M) define (fi)i I Q i I HomR(Mi, M) by setting fi(mi) := f(m i) for each i I. This is a two-sided inverse to the map in the first paragraph. Note here it is NOT true that Hom M R( M i, M) M = Hom R(M i, M) i I if I is infinite. Take the case R is a field F, and let M = M i = F (the one dimensional vector space) and I = N (or any countable set). The left hand side is then the dual vector space F I of a countable dimensional vector space, which is of uncountable dimension. The right hand side is just the vector space F I again, of countable dimension. So they cannot be isomorphic... i I (ii) Suppose we are given an element f Hom Q R(M, i I Mi). Define (fi)i I Q i I HomR(M, Mi) by letting f i(m) = π i(f(m)), the ith coordinate of f(m). Conversely given (f i) i I Q i I HomR(M, Mi) define f Hom Q R(M, i I Mi) by letting f(m) = (fi(m))i I. (iii) This is false in general but I didn t try to find a counterexample, sorry... But if M is finitely generated as an R-module then this is true and you can prove it just like in (ii). Indeed the direct sums here sit inside the direct products from (ii) as the tuples with all but finitely many entries being zero. The restriction of the isomorphism from (ii) to these subgroups gives an isomorphism between the direct sums providing M is finitely generated. i I

3 17 5. Recall that a left R-module M is of finite length if it has a composition series. In that case, the length l(m) the number of non-zero factors appearing in a composition series of M is a well-defined invariant of M by the Jordan-Hölder theorem. (i) Suppose that 0 K M Q 0 is a short exact sequence. Prove that M is of finite length if and only if both K and Q are of finite length, in which case l(m) = l(k) + l(q). (ii) Suppose that M is a module of finite length and let N and L be submodules. Prove that l(n + L) + l(n L) = l(n) + l(l). (iii) Interpret (ii) in terms of dimension if M is a finite dimensional vector space over a field and N and L are subspaces. (i) If K and Q are of finite length, take composition series for them both. Then glue them together using the lattice isomorphism theorem to get a composition series for M. In particular, l(m) = l(k) = l(q). Now suppose that M has finite length. Say is a composition series. Consider 0 = M 0 < M 1 < < M n = M 0 = M 0 K M 1 K M n K = K. I claim that the factors are either simple or zero, hence K has a composition series. Well, by the second isomorphism theorem, M i K/M i 1 K = M i K/(M i 1 K) M i 1 = (Mi K + M i 1)/M i 1. This is a submodule of M i/m i 1 which is simple. simple... In a similar way to look at the series So its either zero or all of M i/m i 1 which is 0 = π(m 0) π(m n) = Q where π : M Q is the epimorphism from the s.e.s. The factors are zero or simple again. So Q has a composition series. (ii) Consider the s.e.s. 0 N L N N/N L 0. Since M has finite length, so do N and L, so by (i) we have that But N/N L = (N + L)/L. Now look at and apply (i) again to see that l(n/n L) = l(n) l(n L). 0 L N + L (N + L)/L 0 l((n + L)/L) = l(n + L) l(l). The right hand sides of our two equations are equal, and rearranging gives the answer. (iii) dim(v + W ) + dim(v W ) = dim V + dim W. 6. True or False? (i) Let V = W X and V = W Y be two decompositions of a left R-module V as a direct sum of submodules. Then X = Y. (ii) Let R be a commutative ring and I, J be two left ideals in R. If the modules R/I and R/J are isomorphic then I = J. (iii) Let R be a ring and I, J be two left ideals in R. isomorphic then I = J. If the modules R/I and R/J are

4 18 (i) False. Take V = Z 2 Z 2 and W = {(0, 0), (1, 1)}, X = {(0, 0), (1, 0)} and Y = {(0, 0), (0, 1)}. (ii) TRUE!!!! Note that Ann R(R/I) = I, Ann R(R/J) = J and if R/I and R/J are isomorphic they have the same annihilators in R. (iii) False. For example take the matrix algebra R = M 2(C). It is a direct sum of two left ideals I 1 I 2 where I 1 consists of matrices with zeros in the second column and I 2 consists of matrices with zeros in the first column. Now R/I 1 = I2 and R/I 2 = I1. It remains to observe that I 1 and I 2 are isomorphic as left R-modules (they are both isomorphic to the module of column vectors C 2 ). 7. True or False? (i) Every finitely generated module over a field has finite length. (ii) Every finitely generated module over a PID has finite length. (i) Of course: finitely generated modules over fields are finite dimensional vector spaces, and they have finite bases... (ii) False. Take R = Z and consider the regular module RR which is cyclic, so certainly finitely generated. If it has a composition series, you get it by picking a maximal submodule, then a maximal submodule of that,... and the process stops in finitely many steps. But a maximal submodule of Z is (2), and as a module that is isomorphic to Z again. So you can go on forever Prove that every short exact sequence of the form 0 K M R R 0 is split. Is it true that every short exact sequence of the form is split? 0 R R M Q 0 Let π : M RR be the epimorphism. We just need to show that it is split. Let m M be any element with π(m) = 1 R. Since R is a free R-module on basis 1 R, there is a unique R-module homomorphism τ : RR M with τ(1 R) = m. Then π τ = id as required... But the second short exact sequence need not be split! Take for example R = Z and the monomorphism i : Z Z given by multiplication by 2. For this to split, 2Z would have to have a complement in Z. But that complement would have to be a submodule of Z isomorphic to Z 2, and there can t be such a thing since Z is an integral domain This exercise is known as the five lemma. Let and 0 A 1 A 2 A B 1 B 2 B 3 0 be short exact sequences of R-modules. Suppose there are vertical maps γ i : A i B i for each i = 1, 2, 3 so that the resulting diagram commutes. Prove that: (i) γ 1, γ 3 monomorphisms γ 2 is a monomorphism; (ii) γ 1, γ 3 epimorphisms γ 2 is an epimorphism; (iii) γ 1, γ 3 isomorphisms γ 2 is an isomorphism. This is an example of diagram chasing which is usually not hard but often tedious. I ll just do (i). Hopefully my notation for the maps is obvious. Take a 2 A 2 such that γ 2(a 2) = 0. We need to show that a 2 = 0. Well, γ 3(π A(a 2)) = π B(γ 2(a 2)) = 0. As γ 3 is mono, this implies that π A(a 2) = 0 already. Hence a 2 ker π A = im i A. Hence a 2 = i A(a 1) for some a 1 A 1. Now γ 2(i A(a 1)) = i B(γ 1(a 1)) = 0. As i B and γ 1 are mono, this implies that a 1 = 0 already. Hence a 2 = i A(a 1) = 0 too.

