Bdtyktl

Definition: An element $p$, not zero and not a unit, is called prime if $p|ab$ implies $p|a$ or $p|b$.

I am having trouble understanding why this definition of prime follows naturally. Could someone provide an intuitive explanation for this definition? For context, the text I am learning from has briefly gone over groups, rings, and integral domains.

Thank you!

Good question!

There are probably several possible answers to this question, but here is my perspective.

Reason 1: Ring Theoretic

Let $R$ be a commutative ring. Let $fin R$, then consider $fR={af : ain R}$.
You can check that this is an ideal of $R$, and it is also often denoted by simply $(f)$ (meaning the ideal of $R$ generated by $f$).

Now we can ask the question

When is $R/fR$ a domain?

It turns out the answer is:

Precisely when $f$ is prime.

Proof:

If $p$ is prime, then if $$abequiv 0 pmod{pR},$$
by definition this means $pmid ab$, which since $p$ is prime implies that $pmid a$ or $pmid b$. However this in turn means
$$aequiv 0!!pmod{pR}quadtext{or}quad bequiv 0!!pmod{pR},$$
which is what it means for $R/pR$ to be a domain.

Conversely, if $R/pR$ is a domain, then if $pmid ab$, $abequiv 0 pmod{pR}$, so either $aequiv 0 pmod{pR}$ or $bequiv 0 pmod{pR}$, which means either $pmid a$ or $pmid b$. Hence $p$ is prime. $blacksquare$

Reason 2: Number Theoretic (kind of)

The other way we come up with this naturally is that it is the condition we need to hold in order to get unique factorizations.

I.e., suppose we have two factorizations of an element $xin R$ into irreducibles
$$x = prod_i p_i = prod_j q_j,$$
with $p_i,q_jin R$ irreducibles,
then when are we guaranteed that some $p_i$ occurring in the first factorization appears somewhere on the left hand side (or an associate of $p_i$, i.e. a unit times $p_i$, since for example in the integers we could have $9=3cdot 3 = (-3)cdot (-3)$)?

Well, we need to have $p_i$ divide one of the $q_j$ (for then they are associates, since $p_i$ and $q_j$ are irreducibles).

The condition that for all multiples $x$ of $p_i$, $p_i$ divides some $q_j$ for any factorization $x=prod_j q_j$ of $x$ into irreducibles is equivalent to $p_i$ being prime (for a Noetherian ring, so that we are guaranteed to have factorizations into irreducibles, otherwise bad things could happen).

Proof:

A note on notation: I'll replace $p_i$ with $p$.

Suppose $p$ is prime, and $pmid x$, and $prod_j q_j$ is a factorization of $x$ into irreducibles, then we induct on the length of the factorization. If $x=q_1$ is irreducible, then $pmid q_1$ by definition, and we are done. Otherwise, since $pmid (q_1cdots q_{n-1})q_n$, then by primality of $p$, either $pmid q_1cdots q_{n-1}$, in which case $pmid q_j$ for some $j$ by the inductive hypothesis, or $pmid q_n$, and we are done.

Conversely, if $p$ has the property discussed above, then if $pmid ab$ for some $a$ and $b$, then let $a=prod_ialpha_i$ and $b=prod_jbeta_j$ be factorizations of $a$ and $b$ into irreducibles (since $R$ is Noetherian. If you aren't familiar with Noetherianness yet, then just take the existence of factorizations into irreducibles as a black box for now). Then $pmid x ab= prod_i alpha_i prod_j beta_j$, so by the property we're assuming $p$ has, either $pmid alpha_i$ for some $i$, or $pmid beta_j$ for some $j$, and thus either $pmid a$ or $pmid b$. Hence $p$ is prime.

Reason 3: (Actually a consequence of reason 2)

There is a theorem, which is relevant here.

A Noetherian domain $R$ is a UFD if and only if every irreducible is prime.

An ideal is prime if and only if its set-theoretical complement is multiplicatively closed. (We want to keep obtaining non-multiples of $p$ when multiplying non-multiples of $p$.)

This allows for constructing a new ring (addition and multiplication) of “fractions” whose “denominators” live only in the complement of the prime ideal (such construction is called “localization”). This new ring is simpler than the original ring, exposing only “local behaviour”.

Welcome!
take 6 for example it is obviously not prime. We have that $6|6=2*3$ but $6nmid 2$ and $6nmid 3$.

So if this definition is not true, that means there must be "a part" of the number hidden as a factor of $a$ and a different part hidden in $b$. Which whould of course mean that the number itshelf has more than one factor so it is not prime in the classical sence