Wikipedia:Reference desk/Archives/Mathematics/2017 May 24

Mathematics desk
< May 23	<< Apr | May | Jun >>	May 25 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

May 24

Irreducibles without primes

Is there an integral domain with an irreducible element but no prime elements? GeoffreyT2000 (talk) 01:26, 24 May 2017 (UTC)

How about $\mathbb {C} [x,y,z]/(y^{2}-x,z^{2}-x)$? y and z are irreducible and are obviously not prime. I'm fairly certain there are no prime elements: given a non-unit polynomial $p(y,z)=y^{m}z^{n}+q(y,z)$, $[y^{m}z^{n}+q(y,z)]\cdot [y^{m}z^{n}-q(y,z)]=[y^{m-1}z^{n+1}+q(y,z)]\cdot [y^{m-1}z^{n+1}-q(y,z)]$. And by degree considerations, if $p(y,z)$ divides either of these factors, it differs from them by a unit. But counting the parity of the number of occurrences of y rules out that possibility.--2406:E006:332E:1:423:6A72:E875:1DD9 (talk) 05:17, 24 May 2017 (UTC)

The ring that you give here is not an integral domain because y and z have equal squares but are neither themselves equal nor additive inverses of each other, so y-z and y+z are nonzero elements with a product of zero. (The usual difference times sum formula for a difference of squares works in any commutative ring.) GeoffreyT2000 (talk) 16:31, 24 May 2017 (UTC)

Any noetherian domain has irreducibles proof here. If you take K[t^2,t^3] (= K[x,y]/(y^2-x^3), the cuspidal cubic) and localize at the singular point, or formally complete to $K[[t^{2},t^{3}]]$, you get a local ring with only 1 prime ideal: (t^2, t^3), which clearly is not principal, clearly has no generator which is prime, while t^2 and t^3 are irreducible. See[1] or [2] also.John Z (talk) 23:58, 24 May 2017 (UTC)

Square root of -7 in 2-adic numbers

Does the Ramanujan–Nagell equation give rise to a square root of -7 (= ...111001) in $\mathbb {Q} _{2}$, the 2-adic numbers? GeoffreyT2000 (talk) 01:29, 24 May 2017 (UTC)

It would if there were an infinite number of solutions, but it doesn't seem that there are. You don't need that though since all you need is

$x^{2}=-7\mod 2^{n}$

not

$x^{2}=2^{n}-7$

Two square roots of -7 are

...0 1100 0000 1011 0101 & ...1 0011 1111 0100 1011. --RDBury (talk) 07:26, 25 May 2017 (UTC)