5 Quite soon in class we will show that if R is a PID then every submodule of a free module is free. Actually we ll only prove this in class for finitely generated modules, but it is true in general. Bearing this in mind, think about the following True or False? (i) If R is a commutative ring, then every submodule of a free module is free. (ii) If R is an integral domain, then every submodule of a free module is free. (iii) If R is commutative and every submodule of a free R-module is free, then R is a PID. (i) False. For example take R = Z 6 and consider the submodule 2Z 6. (ii) False. For example take R = F [x, y], F a field. Consider the submodule (x, y). (iii) True. Its obvious that R has to be an integral domain (else you get counterexamples like in (i)). Now suppose (x, y) is an ideal that is not principal. Then (x, y) is not free (like in (ii)). 11. In linear algebra, you often take linearly independent sets and extend them to bases, and you often take spanning sets and cut them down to a basis. (i) Use Zorn s lemma to prove that any set X of linearly independent vectors in an R-module M is contained in a maximal linearly independent set. (ii) Suppose that D is a division ring. We know already that all D-modules are semisimple, hence free. Prove that every maximal linearly independent set of vectors in an D-module M is a basis. (iii) In general you need to be careful: if M is a free R-module and X is a maximal linearly independent set of vectors in M, then X need not be a basis. For example, suppose that R is an integral domain with the property that every maximal linearly independent set of vectors in a free R-module M is a basis. Prove that R is a field. (iv) Let M be a module over a division ring D and let X M be a subset of vectors that span M as a D-module. Prove that X contains a basis of M. (Hint: use the fundamental lemma about semisimple modules.) (i) Let F be the set of all sets Y of linearly independent vectors M containing X. Note F is non-empty as X F. We just need to prove that F has a maximal element. This follows by Zorn s lemma if we show every chain (Y i) i I in F has an upper bound in F. Well just take Y = S i I Yi. Its an upper bound, but why is it in F? Well if not, there is a dependency r 1y i1 + + r k y ik = 0 for some i 1,..., i k I and r 1,..., r k 0. But then you can find i I such that y i1,..., y ik all already lie in Y i. But that contradicts the linear independence of the set Y i. (ii) Let X be a maximal linearly independent set of vectors in M. Take m M. Since X is maximal linearly independent, X {m} is linearly dependent. Hence we can write m as a linear combination of elements of X. Hence X spans M, so its a basis. (iii) Take M = RR. Take 0 x R. This is a linearly independent set of vectors in M, since R is an integral domain. I claim its maximal. Well, if not you can find another 0 y R such that x, y is still linearly independent. But that s a contradiction: (y)x + ( x)y = 0 is a non-trivial linear relation between x and y. Hence by assumption x is a basis for R. So we can write 1 = yx for some y R. Hence x is a unit, and R is a field. (iv) Note M = P 0 x X Dx. Each Dx is simple. Hence by the fundamental lemma on semisimple modules, there exists Y X such that M = L y Y Dy. That is, y is a basis. 12. Let R and S be rings, let M be a left R-module and let B be an R, S-bimodule. (i) Prove that Hom R (B, M), the abelian group of all left R-module homomorphisms from B to M, can be made into a left S-module by defining the action of s S on f Hom R (B, M) by letting sf be the unique element of Hom R (B, M) with (sf)(b) := f(bs) for all b B. (ii) Prove that Hom R (M, B), the abelian group of all left R-module homomorphisms from M to B, can be made into a right S-module by defining the action of s S on f Hom R (M, B) by letting fs be the unique element of Hom R (M, B) with (fs)(m) = f(m)s for all m M.

