1
$\begingroup$

Suppose $x,y \in R$ where $R$ is a ring (not necessarily with an identity) and suppose $m$ is a positive integer. What is the characteristic of a ring with the property that if $mx = my$, then $x = y$?

This is not true if the characteristic divides $m$.

If the characteristic has a factor in common with $m$, then in some rings of integers modulo $n$ this would not be true either for every $x,y$.

So is it only true in characteristic 0 rings?

$\endgroup$
4
  • $\begingroup$ Your question isn't very clear. Is $m$ fixed? If so, then $mx = my$ implies $x = y$ iff the characteristic is coprime to $m$. $\endgroup$
    – Rob Arthan
    Sep 25, 2021 at 20:04
  • $\begingroup$ I clarified it I think. $\endgroup$
    – E2R0NS
    Sep 25, 2021 at 20:10
  • 2
    $\begingroup$ Watch out for the ring $\mathbb Z[x]/(mx)$. Here, $mx=0$ but $x \ne 0$; however, the ring has characteristic $0$. $\endgroup$
    – Misha Lavrov
    Sep 25, 2021 at 20:16
  • $\begingroup$ @MishaLavrov Ah, so the characteristic of the ring must be $n>0$ but coprime with $m$. $\endgroup$
    – E2R0NS
    Sep 25, 2021 at 20:27

1 Answer 1

Reset to default
1
$\begingroup$

If this property holds and the ring has characteristic $c$, then in particular we must have either $c=0$ or $\gcd(m,c)=1$. Otherwise, let $x = \frac{c}{\gcd(m,c)}a$ for any element $a$ such that $\frac{c}{\gcd(m,c)}a \ne 0$ (this must exist, because $\frac{c}{\gcd(m,c)} < c$), and let $y=0$. Then $my = mx = 0$, but $x \ne y$.

Also, whenever $m = \pm 1$, this property is guaranteed to hold for any ring; when $m=0$, it only holds for the trivial ring.

However, for any other $m$, there are examples of any characteristic $c \ne 1$ where it does not hold. For example, take $\mathbb Z[x]/(mx, c)$. Here, $mx=0$ but $x \ne 0$.

In conclusion, given this property, we can limit the characteristic of the ring; however, the characteristic does not help us determine if this property holds.

$\endgroup$
14
  • 1
    $\begingroup$ You can reduce somewhat to "rings where $mx = 0$ implies $x=0$". A sufficient condition is "unital rings with no zero divisors and characteristic not dividing $m$" but that might be too strong for you. $\endgroup$
    – Misha Lavrov
    Sep 25, 2021 at 20:43
  • 1
    $\begingroup$ The logic is that if $R$ is unital (yes, with an identity) and characteristic not dividing $m$, then $m \in R$. If $mx = my$ but $x \ne y$, then $m$ and $x-y$ are zero divisors, so ruling those out prevents this from happening. $\endgroup$
    – Misha Lavrov
    Sep 25, 2021 at 20:53
  • 1
    $\begingroup$ Those are actually the same. In unital rings with no zero divisors, the characteristic is always prime, so if it does not divide $m$ then the gcd is $1$. No, by "too strong" I meant that there are plenty of rings that have no identity, or that have zero divisors, which also work, I just don't know how to give a nice rule for when they have the property you want. $\endgroup$
    – Misha Lavrov
    Sep 27, 2021 at 21:07
  • 1
    $\begingroup$ It's definitely not necessary. I don't have good go-to examples of rings without an identity in my head, but $\mathbb Z/n\mathbb Z$ with $\gcd(m,n)=1$ has zero divisors - it's just that $m$ isn't one of them, so it still has your property. $\endgroup$
    – Misha Lavrov
    Sep 27, 2021 at 21:40
  • 1
    $\begingroup$ Ah, we have such an $m$ fixed in advance. So take $\mathbb{Z}/12\mathbb{Z}$ and $m=5$. Since $5$ is not a zero divisor in this ring we must have that $x-y$ in $m(x-y)=0$ is also not a zero divisor, so must be zero; hence, property holds. But $\mathbb{Z}/12\mathbb{Z}$ has zero divisors such as $2$ and $6$. $\endgroup$
    – E2R0NS
    Sep 27, 2021 at 23:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .