Page 152 - DMTH403_ABSTRACT_ALGEBRA

Unit 14: Rings

Figure 14.1: The Graphs of f and (–f) over [0, 1] Notes

Now let us see if the axiom R6 holds.
To prove f . (g + h) = f . g + f . h, we consider (f . (g + h) (x) for any x in [0, 1].
Now (f . (g + h)(x) = f(x) (g + h) (x)
= f(x) g(x) +h(x))

= f(x) g(x) + f(x)h(x), since, distributes over + in R.
= (f.g)(x) + (f.h)(x)
Since multiplication is commutative in C [0, 1], the other distributive law also holds. Thus, R6 is
true for C [0, 1]. Therefore, (C [0, 1], +, .) is a ring.
This ring is called the ring of continuous functions on [0, 1].
The next example also deals with functions.

Example: Let (A, +) be an abelian group. The set of all endomorphisms of A is
End A = ( f : A – A | f(a + b) = f(a) + f(b)  a, b  A )

For f, g  End A, we define f + g and f . g as
(f + g) (a) = f(a) + g( a), and ...(1)
(f. g) (a) = fog(a) = f(g(a))  a  A
Show that (End A, +,.) is a ring. (This ring is called the endomorphism ring of A.)

Solution: Let us first check that + and . defined by (1) are binary operations on End A.
For all a, b  A,
(f + g) (a + b) = f(a + b) f g(a + b)
= (f(a) + f(b)) + (g(a) + g(b))

= (f(a) + g(a)) + (f(b) + g(b))
= (f + g) (a) + (f + g) (b), and

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