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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)

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  • $\begingroup$ Is the ring R commutative? If so then $n⋅r⋅1_{R}=0$ follows when you show that $n⋅1_{R}=0$ $\endgroup$ – user758469 Apr 1 at 15:47
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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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  • $\begingroup$ If we consider $\phi \colon \mathbb{Z} \longrightarrow \mathbb{Z}/2\mathbb{Z}$, $\phi(z)=[1]_2$ $z$ is not necessarily $1$ because $\phi(3_\Bbb{Z})=[1]_2$. 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 $n\cdot 1_R=n\cdot\phi(1_{\Bbb{Z}})$? $\endgroup$ – mug_donut Apr 2 at 10:14
  • $\begingroup$ Because $\phi(1_{\Bbb{Z}})=1_R$. $\endgroup$ – Servaes Apr 2 at 10:24
  • $\begingroup$ 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 $ker(\phi)=(n)$ $\endgroup$ – mug_donut Apr 2 at 10:54
  • $\begingroup$ 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 $\phi(1_{\Bbb{Z}})=1_R$. $\endgroup$ – Servaes Apr 2 at 11:47
  • $\begingroup$ Ok, I understand. Thanks! $\endgroup$ – mug_donut Apr 2 at 11:50

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