6 20 Just check some axioms until you are happy. The important thing to remember: right S-module structure on the FIRST ARGUMENT of hom leads to a left S-module structure on the hom space. right S-module structure on the SECOND ARGUMENT of hom leads to a right S-module structure on the hom space. 13. Let R be a ring and M be a left R-module. By 12(i), the abelian group Hom R (R, M) has the structure of a left R-module. By 12(ii), the abelian group Hom R (M, R) has the structure of a right R-module. (i) Taking M = R, these constructions make the space Hom R (R, R) of all left R-module homomorphisms from R to R both into a left R-module and a right R-module. Verify that the left and right actions commute, so that it is actually an (R, R)-bimodule. (ii) Prove using the map f f(1 R ) that Hom R (R, R) is isomorphic to R as an (R, R)- bimodule. (iii) Of course End R (R) = Hom R (R, R) is actually a ring with multiplication fg := f g (composition). Is the map f f(1 R ) a ring isomorphism between End R (R) and R? Be careful... (i) Well we have to check that (rf)s = r(fs). Evaluate on some t R. Then ((rf)s)(t) = (rf)(t)s = f(tr)s and (r(fs))(t) = (fs)(tr) = f(tr)s. Done. (ii) Certainly f f(1 R) is an isomorphism of abelian groups between Hom R(R, R) and R. So consider the image of rfs. We want it to be rf(1 R)s for our map to be a bimodule homomorphism. Well like in (i), (rfs)(1 R) = f(1 Rr)s = f(r)s = f(r1 R)s = rf(1 R)s. Phew. (iii) NO. This is a special case of the result in class when we computed the endomorphism algebra of a free left R-module of rank n. In case n = 1, we showed that End R(R) op = R, and the isomorphism is exactly evaluation at 1. But if R is not commutative, then evaluation at 1 is NOT an isomorphism between End R(R) and R without the op Let D be a division ring and let R be the ring of all 2 2 upper triangular matrices over D. Let e 1 and e 2 be the diagonal matrix units diag(1, 0) and diag(0, 1). (i) Show that e 1 and e 2 are orthogonal idempotents with e 1 +e 2 = 1, so that R R = Re 1 Re 2. (ii) Show that Re 1 is simple but Re 2 is not. (iii) Show that End R (Re 1 ) op = End R (Re 2 ) op = D as rings. Deduce in particular that Re 1 and Re 2 are indecomposable R-modules, and the idempotents e 1 and e 2 are primitive. (i) Re 1 is just the module of column vectors of height 2 with top entry anything from D and bottom entry zero. You can get from any one non-zero such vector to any other by multiplying by D, i.e. by multiplying by the element of R with a scalar in its 11-entry and zeros in all other entries. So Re 1 is irreducible. Re 2 is just the module of column vectors of height 2 with arbitrary entries from D. There is a submodule consisting of all the column vectors with 0 as their bottom coordinate. Hence it is reducible. (ii) Recall that End R( RR) op = R. Hence by lemma from class, End R(Re 1) op = e1re 1. This is just the matrices with anything from D as their 11-entry, and all other entries zero. Clearly this is isomorphic to D. Similarly, End R(Re 2) op is e 2Re 2, which is matrices with anything from D as their 22-entry, all other entries zero. Since D is for sure a local ring, it follows that Re 1 and Re 2 are indecomposable, hence e 1 and e 2 are primitive idempotents. 15. A ring R is SBN (single basis number) if all finitely generated free left R-modules are isomorphic to R R. (This is equivalent to all finitely generated right R-modules being isomorphic

7 21 to R R but the fact that these are equivalent definitions is far from obvious if R is not a commutative ring!) (i) Show that if R is SBN then so is every quotient ring R/I. (ii) Let A = Z N, the free abelian group on the set N. Let R = End Z (A), the ring of all Z-module endomorphisms of A. Prove that R is SBN. (i) L Let M be a finitely generated free left (R/I)-module. Pick a basis x 1,..., x n for M, so M = n i=1 (R/I)xi and each (R/I)xi = R/I. View M now as an R-module. Let P = L n i=1 Rxi be the free R-module on basis x 1,..., x n. Define an R-module homomorphism P M mapping x i to x i for each i = 1,..., n. Note the kernel of this is IP. So, M = P/IP. Now R has IBN, so P = RR. Hence M = P/IP = R/I. Hence R/I has IBN. (ii) Note A is a left R-module by definition. For i N, let e i be the projection of A onto its ith component. Note e i R is an idempotent and Re i = A as a left R-module. (The isomorphism here maps r Re i to r1 i where 1 i is the identity in the ith slot in the direct sum A.). So R = Y Y Re i = A i N i N as a left R-module. But this is a direct product of countably many copies of A. And a finite direct sum (aka direct product) of copies of R is still direct product of countably many copies of A. Hence they re isomorphic. 16. Let A be the abelian group Z r p 1 1 Z r p 2 2 Z p rn n where p 1,..., p n are distinct primes and r 1,..., r n 1. Compute the abelian groups Hom Z (Z, A) and Hom Z (A, Z). Since Z is a free Z-module of rank 1, evaluation at 1 defines an isomorphism between Hom Z (Z, A) and A. Hence the first abelian group is just A again. For the second one, p r 1 1 acts as zero on Z p r 1 1, and the only element in Z with this property is zero. So the only map here is the zero map, and this is the zero module. 17. In class, we proved (using Fitting s lemma) that if M is an indecomposable module of finite length, then End R (M) is a local ring, i.e. the sum of two non-units is a non-unit. Instead prove Schur s lemma: if M is an irreducible module then End R (M) is a division ring, i.e. 0 is the only non-unit. Let f : M M be a non-zero R-module endomorphism with M simple. The kernel of f is not all of M, but it is an R-submodule of M, hence it is zero as M is irreducible. Therefore f is injective. The image of f is not zero, but it is an R-submodule of M, hence it is M as M is irreducible. Therefore f is surjective. Therefore f is invertible. 18. Let e, f R be idempotents, and consider the summands Re and Rf of R R. (i) Prove that Re = Rf if and only if f = e + (1 e)ae for some a R. (Hint: ( ) If f = ae and e = bf then efe = e and 1f1 = f where 1 = e + (1 e).) (ii) Prove that Re = Rf if and only if there exist a, b R with ab = e and ba = f. (iii) Deduce that Re = Rf as left R-modules if and only if er = fr as right R-modules. (i) ( ). If Re = Rf then f = ae and e = bf for some a, b R. Hence ef = bff = bf = e and f(1 e) = ae(1 e) = 0. So f = ((1 e) + e)f((1 e) + e) = efe + (1 e)fe = e + (1 e)fe. ( ). If f = e + (1 e)ae then f Re so Rf Re. And ef = e 2 = e so e Rf so Re Rf. (ii) Note that every element of Hom R(Re, Rf) is gotten just by right multiplication by some element of erf. Indeed, right multiplication by such elements certainly gives R-module maps from Re to Rf. Conversely, given an R-module map from Re to Rf, compose with the projection e from R to Re at the beginning and the inclusion of Rf into R at the end and you get an R-module homomorphism from R to R, i.e. an element of End R(R) op = R which actually lies in erf.

