Advanced Algebra II Homework 3

Advanced Algebra II Homework 3

Yang Ju-Chen B95201057

2007/03/23 1. Determine Z^m ZZⁿ.

Proof.

Let the free Z module on Zm Zn be F; and let the submodule used to create tensor product be K: De…ne the Z module homomorphism f : F ! Z(m;n) by f a; b = ab: Now for any a1; a2; a 2 Z^m; b1; b2; b 2 Zⁿ and r 2 Z;

f (a1+ a2; b) a1; b a2; b = f a1+ a2; b f a1; b f a2; b

= (a1+ a2) b a1b a2b = 0;

f a; b1+ b2 a; b1 a; b2 = f a; b1+ b2 f a; b1 f a; b2

= a (b1+ b2) ab1 ab2 = 0;

f a; rb r a; b = f a; rb rf a; b

= a (rb) rab = 0;

f ra; b r a; b = f ra; b rf a; b

= (ra) b rab = 0;

so K ker f: We may identify F= ker f as a submodule of F=K = Z^m ZZⁿby a; b + ker f ! a; b + K = a _Zb. It’s clear that Im f = Z(m;n); so F= ker f = Z(m;n); hence

jZ^m ZZⁿj jF= ker fj = Z(m;n) = (m; n) :

For any a 2 Zm and b 2 Zn; we have

a _Zb = ab1 _Z1 2 1 Z1 _Z:

Since Z^m ZZⁿ is generated from all those elements like above, we conclude that Z^m ZZⁿ= 1 _Z1

Z: From elementary number theory, there exists x; y 2 Z such that xm + yn = (m; n) : Hence we have

(m; n) 1 _Z1 = xm1 _Z1 + yn1 _Z1 = xm _Z1 + y1 _Zn = 0 ) 1 Z1 j (m; n) ; in particular,

jZm ZZnj = 1 Z1 (m; n) :

Therefore, jZ^m ZZⁿj = 1 Z1 = (m; n) ; Z^m ZZⁿ= 1 _Z1 _Z=Z(m;n):

2. Let M = M₁ M₂. Prove that M is ‡at if and only if both M_i are ‡at. What can you say if M = _i2IMi with general index set I.

Proof.

CautionWhen I is any index set, the symbol _i2IMi actually means all the function ' on I such that ' (i) 2 Mi and ' (i) = 0 for all but …nitely many.

Given any exact 0 ! N1 ! N2; say f_i : N₁ M_i ! N2 M_i; and f : _i2I(N₁ M_i) !

i2I(N2 Mi) de…ned naturally by fi: Claim ker f = _i2Iker f_i:

dGiven ' 2 ker f; then f (') = 0 ) f (') (i) = fi(' (i)) = 0; so ' (i) 2 ker fi for all i: Since ' 2 ker f i2I(N1 Mi) ; there is only …nite i such that ' (i) 6= 0; hence ' 2 i2Iker fi:

Now given ' 2 i2Iker f_i; say 0 6= ' (i) 2 ker fifor …nite i 2 I0; then f_i(' (i)) = f (') (i) = 1

0 for all i 2 I; hence ' 2 ker f:c

By theorem IV.5.9 on Hungerford, we have

N M = _i2I(N Mi)

for any module N: Above claim says that f_i is injective for all i 2 I i¤ f is, that is, 0 ! N¹ M = _i2I(N1 Mi)!^f i2I(N2 Mi) = N2 M is exact i¤ 0 ! N¹ Mi

fi

! N² Mi is exact for all i 2 I: Hence M is ‡at i¤ Mⁱ is for all i:

3. Let R be a local ring and M is a …nitely generated ‡at R-module. Then M is free.

In fact, if fx¹; :::; xng M such that they form a basis in M=mM over R=m, then it forms a basis of M .

Proof.

Since M is …nitely generated, so is for M=mM; hence dimKM=mM = n < 1 for some n 2 N; where K R=m: Let fx¹; :::; xng M such that fx¹; :::; xng forms a basis of M=mM;

and let : Rⁿ! M de…ned by

(r1; :::; rn) =X

rixi; ri2 R:

Corollary 1.4.11 says hx¹; :::; xni = M; so is surjective. Do the right exact operation m on the exact sequence:

0 ! ker ! Rⁿ! M ! 0 to get the following exact commute diagram:

m ker ! m Rⁿ ! m M ! 0

# # #

0 ! ker ! Rⁿ ! M

Say f : m ker ! ker , g : m Rⁿ ! Rⁿ and h : m M ! M; Snake lemma induces the following exact sequence:

ker h ! coker (f) ! coker (g) ! coker (h) : Now since M is ‡at, exact sequence 0 ! m ! R induces that

0 ! m M ! R M = M is exact, that is, ker h = 0: Therefore,

0 ! coker (f) = ker =m ker ! coker (g) = Rⁿ=mRⁿ= (R=mR)ⁿ! coker (h) = M=mM:

But actually (R=mR)ⁿ = M=mM because we takefx¹; :::; xng as basis of M=mM over R=mR;

so we conclude that ker =m ker = 0; that is, ker = m ker : When R is Noetherian, or M is projective, or more generally, M is …nitely presented (see Lang: Lemma XVI.3.9), ker is

…nitely generated. We can therefore apply Nakayama lemma on the relation ker =m ker = 0 to conclude that ker = 0 and Rⁿ= M .

4. Let 0 ! M¹! M² ! M³ ! 0 be an exact sequence of R-modules. If both M¹; M3 are …nitely generated then so is M₂.

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Proof.

