The characteristic of a ring with unity is defined to be the least positive integer $n$ such that $1$ plus itself $n$ times $=0$. How does this make sense? $1$ plus itself $n$ times $=n1=n=0$, but $n$ is defined to be nonzero.

One exercise that is bothering me is:

Let $A$ be a finite integral domain. Prove: Let $a$ be any nonzero element of $A$. If $na=0$, where $n\neq0$, then $n$ is a multiple of the characteristic of $A$.

This doesn't make sense. If $A$ is an integral domain, and $a$ is nonzero, and if $na=0$ where $n\neq0$, then this statement doesn't make sense. The characteristic is defined to be nonzero, and if $n$ is multiple of the characteristic then it is nonzero, and also $a$ is nonzero, but $na=0$. This contradicts being an integral domain.

Any help?

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    The source of your confusion is that $n$ is not in the ring, it is a description of how many times we add $1$ (which *is* in the ring) to itself. Consider $\mathbb{Z}/5\mathbb{Z}$; this is not only a ring, but a field. And yet $1+1+1+1+1=0$. – vadim123 Dec 05 '14 at 18:13
  • If $n$ is the characteristic of $A$, then $(1+1+...+1)$ $=$ $1$ $n$ times $=0$, and $1$ $n$ times $=n1=n$, so $n$ must be in the ring, since the ring is closed w.r.t. to multiplication. – user46372819 Dec 05 '14 at 18:18
  • @JohannFranklin, that is incorrect. The meaning is *exactly* what Vadim told you. Besides this, his simple example gives you a good motivation. – Timbuc Dec 05 '14 at 18:25
  • What is incorrect? How? – user46372819 Dec 05 '14 at 18:25
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    "closed wrt multiplication" means that if $a,b$ are both in the ring, then $ab$ is also in the ring. Multiplying $1$ by a potato does not mean that $1\cdot potato$ is in the ring. – vadim123 Dec 05 '14 at 18:27
  • @JohannFranklin That somehow $\;n\;$ must be in the ring...and yes, it is, but in the sense Vadim told you: $$n:=\overbrace{\cdot1_A+\ldots+1_A}^{N\;\text{times}}$$ – Timbuc Dec 05 '14 at 18:27
  • See also http://math.stackexchange.com/questions/98605/why-characteristic-zero-and-not-infinite-characteristic. – lhf Dec 05 '14 at 18:32
  • If $n$ is the characteristic of A, then $1$ $n$ times equals $0$. But $1$ $n$ times equals $n$. So, $1$ $n$ times equals $0$. So, $n=0$ and since $0$ is in the ring, $n$ is in the ring. Is this incorrect? – user46372819 Dec 05 '14 at 18:39
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    This is incorrect. The ring $\mathbb{Z}/5\mathbb{Z}$ contains exactly five elements: $\{0,1,2,3,4\}$. It does not contain $5$, or potato. Your statement "1 $n$ times equals $n$" is an error, because in this ring $1+4=0$. It may be useful in certain circumstances to *define* $5$ to be a synonym for $0$ -- that is what Timbuc's comment means -- however then your elements represent equivalence classes of numbers rather than numbers. – vadim123 Dec 05 '14 at 19:11
  • Gotcha. So, to answer the proof exercise: Since $A$ is an integral domain, if $na=0$ where $a,n \neq 0$, then $na=(n1)a=0$. Therefore, $n$ is multiple of the characteristic of $A$. Is this correct? – user46372819 Dec 05 '14 at 19:18

1 Answers1


The abstract answer to your question involves the (unique!) ring homomorphism $\mathbb Z\to R$, which exists if $R$ itself has a unit element $1_R$. This homomorphism sends the natural number $1$ to $1_R$, and a positive number $n\in\mathbb Z$ to the result of adding $1_R$ to itself $n$ times. Any homomorphism has a kernel, in this case it’ll be an ideal of the domain $\mathbb Z$, and so generated by a well-defined nonnegative integer $\chi$. This integer is the characteristic of $R$, and it may be any nonnegative integer. Looking through the construction and definition, you see that that characteristic is the smallest positive $n$ such that adding $1_r$ to itself $n$ times gives a result of zero; but if there is no such positive number, the construction gives $\chi=0$.

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  • If $n$ is the characteristic of $A$, then $1$ $n$ times equals $0$. But $1$ $n$ times equals $n$. So, $1$ $n$ times equals $0$. So, $n=0$ and since $0$ is in the ring, $n$ is in the ring. Is this incorrect? – user46372819 Dec 05 '14 at 18:48
  • No, not correct. $1_R$ added to itself $n$ times gives $0_R$. Adding $1_R$ to itself $n$ times does not give $n$ because $n$ is an integer (more exactly a natural number), and $n$ is not in the ring $R$ (unless of course $R$ contains $\mathbb Z$). And this really becomes interesting when $R$ *doesn’t* contain $\mathbb Z$. – Lubin Dec 06 '14 at 05:33