Section 11.4 and 12.1.
Homework #6
Masaya Sato
Section 11.4 3. Let R be any commutative ring with 1, let V be an R-module and let x1 , x2 , . . . , xn ∈ V . V .
Assume that for some A ∈ M n
×n
(R),
x A ... = 0. 1
xn
Prove that (det A)xi = 0, for all i ∈ {1, 2, . . . , n}. Proof. Consider the following two cases. Case 1. Suppose all xi ’s are 0. Then it is obvious that (det A)xi = 0 for all i = {1, 2, . . . , n}. Case 2. Suppose there exist some xi such that xi = 0. Then the equation
x A ... = 0. 1
xn
has a nontrivial solution. Therefore the set of column vectors {a1 , a2 , . . . , an } of the matrix dependent. Hence Hence det A = 0 and thus (det A)xi = 0 for all A = [a1 a2 · · · an] are linearly dependent. i = {1, 2, . . . , n}. 6. (Minkowski’s Criterion) Suppose that A is an n × n matrix with real entries such that
the diagonal elements are all positive, the off-diagonal elements are all negative and the row sums are all positive. Prove that det A = 0. [Consider the corresponding system of equations (x1 , . . . , xn ). If x If xi has the largest absolute AX = 0 and suppose there is a nontrivial solution (x value show that the ith equation equation leads to a contradi contradiction ction.] .] Proof. Suppose by contradiction that det A = 0, i.e. the matrix equation AX = 0 has a
nontrivial solution X = (x1 , . . . , xn), where not all xi ’s are zero. zero. Then Then choose choose xi = 0 such that |xi| is the largest value. For the corresponding system of equations of x1 , . . . , xn the ith equation is given by ai1 x1 + · · · + aii 1 xi −
−1
+ aii xi + aii+1 xi+1 + · · · + ainxn = 0,
where ai1 + · · · + aii 1 + aii + aii+1 + · · · + ain > 0. So even even if xi > 0 or xi < 0 the above equation cannot cannot be equal equal to zero. This This is contra contradic dictio tion n and theref therefore ore det A = 0 as ith equation desired. −
Section 12.1 1. Let M be a module over the integral domain R.
(a) Suppose x is a nonzero torsion element in M . Show that x and 0 are “linearly dependent.” Conclude that the rank of Tor(M Tor( M ) is 0, so that in particular any torsion R-module has rank 0. Abstract Abstract Algebra Algebra by Dummit and Foote 1
Section 11.4 and 12.1.
Homework #6
Masaya Sato
(b) Show that that the rank rank of M of M is the same as the rank of the (torsion free) quotient M/Tor( ). M/ Tor(M M ). Tor(M ), ), there exists a nonzero r ∈ R such that rx = 0. Therefore the Proof. (a) Since x ∈ Tor(M equation rm + r 0 = 0 holds true, where r ∈ R, and hence x and 0 are linearly dependent. Every nonzero x ∈ M has some nonzero r ∈ R such that rx = 0 since R is an integral domain, so the maximum number of linearly independent elements is 0. Thus Tor( M ) is of rank 0. (b) Define an R-module homomorphism ϕ : M → M/Tor( M/ Tor(M M ) by
Tor(M ). ). ϕ(x) = x + Tor(M By its construct construction ion ϕ is surjective. Moreover for every x ∈ ker ϕ ≤ M Tor(M ) ⇒ x + Tor(M Tor(M ) = 0 + Tor(M Tor(M ) ϕ(x) = 0 + Tor(M Tor(M ). ). ⇒ x ∈ Tor(M Now observe that there exists some r1 , . . . , rn ∈ R such that x = r1 x1 + · · · + rnxn and moreover there is some nonzero r so that rx = 0. Then rx = 0 ⇒ r(r1 x1 + · · · + rn xn ) = 0 (rrn)xn = 0 ⇒ (rr 1)x1 + · · · + (rr ⇒ rri = 0 ∀i = 1, · · · , n since {x1 , · · · , xn} is a basis for M . Furthermore rri = 0 implies that ri = 0 since R is an integral integral domain. Therefor Thereforee ker ker ϕ = 0 and hence ϕ is an isomorphism. isomorphism. Thu Thuss M/Tor( M/ Tor(M M ) is also of rank n. 2. Let M be a module over the integral domain R.
