Prime ideals of Z

Question

I must be going crazy...

We know that for an integral domain, $R, a \in R$ is prime if and only if $(a)$ is a prime ideal. So taking $R$ to be the integers and $a=2$. Obviously 2 is prime and looking at $(2) = 2\mathbb{Z} = \{0, \pm 2, \pm 4, \pm 6,...\}$. But for $8 = 2 \times 4$ we have both 2 and 4 $\in (2)$ and therefore (2) is not prime...

What am I missing? Thanks, Tal

Hi Praphulla, I thought that a prime ideal, $P$, is one where if ab $\in P$ then either $a \in P$ OR $b \in P$. — Tal Bashan, Oct 09 '14 at 04:21
That is fine... so, why do you think that is missing here... — , Oct 09 '14 at 04:21
because $8 \in (2)$ but both 2 AND 4 are $\in (2)$. Therefore (2) is not a prime ideal of $\mathbb{Z}$ even though 2 is a prime number... — Tal Bashan, Oct 09 '14 at 04:24
No.. No.. No... $ab\in P$ implies either $a$ or $b$ in $P$ does not say that both should not be there.... — , Oct 09 '14 at 04:25
ahhhhhh.... I've been trying to ascertain that but nowhere does it say that. For me 'or' means one or the other not both.. my mistake. Thanks for your help... — Tal Bashan, Oct 09 '14 at 04:27
It's what logicians and digital circuit designers call "inclusive OR". — Robert Lewis, Oct 09 '14 at 04:32

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

1

I'm not sure that you are understanding the concept of a prime ideal correctly. A prime ideal is a proper ideal $I$ such that whenever $a,b\in R$ and $ab\in I$ one has either $a\in I$ or $b\in I$ already. So, in your case, both $a$ and $b$ are in $(2)$ and this ideal fits the definition perfectly.

I am assuming, based on your statement, that you are under the misconception that a prime ideal is an ideal $I$ such that whenever $ab\in I$ one has $a\not\in I$ and $b\not\in I$ (in other words, exactly the negation of the correct definition). Actually, here's the intuition as to why the definition is as it is: a prime number in $\mathbb{Z}$ is a number $p\ne 1$ so that whenever $p|ab$ one has $p|a$ or $p|b$. This translates naturally to the definition that $ab\in(p)$ implies $a\in(p)$ or $b\in(p)$. (This is for $\mathbb{Z}$, which is a principal ideal domain in which all ideals are generated by one element, and so all ideals are of the form $(n)$ for some $n$.)

This answer may provide further insight into the definition of a prime ideal: Motivation behind the definition of Prime Ideal

edited Apr 13 '17 at 12:21

Community

1

answered Oct 09 '14 at 04:30

Marc

965
4
13

Hi Marc, my misconception (as was highlighted above) is that if $ab \in I$ then either $a \in I$ OR $b \in I$ but not both. I now understand that if both a and b are $\in I$ this is not a violation of the definition of prime ideal. – Tal Bashan Oct 09 '14 at 04:33
@TalBashan I see. It is relatively rare in mathematics when one defines a concept with exclusive disjunction (i.e. X or Y but not both). I believe that usually an author would explicitly state "either X, or Y, but not both" in this case, and you should assume inclusive disjunction whenever you see "either X or Y" by itself. – Marc Oct 09 '14 at 04:36
I'll need to speak to my lecturer about this as this was never brought up in past units and I wasn't aware of this. The wording in our definition was "...either a $\in I$ or $b \in I$. The 'either' made me think that it is exclusive. Thanks for your help... – Tal Bashan Oct 09 '14 at 04:40

Prime ideals of Z

1 Answers1