# Easiest way to prove that a subset of even integers is closed under multiplication?

What's the easiest way of showing that; $2\mathbb{Z}\setminus (4n-2)\mathbb{Z}$ is closed under multiplication?

(I'm trying to show that $(4n-2)$ is a prime element of $2\mathbb{Z}$ by showing $(4n-2)\mathbb{Z}$ is a prime ideal)

What is n? If you let n=5, for instance, you don't have $2\mathbb{Z} \setminus 18 \mathbb{Z}$ closed under multiplication, since $6$ is in this set but $6^2$ is not. – Barry Smith Oct 18 '11 at 11:23
Or did you mean "Let S be the set of integers of the form 4n−2. Show that $2\mathbb{Z}\setminus S$ is closed under multiplication"? If so, the easiest way is to rephrase the statement in a much simpler form. – Barry Smith Oct 18 '11 at 11:31
Thankyou, you have highlighted an error in my reasoning. Clearly all $4n-2$ with integer $n$ are not prime elements. What are the prime elements of $2\mathbb{Z}$? – Bruised Oct 18 '11 at 12:11
The numbers $2p$ as $p$ runs through the odd prime natural numbers. – Barry Smith Oct 18 '11 at 12:21
$nk$ is a prime in $n\mathbb{Z}$ if and only if whenever $nk$ divides a product $(na)(nb) = n^2ab$, then either $nk$ divides $na$ or $nk$ divides $nb$; that means in particular that whenever $k$ divides $nab$, then $k$ divides $a$ or $k$ divides $b$. Hence $k$ must be a prime that does not divide $n$. Conversely, if $k$ is a prime that does not divide $n$, then whenever $nk$ divides $(na)(nb)$, then $k$ divides $nab$, hence $k$ divides $a$ or divides $b$, hence $nk$ divides $na$ or $nb$, so $nk$ is a prime in $n\mathbb{Z}$. – Arturo Magidin Oct 18 '11 at 13:15