Download Solutions to Assignement 1

MATH470 Winter 2016
Solutions to Assignment 1
Due Friday January 22, 2016
1. (a) The center of a ring R is
C = {z ∈ R : zr = rz
for all r ∈ R} ,
i.e. it is the set of all elements which commute with every element of
R. Prove that the center of a ring is a subring that contains the identity.
Prove that the center of a division ring is a field.
(b) Describe the center of the real Hamilton quaternions H.
Prove that {a + bi : a, b ∈ R} is a subring of H which is a field, but it is
not contained in the center.
(c) Let G = {g1 , . . . , gn } be a finite group. Show that the element N =
g1 + . . . + gn is in the center of the group ring RG.
Solutions: (a) It is clear that 1 ∈ C since 1 · r = r = r · 1 for all r ∈ R. Let
z1 , z2 ∈ C. We have to show that z1 + z2 and z1 z2 are in C. For any r ∈ R, we
have
(z1 + z2 )r = z1 r + z2 r = rz1 + rz2 = r(z1 + z2 ),
and z1 + z2 ∈ C. Similarly,
(z1 z2 )r = z1 (z2 r) = z1 (rz2 ) = (z1 r)z2 = r(z1 z2 ),
and z1 z2 ∈ C.
If R is a division ring, we also have that x ∈ C implies that x−1 ∈ C. Indeed,
let x ∈ C, r ∈ R such that xr = rx. Taking inverses (since R is a division ring,
every non-zero element of R has a multiplicative inverse), then
(xr)−1 = r−1 x−1 = (rx)−1 = x−1 r−1
and x−1 commutes with every element s ∈ R (every s 6= 0 ∈ R can be written
as s = r−1 for some r). Then, the center of R is a commutative ring with 1 in
which every element has an inverse, i.e. it is a field.
(b) Let z = a + bi + cj + dk ∈ H. Then
zj = aj + bij + cj 2 + dkj = −c − di + aj + bk
jz = aj + bji + cj 2 + djk = −c + di + aj − bk
Then, z does not commute with j unless d = b = 0, and we are left to consider
z = a + cj. In this case, zi = ai − ck and iz = ai + ck, and this forces c = 0,
and we are left to consider z = a ∈ H. For such z ∈ H, we have that
a(a0 + b0 i + c0 j + d0 k) = aa0 + ab0 i + ac0 j + ad0 k
(a0 + b0 i + c0 j + d0 k)a = aa0 + ab0 i + ac0 j + ad0 k
and z = a commutes with every element of H. This shows that the center of
the quaternion is the set of real quaternions of the form z = a, a ∈ R. Then
S = {a + bi : a, b ∈ R} in not contained in the center of H.
We now show that S = {a + bi : a, b ∈ R} which is a subring of H: it contains
1, it is clearly closed under addition, and
(a + bi)(a0 + b0 i) = aa0 + (ba0 + a0 b)i + bb0 i2
= (aa0 − bb0 ) + (ba0 + a0 b)i ∈ S.
Also, every element z = a + bi 6= 0 in S has an inverse
z −1 =
a − bi
a2 + b 2
in S. Finally, S is commutative, since
(a + bi)(a0 + b0 i) = (aa0 − bb0 ) + (ba0 + a0 b)i = (a0 + b0 i)(a + bi).
This proves that H is a field (it is the field of complex numbers).
(c) Let g be any element of the group G. Then, left multiplication by g is a
bijection of the group G, i.e.
{gg1 , . . . , ggn } = {g1 , . . . , gn } ,
and the same holds for right multiplication by g. Then,
g(g1 + . . . + gn ) = g1 + . . . + gn = (g1 + . . . + gn )g,
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and g1 + . . . + gn commutes with g. By distributivity, we have that for any
element a1 g1 + . . . + an gn ∈ RG,
(g1 + . . . gn )(a1 g1 + . . . + an gn ) = a1 g1 (g1 + . . . gn ) + . . . + an gn (g1 + . . . gn )
= (a1 g1 + . . . + an gn )(g1 + . . . + gn ),
which shows that (g1 + . . . + gn ) is in the center of RG.
2. (a) Let R and S be rings. Prove that the direct product R × S is a ring
under componentwise addition and multiplication. Prove that R × S is
commutative if and only if both R and S are commutative. Prove that
R × S has an identity if and only if both R and S have identity.
(b) Prove that {(r, r) : r ∈ R} is a subring of R × R.
Solutions: (a) Let R × S = {(r, s) : r ∈ R, s ∈ S} with addition and multiplication defined by
(r1 , s1 ) + (r2 , s2 ) = (r1 + r2 , s1 + s2 )
(r1 , s1 ) · (r2 , s2 ) = (r1 r2 , s1 s2 ).
