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$\mathbb{Z}/6\mathbb{Z}= \{ [0],[1],[2],[3],[4],[5]\}$. I'm having a problem especially in finding the prime ideals. it seems to me like the whole ring is a prime ideal, but it can't be, because the definition of prime ideal is that it not to be the whole ring. Any explanations on how should I check it?

egreg
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CnR
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  • It's not hard to write down all the ideals of this ring, and then check each one to see whether it's prime. – Gerry Myerson Nov 01 '13 at 22:23
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    The ring $\mathbb{Z}/6\mathbb{Z}$ has four ideals: $6\mathbb{Z}/6\mathbb{Z}$, $2\mathbb{Z}/6\mathbb{Z}$, $3\mathbb{Z}/6\mathbb{Z}$ and $\mathbb{Z}/6\mathbb{Z}$. The whole ring is never a prime ideal, by definition. – egreg Nov 01 '13 at 22:25

2 Answers2

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As you can calculate either by elementary number theory or by using the homomorphism theorems for rings, the only proper ideals are the zero ideal, $2\mathbb{Z}/6\mathbb{Z} = \{[0], [2], [4]\}$ and $3\mathbb{Z}/6\mathbb{Z} = \{[0],[3]\}$ Now, $[2]\cdot [3]=[0]$, so the zero ideal is not prime in this ring. The other two ideals are maximal and hence also prime.

TBrendle
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  • why 5Z is not a proper integral? and why if they're maximal they're also prime? – CnR Nov 01 '13 at 23:13
  • If you multiply all the elements in $\mathbb{Z}/6\mathbb{Z}$ by 5 you'll get the elements of $\mathbb{Z}/6\mathbb{Z}$ again. So in particular, any ideal containing 5 must be the whole ring. All maximal ideals in any ring are necessarily prime; is this not proved in your text? – TBrendle Nov 02 '13 at 02:16
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By the homomorphism theorems, the ideals in $\mathbb{Z}/n\mathbb{Z}$ are of the form $m\mathbb{Z}/n\mathbb{Z}$ where $m$ is a divisor of $n$. If $n=dm$, then $$ (\mathbb{Z}/n\mathbb{Z})\big/(m\mathbb{Z}/n\mathbb{Z})\cong \mathbb{Z}/d\mathbb{Z} $$ and it should now be easy to answer the general question, so also the case $n=6$.

egreg
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