# What is the characteristic of a ring with the property that if $mx = my$, then $x = y$?

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?

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

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.

• 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. Sep 25, 2021 at 20:43
• 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. Sep 25, 2021 at 20:53
• 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. Sep 27, 2021 at 21:07
• 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. Sep 27, 2021 at 21:40
• 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 $6$. Sep 27, 2021 at 23:53