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Unformatted text preview: 276 6. RINGS Example 6.2.17. The kernel of the ring homomorphism KŒx ! K given
by p 7! p.a/ is the set of all polynomials p having a as a root.
Example 6.2.18. The kernel of the ring homomorphism C.R/ ! C.S /
given by f 7! fjS is the set of all continuous functions whose restriction
to S is zero.
Example 6.2.19. Let K be a ﬁeld. Deﬁne a map ' from KŒx to
Fun.K; K/, the ring of K –valued functions on K by '.p/.a/ D p.a/.
(That is, '.p/ is the polynomial function on K corresponding to the polynomial p .) Then ' is a ring homomorphism. The homomorphism property
of ' follows from the homomorphism property of p 7! p.a/ for a 2 K .
Thus '.p C q/.a/ D .p C q/.a/ D p.a/ C q.a/ D '.p/.a/ C '.q/.a/ D
.'.p/ C '.q//.a/, and similarly for multiplication.
The kernel of ' is the set of polynomials p such that p.a/ D 0 for all
a 2 K . If K is inﬁnite, then the kernel is f0g, since no nonzero polynomial
with coefﬁcients in a ﬁeld has inﬁnitely many roots.
If K is ﬁnite, then ' is never injective. That is, there always exist
nonzero polynomials p 2 KŒx such that p.a/ D 0 for all a 2 K . Indeed,
we need merely take p.x/ D a2K .x a/. Deﬁnition 6.2.20. A ring R with no ideals other than f0g and R itself is
said to be simple.
Any ﬁeld is a simple ring. You are asked to verify this in Exercise
In Exercise 6.2.11, you are asked to show that the ring M of n-by-n
matrices with real entries is simple. This holds equally well for matrix
rings over any ﬁeld.
(a) Let fI˛ g be any collection of ideals in a ring R. Then \ I˛ is ˛ (b) an ideal of R.
Let In be an increasing sequence of ideals in a ring R. Then
In is an ideal of R.
n Proof. Part (a) is an Exercise 6.2.17. For part (b), let x; y 2 I D [ In . n Then there exist k; ` 2 N such that x 2 Ik and y 2 I` . If n D maxfk; `g, ...
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