math202hw8sols

7/24/2019 Math202hw8sols

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Math 202 - Assignment 8 Solutions

Solutions by: Elizabeth Pannell, Toby Stockley, and Wei Yuan

Supplemental Problem

Both maps are impossible. The easier part is that ifm < n, then it is impossible to have a surjectionRm Rn. To see this, we can choose any maximal ideal I R, and note that a surjection Rm Rn

would yield upon reduction modulo Ia surjection (R/I)m (R/I)n of vector spaces over the fieldR/I. This is impossible by basic linear algebra (for instance, by the rank nullity theorem).

Unfortunately, this technique of reducing to a field does not work directly for the other part of theproblem, since modding out by Idoes not preserve injectivity (in the more advanced language from afuture section, this is saying that tensoring with R/I over R is right exact, but not left exact).

We therefore give a direct proof. Suppose we have an R-module injection Rm Rn with m > n.This is equivalent to the existence ofm elements ofRn that areR-linearly independent. We will provethat givenn + 1 elements ofRn, there must exist a non-trivial dependence between them.

Suppose for a contradiction that we have row vectors vi Rn, fori = 1, . . . n + 1, that are linearly

independent over R. LetMbe the (n+ 1) n matrix with entries in R created from the row vectorsvi. We will prove the following statement for 0 t n 1, by induction on t:

The determinant of the (n t) (n t) matrix created by deleting any (t+ 1) rows andanyt columns from Mis equal to zero.

This claim for t = n 1 implies that M = 0, which is clearly a contradiction to the linearindependence of the rows ofM, and completes the proof. It remains to show the claim.

Base case of claim (t= 0): We must show that if we remove one of the rowsvi, the determinantof the n n matrix that remains is equal to 0. Let us call this determinant di. Then we claim thatif we let x:=

n+1i=1(1)

idivi Rn, then x= 0. Indeed, one can check that the j th component ofx,

namelyn+1

i=1(1)idivij , is equal to the determinant of the (n+ 1) (n+ 1) matrix obtained by

concatenating another copy of thej th column toM, and expanding by minors along this last column.Since the determinant of a matrix with a replicated column is zero, we obtain that x = 0 as claimed.Now, the R-linear independence of the vi implies that di= 0 for all i. This gives the base case t = 0.

Inductive step of claim: Let t >0, and delete t rows of the matrix M, yielding an (n+ 1 t) nmatrix N. Let us delete t columns ofN, and call the resulting (n+ 1 t) (n t) matrix L. Foreach i = 1, . . . , n+ 1 t, letdi denote the determinant of the matrix obtained from L by eliminatingthe ith row ofL. Our goal is to show that each di = 0. We claim that

n+1ti=1 (1)

idiwi = 0, wherewe have written wi for the ith row ofN. To see this, note that the jth component of this sum is the determinant of the square matrix obtained by concatenating a copy of the jth column of Nto the matrix L. Each of these (n+ 1 t) (n+ 1 t) matrices either has a duplicated column(and hence has determinant zero) or has determinant zero by the inductive hypothesis. Thereforen+1t

i=1 (1)idiwi= 0 as claimed, implying thatdi= 0 for all i since thewiare R-linearly independent

(as these are a subset of the original vi). This completes the inductive step, hence the proof of theclaim and of the problem.

Exercise 10.4.3.

Clearly C is both a left and right R-module, so by the construction of tensor product in 10.4, CRCis an R-module. Similarily, CC C is a C-module, and since R C, CC C is an R-module as well.

Now, since {1, i} is a basis for C over R, {1 1, 1 i, i i, i 1} is a basis for CRCover R. Also,CC C = C (via a b ab), so CC C has dimension 2 over R.

So CR C has dimension 4 while CC C has dimension 2. Therefore, C R C and CC C are notisomorphic as R-modules.

Exercise 10.4.7.


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Supposex QR N. Thenx can be written as the finite sum of simple tensors, x =k

i=1

(ai/bi) (ni).

Letd = lcm{bi}and di= (d/bi). Then,

x=

k

i=1

(ai/bi) (ni) =

k

i=1

((aidi)/d) (ni)

=

k

i=1

(1/d) (aidini) = (1/d) (k

i=1

(aidini)),

with

k

i=1

(aidini) N sinceaidi R and Nis a left R-module.

Exercise 10.4.8.

First note the misprint: (u, n) (u, n) if and only ifun= un inN should read (u, n) (u, n)if and only ifxun = xun N for some x U. (The relation given in the book is not transitive,hence this fix is needed.)

(a) We show thatU1Nis an abelian group. Suppose (u, n), (u1, n1), (u2, n2),and (u3, n3) U1N.For associativity, we calculate:

((u1, n1) + (u2, n2)) + (u3, n3)

=(u1u2, u2n1+u1n2) + (u3, n3)

=((u1u2)(u3), +u3(u2n1+u1n2) + (u1u2)n3)

=((u1u2)(u3), +(u3u2)n1+ (u3u1)n2+ (u1u2)n3)

=(u1(u2u3), +(u2u3)n1+ (u1u3)n2+ (u1u2)n3)

=(u1(u2u3), +(u2u3)n1+u1(u3n2+u2n3))

=(u1, n1) + (u2u3, u3n2+u2n3)

=(u1, n1) + ((u2, n2) + (u3, n3)).

