Some Properties about the Characteristic of a Ring

abstract-algebraidealsnumber theoryring-homomorphismring-theory

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)

Best Answer

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 $r\in R$ $$n\cdot r=(n\cdot1_R)\cdot r=(n\cdot\phi(1_{\Bbb{Z}}))\cdot r=\phi(n\cdot 1_{\Bbb{Z}})\cdot r=\phi(n)\cdot r=0_R\cdot r=0.$$

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