Modifying the definition of a prime ideal in a commutative ring.

Question

According to JOSEPH A.GALLAIN EDITION 8th

A prime ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that $a, b \in R$ and $ab \in A$ imply $a \in A$ or $b \in A$.

I am confused about what we may conclude after we change the above definition to be as follows: if $a,b \in R$ then $ab \in A$ implies both $\,a,b\in A$. Does this mean that $A$ will not be a proper ideal?

Since $a,b$ are arbitrary then $ab \in A$ implies $a,b \in A$ then $A=R$.

But I am confused if we choose $a,b$ from $A$ then $a,b$ are in $R$ and in $A$ but can I choose $a,b$ in this way !

Please help me to reach the other side of river !

Answer 1

Try and think where the definition of prime ideal comes from.

A prime number $p$ satisfies the property “if $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$”.

You seem to be interested in the set of all positive integers $s$ that satisfy

if $s$ divides $ab$, then $s$ divides $a$ and $b$

Well, this set is quite small, because only $1$ satisfies the requirement: indeed, for every $a$, $p$ divides $ap$.

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Let's generalize: you should be able to see the similarity in the argument.

Let now $I$ be an ideal of a ring $R$ (no unit element required) such that

for all $a,b\in R$, if $ab\in I$ then $a\in I$ and $b\in I$

Take $a\in I$; then, for every $b\in R$, $ab\in I$, forcing $b\in I$.