Find all maximal ideals in Z 8

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Find all maximal ideals in Z8...

Homework Statement



http://gyazo.com/e292522bc3d99584d5abb55826b4a50f

Homework Equations



Some definitions.

http://gyazo.com/f095ef61ecc9806c8f6a95fa99dad6fb

I was also thinking about using a lattice of ideals to show this.

The Attempt at a Solution



Okay, when I draw out the ideal lattice for these, it's obvious to see which ideals are maximal. For part a, b, and c.

a) For Z8, <2> is maximal.
b) For Z10, <2> and <5> are maximal.
c) For Z12, <2> and <3> are maximal.
d) Requires a proof. Lattice fails.

My question is, how would I argue this without the aid of a lattice for a, b and c?
 
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Zondrina said:

Homework Statement



http://gyazo.com/e292522bc3d99584d5abb55826b4a50f

Homework Equations



Some definitions.

http://gyazo.com/f095ef61ecc9806c8f6a95fa99dad6fb

I was also thinking about using a lattice of ideals to show this.

The Attempt at a Solution



Okay, when I draw out the ideal lattice for these, it's obvious to see which ideals are maximal. For part a, b, and c.

a) For Z8, <2> is maximal.
b) For Z10, <2> and <5> are maximal.
c) For Z12, <2> and <3> are maximal.
d) Requires a proof. Lattice fails.

My question is, how would I argue this without the aid of a lattice for a, b and c?

An ideal must be a subgroup. What do the subgroups of Z_n look like? Think about prime divisors of n. Can you characterize a maximal ideal in terms of those?
 


Dick said:
An ideal must be a subgroup. What do the subgroups of Z_n look like? Think about prime divisors of n. Can you characterize a maximal ideal in terms of those?

Hmm... The subgroups of Zn are cyclic and the order of each subgroup divides the order of the group.

In each case then, ideals of the form <p> where p is a prime that divides the order of the group are maximal ideals?

EDIT : So I suppose I could conclude since all ideals of Zn come from the ideals of Z that contain nZ, any maximal ideal of Zn is of the form pZn where p is a prime that divides n.
 
Last edited:


Zondrina said:
Hmm... The subgroups of Zn are cyclic and the order of each subgroup divides the order of the group.

In each case then, ideals of the form <p> where p is a prime that divides the order of the group are maximal ideals?

EDIT : So I suppose I could conclude since all ideals of Zn come from the ideals of Z that contain nZ, any maximal ideal of Zn is of the form pZn where p is a prime that divides n.

Right.