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Explanation: Characteristic Zero Instead of Characteristic Infinite

(related to Definition: Characteristic of a Ring)

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).

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References

Bibliography

  1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013
  2. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013