Show briefly that 6Z is an ideal of 2Z Find addition and mul

Show briefly that 6Z is an ideal of 2Z. Find addition and multiplication table for the factor ring^2Z/6Z. Is the factor ring in part isomorphic to the field^Z_3? Why?

Solution

1(a)   Let J be an ideal in E such that I J. Then there exists x belongs to J such that x does not belong to I. Thus x
is even but not divisible by 6. Thus x = 6k + r for some k belong to Z and r = 2 or r = 4. Now 6k belongs to I, thus
6k belongs to J. So x 6k belongs to J. In other words, r belongs to K. If r = 2, then 2 belongs to J implies J = E. If r = 4, then
4 belongs to J. However, 6 belongs to I, so 6 belongs to J. Thus 6 4 belongs to J. Hence 2 belongs to J and so J = E. Thus in either case
J = E and hence I is maximal and an ideal

b) 6 / 2Z and 6 2belongs to3Z, but 6 does not belong to pZ for any other prime p. In general, in a commutative ring
R, (b) (a) if and only if b belongs to (a) if and only if a divides b (b = ra). So the containments between

principal ideals really react the multiplicative structure. This is what algebraic number theory is
all about.

Multiplication in Z2[x]/6z

c) We can prove this using the fourth isomorphism theorem,We know that there is a y so that (x+M)(y +M) = 1+M, so that xy = 1+m for some
element y belongs R. But m belongs (M [ fxg) and xy belongs (M [ fxg), so 1 belongs to (M [ fxg).
Assume M is maximal. Then let x +M 6= 0 belongs to R=M. That is x =2 M. Thus M + (x) = R. So
1 belongs to M+(x), and we can write 1 = rm+sx for r; s 2 R. Then (s+M)(x+M) = (1-rm)+M = 1+M,
and x +M is a unit. so we can say it is isomorphic for Z3