How do you find the maximal ideals?
-Wooi-Chen Ng
I looked at the example 14 and illstrated by picture in example 15 on page 266 and it appear to be the multiple of n for example z_36 being <2> and <3> are the maximal ideas, the rest of the multiple of 36 are contained in 2 and 3. I am not sure this is correct but that is what I got from it. -Herr-
This is what I got, as well. I see that the smallest ideals are essentially factors of the other ideals. The "maximal" part is that they are only "contained" by the main. At least that's how I think of it. -Tim F
Perhaps you choose the prime ones? -Josh
I think it is the prime divisors
I'm lost...
Think of it this way. What are the prime factors of 36? - 2&3
An ideal I in R is maximal if there is no interesting, in other words can't contain the whole ring, ideal that contains I but does not equal I.
For example, Z_10= {0,1,2,3,4,5,6,7,8,9}
I contains 2 therefore I contains 4,6,8,0
I contains 3 therefore I contains 6,9, but 9 is -1 mod 10 which is 1 which is the unity and thus not interesting
I contains 4 which is I contains 2
I contains 5 which is I contains 0 thus is a maximal
I think that explains the reasoning a little more, but essentially due the fact that ideals are in correspondence to single elements of Z mod n then the maximals are the prime divisors of n as you stated.
--Robertsr 21:48, 29 October 2008 (UTC)
Thanks for the explanation. I used the prime factors, but honestly didn't know why...
This is what I did for part (d): If $ n = p_1^tp_2^t...p_m^t $ then the maximal ideals are $ p_1 $, $ p_2 $,..., $ p_m $ Hope this helps.
