Page 177 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Solution: For any m, n  Z,

f(m + n) = m t n = m  n = f(m) + f(n), and

f(mn) = mn = m  n = f(m) f(n).

Therefore, f is a ring homomorphism.
Now, Kerf = (m  Z | f(m) = 0 |

= {m  Z | m= 01
= (m  Z | m  0 (mod s))
= sz.

Im f = (f(m) | m  Z)
= ( m | m  Z)
= Z , s

showing that f is an epimorphism.

Example: Consider the map f : Z – Z : f(n (mod 6)) = n(mod 3). Show that f is a ring
3
6
homomorphism. What is Ker f?
Solution: Firstly, for any n, m  Z,
f(n(mod 6) + m(mod 6)) = f((n + m) (mod 6)) = (n + m) (mod 3)

= n (mod 3) + m(mod 3)
= f(n (mod 6)) + f(m(mod 6))
You can similarly show that

f(n(mod 6) . m(mod 6)) = f(n(mod 6)) . f(m(mod 6)).
Thus, f is a ring homomorphism.
Ker f = {n(mod.6) | n  0(mod 3)) = {n(mod 6) | n  3Z)

= {0,3}, bar denoting ‘mod 6’.
Before discussing any more examples, we would like to make a remark about terminology. In
future we will use the term ‘homomorphism’ for ‘ring homomorphism’. You may remember
that we also did this in the case of group homomorphisms.
Now let us look at some more examples.

Example: Consider the ring C[0, 1] of all real valued continuous functions defined on the
closed interval [0, 1].
Define  : C[0, 1] + R : (f) = f(1/2). Show that  is a homomorphism.
Solution: Let f and g  C [0, 1]
Then (f + g) (x) = f(x) f g(x) and

(fg) (x) = f(x) g(x) for all x  C[0, 1].

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