8 22 Hence an isomorphism Re Rf is given by right multiplication by some a erf and the inverse is given by right multiplication by some b fre. Then right multiplication by ab gives the identity map from Re to Re, i.e. ab = e, and similarly ba = f. (iii) Clear as the statement in (ii) is left-right symmetric. 19. Suppose that R = M n (F ) for F a field and n > 1. (i) Find a decomposition 1 R = e 1 + +e m of 1 as a sum of primitive orthogonal idempotents, and describe the left R-modules Re i and the right R-modules e i R. (ii) Find another such decomposition. (iii) Prove that R is a simple ring, i.e. it has no non-trivial two-sided ideals. (i) Let e i be the matrix with 1 F in its ii entry and all other entries zero. These are orthogonal idempotents summing to 1. To see that they are primitive, just note that Re i is the left R-module of column vectors, which is irreducible as you can get any non-zero column vector to any other by multiplying by some matrix. Hence Re i is indecomposable. Hence e i is a primitive idempotent by result from class. (ii) Take any unit in R, i.e. an invertible matrix u. Then ue iu 1 also give such things. (iii) Let I be a non-zero two-sided ideal. Take a non-zero matrix m I. Say the ij-entry of m is not zero. Then e ime j is the matrix equal to this entry in the ij-place and all other entries are zero. Hence I contains the ij-matrix unit e i,j. Now left multiplying you get all other matrix units in the jth column, and right multiplying you get all other matrix units in the ith row. Hence we ve got all matrix units. Adding we get any matrix. So I = R. 20. Let A = (a i,j ) be an n n matrix with integer entries and determinant d 0. (i) Let X be the free abelian group on basis x 1,..., x n and let Y by the subgroup generated by y 1,..., y n where n y j = a i,j x i. i=1 Prove that X/Y is a finite abelian group of order d. How would you compute the isomorphism type of this group explicitly? (ii) Compute the isomorphism type of the group X/Y explicitly for the following matrices: [ ] , 1 2 1, What about for the n n matrix following the same pattern? (iii) Do the same for the matrix (i) Say A diag(d 1,..., d n) are the invariant factors of A over Z, and assume all d i 0 by rescaling by 1 if necessary. Note d 1 d n = ±d 0, since the determinant of an invertible matrix over Z is ±1 (the only units in Z). So all d i are non-zero. Then we can pick a basis v 1,..., v n for X so that d 1v 1,..., d nv n is a basis for Y. But then its obvious that X/Y = Z d1 Z dn. In particular its order is d 1 d n = d. (ii) The determinant of the n n matrix following this pattern is (n + 1), and its invariant factors are (1,, 1, n + 1). This is an exercise in determinants... Hence the group X/Y is Z n+1. (iii) Using the J i trick, you see that it has invariant factors (1, 1, 2, 2) and X/Y = Z 2 Z 2..

9 Let R = C[[x]], the ring of formal power series over C (it is a PID). Consider the submodule W of the free module V = Rv 1 Rv 2 generated by and (1 x) 1 v 1 + (1 x 2 ) 1 v 2 (1 + x) 1 v 1 + (1 + x 2 ) 1 v 2. Find a basis v 1, v 2 of V and elements d 1 d 2 R such that d 1 v 1 and d 2 v 2 is a basis for W. Describe V/W. By the way here is why C[[x]], formal power series in an indeterminant x with coefficients in C, is a PID. Elements of this ring consist of infinite series a 0 + a 1x + a 2x 2 + for a 0, a 1, C (possibly infinitely many being non-zero). Multiplication is the obvious thing. The polynomial ring C[x] is a subring of C[[x]]. Observe first that any power series whose constant term is not zero is actually a unit. For instance 1 x is a unit: its inverse is 1 + x + x 2 + x 3 +. If you think for a moment you will see that you can algorithmically work out how to invert any power series with constant term non-zero inductively... and end up with a power series going on forever. Given this, it is easy to see that any non-zero element of C[[x]] is just a power of x, x i, times a unit. Hence the ideals in C[[x]] are just (x i ) for i 0 together with the zero ideal. So its a PID. Now for the problem. Subtract (1 + x) 1 times row 1 from row 2 of the matrix (1 x) 1 (1 + x) 1 «(1 x 2 ) 1 (1 + x 2 ) 1 to get the matrix (1 x) 1 (1 + x) 1 «0 (1 + x 2 ) 1 (1 + x) 2 Now column 2 minus (1 x)(1 + x) 1 times column 1 gives (1 x) (1 + x 2 ) 1 (1 + x) 2 Thus the invariant factors are d 1 = (1 + x) 1 and d 2 = (1 + x 2 ) 1 (1 + x) 2 (or up to units 1 and x would do). Implementing the row operations above as matrix multiplication, you see that «(1 x) (1 + x 2 ) 1 (1 + x) 2 = «««1 0 (1 x) 1 (1 + x) 1 1 (1 x)(1 + x) 1 (1 + x) 1 1 (1 x 2 ) 1 (1 + x 2 ) «Hence Finally, (v 1, v 2) 1 0 = (v 1, v 2) (1 + x) 1 1 «= (v 1 (1 + x) 1 v 2, v 2). V/W = R/((1 x) 1 ) R/((1 + x 2 ) 1 (1 + x) 2 ) = (0) R/(x) = R/(x) = C. 22. Prove that Q is not a free Z-module (even though it is torsion free). Suppose it was a free Z-module on basis v i (i I). Note I cannot be of size 1, because Q is not isomorphic to Z as an abelian group (obviously its not cyclic rational numbers are not integer multiplies of any one fraction!). Let v and v be two different elements from the basis. Then v = a/b and v = c/d are non-zero rational numbers. But then cbv adv = 0, contradicting linear independence. 23. Let R be a PID and V be an R-module. A submodule W of V is called pure if W xv = xw for all x R. If V is finitely generated, prove that W is a pure submodule of V if and only if it is a direct summand of V.

