You might ask, why the characteristic of a ring is defined to be $\operatorname{char}(R)=0$ and not infinite $\operatorname{char}(R)=\infty$ if a minimal number $n$ with the property $$\underbrace{1+\cdots+1}_{n \text{ times}} = 0$$ does not exist. Under this alternative definition, the characteristic of a ring would be a kind of the order of the additive cyclic group $(R, + ).$ generated by $1\in R.$

The key to an explanation is the classification of cyclic groups. Here, we recap why:

By hypothesis, $(R, + )$ is a cyclic group with a generator $1\in R$ with $R=\langle 1\rangle.$ Consider a group homomorphism $f:(\mathbb Z, + )\to (R, + )$ defined by $$f(a):=\underbrace{a+\cdots+a}_{n \text{ times}}$$ for all $a\in (R, + )$ for all $n\in\mathbb Z.$ Note that the kernel $\ker(f)=\{a\in R\mid f(a)=1\}$ is a subgroup of $(Z,+)$. Now, if $R$ has an infinite order $|R|=\infty,$ then $\ker(f)=\{0\}.$ It follows from the isomorphism theorem for groups that $(R, + )$ is isomorphic to $\mathbb Z/\{0\}=\mathbb Z.$ Therefore, $\operatorname{char}(R)=0$, since $\ker(f)=\{0\}$ (the kernel consists of all multiples of $0$ and has therefore only the element $0$).

To even better see the parallel, let consider the finite case. If $R$ has a finite order $|R|=n$ with $n > 0.$ Note that all additive subgroups of integers have the form $(\mathbb Z_n, + )$ for a natural number $n\in\mathbb N.$ In particular, $\ker(f)$ has this form. It follows again from the isomorphism theorem for groups that $(R, + )$ is isomorphic to $(\mathbb Z/\ker(f), + ) = (\mathbb Z_n, + ).$ Therefore, $\operatorname{char}(R)=n$, since $\ker(f)=\{kn\mid k\in Z\},$ (the kernel consists of all multiples of $n$).

| | | | created: 2019-08-10 08:06:32 | modified: 2019-08-11 09:42:05 | by: *bookofproofs* | references: [677], [696]

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[696] **Kramer Jürg, von Pippich, Anna-Maria**: “Von den natürlichen Zahlen zu den Quaternionen”, Springer-Spektrum, 2013

[677] **Modler, Florian; Kreh, Martin**: “Tutorium Algebra”, Springer Spektrum, 2013