### . Find all sub-rings of R. Because rings as defined in the book have

```Let R = ab | a ∈ Z, b = 3n for some n ≥ 0, gcd(a, b) = 1 ⊂ Q.
Find all sub-rings of R.
Because rings as dened in the book have the additive identity and multiplicative identity, we immediately know any subring S of R must contain Z. In
particular, Z is a subring of R. Suppose, however, that the subring S had an
a
element n for some n ≥ 1 and a 6= 0. If necessary we can take its additive
3
inverse and assume a > 0. Then under addition, we'd have
a
a
a
= n + · · · + n ∈ S.
3 |3
{z 3 }
3n−1 times
Now take the greatest integer less than or equal to
jak
3
jak
a
. Then
, denoted
3
3
∈ Z ⊂ S and hence
a jak
∈ S.
−
3
3
It follows that either
1
2
∈ S or ∈ S.
3
3
If 13 ∈ S then S = R. Suppose 23 ∈ S . Then − 13 = 23 − 13 ∈ S and hence 13 ∈ S
(closed under additive inverses). Thus, S = R. In conclusion, subrings of R are
just Z and R itself.
3.2.14:
Suppose there was a eld F of order 6, then the additive subgroup is isomorphic to C6 . Thus there is an element of (additive) order 6 which generates
the additive subgroup, call it g . Thus we have F = {0, g, 2g, 3g, 4g, 5g}.
But then
2g · 3g = (g + g)(g + g + g)
= g2 + g2 + g2 + g2 + g2 + g2
= 6g 2 = 0
where the last equality comes from Lagrange's theorem: the (additive) order of
element divides the order of the (additive) group.
3.2.16b
An additive subgroup of F must have order 2, 4, 8. If it is 2, then it must be
{0, 1}. If it is 8 it must be F . Assume it is 4. Then the multiplicative part is
order 3 and a subgroup of the multiplicative group of F which is order 7. But
3 doesn't divide 7, so no such subeld of order 4 may exist.
3.2.16c
Then we may have 4 or 8 and respectively multiplicative groups of order 3
or 7. 3 does divide 15, whereas 7 does not. Thus the same argument says we
cannot have a subeld of order 8, but we wouldn't know about a subeld of
order 4.
3.2.29c
1
We show every nonzero element a + bω has an inverse. Formally, we might
want to write
1
1 a − bω
=
a + bω
a + bω a − bω
a − bω
= 2
a + b2
= a(a2 + b2 )−1 − b(a2 + b2 )−1 .
Thus, it suces to show that a2 + b2 6= 0. If, however, we could nd a, b ∈ Zp
such that a2 + b2 = 0 then if a = 0 then b = 0, contradicting that a + bω is
nonzero, if b = 0 then a = 0, contradicting that a + bω is nonzero, and nally
if both are nonzero then (ab−1 )2 + 1 = a2 (b−1 )2 + 1 = 0 which contradicts the
corollary that x2 = −1 has no solutions in Zp for prime p with p ≡ 3 (mod 4).
3.3.17
a) When R is commutative, because the center is a subring and hence contains 1. Thus R = Z . If R = Z , then R is a commutative ring.
b) Z is a commutative subring of R. Then for a ∈ Z , a 6= 0, consider the
principal ideal generated by a. Because a ∈ Ra, then Ra is not the 0 ideal
and hence Ra = R, because R is simple. It follows that a has an inverse in R:
ba = ab = 1 for some b ∈ R. We conclude that Z is a eld (when R is simple).
c) Suppose R/Z = ha + Zi for some a. Thus given r, s ∈ R/Z we have
rs = (ja + u)(ka + v) for some integers j, k and some elements u, v ∈ Z . Expanding we have (ja)(ka) + (ja)v + u(ka) + uv . Using the property of Z and
the commutativity of Z we have
(ja)(ka) + (ja)v + u(ka) + uv = (ka)(ja) + v(ja) + (ka)u + vu
and the right-hand side equals sr.
2
```