(a) Suppose that M has rank n and that x1 , x2 , . . . , xn is any maximal set of linearly independent elements of M . Let N = Rx1 + · · · + Rxn be the submodule generated by Prove that that N is isomorphic to Rn and that the quotient M/N is a torx1 , x2 , . . . , xn . Prove sion R-module (equivalently, the elements x1 , . . . , xn are linearly independent and for any y ∈ M there is a nonzero element r ∈ R such that ry can be written as a linear combination r1 x1 + · · · + rn xn of the xi ). (b) Prove that conversely that if M contains a submodule N that is free of rank n (i.e. N ∼ = Rn) such that the quotient M/N is a torsion R-module then M has rank n. [Let y1 , y2 , . . . , yn+1 be any n + 1 elements of M . Use the fact that M/N is torsion to write ri yi as a linear combination of a basis for N for some nonzero elements r1, . . . rn+1 of R. Use an argument as in the proof of Proposition 3 to see that the ri yi , and hence also the linearly dependent. dependent.]] yi, are linearly Abstract Abstract Algebra Algebra by Dummit and Foote 2
Section 11.4 and 12.1.
Homework #6
Masaya Sato
Proof. (a) Think of Rn as an R-module and define an R-module map ϕ : N → Rn by
ϕ(r1 x1 + · · · + rn xn) = (r1 , . . . , rn ). It is easy to see that ϕ is an R-module homomorphism homomorphism and moreove moreoverr surjectiv surjectivee by its construct construction. ion. So for all x ∈ N with some r1 , . . . , rn ∈ R such that x = r1 x1 + · · · + rnxn if x ∈ ker ϕ, then ϕ(x) = 0R ⇒ ϕ(r1 x1 + · · · + rn xn ) = 0R ⇒ (r1 , . . . , rn ) = 0R . n
n
n
Thus ri = 0 for i = 1, . . . , n and x = 0. Therefor Thereforee ker ϕ = 0 and hence ϕ is injective injective.. This proves that ϕ is an isomorphism. isomorphism. Now observe observe that that {x1 , . . . , xn } is a basis for M since M has rank n. So for eve every ry y ∈ M and some nonzero r ∈ R ry ∈ M is expressed as a linear combination of x1, . . . , and xn , i.e. ry = r1 x1 + · · · + rnxn , where ri ∈ R for i = 1, . . . , n. n. This implies that ry ∈ N and ry + N = 0 + N ∈ M/N and thus y is a torsion element in the quotient M/N . (b) Suppose that N is isomorphic to Rn and M/N is a tors torsio ion n modul module. e. Th Then en choos choosee a basis {x1 , . . . , xn} for N and extend this basis {x1 , . . . , xn, xn+1 , . . . , xm} for M . M has rank m ≥ n, so suppose by contradiction that m > n. Observe that xn+1 + N , . . . , xm + N form a basis for M/N . Since M/N is a torsion module, there exist some ri ∈ R such that ri (xi + N ) N ) = ri xi + N = 0 + N ∈ M/N for i = n + 1, . . . , m. m. I.e. ri xi ∈ N . N . Then for every x ∈ M there exist ai ∈ R for i = 1, . . . , m such that x = a1 x1 + · · · + an xn + an+1 xn+1 + · · · + amxm. Now let r = rn+1 · · · rm. So rx ∈ M since M is a R-module. However for i = n + 1, 1, . . . , m r (aixi ) = (rai )xi = (ai r )xi = ai (rxi ) ∈ N since R is an integral domain and N is an R-submodule. -submodule. This contradict contradictss the assumption assumption that M has rank m such that m > n. n . Therefore m = n and thus M is of rank n. 3. Let R be an integral domain and let A and B be R-modules of rank m and n, respectively.
Prove that the rank of A ⊕ B is m + n. [Use the previous exercise.] Abstract Abstract Algebra Algebra by Dummit and Foote 3
Section 11.4 and 12.1.
Homework #6
Masaya Sato
Proof. Observe that A ⊕ B is finite dimensional since A and B are both finite dimensional. Then by the Second Isomorphism Theorem for modules
(A + B )/B ∼ A/(A ∩ B ). = A/( Moreover since A ∩ B = 0,
(A ⊕ B )/B ∼ = A.
Therefore dim(A dim(A ⊕ B )/B = dim A ⇒ dim A ⊕ B − dim B = dim A ⇒ dim A ⊕ B = dim A + dim B = m + n and thus A ⊕ B is of rank m + n. 4. Let R be an integral domain, let M be an R-module and let N be a submodule of M .
