(1) Proof:
By the Euclidean (division) algorithm there exist numbers q, r ∈ Z
Such that 0 < r
< n, and a is of the form:
a = q n + r
By definition of [a ] mod n = [a]
r ∈ [a] , And: r = a + (-q) n
(2) Let Z be the
integers. The ideal:
(5) = {5 j: j ∈ Z }
Show all the congruence classes with respect to this ideal.
The
congruence classes are:
[a] = {a
+ j: j ∈ (5)}
= {a + 5j: j ∈
Z }
[0] = {0 +
5 j: j ∈ Z}
= { 5j: j ∈
Z } =
(5)
[1]
= {1 + 5 j: j
∈ Z} = {1,
6, -4, 11, -9}
[2] =
{2 + 5 j: j ∈
Z} = {2, 7, -3, 12, -8}
[3] =
{3 + 5 j: j ∈
Z} = {3, 8, -2, 13, -7}
[4] =
{4 + 5 j: j ∈
Z} = {4, 9, -1, 14, -6}
[5] = {5 + 5j: : j
∈ Z} = 5 (j+
1) : j ∈ Z} = [5]
= [0]
3) Using set notation we may define:
A · B = {x · y: x ∈ A, y ∈ B}
A + B = = {x + y: x ∈ A, y ∈ B}
4) Take S as the set of
integers, Z. Let the ideal I = (2) so that S / I = Z 2
Thence or otherwise, find:
a (a) [0]
b) [1] c) S/ I =
Z 5
Solutions:
Take S = Z. I = (2)
so that S / I = Z 2
a) [0] = {0, +
2, + 4, + 6…..} = I = (2)
b) [1] = {1, 3, -1, 5, -3, 7….}
c)If S / I = Z 5
S/I = {[0], [1], [2],
[3], [4]}
5 ) Show every ring S has two ideals: S itself and {0}.
Every ring S always has at least two subrings,
namely S and the zero ring, S 0.
Further, if S is also a field then the only ideals in S are S and {0}. Further:
if 1 ∈ I then I = S. (Let I = S be an ideal,
I = {0} is also an ideal. )
(In a field F, the
only ideals are 0 and F)
Basically, an ideal of a ring S is an additive
subgroup a of S with the property that:
a ∈ S and a ∈ a imply ra ∈ a . Clearly then the set containing the single
element 0 and the set containing the whole ring S are ideals.
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