10 24 (Thanks to J.B. for this solution.) Suppose that W is a summand of V, say V = W U. Take x R. We need to show that W xv xw. Well, say we have some v V so that xv W. Write v = w + u for w W, u U. Then xv = xw + xu W, hence xu = xv xw W U = {0}. This shows that xv = xw so it lies in xw. Conversely, suppose that W is a pure submodule of V. The quotient V/W is finitely generated so by structure theorem it is a direct sum of cyclic modules. Say V/W = R(v 1 + W ) R(v n + W ) for v 1,..., v n V. Suppose that R(v i + W ) is cyclic of order d i. If d i = 0 set u i = v i. If d i 0, then d iv i W hence by purity d iv i = d iw i for some w i W. Set u i = v i w i in this case. Note in both cases that v i + W = u i + W, hence V/W = R(u 1 + W ) R(u n + W ) and each R(u i + W ) is also cyclic of order d i. The advantage now is that we actually have d iu i = 0 (whereas we only had d iv i W...). Finally set U = Ru Ru n. We claim that V = W U, hence W is a summand of V as required. Clearly V = W + U because (U + W )/W = V/W. So we just need to check that W U = {0}. Take r 1u r nu n W. Then r 1(u 1 + W ) + + r n(u n + W ) = 0 in V/W. Hence as V/W = R(u 1 + W ) R(u n + W ), we have that r i(u i + W ) = 0 in V/W for each i. But this means that d i r i, i.e. r i = d is i for some s i, hence r iu i = s id iu i = 0. Hence r 1u r nu n = 0 which is what we wanted Let R be a commutative ring. Is it true that the left regular R-module is indecomposable? What if R is a PID? No. Take for example the ring Z 6. By CRT, it is isomorphic to Z 2 Z 3. So it is decomposable. But if R is an integral domain, then the left regular R-module is indecomposable. Proof: Suppose R = A B is decomposable as a left R-module. Take 0 a A and 0 b B. Consider ab = ba. It lies in A and in B. Hence its zero. This contradicts R being an integral domain. 25. (i) How many abelian groups are there of order up to isomorphism? (ii) How many isomorphism classes of abelian groups are there of order 108 having exactly 4 subgroups of order 6? (i) It is the number of partitions of 6 times the number of partitions of 5. For 6 there are 11: (6), (5, 1), (4, 2), (4, 1, 1), (3, 3), (3, 2, 1), (3, 1, 1, 1), (2, 2, 2), (2, 2, 1, 1), (2, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1). For 5 there are 7: (5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1). So there are 77. I hope. (ii) The abelian groups of order 108 are as follows: Z 27 Z 4, Z 27 Z 2 Z 2, Z 9 Z 3 Z 4, Z 9 Z 3 Z 2 Z 2, Z 3 Z 3 Z 3 Z 4, Z 3 Z 3 Z 3 Z 2 Z 2. A subgroup of order 6 is Z 3 Z 2, and Z 3, Z 2 must be in the unique Sylow subgroups of order 27 or 2. Now Z 27 has one subgroup of order 3, while Z 9 Z 3 has four and Z 3 Z 3 Z 3 has 13. Similarly, Z 4 has one subgroup of order 2 and Z 2 Z 2 has three. So the only possibility is Z 9 Z 3 Z True or False? (i) An n n nilpotent matrix with entries in a field has characteristic polynomial x n. (ii) There are exactly 3 similarity classes of 4 4 matrices A over the field F 2 satisfying A 2 = 1. (iii) The number of similarity classes of n n nilpotent matrices over a field F is independent of the particular field F chosen (iv) The matrices and are similar over Q (i),(iii) Both true. The minimal polynomial of a nilpotent matrix divides x n so all roots of it are zero, so all its eigenvalues are zero, so up to similarity it is just a bunch of Jordan block J d1 (0),..., J dk (0) down the diagonal with d 1 d k > 0 and d d k = n. Obviously the characteristic

11 25 polynomial is x n. And the number of similarity classes is equal to the number of partitions of n, independent of the field F. (ii) True. The minimal polynomial divides x 2 1 = (x 1) 2. So it is just J 1(1) and J 2(1) s down the diagonal in JNF. Now 4 = = = (how to put the Jordan blocks down the diagonal). So there are 3 similarity classes. (iv) Well this was typo, I meant to have 4 by 4 matrices here! But as it is its a very easy question, the answer is false I guess: it doesn t even make sense to say that non-square matrices are similar at all. 27. How many similarity classes are there of 2 2 matrices over F 2? Possible characteristic polynomials: x 2, x 2 +1 = (x+1) 2, x 2 +x = x(x+1), x 2 +x+1. The first three give classes diag(0, 0), J 2(0), diag(1, 1), J 2(1), diag(0, 1). The final one is irreducible polynomial, with rational normal form» So I make it 6. Note I often use Jordan normal form even if the field is not algebraically closed. PROVIDING all the roots of the characteristic polynomial lie in the field F, it makes sense to work with JNF. Its just that there are some characteristic polynomials which don t have roots lying in F, and for such ones you have to resort to the rational normal form. 28. Let V be a 7 dimensional vector space over Q. (i) How many similarity classes of linear transformations on V are there with characteristic polynomial (x 1) 4 (x 2) 3? (ii) Of the similarity classes you listed in (i), how many have minimal polynomial (x 1) 2 (x 2) 2? (iii) Let φ be a linear transformation from V to V having characteristic polynomial (x 1) 4 (x 2) 3 and minimal polynomial (x 1) 2 (x 2) 2. Find dim ker (φ 2 id). (i) Partitions of 4: 4, 3 + 1, 2 + 2, , , 5 in total. Partitions of 3: 3, 2 + 1, , 3 in total. So there are 15 similarity classes of such linear transformations. (ii) Need biggest Jordan blocks of size 2. So just partitions or for eigenvalue 1 and for eigenvalue 2, giving just 2 classes in total. (iii) We need to compute the dimension of the 2-eigenspace. There are exactly two Jordan blocks of eigenvalue 2. So this dimension is Let V be a finite dimensional vector space over Q and θ : V V be a linear transformation having characteristic polynomial (x 2) 4. (i) Describe the possible Jordan normal forms for θ and for each one give the minimal polynomial of θ. (ii) For each of the possibilities in (i) compute the dimension of the 2-eigenspace of θ. (iii) Assume that θ leaves stable only finitely many subspaces of V. What can be said about the Jordan normal form of θ? (i) Note we re looking for a 4 by 4 matrix with all eigenvalues equal to 2: diag(2, 2, 2, 2), diag(j 2(2), 2, 2), diag(j 2(2), J 2(2)), diag(j 3(2), 2) and J 4(2). Their minimal polynomials are (x 2), (x 2) 2, (x 2) 2, (x 2) 3 and (x 2) 4 respectively. (ii) This is the number of Jordan blocks, so 4, 3, 2, 2, 1 respectively. (iii) If the 2-eigenspace is more than 1 dimensional, there are infinitely many 1-subspaces of the 2-eigenspace, all of which are θ-stable. So we must have just one Jordan block of eigenvalue 2, i.e. θ has JNF J 4(2).

