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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 $ℂ[x,y,z]/(y^2-x,z^2-x)$? 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

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

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)