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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.  
 
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, <math>Z_"10"</math> = {0,1,2,3,4,5,6,7,8,9}
+
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 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 3 therefore I contains 6,9, but 9 is -1 mod 10 which is 1 which is the unity and thus not interesting

Revision as of 16:48, 29 October 2008

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


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)

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