12 Let θ : V V be an endomorphism of a finite dimensional vector space V. Recall for λ F that the λ-eigenspace of V is V λ := ker (θ λ id), and elements of it are called λ-eigenvectors. Also θ is called diagonalizable if and only if V has a basis in which the matrix of θ is a diagonal matrix, in other words, a basis consisting of eigenvectors. (i) Prove that the sum λ F V λ is direct, i.e. eigenvectors with different eigenvalues are linearly independent. (ii) Prove that θ is diagonalizable if and only if V = λ F V λ. (i) Suppose not. Then there exists a linear relation 0 = v v n be a linear relation with 0 v i V λi, distinct λ 1,..., λ n and n > 0. Choose one with n as small as possible. Act with θ to get that 0 = λ 1v λ nv n. Subtracting λ 1 times the first equation we get that 0 = (λ 2 λ 1)v (λ n λ 1)v n. This is a linear relation of strictly smaller length, contradicting the minimality of n. (ii) If θ is diagonalizable, then there exists a basis of eigenvectors, hence V is spanned by eigenvectors, hence V = P λ F V λ. We already showed in (i) that this sum is direct. Coversely suppose V = L λ F V λ. Pick a basis for each V λ and glue together to get a basis for V consisting of eigenvectors. The matrix of θ in this basis is diagonal. 31. Recall from class that θ : V V is diagonalizable if and only if its minimal polynomial is a product of distinct linear factors. Actually we only observed this in the case that the ground field was algebraically closed (since we deduced it from the Jordan normal form). Prove it now for any ground field. If θ is diagonalizable, let λ 1,..., λ n be the distinct eigenvalues. Then the minimal polynomial is (x λ 1) (x λ n), i.e. a product of distinct linear factors. Conversely, suppose that the minimal polynomial is a product of distinct linear factors, (x λ 1) (x λ n). View V as an F [x]-module so that x acts as θ. By the primary decomposition, V = V 1 V m where V i is cyclic of prime power order. Say V i is of order p i(x) m i for p i(x) F [x] prime. The minimal polynomial acts as zero on V. Hence p i(x) m i (x λ 1) (x λ n) for each i. But this means each m i = 1 and each p i(x) is of degree 1. Hence V is diagonalizable Let θ i (i I) be a family of endomorphisms of a finite dimensional vector space V. Prove that they are simultaneously diagonalizable, i.e. there exists a basis for V with respect to which the matrices of all of the θ i s are diagonal, if and only if each θ i is diagonalizable and θ i θ j = θ j θ i for all i, j I. If the θ i s are simultaneously diagonalizable, they are obviously diagonalizable and commute. Now consider the converse. Assume the θ i s are all diagonalizable and that they pairwise commute. We show that they are simultaneously diagonalizable by induction on dim V. If dim V = 1 it is clear (everything is diagonal!). Now in general, if each θ i is just multiplication by a scalar c i, then in any basis each θ i looks like c ii n and they re simultaneously diagonalizable for free. So we can find some θ i, wlog θ 1, having distinct eigenvalues λ 1,..., λ n with n > 1. Let V i = ker (θ 1 λ i), so dim V i < n. Note all the other θ j leave V i invariant: if v V i then θ 1(θ j(v)) = θ j(θ 1(v)) = λ iθ j(v) hence θ j(v) V i still. I claim that the restrictions θ j Vi of each θ j to V i are still diagonalizable. To prove this, note that that the minimal polynomial of θ j is a product of distinct linear factors. The minimal polynomial of θ j Vi divides this, so it too is a product of distinct linear factors. Hence θ j Vi is diagonalizable. Now we can apply induction to see for each i that the {θ j Vi j I} are simultaneously diagonalizable. Pick a basis for each V i consisting of simultaneous eigenvectors, then glue these bases together to get a basis of simultaneous eigenvectors for all of V. 33. Let V and W be vector spaces over a field F. Construct a natural injective homomorphism from V W to (V W ), and prove that it is an isomorphism if V and W are both finite dimensional. Is it an isomorphism if they are infinite dimensional?

