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Math 202, Homework #8 Solutions

Elizabeth Pannell, Toby Stockley, Wei Yuan

Exercise 10.4.19.

Show that 2 x x 2 is a torsion element:

2(2 x x 2) = 4 x 2x 2 = 2 2x 2 2x= 0

Show that 2 x x 2 is annihilated by both 2 and x:

2 : 2(2 x x 2) = 0 (from the calculation above)

x: x(2 x x 2) = 2x x x2 2 =x 2x x 2x= 0

The submodule ofIR Igenerated by 2xx2 is by definition cyclic, and hence has the form R/JwhereJ= Ann(2 x x 2). We have just shown that J contains the maximal ideal I (note that R/I= Z/2Zvia f(x) f(0) (mod 2)). Therefore, it remains to show thatJ=I, or equivalently that J=R. In otherwords, we must show that 2 x x 2 = 0 in IRI.

To this end, let : I I Z/2Z be the map (p(x), q(x)) p(0)2 q(0) (mod 2).

Claim: is anR-balanced map.

Proof:

Suppose p(x), q(x), m(x), n(x) Iandr R. Then,

(1)(p(x) +q(x), m(x)) = (p(0) +q(0))m(0)/2 (mod 2) =p(0)m(0)/2 (mod 2) +q(0)m(0)/2 (mod 2) =(p(x), m(x)) +(q(x), m(x))

(2)(p(x), m(x) + n(x)) = p(0)(m(0) + n(0))/2 (mod 2) =p(0)m(0)/2 (mod 2) +p(0)n(0)/2 (mod 2) =(p(x), m(x)) +(p(x), n(x))

(3)(p(x), rm(x)) = p(0)(rm(0))/2 (mod 2) = (p(0)r)n(0)/2 (mod 2) = (p(x)r, m(x)).

Therefore is anR-balanced map and hence induces a map : II Z/2Z We evaluate(2xx2) =(2 x) (x 2) = 2/2 0 (mod 2) = 1 (mod 2).

Therefore 2 x x 2 is nonzero as desired.

Exercise 10.4.20.

Let : I I Ibe the map such that (p, q) pq.

Claim: is an R-balanced map.

Proof:

Suppose p,q,m, n Iandr R. Then,

(1)(p+q, m) = (p+q)m= pm+qm = (p, m) +(q, m)

(2)(p, m+n) = p(m+n) = pm+pn= (p, m) +(p, n)

(3)(p, rm) = p(rm) = (pr)n= (pr, m)

Therefore, is an R-balanced map.

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So, by the universal property of the tensor product, induces the group homomorphism : I I I.

Now, suppose 2 2 +x x= a bfor some a, b I.

Then,

(2 2 +x x) = (a b)

(2 2) +(x x) = (a b)22 +x2 =ab.

But 22 +x2 is irreducible in Z[x], so a or b must be a unit. ButIdoes not contain any units, which is acontradiction.

Therefore, there does not exista, b Isuch that 2 2 +x x= a b.

Exercise 10.4.24.

Let us consider the map : Z[i] Z R C where((a+bi) r) = ra +rbi for a, b Z and r R.

First we must show this is a homomorphism:

([(a +bi) + (a+bi)]r) = ra +rbi+ra+rbi = (a +bi +r) +(a+bi r)

(a+bi (r+r)) = (r+r)a+ (r+r)bi= r(a+bi) +r(a+bi) = (a+bi r) +(a+bi r)

([a + bir] [a+ bir]) = rr (aabb)+ rr(ab+ ab)i= r(a + bi)r(a+ bi) = (a + bir)(a+ bir)

[In the 3rd step, the multiplication of simple tensors is well-defined by Proposition 21.]

Now all that remains is to show that is a bijection.

It is clear from properties of the tensor product that every element x Z[i]ZRcan be written in the formx = (1 c) + (i d) for some c, d R. This element maps under to (x) = c +id; this is equal to zeroonly ifc = d = 0, in which case x = 0. Thus is injective.

Similarly, for any c +di C, we note that (1 c) +(i d) = c+di, so is surjective.

Exercise 10.4.26.

We define a map f : S R[x1, , xn]/I S[x1, , xn]/IS[x1, , xn] by f(s, g) = sg. Here the bardenotes reduction mod I. It is easy to see that the mapfis well-defined (i.e. that the equivalence class ofsg modulo IS[x1, , xn] depends only on the equivalence class ofg modulo I) and that f is R-balanced.We therefore get an induced map

: SRR[x1, , xn]/I S[x1, , xn]/IS[x1, , xn]

ofS-modules. To check that is furthermore a map ofS-algebras (i.e. of rings), we note that

((s g) (s g) = ((ss) gg) = ssgg,

which is equal to (s g) (s g) = sg sg.

