If $R$ is isomorphic to a ring $S$, show that $S$ has this same property: $n*1_{S}=0$

Question

I am having some trouble with a homework problem... Hoping someone may be able to provide some insight and a bump in the right direction!

Suppose $R$ is a commutative ring with unity $1_{R}$, and has the property that $5$ is the smallest positive integer $n$ so that $n*1_{R}=1_{R}+1_{R}+...+1_{R}=0$. If $R$ is isomorphic to a ring $S$, show that $S$ has this same property: $5*1_{S}=0$, and $n$ is the smallest positive integer $n$ such that $n*1_{S}=0$.

My professor provided a couple hints: Don't assume that either $R$ or $S$ has to be $\mathbb{Z}$. Show this using the isomorphism - don't just say it it's because they are isomorphic.

Answer 1

Let $\phi: R\rightarrow S$ be an isomorphism. Since $\phi$ is a ring homomorphism, we must have $\phi(1_{R})=1_{S}$ and $\phi(0_{R}) = 0_{S}$. Then

$\phi(0) = \phi(1_{R}+1_{R}+1_{R}+1_{R}+1_{R}) = \phi(1_{R})+\phi(1_{R})+\phi(1_{R})+\phi(1_{R})+\phi(1_{R}) = 1_{S}+1_{S}+1_{S}+1_{S}+1_{S} = 0_{S}$. Hence, $5*1_{S} = 0_{S}$.

Can you complete the proof by showing that $5$ is the least such integer for which this is true? (Hint: suppose $k*1_{S} = 0$ for some $0 < k < 5$. What would be true of $k*1_{R}$?)

Answer 2

A high level view of the characteristic of a ring with unity can be obtained by considering that, for every ring $R$ (commutative or not) with unity, there is a unique ring homomorphism $$ \varepsilon_R\colon\mathbb{Z}\to R $$ It's just the unique group homomorphism that sends $1\in\mathbb{Z}$ to $1_R$. It can be checked that this is indeed a ring homomorphism: $$ \varepsilon_R(m0)=01_R=0_R=\varepsilon_R(m)\varepsilon_R(0) $$ If we assume $n\ge0$ and $\varepsilon_R(mn)=\varepsilon_R(m)\varepsilon_R(n)$, we have \begin{align} \varepsilon_R(m(n+1))=\varepsilon_R(mn+m) &=\varepsilon_R(mn)+\varepsilon_R(m)\\ &=\varepsilon_R(m)\varepsilon_R(n)+\varepsilon_R(m)\\ &=\varepsilon_R(m)(\varepsilon_R(n)+1)\\ &=\varepsilon_R(m)\varepsilon_R(n+1) \end{align} and, by induction, we have the proof, for $n\ge0$, that $\varepsilon_R(mn)=\varepsilon_R(m)\varepsilon_R(n)$. For $n<0$, it's easy to apply the rule of signs and, finally, we have that $$ \varepsilon_R(mn)=\varepsilon_R(m)\varepsilon_R(n) $$ for all $m,n\in\mathbb{Z}$.

Now, $\ker\varepsilon_R$ has a special significance: if $\ker\varepsilon_R=\{0\}$, then $n1_R\ne0_R$ whenever $n\ne0$. If instead $\ker\varepsilon_R=k\mathbb{Z}$ and $k>0$, then $k1_R=0_R$ and, consequently, $kr=0_R$ for all $r\in R$ and $k$ is the minimum positive integer with this property.

The unique non negative integer $k$ such that $\ker\varepsilon_R=k\mathbb{Z}$ is called the characteristic of $R$. The properties above characterize it.

If $\varphi\colon R\to S$ is an isomorphism of rings, then $$ \varphi\circ\varepsilon_R=\varepsilon_S $$ by the uniqueness, so $$ \ker\varepsilon_R\subseteq\ker\varepsilon_S $$ The converse inclusion follows by considering $\varphi^{-1}$. Thus

Isomorphic rings $R$ and $S$ have the same characteristic, because $\ker\varepsilon_R=\ker\varepsilon_S$.