Q:

Can't we say that on a commutative ring every ideal is prime?

On a commutative ring K, an ideal I is prime when:
$$
    I \text{ is a subgroup of }K,\text{ }  I \neq K, \text{and if } x,y \in K, \text{the relation } xy \in I \text{ implies } x \in I \text{ or }y\in I.
$$
But is this not always the case in a commutative ring K?
Because by definition, if  I is an ideal of K, we have $ax \in I$ for any $a \in K$ and $x \in I$. 
Where my understanding went wrong?

A:

Take $\;2\cdot 3\in6\Bbb Z\;$ . Yet $\;2,\,3\notin 6\Bbb Z\;$ 
I think the understanding problem here is: a prime ideal certainly fulfills the condition$\;xy\in I\implies (x\in I\;\;or\;\;y\in I\,$) , and this is not the same as the very definition of ideal, as the example above shows: observe that
$$a\in6\Bbb Z\implies a=6m,\,m\in\Bbb Z\implies\;\;\forall\,y\in\Bbb Z\;,\;\;ay=6my\in\Bbb Z$$