Let f : M₁ ! M2 and g : M₂ ! M3; then Im f = ker g is …nitely generated, say fxig ; because M₁ is. Let fzjg be a generating set of M3; surjection g permits us to choose one yj 2 g ¹fz^jg for each j; consider X = fxⁱ; yjg : For each x 2 M²; we have for some sj 2 R

g (x) =X

j

sjzj =X

j

sjg (yj) = g 0

@X

j

sjyj

1

A ) x X

j

sjyj 2 ker g;

hence for some ri2 R;

x =X

i

rixi+X

j

sjyj 2 hXi ; so M₂ is …nitely generated.

5. * Tensor product commutes with direct limit. That is, lim

!(Mi N ) = (lim

!Mi) N:

Proof.

See Exercise 14 to 20 in Atiyah chapter 2.

6. A ring S is said to be an R-algebra if there is a ring homomorphism R ! S. Show that S is an R-module.

Proof.

Let the ring homomorphism be f; de…ne the operation : R S ! S by r s f (r) s:

For all a; b 2 R, x; y 2 S, we have

a (x + y) = f (a) (x + y) = f (a) x + f (a) y = a x + a y;

(a + b) x = f (a + b) x = f (a) x + f (b) x = a x + b x;

a (b x) = a (f (b) x) = f (a) f (b) x = f (ab) x = (ab) x;

1_R x = f (1_R) x = 1_Sx = x;

so S is an R module.

7. Let S be a ‡at R-algebra and M be a ‡at S-module. Then M is a ‡at R-module.

Proof.

We …rst prove the following useful lemma:

Lemma 1 (Atiyah Exercise 2.15) Let A; B be rings, let M be an A module, P a B module and N an (A; B) bimodule. Then M AN is naturally a B module, N BP an A module, and we have

(M _AN ) _BP = M A(N _BP ) : Proof of Lemma.

For simply, we always use x; y; z; a; b to denote elements in M; N; P; A; B respectively. Any original product will denote without any speci…c symbol. M AN is an A module under A tensor construction, and is module B under new de…ned _{M AN B} :

(x _Ay) _{M AN B}b x _A(yb) : Moreover, we have

(a (x _Ay)) _{M AN B}b = ((ax) _Ay) _{M AN B}b = (ax) _A(yb)

= a (x _A(yb)) = a ((x _Ay) _{M AN B}b) ; 3

so M AN is an (A; B) bimodule. Similarly, N BP is an (A; B) bimodule under B tensor construction and _{A N BP} :

a _{A N BP} (y _Bz) (ay) _Bz:

(M _AN ) _BP is an (A; B) bimodule under B tensor construction and _{A (M AN ) BP} : a _{A (M AN ) BP} (x _Ay) _Bz (a (x _Ay)) _Bz:

M _A(N _BP ) is an (A; B) bimodule under A tensor construction and _{M A(N BP )}: x _A(y _Bz) _{M A(N BP )}b x _A((y _Bz) b) :

Fix an z 2 P: The mapping (x; y) ! x A(y _Bz) is clearly A linear in x, and x _A((ay) _Bz) = x _A(a _{A N BP} (y _Bz)) = a (x _A(y _Bz)) ;

where the last equality holds by A linearity of _A: Hence (x; y) ! x A(y _Bz) is A linearity in both x and y; universal property induces a A module homomorphism f_z : M _AN ! M A(N BP ) such that

fz(x _Ay) = x _A(y _Bz) :

Next, consider the mapping (t; z) ! f^z(t) from (M AN ) P into M A(N BP ) ; f_z((x _Ay) _{M AN B}b) = f_z(x _A(yb)) = x _A((yb) _Bz)

= x _A((y _Bz) b) = (x _A(y _Bz)) _{M A(N BP ) B}b

= f_z(x _Ay) _{M A(N BP ) B}b;

f_bz(x _Ay) = x _A(y _B(bz)) = x _A((y _Bz) b) = f_z(x _Ay) _{M A(N BP ) B}b;

so (t; z) ! f^z(t) is B linear in both t and z (summation homomorphic property is easy to verify). Universal property induces a B module homomorphism f : (M _AN ) _BP ! M A

(N _BP ) such that

f ((x Ay) Bz) = x A(y Bz) : Moreover,

f a _{A (M AN ) BP} ((x _Ay) _Bz) = f ((a (x _Ay)) _Bz) = f (((ax) _Ay) _Bz)

= (ax) _A(y _Bz) = a (x _A(y _Bz))

= af ((x _Ay) _Bz) ;

hence f is an (A; B) bimodule homomorphism. Similarly, we may construct an (A; B) bimodule homomorphism g : M _A(N _BP ) ! (M AN ) _BP such that

g (x _A(y _Bz)) = (x _Ay) _Bz:

Clearly f g and g f are identity maps, hence f and g are (A; B) bimodule isomorphism.

Since S is a R algebra, in particular, S is a (R; S) bimodule. Now given any exact R module sequence 0 ! N1 ! N2; that S is R ‡at implies 0 ! N1 RS ! N2 RS is exact. Ni RS is nothing but extension of scalars from R to S; hence S ‡atness of M implies 0 ! (N¹ RS) _SM ! (N² RS) _SM is exact. By previous lemma and Proposition 1.5.7,

(Ni RS) SM = Ni R(S SM ) = Ni RM;

that is, 0 ! N^{1 R}M ! N^{2 R}M is exact. Therefore, M is a ‡at R-module.

8. * Complete the exercises and incomplete proofs in the note.

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