Suppose that M has rank n, N has rank r and the quotient M/N has rank s. Prove Prove that that [Let x1 , x2 , . . . , xs be elements of M whose images in M/N are maximal set of n = r + s. [Let independent elements and let xs+1 , xs+2 , . . . , xs+r be a maximal set of independent elements in N . Prove Prove that that x1 , x2 , . . . , xs+r are linearly independent in M and that for any element y ∈ M there is a nonzero element r ∈ R such that ry is a linear combination of these elements. Then use Exercise 2.] Proof. Let {x1 , . . . , xr } be a basis for N and extend this to a basis {x1 , . . . , xr , xr+1 , . . . , xn
for M . Observe that xi + N = 0 + N ∈ M/N for i = 1, . . . , r since each is a basis element for N . Now claim that
{xr+1 + N , . . . , xn + N } form a basis for M/N . Then for every x ∈ M there exist a1 , . . . , an ∈ R such that x = a1 x1 + · · · + an xn ⇒ x + N = (ar+1 xr+1 + · · · + an xn) + N ⇒ x + N = ar+1 (xr+1 + N ) + · · · + an (xn + N ) N ) ∈ M/N . So the set {xr+1 + N , . . . , xn + N } generates M/N . Moreover for ar+1 , . . . , an ∈ R ar+1 (xr+1 + N ) + · · · + an (xn + N ) = 0 + N ⇒ ar+1 xr+1 + · · · + an xn ∈ N . Thus ar+1 xr+1 + · · · + an xn = 0 and ar+1 = · · · = an = 0 because otherwise ar+1 xr+1 + · · · + an xn would not be expressed uniquely. Therefore
{xr+1 + N , . . . , xn + N } is linearly independent and then the set form a basis for M/N . Hence M/N has rank n − r and n−r = s⇒n=r+s as desired. Abstract Abstract Algebra Algebra by Dummit and Foote 4
Section 11.4 and 12.1. 5. Let R =
Homework #6
Masaya Sato
Z[x]
and let M = (2, (2, x) be the ideal generated by 2 and x, considered as a submodule of R. Sh Shoow tha thatt {2, x} is not a basis of M . [Fin [Find d a nontr nontriv ivia iall R-linear dependence dependence between between these two elements elements.] .] Sho Show w that the rank of M is 1 but that M is not free of rank 1 (cf. Exercise 2). of M since {2, x} is linearly dependent i.e. the equation in terms Proof. {2, x} is not a basis of M of r1 and r2 r1 2 + r2 x = 0 has a nontrivia nontriviall solution solution r1 = −x and r2 = 2 because multiplication is commutative. So the maximum number of a linearly independent set is 1 and thus M is of rank 1. However M is not generated by any nonzero single element since M = (2, (2, x), which is a nonprincipal ideal of R. 6. Show that if R is an integral domain and M is any nonprincipal ideal of R then M is
torsion free of rank 1 but is not a free R-module. Proof. Observe first that M is an R-submodule of R of R. For all nonzero x, y ∈ M the equation
in terms of x and y r1 x + r2 y = 0, where r1 and r2 are elements in R, has a nontrivial solution r1 = −y and r2 = x because multiplication is commutative. So the maximum number of linearly independent set is 1 and thus M is of rank 1. Moreover for every nonzero x ∈ M rx = 0 for every nonzero r ∈ R since R is an integ integral ral domain. domain. Th Thus us M is a torsion free module, but M is not free because M is nonprincipal ideal of R, i.e. M cannot be generated by any nonzero single element in M . 7. Let R be any ring, let A1 , A2 , . . . , Am be R-modules and let Bi be submodules of Ai ,
1 ≤ i ≤ m. Prove that (A1 ⊕ A2 ⊕ · · · Am)/(B1 ⊕ B2 ⊕ · · · ⊕ Bm) ∼ = (A1 /B1 ) ⊕ (A2 /B2) ⊕ · · · ⊕ (Am/Bm). Proof. Define an R-module homomorphism ϕ : A1 ⊕ A2 ⊕ · · · ⊕ Am → (A1 /B1 ) ⊕ (A2 /B2 ) ⊕
· · · ⊕ (Am/Bm) by ϕ(a1 ⊕ a2 ⊕ · · · ⊕ am) = (a1 + B1) ⊕ (a2 + B2 ) ⊕ · · · ⊕ (an + Bn). ϕ is well-defined and moreover surjective by its construction. Then the kernel ker ϕ is given by all elements x1 ⊕ x2 ⊕ · · · ⊕ xm ∈ A1 ⊕ A2 ⊕ · · · ⊕ Am such that ϕ(x1 ⊕ x2 ⊕ · · · ⊕ xm) = 0 + B1 ⊕ 0 + B2 ⊕ · · · ⊕ 0 + Bm. So x1 ⊕ x2 ⊕ · · · ⊕ xm ∈ (A1 ⊕ A2 ⊕ · · · ⊕ Am) ∩ (B1 ⊕ B2 ⊕ · · · ⊕ Bm) Abstract Abstract Algebra Algebra by Dummit and Foote 5
Section 11.4 and 12.1.
