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Suppose by absurd that the characteristic of an integral domain is an integer $n$ not prime, say $n=n_1 \cdot n_2$.

Now we have $$na=(n_1 \cdot n_2)a=0 \implies (n_1 \cdot a) \cdot (n_2 \cdot a)=0 \implies n_1 \cdot a=0 \space \lor n_2 \cdot a=0$$

In my book it is considered a contradiction (for the minimality of $n$) but a priori it could be possible that $n_1$ and $n_2$ don't satisfy characteristic property.

Adopted definition of characteristic of an integral domain (with $charD \neq 0$):

$$charD := min \{n \in \mathbb{N}| \space n \cdot a=0 \space \forall a \in D\}$$

In my book an integral domain is a commutative ring without non-zero zero divisors. I do not assume identity.

rschwieb
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ictibones
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  • I don't really get what is your question, sorry. Could you enlighten me? – RGS Jan 24 '18 at 18:23
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    @RGS In the definition you have "for all a". – Arnaud Mortier Jan 24 '18 at 18:27
  • Sure: I don't see the contradiction... neither $n_1$ nor $n_2$ have the characteristic property (if $m$ is the characteristic then $m \cdot a=0$ $\forall$ a in D) – ictibones Jan 24 '18 at 18:27
  • @LeonardoVannini Are you saying your definition of integral domain does not include multiplicative identity? Because that is a rare assumption for integral domains. – rschwieb Jan 24 '18 at 18:36
  • @rschwieb Yes, for my book an integral domain is a commutative ring without non-zero zero divisors – ictibones Jan 24 '18 at 18:40
  • @LeonardoVannini What you wrote does no answer my question clearly. Do you mean yes, I do not assume identity? – rschwieb Jan 24 '18 at 18:42
  • Sorry, I do not assume identity – ictibones Jan 24 '18 at 18:43
  • @LeonardoVannini Next time, directly edit in your assumptions into the question from the start. Putting them in the comments later is helpful... but you really ought to refine the question with them too. – rschwieb Jan 24 '18 at 19:04

2 Answers2

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You still have proved that any vector will be killed by either one of $n_1$ and $n_2$. What you want is to be able to take the same $n_i$ for all vectors.

So assume that there is a vector $a_1$ annihilated only by $n_1$ and a vector $a_2$ annihilated only by $n_2$.

Who kills $a_1+a_2$ then? If it is $n_1$ then $n_1(a_1+a_2)=0$ together with $n_1a_1=0$ proves that $n_1a_2=0$, which is a contradiction.

Arnaud Mortier
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You've shown that for every element $r\in R$, $n_1r=0$ or $n_2r=0$.

You can check that $I_1=\{x\mid n_1x=0\}$ and $I_2=\{x\mid n_2x=0\}$ are additive subgroups of $R$, and moreover by your discovery above, $R=I_1\cup I_2$.

But a group is never a union of two proper subgroups, so one of them is in fact all of $R$. That contradicts the minimality of $n$.

In my book it is considered a contradiction (for the minimality of $n$) but a priori it could be possible that $n_1$ and $n_2$ don't satisfy characteristic property.

If your book considers it a contradiction without resolving this detail, then I think it is highly likely your book assumes an integral domain has an identity. In that case, you can conclude the proof much more simply. Suppose $n$ is the additive order of $1$. Then obviously $n\cdot x=(n\cdot 1)\cdot x=0$ for every $x\in R$. If $n$ is were composite, $n_1n_2\cdot 1=(n_1\cdot 1)(n_2\cdot 1)=0$, at which point we would have to conclude one of the factors is zero... but that would contradict the minimality of $n$ as the additive order of $1$.

rschwieb
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