Im soooooo close to solving this problem (Rings)

  • Thread starter Thread starter pureouchies4717
  • Start date Start date
  • Tags Tags
    Rings
pureouchies4717
Messages
98
Reaction score
0
Let R be a ring of characteristic m > 0, and let n be any
integer. Show that:

if 1 < gcd(n,m) < m, then n · 1R is a zero divisor



heres what i got out of this:

Let gcd(n,m) = b

1< d < m so m/d = b < m
and d | n


Also, m * 1_R = 0

can someone please offer some insight?
thanks,
nick
 
Physics news on Phys.org
You know m 1_R=0_R. You need to show that there are some x, y in R with x\cdot (n 1_R) = 0_R and (n 1_R) \cdot y = 0_R.

I suggest trying y=x = \frac{m}{(n,m)}1_R. :smile:
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
Back
Top