To see the existence of an identity, let e = (1, 0). Then,

(u, n) + (1, 0) = (u1, 1n+u0) = (u, n)

and(1, 0) + (u, n) = (1u, u0 + 1n) = (u, n).

For commutativity, note:

(u1, n1) + (u2, n2) = (u1u2, u2n1+u1n2) = (u2u1, u1n2+u2n1) = (u2, n2) + (u1, n1).

To see the existence of additive inverses, note that for any (u, n) U1N, we have

(u, n) + (u, n) = (u2,un+un) = (u2, 0) = (1, 0).

Therefore,U1Nis an abelian group. Next we check that U1N is an R-module. TheR action isgiven byr(u, n) = (u,rn). We check this has the desired properties:

1.

(r+s)(u, n) = (u, (r+s)n) = (u,rn+sn)

= (u2, u(rn+sn)) = (u,rn) + (u,sn)

=r(u, n) +s(u, n).


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2.(rs)(u, n) = (u, (rs)n) = (u, r(sn)) = r(u,sn) = r(s(u, n)).

3.

r((u, n) + (u1, n1)) = r(uu1, u1n+un1) = (uu1, r(u1n+un1))= (uu1, ru1n+run1) = (uu1,rnu1+urn1)

= (u,rn) + (u1, rn1)) = r(u, n) +r(u1, n1)

4.1(u, n) = (u, 1n) = (u, n).

ThereforeU1N is an R-module.

(b) Let h : Q NU1Nbe the map that sends (a/b,n) to (b,an). We will show thath is anR-balanced map. For a/b, c/d Q,m,n N, and r R we have

1.

h(a/b+c/d, m) = h((ad+cb)/(bd), m) = (bd, (ad+cb)m)

= (b,am) + (d,cm) = h(a/b,m) +h(c/d,m).

2.

h(a/b,m+n) = (b, a(m+n)) = (b2, ba(m+n))

= (b2,bam+ban)) = (b,am) + (b,an)

=h(a/b,m) +h(a/b, n).

3.h(a/b, rm) = (b, a(rm)) = (b, (ar)m) = h(ar/b, n).

Thereforeh is anR-balanced map, and hence by the universal property of tensor products there isa unique homomorphism f :Q RN U

1N such thatf(q n) = h(q, n).Next we claim that the map g :U1NQ RN given by g((u, n)) = (1/u) n is well defined.

To see this, suppose (u1, n1) = (u2, n2). Then there exists anx U such that xu1n2 = xu2n1. Wehave:

(1/u1) n1 =((1/u1)(1/(xu2))) (xu2n1)

=((1/u1)(1/(xu2))) (xu1n2)

=(1/u2) n2,

s g is well-defined.It is easy to see that f andg are inverses:

f(g((u, n))) =f((1/u) n) = (u, 1 n) = (u, n)

andg(f((a/b) n)) = g((b,an)) = (1/b) an= (a/b) n.

Therefore,U1Nis isomorphic to Q RNas an R-module.


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(c) Since U1Nis isomorphic to Q RN, we have

(1/d) n= 0 f((1/d) n) = (d, n) = 0 = (1, 0)

there exists r R {0}such that rn= 0.

(d) Suppose A is an abelian group. We use the results of parts (a), (b), and (c), with Q = Q,R = Z, and N =A. We see that every element ofQ ZA can be written (1/d) a for some a A,and this tensor element is 0 if and only na = 0 from some n Z, n = 0. In other words, QZA = 0if and only ifA is a torsion group.

Exercise 10.4.11.

Suppose v, w V such that e1 e2+e2 e1 =v w. Writev = ae1+be2 and w = ce1+de2for some a, b, c, d R. Then we calculate

(ae1+be2) (ce1+de2) = (ae1) (ce1) + (ae1) (de2) + (be2) (ce1) + (be2) (de2)

= (ac)(e1 e1) + (ad)(e1 e2) + (bc)(e2 e1) + (bd)(e2 e2).

Since the 4 vectors ei ej for i, j = 1, 2 form a basis ofVRV, this equals e1 e2+e2 e1 if andonly ifac = 0, ad = 1, bc= 1, and bd = 0. The first equation impliesa = 0 or c = 0. But ifa = 0,thenad = 0 = 1 and if c=0, then bc = 0 = 1. This is a contradiction. Therefore, there does not existv, w V such that e1 e2+e2 e1 = v w.

Exercise 10.4.12.

For the reverse direction, suppose thatv = av for some a F. Then

v v = (av) v =v (av) = v v.

For the forward direction, note that since v and v are nonzero, and there exists no scalar a Fsuch that v = av , it follows that v and v are linearly independent. We can therefore choose linearfunctionals, : V F such that

(v) = 1, (v) = 0, (v) = 1, (v) = 0.

The fact that such functionals exist relies essentially on the fact that we are dealing with vector spacesover a field; for example, we may complete {v, v} to a basis ofV and let , be the correspondingelements of the dual basis. One easily checks that the map VV Fgiven by (a, b) (a)(b) isan F-bilinear map, and hence induces an F-vector space map : VFV F. The image ofv v

under is 1, whereas the image ofv v under is 0. Therefore, v v =v v in VFV.

References

[DF] Dummit, David and Foote, Richard. Abstract Algebra, 3rd edition.

math202hw8sols

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