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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.

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  • $\begingroup$ I don't really get what is your question, sorry. Could you enlighten me? $\endgroup$
    – RGS
    Jan 24, 2018 at 18:23
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    $\begingroup$ @RGS In the definition you have "for all a". $\endgroup$ Jan 24, 2018 at 18:27
  • $\begingroup$ 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) $\endgroup$
    – ictibones
    Jan 24, 2018 at 18:27
  • $\begingroup$ @LeonardoVannini Are you saying your definition of integral domain does not include multiplicative identity? Because that is a rare assumption for integral domains. $\endgroup$
    – rschwieb
    Jan 24, 2018 at 18:36
  • $\begingroup$ @rschwieb Yes, for my book an integral domain is a commutative ring without non-zero zero divisors $\endgroup$
    – ictibones
    Jan 24, 2018 at 18:40

2 Answers 2

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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 $1$.

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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.

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  • $\begingroup$ $n_1 n_2=n$, which I have assumed to be the characteristic of D... $\endgroup$
    – ictibones
    Jan 24, 2018 at 18:38
  • $\begingroup$ @LeonardoVannini I have edited to make it clearer. $\endgroup$ Jan 24, 2018 at 18:54
  • $\begingroup$ @rschwieb Yes it does, because in that case there is nothing to prove since the OP has already proved that for any vector $a$, either $n_1a=0$ or $n_2a=0$. $\endgroup$ Jan 24, 2018 at 19:00
  • $\begingroup$ @ArnaudMortier Or rather that case falls out before your argument is necessary, I see. Yes, it works, with a little grease :) $\endgroup$
    – rschwieb
    Jan 24, 2018 at 19:01
  • $\begingroup$ Yes. In fact this is how you prove the property that you have mentioned. $\endgroup$ Jan 24, 2018 at 19:03

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