What's so special about a prime ideal? An ideal is defined something like follows:

Let $R$ be a ring, and $J$ an ideal in $R$. For all $a\in R$ and $b\in J$, $ab\in J$ and $ba\in J$.

Now, $J$ would be considered a prime ideal if 

For $a,b\in R$, if $ab\in J$ then $a\in J$ or $b\in J$.

To my (admittedly naive) eyes, this isn't saying much. More or less, I guess it just sounds like a backwards way of describing a regular ideal.


*

*$a,b$ are always elements of $R$, though the prime ideal definition doesn't specify that one has to be in $J$...

*... but the definition of a normal ideal already tells that the product is in $J$ if one of the elements is in $J$.


So, in both cases, the product is in $J$, and either of the elements is in $J$, making them seem like incredibly similar statements to me, and not saying much about the interesting "prime-like" properties of a prime ideal.
What makes these two different?

The easiest way to answer this, I think, is with an example.  Let $R=\mathbb{Z}$, and let's consider the ideal $I=(6)$.  This is the set of all integers that are multiples of $6$.  You can see that it's an ideal because if you take any multiple of $6$ and multiply it by any other integer, the result is still a multiple of $6$.  So $I$ is closed under multiplication by any element of the ring.
But $I=(6)$ is not a prime ideal.  You can see this because $2 \notin (6)$ and $3 \notin (6)$, but $2 \cdot 3 \in (6)$.
On the other hand, the ideals $(3)$ is prime:  If you multiply two numbers together and the result is a multiple of $3$, then at least one of the two numbers you began with must also be a multiple of $3$.
If you ponder this example, you will also understand why the word "prime" is used for this property.
