# How can a finite integral domain have a non-zero characteristic?

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?

• 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. – Tobias Kildetoft May 23 '13 at 17:44
• If $n*1_R=m*1_R$ then $n=m$. – Angela Pretorius May 23 '13 at 17:46
• @AngelaRichardson- sorry I stand corrected. You are right. Quite obvious, in fact. I was making the erroneous assumption that $\underbrace{1+1+\dots}_{\text{n times}}=n$. – fierydemon May 24 '13 at 4:48

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 $na=0$.

• Oh. So the basic argument is $\underbrace{1+1+1\dots}_{n times}\neq n$. The notation is very confusing then. – fierydemon May 23 '13 at 18:02
• 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 $p$". – Zev Chonoles May 23 '13 at 18:08
• 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? – fierydemon May 23 '13 at 18:38
• 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 $(\underbrace{1+1+\dots}_{p times})=0$. – fierydemon May 23 '13 at 18:44
• @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. – vonbrand May 23 '13 at 19:35

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 $R$.

• You contradict yourself. Your own example $\mathbb{Z}/3 \mathbb{Z}$ does not contain a "copy of the integers". – vonbrand May 23 '13 at 19:36
• @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? – egreg May 23 '13 at 19:46
• 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? – fierydemon May 24 '13 at 5:46
• Please ignore the comment above I got it :). Thanks a lot for your help – fierydemon May 24 '13 at 5:52