indivisible elements

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15 events

when toggle format	what		by	license	comment
May 11, 2015 at 4:45	answer	added	CPM		timeline score: 5
Jan 25, 2015 at 16:00	vote	accept	GreginGre
Jan 25, 2015 at 11:25	answer	added	YCor		timeline score: 8
Jan 25, 2015 at 10:05	vote	accept	GreginGre
Jan 25, 2015 at 16:00
Jan 25, 2015 at 10:03	history	edited	GreginGre	CC BY-SA 3.0	added 1 character in body
Jan 25, 2015 at 9:53	vote	accept	GreginGre
Jan 25, 2015 at 10:04
Jan 23, 2015 at 21:48	answer	added	Pace Nielsen		timeline score: 7
Jan 21, 2015 at 6:01	comment	added	GreginGre		thanks for your answer. Your example for $Q1$ does not work, since $x$ is certainly indivisible, but it is prime. iN fact, any noetherian boolean ring won't work , because it will be finite, hence isomorphic to $\mathbb{F}_2^n$ for some $n\geq 1.$ for the later ring, there is no irreducible elements, and all indivisible elements are prime. However, your candidate for Q3 seems to work, thanks a lot !
Jan 21, 2015 at 1:41	comment	added	Pace Nielsen		On question 3, take the ring $A=K[[t^2,t^3]]$. The only non-zero prime ideal in this ring is the maximal ideal $(t^2,t^3)$, which is not principal. Thus, there are no prime elements. It is noetherian, according to this wiki: en.wikipedia.org/wiki/Cohen%E2%80%93Macaulay_ring
Jan 20, 2015 at 21:02	comment	added	Arturo Magidin		I misread the condition for "indivisible".
Jan 20, 2015 at 21:02	comment	added	GreginGre		Hi,sorry , I meant "indivisible" in Q1, and I forgot to get rid of the trivial case of a field. I've modified the text now. However, I don't understand how you can say that $p\mid ab\Rightarrow p\mid a$ or $p\mid b$ when $p$ is indivisible.
Jan 20, 2015 at 20:54	history	edited	GreginGre	CC BY-SA 3.0	added 42 characters in body
Jan 20, 2015 at 20:52	comment	added	Arturo Magidin		In a commutative ring, "indivisible" and "prime" are the same thing for nonzero elements. For in a commutative ring, prime and completely prime are equivalent for ideals. If $p$ is prime, and $p\|ab$, then $ab\in (p)$, hence $a\in (p)$ or $b\in (p)$, hence $p\|a$ or $p\|b$. Conversely, if $p$ is indivisible and $ab\in (p)$, then $p\|ab$, hence $p\|a$ or $p\|b$, hence $a\in (p)$ or $b\in (p)$, thus $(p)$ is completely prime, and hence prime; therefore, $p$ is prime.
Jan 20, 2015 at 20:50	comment	added	Arturo Magidin		Q2 and Q3: Fields are noetherian domains and have no primes and no irreducible elements. However, they have no zero divisors, so your parenthetical comment is false. In particular, it is false that Noetherian domains must have indivisble elements.
Jan 20, 2015 at 20:32	history	asked	GreginGre	CC BY-SA 3.0