# Some Properties about the Characteristic of a Ring

Define characteristic of a ring $$R$$ as the natural number n such that $$n\mathbb{Z}$$ is the kernel of the unique ring homomorphism from $$\mathbb{Z}$$ to $$\mathbb{R}$$, which is given by $$\begin{array}{rccl} \phi \colon & \mathbb{Z} & \longrightarrow & R\\ &z & \longmapsto & z \cdot 1_R, \end{array}$$ where $$R$$ is a commutative ring with unity.

I want to prove that $$n \cdot 1_R = 0$$ and $$n \cdot r \cdot 1_R = 0$$ where $$r \in R$$ and $$n$$ is the characteristic of $$R$$.

My try is:

1. Let $$k$$ be an element in the kernel, then (as $$\ker\phi=n\mathbb{Z}$$) there exist an interger $$z$$ such that $$k = n \cdot z$$. By definition of homomorphism and kernel $$\phi(k)=\phi(n\cdot z)=\underbrace{1_R+\cdots+1_R}_{n\cdot k}\; \forall k \in \mathbb{Z},$$ and in particular this holds for $$k = 1$$. Therefore $$\underbrace{1_R+\cdots+1_R}_{n} = n \cdot 1_R = 0.$$

2. Let $$r\in R$$, then $$n \cdot r = \underbrace{r + \cdots + r}_{n}= r \cdot \underbrace{(1_R + \cdots + 1_R)}_{n}=r \cdot0_R=0_R.$$ Hence $$n \cdot r \cdot 1_R = 0$$.

But I don't know if my proof is correct. What do you think? (I'd like to make sure my task is correct before I handle it)

• Is the ring R commutative? If so then $n⋅r⋅1_{R}=0$ follows when you show that $n⋅1_{R}=0$ – user758469 Apr 1 at 15:47

There are some small errors in your proof, and the proof can be made much more direct:

1. In the expression $$\phi(k)=\phi(n\cdot z)=\underbrace{1_R+\cdots+1_R}_{n\cdot k}\; \forall k \in \mathbb{Z},$$ you should have $$n\cdot z$$ below the brace, not $$n\cdot k$$. Also the last part ($$\forall k\in\Bbb{Z}$$) makes no sense here; you have already specified $$k$$ to be an element of $$\ker\phi$$.

2. There is no need to consider a general element in the kernel. Instead you can argue as follows:

Let $$n$$ be the characteristic of $$R$$. Then $$\phi(n)=0_R$$ and hence $$n\cdot 1_R=n\cdot\phi(1_{\Bbb{Z}})=\phi(n\cdot1_{\Bbb{Z}})=\phi(n)=0_R.$$

3. For the converse you have by definition of the identity element of $$R$$ that for all $$r\in R$$:

$$n\cdot r=n\cdot(1_R\cdot r)=(n\cdot 1_R)\cdot r=0_R\cdot r = 0_R.$$

In general I would prefer to avoid describing a sum with '$$\cdots$$' and and underbrace, as in $$\underbrace{1_R+\cdots+1_R}_{n\cdot k},$$ when you can just as well describe this sum as $$n\cdot k\cdot 1_R.$$ In fact, your proof can be condensed to the following:

If $$R$$ is a commutative ring with unity of characteristic $$n$$, then for all $$r\in R$$ $$n\cdot r=(n\cdot1_R)\cdot r=(n\cdot\phi(1_{\Bbb{Z}}))\cdot r=\phi(n\cdot 1_{\Bbb{Z}})\cdot r=\phi(n)\cdot r=0_R\cdot r=0.$$

• If we consider $\phi \colon \mathbb{Z} \longrightarrow \mathbb{Z}/2\mathbb{Z}$, $\phi(z)=_2$ $z$ is not necessarily $1$ because $\phi(3_\Bbb{Z})=_2$. So why can we do this step: $$n\cdot 1_R=n\cdot\phi(1_{\Bbb{Z}})=\phi(n\cdot1_{\Bbb{Z}})=\phi(n)=0_R.$$ Why $n\cdot 1_R=n\cdot\phi(1_{\Bbb{Z}})$? – mug_donut Apr 2 at 10:14
• Because $\phi(1_{\Bbb{Z}})=1_R$. – Servaes Apr 2 at 10:24
• Yeah, I know by definition $\phi(1_{\Bbb{Z}})=1_R$, then let $\phi(z)=r$, If $z=1$ for sure implies $r=1_R$. But I'm not sure about the fact $r=1_R$ implies $z=1$. I mean, let $\phi \colon \mathbb{Z} \longrightarrow \mathbb{Z}/2\mathbb{Z}$ then: $$n\cdot 1_R=n\cdot\phi(3_{\Bbb{Z}})=\phi(n\cdot3_{\Bbb{Z}})=0_R.$$ So we get to the resoult anyways because $ker(\phi)=(n)$ – mug_donut Apr 2 at 10:54
• It is not true that $\phi(z)=1_R$ implies $z=1$, but this is not at all what the proof uses. It just uses that $\phi(1_{\Bbb{Z}})=1_R$. – Servaes Apr 2 at 11:47
• Ok, I understand. Thanks! – mug_donut Apr 2 at 11:50