Page 183 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes (iii)  is 1 – 1. To show this let x, y  R such that
(x + I) = (y + I). Then f(x) = f(y),

so that f(x – y) = 0, i.e., x – y  Ker f = I.
i.e., x + I = y + I.
Thus,  is I - I
So, we have shown that R/Ker f  1m f.

Thus, iff is onto, then Im f = S and R/Ker f  S.
Note that this result says that f is the composition o , where  is the canonical homomorphism:
R – R/I : (a) = a + I. This can be diagrammatically shown as

Let us look at some examples of the use of the Fundamental Theorem.

Consider p : Z – Z : p(n) = n.p is an epimorphism and Ker p = {n|n = 0 in Z,,,} = mZ.
m
Therefore, Z/mZ  Z,
(Note that we have often used the fact that Z/mZ and Z are the same.)
m
As another example, consider the projection map
p : R × R  R : p(a, b) = a, where R and R are rings. Then p is onto and its kernel is ((0, b) |
1
1
2
1
2
b  R }, which is isomorphic to R . 2
2
Therefore, (R × R )/R  R .
1
2
2
1
Let us now apply Theorem 9 to prove that any ring homomorphism from a ring R onto Z is
uniquely determined by its kernel. That is, we can’t have two different ring homomorphisms
from R onto Z with the same kernel. (Note that this is not true for group homomorphisms. In
fact, you know that I and – I are distinct homomorphisms from Z onto itself with the same
z
z
kernel, {0}. To prove this statement we need the following result.
Theorem 10: The only non-trivial ring homomorphism from Z into itself is I .
z
Proof: Let f : Z - Z be a non-trivial homomorphism. Let n be a positive integer.
Then n = 1 + 1 + ..... + 1 (n times). Therefore, . ,
f(n) = f(1) + f(1) + ..... -1 f(1) (n times) = n f(1).
On the other hand, if n is a negative integer, then –n is a positive integer. Therefore, f(–n) = (–n)
f(l), i.e., –f(n) = – nf(1), since f is a homomorphism. Thus, f(n) = n f(1) in this case too.
Also f(0) = 0 = of(1).

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