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For every natural number n, i have to give a finite and a infinite ring with characteristic n.

For the finite case is it simple: for all n $\in\mathbb{N}$, the ring $\mathbb{Z}_n$ is a finite ring with characteristic n.
But in the infinite case i can't think of a ring...

The only infinite ring with a finite characteristic that is not 0, is the ring of all subsets of a infinite set X. With the operations of symmetric difference and intersection. Here is the characteristic equal to 2.

Robbe Motmans
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  • The characteristic is always $0$ or a prime number. So there's no way you will have something of characteristic $4$... – Olórin Mar 16 '15 at 18:49
  • @user10000100_u: Characteristic of a ring $A$ is the generator of the kernel of the natural map $\mathbb Z \to A, 1 \mapsto 1.$ – Krish Mar 16 '15 at 18:51
  • Let me put it differently : the notion of characteristic is not pertinent for all rings. Because honestly, we really don't want to say that $\mathbf{Q}$, as well as $\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/3\mathbf{Z}\times\ldots$, are of characteristic $0$, do we ?... You see what I mean ?... – Olórin Mar 16 '15 at 18:57
  • @user10000100_u: Yes! I think I understood what you mean. But that's how it's defined for rings. – Krish Mar 16 '15 at 19:14

2 Answers2

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Consider the ring $\Bbb Z_n[x]$ of polynomials in one variable $x$ with coefficients in $\Bbb Z_n$, where $n$ is prime. It is an infinite ring since $x^m \in \Bbb Z_n[x]$ for all positive integers $m$, and $x^{m_1} \ne x^{m_2}$ for $m_1 \ne m_2$. But the charactetistic of $\Bbb Z_n[z]$ is clearly $n$.

Community
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Robert Lewis
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$A_n:= \prod \mathbb Z/n\mathbb Z$ (infinite product, fixed $n$). This is an infinite ring with characteristic $n.$

Krish
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  • oke thnx! is the infinite product $\prod \mathbb{Z}_n$ also an infinte ring with characteristic n? – Robbe Motmans Mar 16 '15 at 18:59
  • @RobbeMotmans: Do you mean the product is over $n$? In this case it will be of characteristic zero, since there is no $m \in \mathbb N$ such that $m\cdot 1=0.$ – Krish Mar 16 '15 at 19:11
  • i meant an infinite cartesian product, so that you have elements of the form (a,b,c,d,...) with a,b,c,d $\in\mathbb{Z}_n$. So you will have a set with an infinite number of elements. Is the characteristic still n in this case? – Robbe Motmans Mar 16 '15 at 19:15
  • Yes! That's the example I gave. (By $\mathbb Z/n\mathbb Z$ I mean "integer modulo $n$" which is $\mathbb Z_n$ in your notations) – Krish Mar 16 '15 at 19:16
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    In some books (e.g. I. N. Herstein) the characteristic is defined only for Integral Domains, so for that $A_n$ as you defined fails to be an example (even if $n$ is prime) – Peeyush Kushwaha Apr 17 '18 at 08:30