Im soooooo close to solving this problem (Rings)

[pureouchies4717]

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

 

[Data]

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 $y=x = \frac{m}{(n,m)}1_R$.

 

[tacman]

Great job on your progress so far! Here is a bit more explanation to help you fully understand the problem and its solution.

First, let's define a ring R of characteristic m > 0. This means that for any element a in R, m * a = 0. This is because the characteristic is the smallest positive integer such that m * a = 0 for all a in R.

Next, let n be any integer and let b = gcd(n,m). This means that b is the greatest common divisor of n and m. Since 1 < b < m, this means that b is a proper divisor of m.

Now, let's consider the element n * 1_R in our ring R. Since b is a proper divisor of m, we know that there exists some integer k such that m = b * k. Therefore, we have:

n * 1_R = n * (b * k) * 1_R = (n * b) * (k * 1_R) = m * (k * 1_R) = 0 * (k * 1_R) = 0

This shows that n * 1_R is equal to 0, which means that it is a zero divisor in our ring R. This is because there exists a non-zero element k * 1_R that when multiplied by n * 1_R, gives us 0.

Therefore, we have shown that if 1 < gcd(n,m) < m, then n * 1_R is a zero divisor in our ring R. This is an important result because it helps us understand the properties of rings with non-zero characteristics. Keep up the good work!