indivisible elements

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Feb 26, 2015 at 21:33	comment	added	Pace Nielsen		@GreginGre: Okay, I understand now. In the sketch $a$ is assumed to be indivisible but not irreducible. Since it is not irreducible, write $a=xy$ for some $x,y$ non-units. By indivisibility, we may assume $a\|x$, so write $x=a\cdot r$. Setting $d=ry$, which is not a unit since $y$ is not a unit, we have $a=ad$.
Feb 26, 2015 at 21:11	comment	added	GreginGre		Ok, the way i formulated my question was confusing. I was talking about your sketch of proof of the fact that in a finite ring, any indivisible element is prime or irreducible. If $a$ is indivisible but not prime nor irreducible, your first step is to pick a nonzero and nonunit element $d$ such that $ad=a$. I don't understand how you find such a $d$, because for an arbitrary finite ring and an arbitrary $a$, this is not true (take $A=\mathbb{Z}/4\mathbb{Z}, a=\bar{2}$. Then, if $a=ad$, we necessarily have $d=\bar{1}$ or $\bar{3}$, which are units).
Feb 26, 2015 at 18:33	comment	added	GreginGre		Thanks! I wouldn't mind having a detailed proof for the existence of $d$. Clearly, this is not true for any finite ring and any element: for example, if $A=\mathbb{Z}/4\mathbb{Z}$ and $a=\bar{2}$, then all the elements $d$ satisfying $ad=a$ are units. So, it probably uses the fact that $a$ is not prime or irreducible, but I can't see the argument.
Jan 25, 2015 at 10:05	vote	accept	GreginGre
Jan 25, 2015 at 16:00
Jan 25, 2015 at 9:53	vote	accept	GreginGre
Jan 25, 2015 at 10:04
Jan 24, 2015 at 17:31	comment	added	Pace Nielsen		@GreginGre, in a finite ring, indivisible elements are either prime or irreducible. Sketch: Work by contradiction, with $a$ indivisible but not prime or irreducible. Get $a=ad$ for some nonzero, non-unit $d$. Then $d$ is a zero-divisor (since $A$ is finite) so fix $d'\neq 0$ with $dd'=0$. Consider $a=(a+d')\cdot d$. Case 1: $a\|d'$, then $d'=ax$ and $0=dd'=dax=ax=d'$, a contradiction. Case 2: $a\|d$, then $d=ax$ and $a=ad=a^2x$. So $e:=ax=e^2$. But $R/(a)=R/(e)\cong (1-e)R$ is not a domain. Fixing $st=0$, $s,t\in (1-e)R\setminus\{0\}$, we have $(a+s)(e+t)=a$. But $a$ divides neither factor.
Jan 24, 2015 at 5:50	history	edited	Pace Nielsen	CC BY-SA 3.0	Fixed a couple problems in the previous solution
Jan 23, 2015 at 21:48	history	answered	Pace Nielsen	CC BY-SA 3.0