It remains to show that is bijective. We first make the following note: every element z S[x1, . . . , xn]can be written in the form z =

sigi where the gi are distinct simple monomials, i.e. elements of the form

xe11 xenn . In particular, each gi lies in R[x1, . . . , xn], and we see that (

si gi) = z, so is surjective.

For injectivity, it is convenient to first prove the caseI= 0. By properties of the tensor product, every elementofSRR[x1, . . . , xn] can be written in the form

si gi where the gi are distinct simple monomials. This

element maps under to

sigi; for this to be zero, we must have that each si is 0, since the gi are distinct

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simple monomials (and by the definition of S[x1, . . . , xn]. But if each si = 0, then the original elementsi gi is zero, giving the desired injectivity in the case I= 0.

For general I, suppose that

si gi SRR[x1, . . . , xn]/I maps to zero under . Then

sigi is anelement of IS[x1, . . . , xn]. Every element of IS[x1, . . . , xn] can be written in the form

hi s

i g

i wherehi I, s

i S and g

i is a simple monomial. Since hi g

i I by the definition of ideal, we can rewritethis as

sih

i with h

i I , s

i S. This says that

si gi and

si h

i have the same image under theisomorphism

SRR[x1, . . . , xn] =S[x1, . . . , xn],

corresponding to the case I= 0, which we have already proven. Since the map is an isomorphism, we havethat

si gi =

si h

i in SRR[x1, . . . , xn]. This implies that both elements have the same imagein SRR[x1, . . . , xn]/I. However,

si h

i has image 0 in SRR[x1, . . . , xn]/I, since each h

i I. Thisproves that

si gi = 0 as desired, completing the proof.

Exercise 10.4.27.

(a) Writing 1 = 1 1, e2 = 1 i, e3 = i 1, and e4= i i, we have

(a 1 +b e2+c e3+d e4)(a 1 +b e2+c

e3+d e4)

=(aa 1 +ab e2+ac e3+ad

e4)

+ (ba e2 bb 1 +bc e4 bd

e3)

+ (ca e3+cb e4 cc

1 cd e2)

+ (da e4 db e3 dc

e2+dd 1)

=(aa bb cc +dd) + (ab +ba cd dc)e2

+ (ac bd +ca db)e3+ (ad +bc +cb +da)e4.

(b) Let 1 = 12(1 1 +i i) =

12(1 +e4) and 2 =

12(1 1 i i) =

12(1 e4). It is clear that 1+2 = 1,

and the formulas 21 = 1, 22 = 2, 12 = 0 follow from part (a). We can then define a ring isomorphism

A A1A2 by the map a (a1, a2), with inverse map given by (x, y) x + y. One easily checks fromthe fact that 1, 2 are orthogonal idempotents that these maps are ring homomorphisms and inverse to oneanother. (Example: to see that (x, y) x + y is multiplicative, note that (x, y) (x, y) xx + yy.On the

other hand, (x+y)(x +y) =xx +yx +xy +yy . However, y = a2 andx =a1 for some a, a A, soyx =aa12 = 0. Similarly xy

= 0, so xx +yy = (x+y)(x +y). The other verifications are similar.)

(c) This is a trivial verification; for example, (z1 + z

1, z2) = ((z1 + z

1)z2, (z1 + z

1)z2) = (z1z2, z1z2) +(z1z2, z

1z2) = (z1, z2) +(z

1, z2), and (rz1, z2) = (rz1z2, rz1, z2) = r(z1, z2).

(d) We have (1) = 12((11)+(ii)) =

12((1, 1)+(1, 1)) = (0, 1). Similarly (2) =

12((1, 1)(1, 1)) =

(1, 0). To see C-linearity, we note that (z (z1 z2)) = ((zz1) z2) = (zz1z2, zz1z2) = z (z1z2, z1z2) =z(z1 z2). This implies surjectivity since (a1+b2) =a(1) +b(2) = (a, 0) + (0, b) = (a, b) for anya, b C. It is easy to see from the definition of that it is a ring homomorphism. Combined with the factthat it is a C homomorphism, we see that it is a C-algebra homomorphism. Since it is surjective and bothsides have dimension 4 over R, it must be a C-algebra isomorphism.

Written problem 1.

Let M = Q ZZ and let N =

(Q Z Z) =

Q. As we saw in Problem 7 above, any element ofM

can be written as (1/d) (ai)

i=1 with ai Z. We can map this element to (ai/d)

i=1 N. This is clearlyan injective map of abelian groups. However, it is not surjective; the image ofM is the set of tuples withbounded denominator. For example, (1/n)n=1 is not in the image ofM.

Written problem 2.

LetM= QZZ/2iZ and letN=

(QZ Z/2iZ).

Each Z/2iZ is a torsion group, so Q Z/2iZ = 0; therefore N= 0.

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However, M is non-zero since

Z/2iZ contains non-torsion elements; for example, it is easy to see that(1, 1, 1, . . . ) is a non-torsion element of

Z/2iZ and hence has non-zero image in M.

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