Prime-like elements of rings - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:54:54Z http://mathoverflow.net/feeds/question/104994 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104994/prime-like-elements-of-rings Prime-like elements of rings Owen Biesel 2012-08-18T18:13:04Z 2012-08-19T02:08:40Z <p>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 $p$. </p> <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?</p> <p>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 $x^2 + bx + c$.</p> <p>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.</p> http://mathoverflow.net/questions/104994/prime-like-elements-of-rings/105008#105008 Answer by Florian Eisele for Prime-like elements of rings Florian Eisele 2012-08-18T22:00:07Z 2012-08-19T02:08:40Z <p>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. </p> <p>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> <blockquote> <p>$p$ is "prime-like" if and only if $pR=P^k$ for some prime ideal $P$ of $R$ and some $k\in \mathbb N$</p> </blockquote> <p>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> <blockquote> <p>$p^2\mid ab$ 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 $\nu_P(b) \geq \nu_P(p)$</p> </blockquote> <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).</p> <p>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).</p>