Download Math 110B HW §5.3 – Solutions 3. Show that [−a, b] is the additive

Math 110B HW §5.3 – Solutions
3. Show that [−a, b] is the additive inverse of [a, b] in Q.
Proof. The zero element in Q is [0, c] where c is any non-zero element R. Let [a, b] ∈ Q.
[−a, b] + [a, b] = [−ab + ba, b2] = [−ab + ab, b2] = [0, b2]
Thus [−a, b] is the additive inverse of [a, b].
6. Prove the right distributive property in Q.
Proof. Let [x, y], [z, w], [u, v] ∈ Q.
([x, y] + [z, w])[u, v] = [xw + yz, yw][u, v]
= [(xw + yz)u, ywv]
= [xwu + yzu, ywv]
= [(xwu + yzu)v, (ywv)v]
= [xuwv + yvzu, yvwv]
= [xu, yv] + [zu, wv]
= [x, y][u, v] + [z, w][u, v]
Thus the right distributive property in Q holds.
8. Assume that the ring R is isomorphic to the ring R0 . Prove that if R is commutative, then
R0 is commutative.
Proof. Let φ : R → R0 be an isomorphism and assume R is commutative. Let a0 , b0 ∈ R0.
Since φ is onto, there exists a, b ∈ R such that φ(a) = a0 and φ(b) = b0. We then have the
following.
a0 b0 = φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a) = b0a0
Thus R is commutative.
9. Let W be the ring defined by the set of all ordered pairs (x, y) of integers x and y. Equality,
addition and mulitplication are defined as follows:
(x, y) = (z, w) ↔ x = z and y = w in Z
(x, y) + (z, w) = (x + z, y + w)
(x, y) · (z, w) = (xz − yw, xw + yz)
Let R be the set of all matrices of the form
a −b
b a
where a and b are integers.
Given that W and R are isomorphic rings, define an isomorphism from W to R and prove
that your mapping is an isomorphism.
Proof. Define φ : W → R by φ(a, b) =
a −b
b a
. First let’s prove φ is a homomorphism.
Let (a, b), (c, d) ∈ W .
φ ((a, b)(c, d)) = φ(ac − bd, ad + bc)
ac − bd −(ad + bc)
=
ad + bc
ac − bd
a −b
c −d
=
b a
d c
= φ(a, b)φ(c, d)
φ ((a, b) + (c, d)) = φ(a + c, b + d)
a + c −(b + d)
=
b+d
a+c
a −b
c −d
=
+
b a
d c
= φ(a, b) + φ(c, d)
Thus φ is a homomorphism.
Now let’s show φ is onto. Let
a −b
b a
∈ R. So (a, b) ∈ W and φ(a, b) =
Thus φ is onto.
a −b
Now let’s show φ is one-to-one. Assume φ(a, b) = φ(c, d). Thus
b a
Hence a = c and b = d. Thus (a, b) = (c, d). Therefore φ is one-to-one.
=
a −b
b a
c −d
d c
.
.
Thus φ is an isomorphism.
13. Prove that if D is a field to begin with, the field of quotients Q is isomorphic to D.
Proof. Let D be a field and let Q be the field of quotients. Define φ : Q → D by φ([a, b]) =
ab−1.
Let’s first check well-defined. Suppose [a, b] = [c, d]. So ad = bc which implies b−1a = cd−1
and by commutativity this gives ab−1 = cd−1. Thus φ([a, b]) = ab−1 = cd−1 = φ([c, d]).
Hence φ is well-defined.
Let’s check the ring structure.
φ([a, b] + [c, d]) = φ([ad + bc, bd])
= (ad + bc)(bd)−1
= ad(bd)−1 + bc(bd)−1
= add−1b−1 + bb−1 cd−1
= ab−1 + cd−1
= φ([a, b]) + φ([c, d])
φ([a, b][c, d]) = φ([ac, bd])
= (ac)(bd)−1
= acd−1 b−1
= ab−1 cd−1
= φ([a, b])φ([c, d])
Suppose φ([a, b]) = φ([c, d]). So ab−1 = cd−1 which implies b−1 a = cd−1 and this implies
ad = bc, which gives [a, b] = [c, d]. Thus φ is one-to-one.
Let d ∈ D. Then φ([d, e]) = d. Hence φ is onto.
Thus φ is an isomorphism and hence D is isomorphic to Q.