A negation of definition

Question

Let $R$ a ring. We know that an ideal $I$ of $R$ is said prime if for all $a,b\in R$ $$ab\in I\Rightarrow a\in I\quad\text{or}\quad b\in I.$$

When an ideal is not prime? That is, what is the negation of this definition formally?

EDIT

I understood thanks to your comments that an ideal is not prime if $\exists a,b \in R$ such that $a\notin I$ and $b\notin I$. From here can I say that an ideal is not prime if for all $a,b\in R\setminus I$, $ab\in I$?

Thanks

Regarding edit! I afraid you can't – Cloud JR K Nov 28 '18 at 19:00 — Cloud JR K, Nov 28 '18 at 19:00
Your edit is not quite right – Prince M Nov 29 '18 at 07:48 — Prince M, Nov 29 '18 at 07:48

score 4 · Answer 1 · answered Nov 28 '18 at 18:28

4

An Ideal $l$ is not prime if there exists $a,b\in R$ such that

$ab\in I , a\notin I $and $b\notin I$

(Note : negation of $P\implies Q$ is $P$ and $\neg Q$

answered Nov 28 '18 at 18:28

Cloud JR K

2,466

score 2 · Accepted Answer · answered Nov 28 '18 at 18:26

2

$\exists a,b\in R$ with $ab\in I, a\notin I, b\notin I$

For example, $2\cdot3\in 6\mathbb{Z}, 2\notin 6\mathbb{Z}, 3\notin 6 \mathbb{Z},$ so $6 \mathbb{Z}$ is not a prime ideal of $\mathbb{Z}.$

answered Nov 28 '18 at 18:26

saulspatz

53,131

Thanks for your ansewer. But it's corret said that $I$ is not prime if exist $a,b\in R$ such that if $ab\in I$, then $a\notin I$ and $b\notin I$, therefore if and only if for all $a,b\in R\setminus I$ $ab\in I$? – Jack J. Nov 28 '18 at 18:38
@JackJ. No, that's wrong. – egreg Nov 28 '18 at 18:46
and what is the correct formulation? – Jack J. Nov 28 '18 at 18:46
@Jack J , what he written is absolutely correct ...negation of if... Then 'is not' if ... Then not – Cloud JR K Nov 28 '18 at 18:47
negation of $P\implies Q$ is $P$ and $\neg Q$. I added this in my answer too – Cloud JR K Nov 28 '18 at 18:48

A negation of definition

2 Answers2