Then, 0R×S = (0, 0) and −(r, s) = (−r, −s) is the additive inverse of (r, s).
Also, addition is commutative and associative because it is in each component.
Similarly, multiplication is associative because it is in each component. The
same holds for distributivity on the left,
(r1 , s1 ) ((r2 , s2 ) + (r3 , s3 )) = (r1 (r2 + r3 ), s1 (s2 + s3 )) = (r1 r2 + r1 r3 , s1 s2 + s1 s3 )
= (r1 r2 , s1 s2 ) + (r1 r3 , s1 s3 ) = (r1 , s1 )(r2 , s2 ) + (r1 , s1 )(r3 , s3 ),
and similarly for distributivity on the right. This shows that R × S is a ring.
Since for any r1 , r2 ∈ R, s1 , s2 ∈ S,
(r1 , s1 )(r2 , s2 ) = (r2 , s2 )(r1 , s1 )
⇐⇒
(r1 r2 , s1 s2 ) = (r2 r1 , s2 s1 )
⇐⇒
r1 r2 = r2 r1 and s1 s2 = s2 s1 ,
we have R × S is commutative if and only if R, S are.
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If R and S have an identity, then clearly 1R×S = (1R , 1S ). Conversly, suppose
that (u, v) is an identity in R × S. Then, for all r ∈ R, s ∈ S,
(u, v)(r, s) = (ur, vs) = (r, s) =⇒ ur = r, vs = s
(r, s)(u, v) = (ru, vs) = (r, s) =⇒ ru = r, sv = s,
and u, v are the identity for R, S respectively.
(b) Since T = {(r, r) : r ∈ R} ⊆ R × R, it suffices to show that T is closed
under substraction and multiplication. We have that
(r1 , r1 ) − (r2 , r2 ) = (r1 − r2 , r1 − r2 ) ∈ T
and
(r1 , r1 ) · (r2 , r2 ) = (r1 r2 , r1 r2 ) ∈ T,
which shows that T is a subring of S.
3. Let
I = {a + bi + cj + dk : a, b, c, d ∈ Z}
be the ring of integral quaternions and define the norm
N :I → Z
a + bi + cj + dk 7→ a2 + b2 + c2 + d2 .
(a) Prove that N (α) = αα for all α ∈ I, where if α = a + bi + cj + dk then
α = a − bi − cj − dk.
(b) Prove that N (αβ) = N (α)N (β) for all α, β ∈ I.
(c) Prove that an element in I is a unit if and only if it has norm ±1. Show
that I x is a group with 8 elements.
Hint: Recall that the inverse in the ring of rational quaternions of an element
α is α/N (α).
Solutions: We first compute the product αβ for two general elements α, β ∈ H.
αβ = (a + bi + cj + dk) · (a0 + b0 i + c0 j + d0 k)
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= aa0 + ab0 i + ac0 j + ad0 k + ba0 i + bb0 i2 + bc0 ij + bd0 ik
+ca0 j + cb0 ji + cc0 j 2 + cd0 jk + da0 k + db0 ki + dc0 kj + dd0 k 2
= (aa0 − bb0 − cc0 − dd0 ) + (ab0 + ba0 + cd0 − dc0 )i + (a0 c + ca0 − bd0 + db0 )j
+(ad0 + da0 + bc0 − b0 c)k.
(1)
(a) Using (1) with β = α, i.e. a0 = a, b0 = −b, c0 = −c, d0 = −d, we have
αα = (a2 + b2 + c2 + d2 ) = N (α).
(b) We show that αβ = βα, which implies the result since
N (αβ) = αβαβ = αββα = ααN (β) = N (α)N (β).
Using (1) with β and α, i.e. replacing a, a0 b, b0 , c, c, d, d0 by a0 , a, −b, −b, −c0 , −c, −d0 , −d
respectively, we get
βα = (aa0 − bb0 − cc0 − dd0 ) + (−ab0 − ba0 − cd0 + dc0 )i + (−a0 c − ca0 + bd0 − db0 )j
+(−ad0 − da0 − bc0 + b0 c)k,
and also
αβ = (aa0 − bb0 − cc0 − dd0 ) − (ab0 + ba0 + cd0 − dc0 )i − (a0 c + ca0 − bd0 + db0 )j
−(ad0 + da0 + bc0 − b0 c)k.
= (aa0 − bb0 − cc0 − dd0 ) + (−ab0 − ba0 − cd0 + dc0 )i + (−a0 c − ca0 + bd0 − db0 )j
+(−ad0 − da0 − bc0 + b0 c)k.