13 27 Define a map from V W to (V W ) mapping (f, g) to the linear map f g : V W F that sends v w to f(v)g(w) (constructed in class). This is bilinear so we get induced a map V W (V W ). I claim that this map is injective. Take something x in the kernel. We can write it as x = f 1 g f n g n for linearly independent functions f 1,..., f n V and some other (not necessarily independent) functions g 1,..., g n W. It is in the kernel of our map, which means for every v V and w W that (f 1 g 1 + +f n g n)(v w) = f 1(v)g 1(w)+ +f n(v)g n(w) = 0. Since f 1,..., f n are linearly independent, we can find v 1,..., v n so that f i(v j) = δ i,j. Then for all i and w W f 1(v i)g 1(w) + + f n(v i)g n(w) = g i(w) = 0. Hence g i = 0 for all i. Hence x = 0. If both are finite dimensional then dim V W = (dim V )(dim W ) = dim(v W ). So the injective map we ve constructed must actually be an isomorphism. If they re both infinite dimensional, it definitely won t be an isomorphism. 34. Recall a bilinear form on a vector space V is a bilinear map from V V to the ground field F. Given any two such bilinear forms b, b, you can add them together to get a new form b + b with (b + b )(v, w) = b(v, w) + b (v, w) and you can multiply by scalars too. This makes the space of bilinear forms on V into a vector space in its own right. (i) Prove that the vector space of bilinear forms on V is naturally isomorphic to the vector space (V V ). (ii) Assuming that V is finite dimensional, prove that (V V ) is naturally isomorphic to Hom F (V, V ). (Hence giving a bilinear form on V is equivalent to giving a linear map from V to V.) (i) Take a bilinear form b. This means a bilinear map from V V to F. By the universal property of tensor product, this induces a unique linear map from V V to F, i.e. an element of (V V ). Conversely given an element f (V V ) you get a bilinear form (v, w) = f(v w)... (ii) If V is finite dimensional then by (i) and the previous question the space of bilinear forms is isomorphic to V V. But from class, V W = Hom F (V, W ) for any finite dimensional V. Hence V V = HomF (V, V ). 35. Let A and B be abelian groups. (i) Prove that A Z Z m = A/mA. (ii) Prove that Z m Z Z n = Zk where k = GCD(m, n). (iii) Assuming that A and B are finitely generated, describe A Z B in general. (i) Define a map A Z m A/mA by (a, z) az +ma. Note this is well-defined: if z z (mod m) then az az (mod ma). Also it is balanced. Hence it induces a map A Z Z m A/mA such that a 1 a + ma. It is clearly onto. Conversely, define a map A A Z Z m mapping a to a 1. Under this map, ma ma 1 = a m = 0, so ma lies in the kernel. Hence our map factors through the quotient A/mA to induce a well-defined map A/mA A Z Z m under which a + ma a 1. This is a two-sided inverse to the map in the preceeding paragraph. (ii) Define a map Z m Z n Z k mapping ([a] m, [b] n) to [a] k [b] k. This is balanced, so induces a map Z m Z Z n Z k such that [a] m [b] n [a] k [b] k. Note it is onto since 1 1 maps to 1 which generates Z k. On the other hand, Z m Z Z n is cyclic generated by the vector [1] [1] (any pure tensor looks like [a] [b] = a[1] b[1] = ab[1] [1]; any tensor is a sum of pure tensors hence a multiple of [1] [1] too). And now if you write k = am + bn you have that k[1] [1] = am[1] [1] + bn[1] [1] = a[m] [1] + b[1] [n] = 0. Hence Z m Z Z n is of order at most k. So the epimorphism from the previous paragraph must be onto. OR: you can use (i) which shows that Z m Z Z n = Zm/(nZ m), which is (Z/mZ)/((nZ + mz)/mz) = Z/(m, n) = Z/(k). (iii) Just write A and B in terms of the primary decompostion and use the fact that commutes with...

14 True or False? (i) If V is a torsion Z-module then V Z Q = 0. (ii) Q Z Q = Q. (iii) If R is a commutative ring and M, N are R-modules, then m n = m n in M R N implies that m = m and n = n. (iv) If R is a commutative ring and M, N are free R-modules, then M R N = M Z N as abelian groups. (i) TRUE. It suffices to show every pure tensor is zero. Take v a. Since v is torsion, mv = 0 for b some m 0. But then v a = mv a = 0. b mb (ii) TRUE. Define a map Q Q Q, ( a, c ab ). It is balanced, hence it induces a map f : Q b d bd ZQ Q. It is an isomorphism, because I can write down a two-sided inverse map: g : a 1 a. Clearly b b f g is the identity, but its not quite clear that g f is... Consider pure tensor a c = b a c = a c = 1 ac. It follows that any element of Q b d b bd bd bd Z Q (not just pure tensors) can be written as 1 a. Now on such an element it is clear that g f is the b identity. (iii) FALSE. I mean rm n = m rn and you neededn t have rm = m... (iv) FALSE. For instance take R = Z Z. Then R R R = R = Z Z = Z 2 as abelian group. But (Z Z) Z (Z Z) = Z 4 as abelian group. 37. Let R and S be rings. We have the tensor functor? R? from mod-r R-mod to abelian groups and the tensor functor? S? from mod-s S-mod to abelian groups. Composing them in two different ways gives two different functors F = (? R?) S? and G =? R (? S?) from mod-r R-mod-S S-mod to abelian groups. (i) Explain what it means to say that the functors F and G are isomorphic functors. (ii) Now use the universal property of tensor product to prove that F and G are indeed isomorphic. (This is associativity of tensor product ). (i) It means for any triple M, N, P of modules there is an isomorphism η M,N,P : (M N) P M (N P ) such that for any morphisms f : M M, g : N N and h : P P, we have that η M,N,P ((f g) h) = (f (g h)) ηm,n,p. (ii) Of course we just need a hommorphism (M N) P M (N P ) such that the pure tensor (m n) p maps to m (n p). The diagram in (i) will then certainly commute. But we need to prove that there exists such a map using the universal property. For each p P, define a map f p : M N M (N P ) sending (m, n) to m (n p). Its balanced so induces a unique map f p : M N M (N P ) sending m n to m (n p). Now given this define a balanced map (M N) P M (N P ) mapping (x, p) to f p(x). Its balanced so induces unique map (M N) P M (N P ) such that (m n) p f p(m n) = m (n p). Phew. 38. Let n > 1. Prove that Z n is both a projective and an injective Z n -module. Clearly its projective as its free. For injectivity, let s apply the criterion for injectivity. We just need to show that if I is an ideal of Z n and α : I Z n is a module homomorphism then α extends to a map Z n Z n. We know I = dz n for some d n. Let k be such that α([d]) = [k]. Since (n/d) times [d] equals zero in Z n, we must have that (n/d) times [k] equals zero in Z n, i.e. n divides nk/d, i.e. d divides k. Let e = k/d. Now define β : Z n Z n by β([x]) = [xe]. Since β([d]) = [k] = α([d]) we get that β extends α. Done. 39. Let R be commutative. True or false? (a) Every submodule of a projective R-module is projective. (b) Every submodule of an injective R-module is injective.