Homework #6
Masaya Sato
and x1 ⊕ x2 ⊕ · · · ⊕ xn ∈ (A1 ∩ B1) ⊕ (A2 ∩ B2 ) ⊕ · · · ⊕ (Am ∩ Bm) and hence x1 ⊕ x2 ⊕ · · · ⊕ xn ∈ B1 ⊕ B2 ⊕ · · · ⊕ Bm. Therefore ker ϕ = B1 ⊕ B2 ⊕ ··· ⊕ Bm and thus ϕ induces the isomorphism ϕ between (A1 ⊕ A2 ⊕ · · · Am)/(B1 ⊕ B2 ⊕ · · · ⊕ Bm) and (A (A1 /B1 ) ⊕ (A2 /B2 ) ⊕ · · · ⊕ (Am/Bm). 9. Give an example of an integral domain R and a nonzero torsion R-module M such that
Ann(M Ann(M ) = 0. Prove Prove that that if N is finitely generated torsion R-module then Ann(N Ann(N ) = 0. Proof. For F (R, R), the set of all functions from
to R, let M be a subset of F (R, R) such that every element f ∈ M is a functi function on with with compac compactt support support.. Then Then M is a module over itself. itself. Moreov Moreover, er, M is a nonzero torsion module since for every f ∈ M there exists some g ∈ M , whose compact support is distinct from the one of f , f , such that R
gf = o, where o : R → R denotes the zero function. However any nonzero g ∈ M does not annihilate annihilate all f ∈ M since some f has distinct distinct compact compact support. support. Now suppose that a torsion R-module N is generated by x1 , . . . , xn , i.e. N = Rx1 ⊕ · · · ⊕ Rxn. So for every nonzero x ∈ N there exist unique a1 , . . . , an ∈ R such that x = a1 x1 + · · · + an xn. Note that for each xi there is a nonzero ri ∈ R so that ri xi = 0 since N is a torsion module. Then for nonzero r = r1 . . . rn (ran )xn rx = r(a1 x1 + · · · + an xn ) = (ra1 )x1 + · · · + (ra = (a1 · · · rn)(r )(r1 x1 ) + · · · + (a (an · · · rn 1 )(r )(rnxn ) = 0+ ··· +0 =0 −
because R is an integral domain. Therefore r ∈ Ann(M Ann(M ) and thus Ann(M Ann(M ) = 0. If M is finitely generated module over the P.I.D. R, describe the structure of M/Tor( ). 13. If M M/ Tor(M M ). Solution: Let M be an R-module of rank n. Then observe that M/Tor( M/ Tor(M M ) is a torsion free quotient module. By existence and uniqueness of the Fundamental Theorem ,
M = R ⊕ · · · ⊕ R.
Abstract Abstract Algebra Algebra by Dummit and Foote 6
Section 11.4 and 12.1.
Homework #6
Masaya Sato
15 Prove that if R is a Noetherian ring then Rn is a Noetherian R-module. [Fix a basis of
If M is a submodule of R of Rn show that the collection of first coordinates of elements of M Rn . If M is a submodule of R of R hence hence is finitely finitely generated. generated. Let m1 , m2 , . . . , mk be elements of M whose first coordinates coordinates generate generate this submodule submodule of R. Show that any element of M can be written as an R-linear combination of m1 , m2 , . . . , mk plus an element of M whose first coordinates is 0. Prove Prove that that M ∩ Rn 1 is a submodule of Rn 1 is the set of elements of Rn with first coordinate 0 and then use induction on n. −
−
-module.. Then Then obser observe ve also that for a Proof. Observe first that Rn = R ⊕ · · · ⊕ R is an R-module direct sum of ideals of Rn I 1 ⊕ · · · ⊕ I n ⊆ J 1 ⊕ · · · ⊕ J n if and only if I i ⊆ J i for all i = 1, . . . , n. consider an infinite infinite ascending ascending sequence sequence of n. Now consider ideals I 11 ⊕ · · · ⊕ I 1n ⊆ · · · ⊆ I k1 ⊕ · · · ⊕ I kn ⊆ I k1+1 ⊕ · · · ⊕ I kn+1 ⊆ · · · . The for each i there exists ki such that I ki = I ki +1 because R is Noeth Noether eria ian. n. Let Let l = max {ki |i = 1, · · · , n}. Therefore i
i
I l1 ⊕ · · · ⊕ I ln = I l1+1 ⊕ · · · ⊕ I ln+1 and hence Rn is a Noetherian R-module by the Ascending Chain Condition .
Abstract Abstract Algebra Algebra by Dummit and Foote 7
