5.4
5.4
J.A.Beachy
1
Quotient Fields
from A Study Guide for Beginner’s by J.A.Beachy,
a supplement to Abstract Algebra by Beachy / Blair
15. Let F be a field. Explain why Q(F ) is isomorphic to F . Why can’t we just say that
Q(F ) = F ?
Solution: Formally, Q(F ) is a set of equivalence classes of ordered pairs of elements
of F , so it is not simply equal to the original set F .
In the general construction, we identified d ∈ D with the equivalence class [d, 1], and
used this to show that D is isomorphic to a subring of Q(D). When D is a field, the
equivalence class [a, b] is equal to the equivalence class [ab−1 , 1] since a · 1 = b · ab−1 .
This shows that when D is a field we have Q(D) ∼
= D.
16. Find the quotient field of Z2 [x].
Solution: We need to formally invert all nonzero polynomials in Z2 [x]. This produces
what look like rational functions, though we do not think of them as functions taking
values in Z2 since as functions things would collapse. (Remember that as functions
f (x)
we have x2 = x.) The set of fractions of the form
such that f (x), g(x) ∈ Z2 [x],
g(x)
where g(x) is not the zero polynomial, is denoted by Z2 (x), and is called the rational
function field over Z2 .
17. Prove that if D1 and D2 are isomorphic integral domains, then Q(D1 ) ∼
= Q(D2 ).
Solution: Let D1 and D2 be isomorphic integral domains, with isomorphism α :
D1 → D2 and β = α−1 . Let φ1 : D1 → Q(D1 ) and φ2 : D2 → Q(D2 ) be the standard
mappings defined in Theorem 5.4.6.
Recall that we have the following diagram from Theorem 5.4.6, in which any one-toone ring homomorphism θ : D → F , where F is a field, gives rise to a unique ring
b = θ.
homomorphism θb : Q(D) → F such that θφ
D
φ-
Q(D)
@
θ @
R
@
θb
?
F
In the above diagram, we first let D = D1 , φ = φ1 , F = Q(D2 ), and θ = φ2 α. We
d
therefore get φd
2 α : Q(D1 ) → Q(D2 ) with φ2 αφ1 = φ2 α. Similarly, letting D = D2 ,
d
φ = φ2 , F = Q(D1 ), and θ = φ1 β we get φd
1 β : Q(D2 ) → Q(D1 ) with φ1 βφ2 = φ1 β. In
following this part of the argument, it may help to draw the corresponding diagrams.
d
We claim that φd
2 α is the isomorphism we are looking for, with inverse φ1 β. To show
this, in the original diagram take D = D1 , φ = φ1 , F = Q(D1 ), and θ = φ1 . Then the
5.4
J.A.Beachy
2
c1 , since 1Q(D ) φ1 = φ1 , and it
identity mapping from Q(D1 ) to Q(D1 fits the role of φ
1
is the unique ring homomorphism for which this is true.
d
dd
On the other hand, consider φd
1 β φ2 α : Q(D1 ) → Q(D2 ). We have (φ1 β φ2 α)φ1 =
φd
1 β(φ2 α) = (φ1 β)α = φ1 (βα) = φ1 , and so the uniqueness guaranteed by Theod
rem 5.4.6 implies that φd
1 β φ2 α is the identity mapping. A similar argument shows
d
that φd
2 αφ1 β : Q(D2 ) → Q(D1 ) is the identity mapping. This completes the proof
that Q(D1 ) ∼
= Q(D2 ).
Comment: It is also possible to directly define an isomorphism θ : Q(D1 ) → Q(D2 )
by setting θ([a, b]) = [α(a), α(b)]. The argument given above is a standard one for any
“universal” construction that guarantees the existence of a unique mapping. If you
continue your study of mathematics, you are likely to encounter arguments like this
in more general situations.
ANSWERS AND HINTS
√
3i].
18. Find the quotient
field
of
Z[
√
Answer: Q( 3i)
m n 19. Find the quotient field of D =
m, n ∈ Z .
−3n m a b a, b ∈ Q . To apply Corollary 5.4.7, show that every
Hint: Let F =
−3b a −1
m n
d 0
element of F can be written in the form
for some m, n, d ∈ Z.
−3n m
0 d
Answer: Q(D) ∼
= F.
m n 20. Find the quotient field of D =
m, n ∈ Z .
−n m a b a, b ∈ Q
Answer: Q(D) =
−b a