Unformatted text preview: 6.4. INTEGRAL DOMAINS 291 any injective homomorphism of R into a ﬁeld F extends to an injective
homomorphism of Q.R/ into F :
'
F
R
qqqqq
qqqq
qqq
qqqqq qqqqqqqqqq
qqq
q Â '
Q qqqqq qqqqq
qqqqqqqq Q.R/ The main steps in establishing these facts are the following:
1. Q.R/ is a ring with zero element 0 D Œ0=1 and multiplicative
identity 1 D Œ1=1. In Q.R/, 0 ¤ 1.
2. Œa=b D 0 if, and only if, a D 0. If Œa=b ¤ 0, then Œa=bŒb=a D
1. Thus Q.R/ is a ﬁeld.
3. a 7! Œa=1 is an injective homomorphism of R into Q.R/.
4. If ' W R ! F is an injective homomorphism of R into a ﬁeld F ,
then ' W Œa=b 7! '.a/='.b/ deﬁnes an injective homomorphism
Q
of Q.R/ into F , which extends ' .
See Exercises 6.4.11 and 6.4.12.
Proposition 6.4.4. If R is an integral domain, then Q.R/ is a ﬁeld containing R as a subring. Moreover, any injective homomorphism of R into
a ﬁeld F extends to an injective homomorphism of Q.R/ into F .
Example 6.4.5. Q.Z/ D Q.
Example 6.4.6. Q.KŒx/ is the ﬁeld of rational functions in one variable.
This ﬁeld is denoted K.x/.
Example 6.4.7. Q.KŒx1 ; : : : ; xn / is the ﬁeld of rational functions in n
variables. This ﬁeld is denoted K.x1 ; : : : ; xn /.
We have observed in Example 6.2.3 that in a ring R with multiplicative
identity 1, the additive subgroup h1i generated by 1 is a subring and the
map k 7! k 1 is a ring homomorphism from Z onto h1i Â R. If this ring
homomorphism is injective, then h1i Š Z as rings. Otherwise, the kernel
of the homomorphism is nZ for some n 2 N and h1i Š Z=nZ D Zn as
rings.
If R is an integral domain, then any subring containing the identity is
also an integral domain, so, in particular, h1i is an integral domain. If h1i
is ﬁnite, so ring isomorphic to Zn for some n, then n must be prime. (Zn
is an integral domain if, and only if, n is prime.) ...
View
Full
Document
This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
 Fall '08
 EVERAGE
 Algebra

Click to edit the document details