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Proposition: Characteristic of an integral domain must be either $0$ or prime number.

I'm confused by this proposition.

I think the characteristic of an integral domain should be always $0$. Suppose it has characteristic $n$. Then $n * a = 0$ for all a of the integral domain.

since n is not $0$ and, if $c * d = 0$ in integral domain, it means $c=0$ or $d=0$, a should be $0$. Hence $n * a$ is not $0$ when $a$ is nonzero. Therefore, characteristic should be always $0$.

What's wrong with my thought?

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If the characteristic is $n$ then this means that $n$ is 0 (in the ring) by definition. –  Tobias Kildetoft Jul 31 '13 at 8:29
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$na$ is not a product of two elements of the integral domain; it's the sum of $n$ copies of $a$. –  Gerry Myerson Jul 31 '13 at 8:35
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4 Answers 4

up vote 3 down vote accepted

In a ring $R$ we define $p*x= {(1_R+1_R+1_R+\cdots+1_R)}x= \sum\limits_{i=1}^p x $, so even if $p= 1_R+1_R+1_R+\cdots+1_R=0 $ as a ring element, it is not necessarily $0$ in $\Bbb Z$.

$ \mathbb F_p$ (the integers modulo $p$ a prime, see here) is an integral domain with characteristic $p$. If $R$ was a ring with characteristic $mn$ then $m \ne 0$ and $ n \ne 0$ but $mn$=0, so $R$ could not be an integral domain.

Note in $\Bbb F_p$ the equivalence class of an integer $n \equiv 0 \mod p$ if and only if $n=pm$ for some $m \in \Bbb Z$, now if $xy \equiv 0 \mod p $ then $p|xy$. By primality of $p$ then $p|x$ or $ p|y$. So either $x$ or $y \equiv 0 \mod p$.

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Hint $\ $ Whenever you have problems understanding such an abstract statement you should look at concrete instances. For example $\,\rm \Bbb Z\ mod\ 3\ $ has characteristic $3$ because $ 3n := n + n + n \equiv 0\ $ for $ n\equiv 0,1,2.\:$ See how you argument breaks down in this simple concrete case, then generalize.

Note that $\,m\cdot a\,$ does not denote an element obtained by applying the ring multiplication to two elements of the ring. Rather, the $m$'th multiple $\,m\cdot a\,$ is the additive analog of the $m$'th power $a^m.$ In the first case we add $m$ copies of $a$ to obtain $\,m\cdot a\,$ and the second we multiply them to get $\,a^m.\,$ They are both well-defined operations in any ring.

One may rigorously define such operations by recursion, viz.

$$\begin{eqnarray} 0\cdot a \,&=&\, 0 \\ (1+n)\cdot a\,&=&\, a + n\cdot a\end{eqnarray}$$

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This is really interesting advice, and a great way of looking at it. Thanks a ton for your help! –  Roger Tynes Dec 15 '13 at 1:30
    
You forgot to \rm your edit. ;) –  tomasz Dec 15 '13 at 2:26
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@tomasz Thanks, Rome has fallen! –  Bill Dubuque Dec 15 '13 at 2:35
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The integer is not an element of $D$. You have the canonical $\mathbb{Z}$-action ($0_\mathbb{Z}a = 0_D$, and $(k+1)a = ka + a$) on the abelian group $(D,+)$, and for that $\mathbb{Z}$-action, you have $ma = 0$ for all $a \in D$.

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This is because any ring is canonically a ${\bf Z}$-algebra (and vice versa), much like any abelian group is canonically a ${\bf Z}$-module (and vice versa). –  tomasz Dec 15 '13 at 2:28
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You are correct - $m$ is not an element of the integral domain D. $m$ is a natural number.

But then of course a fair question to consider is how $m \cdot a$ is defined. Somewhere in your text it should define $m \cdot a$ in which $m$ is a natural number and $a$ is an element of an integral domain D as $a$ added to itself $m$ times.

Consider the integral domain $\mathbb{Z}_5$ with addition and multiplication modulo $5$. $\mathbb{Z}_5$ is finite characteristic because each element may be added to itself a number of times (modulo $5$) to reach $0$.

  • $0 \equiv 0$ (mod $5$)
  • $1+1+1+1+1 = 5 \equiv 0$ (mod $5$)
  • $2+2+2+2+2 = 10 \equiv 0$ (mod $5$)
  • $3+3+3+3+3 = 15 \equiv 0$ (mod $5$)
  • $4+4+4+4+4 = 20 \equiv 0$ (mod $5$)
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So by definition M is not an element. It's not optional, it's that M is just not in D correct? –  Roger Tynes Dec 15 '13 at 0:44
    
Yes $m\cdot a $ only means $\underbrace{a+a+a+\ldots+a}_{m \text{ times}}$. As everybody said $m\in \mathbb{N}$ and $\mathbf{m\not\in D}$. –  Sergio Parreiras Dec 15 '13 at 0:54
    
I see. Thanks so much for that clarification, I really apreciate it. –  Roger Tynes Dec 15 '13 at 1:26
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