Question

An element $p$ of a commutative ring $R$ is called "prime" if, for any $a,b\in R$, whenever $ab$ is a multiple of $p$, either $a$ or $b$ is a multiple of $p$.

Is there a word for the "prime-like" property that, whenever $ab$ is a multiple of $p^2$, either $a$ or $b$ is divisible by $p$? Or another, more usual concept in ring theory that this is connected to?

I ask because the "prime-likeness" of $2$ in $R$ seems to control whether the quadratic formula can be made to work for monic polynomials over $R$ (as long as $2$ is also not a zero-divisor). This is because, if the discriminant of $x^2 + bx + c$ is a square $b^2 - 4c = d^2$, then $(-b+d)(-b-d) = 4c$, so at least one (and hence both) of $(-b+d)$ and $(-b-d)$ are multiples of $2$ in $R$. Their halves are the two roots of $x^2 + bx + c$.

For example, $2$ is "prime-like" in $\mathbb{Z}[\sqrt{2}]$, which is easy to verify elementarily. Hence a monic quadratic over $\mathbb{Z}[\sqrt{2}]$ factors iff its discriminant is a square. But $2$ is not "prime-like" in $\mathbb{Z}[\sqrt{5}]$, since $(\sqrt{5}-1)(\sqrt{5}+1) = 4$. And indeed, the discriminant of $x^2 -x-1$ is a square in $\mathbb{Z}[\sqrt{5}]$, but the polynomial doesn't factor there.

Answer 1

I assume (based on your example) that you're primarily interested in the case where $R$ is the ring of integers in an algebraic extension of $\mathbb Q$. Then your property of being "prime-like" is equivalent to the property of generating a primary ideal.

So assume $R$ is a Dedekind domain (in particular ideals factor uniquely into products of prime ideals), and let $p\in R$ be an arbitrary element. Then

$p$ is "prime-like" if and only if $pR=P^k$ for some prime ideal $P$ of $R$ and some $k\in \mathbb N$

i.e. "prime-like" in Dedekind domains is the same as "primary". An element $x$ in the fraction field of $R$ lies in $R$ iff the valuations $\nu_Q(x)$ are non-negative for all prime ideals $Q$ of $R$. So let $a,b\in R$ and assume $p^2\mid ab$. Then $\nu_Q(a/p)=\nu_Q(a)\geq 0$ and $\nu_Q(b/p)=\nu_Q(b)\geq 0$ for all primes $Q\neq P$. So to see that either $a/p$ or $b/p$ lies in $R$, one just has to check that one of them has positive $P$-valuation. But

$p^2\mid ab$ implies $\nu_P(a)+\nu_P(b)=\nu_P (ab) \geq \nu_P(p^2)=2\nu_P(p)$ which implies $\nu_P(a) \geq \nu_P(p)$ or $\nu_P(b) \geq \nu_P(p)$

so either $\nu_P(a/p)\geq 0$ or $\nu_P(b/p)\geq 0$. On the other hand, if the ideal $pR$ isn't primary then $p$ is not prime-like (the construction in Julian's comment can be generalized).

Of course I'm not sure what exactly you're looking for, but at least this clears up what is going on in your last example: while $2$ isn't "prime-like" in $\mathbb Z[\sqrt{5}]$, it is prime like in its integral closure $\mathbb Z[\frac{1+\sqrt{5}}{2}]$ (which is however unspectacular because it remains a prime in that ring).