This proves that αβ = βα.
(c) Suppose that u ∈ I is a unit. Then, 1 = N (1) = N (uu−1 ) = N (u)N (u−1 ).
Since for any z 6= 0 ∈ I, the norm is a positive integer, this implies that
N (u) = N (u−1 ) = 1. Conversly, suppose that N (u) = 1. Then
u−1 =
u
= u ∈ I,
N (u)
and u−1 ∈ I, so u is a unit. Then,
I x = a + bi + cj + dk : a, b, c, d ∈ Z, a2 + b2 + c2 + d2 = 1
= {±1, ±i, ±j, ±k}
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4. Let K be a field. A discrete valuation on K is a function ν : K x → Z satisfying
(i) ν(xy) = ν(x) + ν(y)
(ii) ν is surjective
(iii) ν(x + y) ≥ min (ν(x), ν(y)), for all x, y ∈ K x with x + y 6= 0.
The set
R = {x ∈ K x : ν(x) ≥ 0} ∪ {0}
is called the valuation ring of ν.
(a) Prove that R is a subring of K containing the identity.
(b) Prove that for each non-zero element x ∈ K, either x ∈ R or x−1 ∈ R.
(c) Prove that x is a unit in R if and only if ν(x) = 0.
Solution: (a) We first remark that R contains 1 since
ν(1) = ν(1 · 1) = ν(1) + ν(1) =⇒ ν(1) = 0.
Also,
0 = ν(1) = ν(−1) + ν(−1) = 2ν(−1) =⇒ ν(−1) = 0.
Then, if ν(−x) = ν(−1) + ν(x) = ν(x), and R contains additive inverses. We
have to show that R is closed under addition and multiplication. Let x, y ∈ R.
Then ν(x + y) ≥ min (ν(x), ν(y)) ≥ 0 since ν(x), ν(y) ≥ 0, and x + y ∈ R.
Similarly, ν(xy) = ν(x) + ν(y) ≥ 0 since ν(x), ν(y) ≥ 0, and xy ∈ R.
(b) Let x ∈ K. If ν(x) ≥ 0, then x ∈ R. If ν(x) < 0, let x−1 be the multiplicative
inverse of x (which exists since K is a field). Since
0 = ν(1) = ν(xx−1 ) = ν(x) + ν(x−1 ),
and ν(x) < 0, we have that ν(x−1 ) ≥ 0, and x−1 ∈ R.
(c) Let x ∈ R such that there exits x−1 ∈ R with xx−1 = 1. Then, since
0 = ν(1) = ν(x) + ν(x−1 ),
ν(x), ν(x−1 ) ≥ 0,
we have that ν(x) = ν(x−1 ) = 0.
Conversly, if ν(x) = 0, then since 0 = ν(x) + ν(x−1 ), we have that ν(x−1 ) = 0
also, and x−1 ∈ R, which shows that x is a unit in R.
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5. A specific example of a discrete valuation ring is obtained by taking K = Q.
Let p a prime, and for any a/b ∈ Q, we write
c
a
= pα , where (c, p) = (d, p) = 1.
b
d
We then define
a
= α.
νp
b
(a) Show that νp is a valuation.
(b) Show that the valuation ring is the set of all fractions whose denominators
are relatively prime to p. Describe the units in this valuation ring.
Solutions: (a) Let x, y 6= 0 ∈ Q and we write
a
c
x = pα and y = pβ ,
b
d
for unique α, β ∈ Z, a, b, c, d ∈ Z/{0} (up to ±) with p - abcd. Then,
α+β ac
νp (xy) = νp p
= α + β = νp (α) + νp (β).
bd
Assume without lost of generality that α ≤ β. Then,
a
c
+ pβ−α
,
x + y = pα
b
d
and νp (x + y) ≥ α. We remark that νp (x + y) is not necessarily equal to α, as
we can have that
a
c
νp
+ pβ−α
≥ 1.
b
d
(This can only happen when α = β. For example, ν5 (10) = 1, ν5 (15) = 1, but
ν5 (10 + 15) = ν5 (25) = 2 > 1.)
Also, νp is surjective as νp (pn ) = n for any n ∈ Z/{0}.
(b) We have that
a
νp pα
≥ 0 ⇐⇒ α ≥ 0,
b
and
pα a
b
is a fraction without any power of p in the denominator. Conversly, for such
a fraction a/b, we have that νp (a/b) = α where a = pα a0 with (a, p) = 1, and
α ≥ 0.
x=
The units are the fractions a/b with a, b ∈ Z, and p - ab.
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