4. Let R be an integral domain. Show that there exists an R-algebra (S, φS )
such that every non-zero element of im φS has a multiplicative inverse in S and
which satisfies the following universal property: if (S 0 , φS 0 ) is an R-algebra such
that every non-zero element of im φS 0 has a multiplicative inverse in S 0 , then
there exists a unique R-algebra homomorphism S → S 0 . [Hint: what is another
name for this S?]
Solution: Recall the definition of fraction field of an integral domain R:
na
o
| a, b ∈ R, b 6= 0 / ∼,
Frac R =
b
where ∼ is the equivalence relation
a
c
∼
b
d
⇐⇒
ad − bc = 0 (in R).
Frac R acquires the structure of a field under the usual multiplication and addition operations.
Set S = Frac R, which is naturally an R-algebra via the ring homomorphism
φS : R → S given by φS (r) = 1r . Note that every non-zero element in im φS
has the form 1r , which has as multiplicative inverse 1r . We claim this (S, φS )
satisfies the universal property.
Suppose (S 0 , φS 0 ) is another R-algebra in which every non-zero element of
im φS 0 has a multiplicative inverse. Define ψ : S → S 0 via
a
= φS 0 (a)φS 0 (b)−1 .
ψ
b
One checks that this formula gives a well-defined R-algebra homomorphism
S → S 0 . It remains to show that this ψ is the only such, so suppose χ : S → S 0
is another R-algebra homomorphism. By definition of R-algebra homomorphism, it must be that χ( 1r ) = χ ◦ φS (r) = φS 0 (r) for all r ∈ R. Because ring
homomorphisms preserve inverses, it must also be the case that χ( 1r ) = φS 0 (r)−1
for all non-zero r ∈ R. Thus, for any ab ∈ S, we have
a
a 1
a 1
χ
=χ
·
=χ
χ
= φS 0 (a)φS 0 (b)−1 ,
b
1 b
1
b
i.e., χ = ψ, which shows uniqueness.