Unformatted text preview: 6.2. HOMOMORPHISMS AND IDEALS 279 Ideals in Z and in KŒx
In the ring of integers, and in the ring KŒx of polynomials in one
variable over a ﬁeld, every ideal is principal:
(a) For a subset S Â Z, the following are equivalent:
(i) S is a subgroup of Z.
(ii) S is a subring of Z.
(iii) S is an ideal of Z.
(b) Every ideal in the ring of integers is principal.
(c) Every ideal in KŒx, where K is a ﬁeld, is principal. Proof. Clearly an ideal is always a subring, and a subring is always a
subgroup. If S is a nonzero subgroup of Z, then S D Zd , where d is the
least positive element of S , according to Proposition 2.2.21 on page 98. If
S D f0g, then S D Z0. In either case, S is a principal ideal of Z. This
proves (a) and (b).
The proof of (c) is similar to that of Proposition 2.2.21. The zero ideal
of KŒx is clearly principal. Let J be a nonzero ideal, and let f 2 J be
a nonzero element of least degree in J . If g 2 J , write g D qf C r ,
where q 2 KŒx, and deg.r/ < deg.f /. Then r D g qf 2 J . Since
deg.r/ < deg.f / and f was a nonzero element of least degree in J , it
follows that r D 0. Thus g D qf 2 KŒxf . Since g was an arbitrary
element of J , J D KŒxf .
I Direct Sums
Consider a direct sum of rings R D R1 ˚
˚ Rn . For each i , set
Ri D f0g ˚ ˚ f0g ˚ Ri ˚ f0g ˚ ˚ f0g. Then Ri is an ideal of R.
How can we recognize that a ring R is isomorphic to the direct sum
of several subrings A1 ; A2 ; : : : ; An ? On the one hand, according to the
previous example, the component subrings must actually be ideals. On the
other hand, the ring must be isomorphic to the direct product of the Ai ,
regarded as abelian groups. These conditions sufﬁce.
Proposition 6.2.28. Let R be a ring with ideals A1 ; : : : As such that R D
A1 C C As . Then the following conditions are equivalent:
(a) .a1 ; : : : ; as / 7! a1 C
C as is a group isomorphism of
As onto R.
(b) .a1 ; : : : ; as / 7! a1 C
C as is a ring isomorphism of A1 ˚
˚ As onto R. ...
View Full Document