ABSTRACT ALGEBRAIC STRUCTURES OPERATIONS AND ...

A subset I is said to be et ideal of the ring R, if I is a subgroup of the underlying
additive structure and is closed wrt to multiplication by any element in the ring.
This can be expressed by
I + I = I
RI = I
IR = I
126. Theorem: The kernel is an ideal.
Let f be a homomorphism of the ring R1 into the ring R2. Then the kernel of
f is an ideal
Proof : Let K denote the kernel of f. Then f a homomorphism between the
underlying groups with kernel K, which therefore is a subgroup of the underlying
group in R1.
This means that the first condition in the definition of ideals is fulfilled.
The second condition: Let k ∈ K, r ∈ R then f(kr) = f(k)f(r) = 0f(r) = 0.
Consequently kr ∈ I. Analogously for rk ∈ I.
127. Theorem: The quotient wrt to an ideal is a ring
Let I be an ideal in the ring R. Then I is a normal subgroup of the underlying
additive group R. Let R/I denote the quotient group. Then the equivalence
relation associated with I is compatible with the ring structure on R and the
quotient structure R/I is a ring.
Proof : We must show that the multiplication is compatible with the equivalence
relation associated with I. Therefore let x1 ∼ y1 and x2 ∼ y2. This
means per definition that x1 − y1 ∈ I and that x2 − y2 ∈ I. We use the good
old trick to express a difference between two products by the difference of the
factors to get
x1x2−y1y2 = x1x2−x1y2+x1y2−y1y2 = x1(x2−y2)+(x1−y1)y2 ∈ IR+RI
and by the properties of an ideal we have that RI + IR = I + I = I and
so we have proved that x1x2 ∼ y1y2. Therefore the quotient structure is well
defined and the product is given by the formula [x][y] = [xy] or in other words
(x + I)(y + I) = (xy) + I. It is easy to check that this product makes the
quotient a monoid with [1] as neutral element for multiplication. To check that
this multiplication is distributive wrt addition is again routine. So the quotient
is a ring.
65