15 29 (c) Every quotient of a projective R-module is projective. (d) Every quotient of an injective R-module is injective. They are all false in general. For counterexamples: (a),(d) Take R = Z 4 and M = Z 2. Note R is both projective and injective (by the preceeding question). And M is both a submodule and a quotient module of R, indeed there is a short exact sequence 0 Z 2 Z 4 Z 2 0 (first map being multiplication by 2). It doesn t split since Z 4 is indecomposable. So Z 2 cannot be injective (since anything starting with an injective does split) and Z 2 cannot be projective (since anything ending with a projective does split). (b) Take R = Z, M = Q which is injective as its divisible, and consider the submodule Z of Q. It is not injective since its not divisible. (c) Take R = Z and M = Z/2Z. Its not projective. 40. Prove that the following conditions on a ring R are equivalent: (a) every left R-module is projective; (b) every left R-module is semisimple; (c) every left R-module is injective. We have shown that every short exact sequence splits 0 K M Q 0 if and only if K is injective, if and only if M is semisimple, if and only if Q is projective. 41. Suppose that 0 K i M π Q 0 is a split short exact sequence of left R-modules. Let B be a right R-module. Prove that 0 B R K id B i B R M id B π B R Q 0 is exact. Give a counterexample to show that this need not be true if you remove the word split above. Since tensor is right exact, everything is fine except that the first map id B i might not be injective. But if the ses is split, then there exists a splitting σ : M K such that σ i = id K. Then id B σ : B M B K satisfies (id B σ) (id B i) = id B (σ i) = id B id K. Hence id B i is injective as required. For a counterexample, take 0 Z Z Z 3 0 where the first map is multiplication by 3. Apply Z 3 Z?. The first map becomes the map Z 3 Z 3 given by multiplication by 3 = 0 That is not injective! 42. If f : V V and g : W W are surjective maps of right and left R-modules, respectively, prove that f g : V R W V R W is surjective. Give a counterexample to show that this need not be true if you replace the word surjective with the word injective everywhere. It suffices to show any pure tensor of the form v w lies in the image, since pure tensors span V R W. But f is onto so there exists v V such that f(v) = v. Similarly there exists w W such that g(w) = w. Then (f g)(v w) = f(v) g(w) = v w. Now replace the word surjective with injective. Take f : Z Z being multiplication by 3, and g : Z 3 Z 3 being the identity map. Then f g : Z Z Z 3 Z Z Z 3 is the map Z 3 Z 3 given by multiplication by 3 = 0. Hence it is the zero map. Its not injective. 43. Describe the abelian group C Z Z n. By general result from last week s homework, for any commutative ring R, M R (R/I) = M/IM. So this is C/nC. But nc = C again. So this is just the trivial abelian group True or false? (a) Every submodule of a projective Z-module is projective. (Hint: every submodule of a free Z-module is free.)

16 30 (b) Every submodule of an injective Z-module is injective. (c) Every quotient of a projective Z-module is projective. (d) Every quotient of an injective Z-module is injective. (a) True. This follows from the hint because every projective Z-module is free. To prove the latter, let P be projective, take a short exact sequence 0 K F P 0 with F free. Then it splits as P is projective, hence F = P K. But this shows P is isomorphic to a submodule of a free module, hence it is free by the hint. (b) False. Take Z inside Q like last week. (c) False. Take Z mapping onto Z 2, the latter is not projective. (d) True. Quotients of divisible abelian groups are divisible. 45. A right R-module M is called flat if the functor M R? is exact. (i) Prove that a direct sum of modules is flat if and only if each of the modules is flat. (ii) Prove that the regular module R R is flat. (iii) Prove that any projective module is flat. (i) Let f : A B be a monomorphism. Applying the functor (M N) R? and using the fact that tensor commutes with direct sum (which remember is actually some natural isomorphism of functors...), we get the following commutative diagram: (M N) R A? y id M N f (M N) R B? y M R A N R A (id M f,id N f) M R B N R B where the vertical maps are isomorphisms. Now the top map is injective if and only if the bottom map is injective, which is if and only if both id M f and id N f are injective. This is exactly what we needed...m N is flat if and only if both M and N are flat. (ii) The functor R R? is isomorphic to the identity functor. The latter is exact, hence so is R R?. (iii) Let P be projective. Its a summand of a free module. Hence by (i) we are done if we can prove that any free module is flat. But free modules are direct sums of the regular module. So by (i) again (which holds even for infinite direct sums... though I only wrote down the proof for two) we are done if we can prove that the regular module is flat. But that is true by (ii) 46. Let M be a left R-module. I explained in class by general nonsense (properties of adjoint functors) that the functor Hom R (M,?) is left exact. Give a direct proof of this statement, then give an example to show that it is need not be exact. Take an exact sequence 0 A f B g C. Apply the functor Hom R(M,?) to get 0 Hom R(M, A) f HomR(M, B) ḡ Hom R(M, C). The map f sends θ to f θ, the map ḡ sends φ to g φ. Note that g f = 0, hence applying the functor Hom R(M,?) we get at once that ḡ f = 0. This shows for free that im f ker ḡ. Now we show that f is injective. Say θ : M A has f(θ) = 0. This means that f(θ(m)) = 0 for all m M. But f is injective, hence in fact θ(m) = 0 for all m M, i.e. θ = 0. Finally we show that im f ker ḡ. Take φ : M B such that ḡ(φ) = 0. This means that g(φ(m)) = 0 for all m M. We need to show that there exists some θ : M A such that ( f)(θ) = φ, i.e. f(θ(m)) = φ(m) for all m M. Take m M. Since g(φ(m)) = 0 we have that φ(m) ker g = im f, hence there exists a unique element θ(m) A such that f(θ(m)) = φ(m). This defines a map θ : M A. It is an R-module homomorphism because f(θ(rm)) = φ(rm) = rφ(m) = rf(θ(m)) = f(rθ(m)), hence θ(rm) = rθ(m).