Verified answer

well, let's look at the possible ideals of Z8 and Z9.

since an ideal has to be a subgroup of the additive group, any ideal has to be principal. this narrows down the possible ideals considerably, we have:

(0) = {0}

(1) = {0,1,2,3,4,5,6,7} = Z8

(2) = {0,2,4,6}

(4) = {0,4}

clearly, (2) is the sole maximal ideal.

similarly, in Z9 we have:

(0) = {0}

(1) = Z9

(3) = {0,3,6} <---maximal ideal

(for any unit u of Z9, we have (u) = (1), and (6) = (3)).

now let's look at ideals of Z10:

(0) = {0}

(1) = Z10

(2) = {0,2,4,6,8} <---maximal ideal

(5) = {0,5} <---another maximal ideal

if an ideal J of Z10 contains a unit, say u, then 1 = (u^-1)u is in this ideal, hence J = Z10. so the only possible PROPER ideals of Z10, are of the form (k) where gcd(k,10) ≠ 1.

well, we have 3 choices, gcd(k,10) = 2, gcd(k,10) = 5, or gcd(k,10) = 10.

the last choice is the 0-ideal, the first choice forces k = 5, the second choice gives us k = 2,4,6, or 8.

note that (3)(4) = 2 (mod 10), so (2) is contained in (4),

(2)(6) = 2 (mod 10), so (2) is contained in (6),

(4)(8) = 2 (mod 10), so (2) is contained in (8).

the containment of (4), (6) and (8) in (2) is obvious, so all these ideals are equal.

similar reasoning shows (3) and (5) are maximal ideals in Z15, which is easy to show because we can explicitly list ALL the ideals of Z15:

(0) = {0}

(1) = (2) = (4) = (7) = (8) = (11) = (13) = (14) = Z15

(3) = (6) = (9) = (12) = {0,3,6,9,12}

(5) = (10) = {0,5,10}

as an easy extension of the logic used here, we see that the ONLY proper ideal of Zp, where p is a positive prime integer, is {0}.