(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.

## No comments:

Post a Comment