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3.23 (i) Show that F = {a + bi : a , b ∈ Q} is a ﬁeld.
Solution. It is straightforward to check that F is a subring of C,
and so it is a commutative ring; it is a ﬁeld because the inverse of
a + bi is r −1 (a − bi ) ∈ F , where r = a 2 + b2 . (ii) Show that every u ∈ F has a factorization u = αβ −1 , where
α, β ∈ Z [i ]. (See Exercise 3.50.)
Solution. Write
a + bi = ( p /q ) + (r /s )i = ( ps + qri )(qs )−1 ,
where p , q , r, s ∈ Z.
3.24 If R is a commutative ring, deﬁne a relation ≡ on R by a ≡ b if there is a
unit u ∈ R with b = ua .
(i) Prove that ≡ is an equivalence relation.
Solution. Absent.
(ii) If a ≡ b, prove that (a ) = (b), where (a ) = {ra : r ∈ R }.
Conversely, prove that if R is a domain, then (a ) = (b) implies
a ≡ b.
Solution. Assume that a ≡ b, so there is u , v ∈ R with a = ub
and u v = 1. We claim that (a ) = (b). If x ∈ (a ), then x = ra =
r ub ∈ (b), and so (a ) ≤ (b). For the reverse inclusion, if y ∈ (b),
then y = sb = s v a ∈ (a ). Hence (a ) = (b).
Conversely, assume that (a ) = (b). If a = 0, then b = 0 and
so a ≡ b. We now assume that a = 0. Since a ∈ (b), there is
r ∈ R with a = r b; since b ∈ (a ), there is s ∈ R with b = sa .
Therefore, a = r b = r sa , and so (1 − r s )a = 0. Since R is a
domain, 1 = r s and r is a unit. Therefore, a ≡ b.
3.25 If R is a domain, prove that there is no subﬁeld K of Frac( R ) such that
R⊆K Frac( R ). Solution. Absent.
3.26 Let k be a ﬁeld, and let R be the subring
R = n · 1 : n ∈ Z ⊆ k.
(i) If F is a subﬁeld of k , prove that R ⊆ F .
Solution. Absent. (ii) Prove that a subﬁeld F of k is the prime ﬁeld of k if and only if it is
the smallest subﬁeld of k containing R ; that is, there is no subﬁeld
F with R ⊆ F
F.
Solution. Absent. ...
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 Fall '11
 KeithCornell

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