Im soooooo close to solving this problem (Rings)

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SUMMARY

The discussion focuses on proving that if \(1 < \text{gcd}(n,m) < m\) for a ring \(R\) of characteristic \(m > 0\), then \(n \cdot 1_R\) is a zero divisor. The user identifies that \(\text{gcd}(n,m) = b\) implies \(d | n\) and that \(m \cdot 1_R = 0_R\). A suggestion is made to demonstrate the existence of elements \(x\) and \(y\) in \(R\) such that \(x \cdot (n \cdot 1_R) = 0_R\) and \((n \cdot 1_R) \cdot y = 0_R\), with a specific choice of \(y = x = \frac{m}{\text{gcd}(n,m)}1_R\).

PREREQUISITES
  • Understanding of ring theory and zero divisors
  • Familiarity with the concept of characteristic of a ring
  • Knowledge of greatest common divisor (gcd) and its properties
  • Basic algebraic manipulation within rings
NEXT STEPS
  • Study the properties of zero divisors in ring theory
  • Explore examples of rings with non-trivial characteristics
  • Learn about the implications of gcd in algebraic structures
  • Investigate the role of elements in a ring that satisfy \(x \cdot (n \cdot 1_R) = 0_R\)
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Mathematicians, algebra students, and anyone studying abstract algebra, particularly those interested in ring theory and zero divisors.

pureouchies4717
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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
 
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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:
 

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