3
$\begingroup$

I know that by virtue of being a finite integral domain, $n\cdot 1_{R}=m\cdot 1_{R}$. Hence, $(n-m)\cdot 1_{R}=0$. The smallest positive element in the set of all possible such $(n-m)$'s is the characteristic.

But then again, in an integral domain, if $a\cdot b=0$, then either $a=0$ or $b=0$. Here, neither $n-m=0$ (by definiton, we have a non-zero characteristic), nor $1_{R}=0$ (we're assuming this is not a zero ring). Isn't this a contradiction?

$\endgroup$
3
  • 2
    $\begingroup$ Non-zero characteristic means precisely that there are distinct $m$ and $n$ with $m-n = 0$ in the ring. It is characteristic $0$ where this cannot happen. $\endgroup$ – Tobias Kildetoft May 23 '13 at 17:44
  • 1
    $\begingroup$ If $n*1_R=m*1_R$ then $n=m$. $\endgroup$ – Angela Pretorius May 23 '13 at 17:46
  • $\begingroup$ @AngelaRichardson- sorry I stand corrected. You are right. Quite obvious, in fact. I was making the erroneous assumption that $\underbrace{1+1+\dots}_{\text{n times}}=n$. $\endgroup$ – fierydemon May 24 '13 at 4:48
3
$\begingroup$

Consider what you mean when you write $na=0$ where $n$ is an integer: $$\underbrace{a+\cdots+a}_{n\text{ times}}=0$$ It very well can be the case that this occurs for a non-zero $n$. This is not a contradiction because the expression $na$ does not mean a product of elements of the ring $R$; it is $\mathbb{Z}$ acting on the ring $R$.

Alternatively, you can think of "$n$" as being a shorthand for $$\underbrace{1_R+\cdots+1_R}_{n\text{ times}}$$ If you prefer this approach, you very well can have that $n=0$ in the ring $R$, and therefore it is not a contradiction to have $na=0$.

$\endgroup$
5
  • $\begingroup$ Oh. So the basic argument is $\underbrace{1+1+1\dots}_{n times}\neq n$. The notation is very confusing then. $\endgroup$ – fierydemon May 23 '13 at 18:02
  • 1
    $\begingroup$ We usually don't bother writing $1_R$ instead of $1$; we just say "let $1$ be the multiplicative identity in $R$". Similarly, we usually don't both writing $n\cdot 1_R$ instead of $n$. It's an abuse of notation like any other; I find it extremely clear. I think that most mathematicians would be perfectly happy with the statement "$p=0$ in a field of characteristic $p$". $\endgroup$ – Zev Chonoles May 23 '13 at 18:08
  • $\begingroup$ But hang on. My text says if $\mathbb{F}$ is of characteristic $p$, then the derivative of $x^{p}\in \mathbb{F[x]}$, which is $px^{p-1}$, equals $0$. Here, we're talking about the element $p$, and not $\underbrace{1+1+\dots}_{p\\times}$. How is this possible then? $\endgroup$ – fierydemon May 23 '13 at 18:38
  • $\begingroup$ The only way I can imagine this would be true is if $x^{p}$ is written as $\underbrace{x*x*\dots}_{p times}$. Now, to differentiate, we use product rule. This would give $\underbrace{x^{p-1}+x^{p-1}+\dots}_{p times}=(\underbrace{1+1+\dots}_{p times})x^{p-1}$, and $(\underbrace{1+1+\dots}_{p times})=0$. $\endgroup$ – fierydemon May 23 '13 at 18:44
  • 1
    $\begingroup$ @AyushKhaitan, that is a formal drivative, defined for formal polynomials as the sum (in the ring) of $n a x^{n - 1}$ for the term $a x^n$. The $n a$ here is $a + a + \ldots + a$ ($n$ times). It just so happens it shares many useful properties with "real" derivatives. $\endgroup$ – vonbrand May 23 '13 at 19:35
3
$\begingroup$

How can a finite integral domain have zero characteristic? It would contain a copy of the integers, which is a contradiction. So a finite integral domain must have non zero characteristic.

When you write $na$, for $n\in\mathbb{Z}$ and $a\in R$ you're not doing a product in $R$, but rather $\underbrace{a+a+\dots+a}_n$ (when $n>0$) or $-((-n)a)$ when $n<0$.

So there's no problem in having $3a=0$ in a domain. This just means that $a+a+a=0$, which happens in $\mathbb{Z}/3\mathbb{Z}$, for instance, which is a field.

I usually consider the unique ring homomorphism $\varepsilon\colon\mathbb{Z}\to R$, defined by $\varepsilon(n)=n1_R$.

In case the ring $R$ is finite, this homomorphism has a non zero kernel, say $k\mathbb{Z}$, so easily $k1_R=0$ and $k$ is the characteristic of $R$.

$\endgroup$
4
  • $\begingroup$ You contradict yourself. Your own example $\mathbb{Z}/3 \mathbb{Z}$ does not contain a "copy of the integers". $\endgroup$ – vonbrand May 23 '13 at 19:36
  • 2
    $\begingroup$ @vonbrand If a ring has characteristic zero it is infinite. Where did I say that $\mathbb{Z}/3\mathbb{Z}$ contains a copy of the integers? $\endgroup$ – egreg May 23 '13 at 19:46
  • $\begingroup$ What I'm not clear on is even if $\underbrace{1+1+\dots}_{\text{n times}}\neq n$, as the field is closed under addition, $\underbrace{1+1+\dots}_{\text{n times}}$ has to be equal to some element in $\mathbb{F}$! Let that element be $a\in \mathbb{F}$. We are still saying that $a=0$. $a=a*1$. Hence, $a*1=0$ when $a,1\neq 0$! Does this not contradict the definition of an integral domain? $\endgroup$ – fierydemon May 24 '13 at 5:46
  • $\begingroup$ Please ignore the comment above I got it :). Thanks a lot for your help $\endgroup$ – fierydemon May 24 '13 at 5:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.