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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!

Asaf Karagila
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Wilson
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3 Answers3

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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 $f\in R$, then consider $fR=\{af : a\in R\}$. 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. (Or $f=0$)

Proof:

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

Conversely, if $R/pR$ is a domain, then if $p\mid ab$, $ab\equiv 0 \pmod{pR}$, so either $a\equiv 0 \pmod{pR}$ or $b\equiv 0 \pmod{pR}$, which means either $p\mid a$ or $p\mid b$. Hence $p$ is prime. $\blacksquare$

Note:

As rschwieb points out in the comments, I should have been a bit more careful when originally writing this. We usually exclude $0$ from the definition of prime (as you've done above). However $R/0R=R/0=R$ is certainly a domain. I suspect that the reason for excluding $0$ is a function of the other motivations for this definition discussed below. Since if we allow $0$ to be prime, then it complicates the statment of unique prime factorization, since after all, $0=0\cdot 3^2=0\cdot 101=0\cdot (-17)$, so how can $0$ have a unique prime factorization?

For more on this and a different perspective, I recommend rschwieb's excellent answer here (same link as in the comments).

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 $x\in R$ into irreducibles $$x = \prod_i p_i = \prod_j q_j,$$ with $p_i,q_j\in R$ 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 $9=3\cdot 3 = (-3)\cdot (-3)$)?

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 $p\mid x$, 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 $p\mid q_1$ by definition, and we are done. Otherwise, since $p\mid (q_1\cdots q_{n-1})q_n$, then by primality of $p$, either $p\mid q_1\cdots q_{n-1}$, in which case $p\mid q_j$ for some $j$ by the inductive hypothesis, or $p\mid q_n$, and we are done.

Conversely, if $p$ has the property discussed above, then if $p\mid ab$ for some $a$ and $b$, then let $a=\prod_i\alpha_i$ and $b=\prod_j\beta_j$ 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 $p\mid x ab= \prod_i \alpha_i \prod_j \beta_j$, so by the property we're assuming $p$ has, either $p\mid \alpha_i$ for some $i$, or $p\mid \beta_j$ for some $j$, and thus either $p\mid a$ or $p\mid b$. 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.

jgon
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  • Could you clarify your choice of representation of factorization? Why is there a factor of $u$ and $v$ separate from the product of the rest of the irreducible factors? Also, could you expand upon this "That $p_{i}$ divides some $q_{j}$ for all factorizations of multiples of $p_{i}$ is equivalent to $p_{i}$ being prime?" I am a little lost. Thank you! – Wilson Jan 08 '19 at 03:20
  • @Wilson, sorry yes. I'll edit in a moment, but $u$ and $v$ are units, though I could have omitted them. – jgon Jan 08 '19 at 03:21
  • Also, what does the notation $R/(p)$ and $R/(f)$ denote in this context? Sorry, I am a little new to these topics. Thank you again. – Wilson Jan 08 '19 at 03:23
  • @Wilson No worries. I'm back to edit now – jgon Jan 08 '19 at 03:26
  • @Wilson I've edited. Hopefully it will be more clear now :) – jgon Jan 08 '19 at 03:42
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    The "ring theoretic" argument isn't completely accurate. $f=0$ is not ever considered to be prime, but in a domain $R/(0)$ will be a domain. I tried to give a little intuition on what makes $0$ different from primes in this solution. Actually, as others have said, it has more to do with actual factorization than some property that a principal ideal has (and so that solution is about irreducibles in a UFD, which are exactly the primes in a UFD.) – rschwieb Jan 08 '19 at 14:38
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    Edge case notwithstanding, I realize that $0$ does satisfy the $p|ab$ definition in a domain. It's just that it's not enough to warrant calling $0$ a prime. Perhaps though, there will come a time when we care about the connection to factorization and irreducibles less, and then we can consider using this inclusive definition. I'm game for that sort of thing, but I just wanted to echo what the current state of literature probably says. – rschwieb Jan 08 '19 at 14:40
  • @rschwieb Fair point, I do agree that the elementwise notion of primeness is primarily of interest for studying factorization, and $0$ being prime would require complicating the statement of unique prime factorization slightly. However, I'm not sure it's that important whether we consider $0$ prime or not, since you could restrict unique prime factorizations to nonzero elements of the ring. – jgon Jan 08 '19 at 14:42
  • For example, something like "$\mathbb Z/(p)$ is a field exactly when $p$ is prime" relies on this convention. – rschwieb Jan 08 '19 at 14:43
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    @rschwieb That's certainly true, I agree that we don't usually consider $0$ a prime. I'm planing to edit in a note to that effect. – jgon Jan 08 '19 at 14:43
  • @rschwieb Also I like your linked answer, I've added a link to it in my answer. – jgon Jan 08 '19 at 14:50
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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”.

ir7
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Welcome! take 6 for example it is obviously not prime. We have that $6|6=2*3$ but $6\nmid 2$ and $6\nmid 3$.

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

George Ntoulios
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