Page 153 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes (f. g) (a + b) = f(g(a + b))
= f(g(a) + g(b))

= f(g(a)) + f(g(b))
= (f . g)(a) + (f . g) (b)
Thus, f + g and f . g  End A.
Now let us see if (End A, +, .) satisfies Rl-R6.

Since + in the abelian group A is associative and commutative, so is + in End A. The zero
homomorphism on A is the zero element in End A. (– f) is the additive inverse of f E End A. Thus,
(End A, +) is an abelian group.
You also know that the composition of functions is an associative operation in End A.
Finally, to check R6 we look at f . (g + h) for any f, g, h  End A. Now for any a  A,
[f . (g + h)l (a) = f((g + h) (8))

= f(g(a) + h(a))
= f(g(a)) + f(h(a))
= (f . g) (a) + (f . h) (a)
= (f . g + f . h) (a)

 f.(g + h) = f . g + f . h.
We can similarly prove that (f + g) . h = f . h + g . h.
Thus, R1-R6 are true for End A.
Hence, (End A, +, .) is a ring.

Note It is not commutative since fog need not be equal to gof for f, g  End A.

Now, let us look at the Cartesian product of rings.

Example: Let (A, +,.) and (B, + , ) be two rings. Show that their Cartesian product
A X B is a ring with respect to  and * defined by
(a, b)  (a’, b’) = (a + a’, b + b’), and
(a, b) * (a’, b’) = (a . a’, b b’)
for a11 (a, b), (a’, b’) in A X B.

Solution: We have defined the addition and multiplication in A X B componentwise. The zero
element of A X B is (0, 0). The additive inverse of (a, b) is (–a, – b), where – b denotes the inverse
of b with respect to – .
Since the multiplications in A and B are associative, * is associative in A × B. Again, using the fact
that R6 holds for A and B, we can show that R6 holds for A × B. Thus, (A × B, 0, *) is a ring.

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