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I am reading Herstein's "Topics in Algebra" and I've encountered with the following problem:

If $D$ is an integral domain and $D$ is of finite characteristic, prove that the characteristic of $D$ is a prime number.

The definition of finite characteristic given by Herstein is: "An integral domain $D$ is said to be of finite characteristic if there exists a positive integer $m$ such that $ma=0$ for all $a \in D$." His definition of characteristic is "the smallest positive integer $p$ such that $pa=0$ for all $a \in D$."

By the way, he doesn't assume that $D$ has a unity. I got stuck at some point of the proof. I'll write what I've done.

First of all, I would like to show the existence of a smallest positive integer that kills all the elements. If I define $S=\{n \in \mathbb N: na=0 \space {\rm for \ all \ } a \in D\}$, then $m \in S$, and since $S$ is nonempty, it must have a minimum element. If I call this element $p$, I want to show that $p$ is prime. Suppose $p$ is not prime, so $p$ can be written as $p=st$ with $s,t>1$. So for any $a \in D$, we have $$pa=(sa)(ta)=0.$$ Since $D$ is an integral domain, then $sa=0$ or $ta=0$ for all $a \in D$. Here comes my silly concern: I have that for some elements in $D$, $t$ kills them and $t<p$, and the same goes for $s$. But I am not so sure if this contradicts any of the statements. It doesn't necessarily happen that $t$ kills ALL of the elements, and similarly for $s$, so maybe $p$ is still the smallest positive integer that kills all the elements. Any explanations would be appreciated.

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    $\begingroup$ It is not necessarily true that $pa=(sa)(ta)$. Note that $(sa)(ta)=sta^2$. $\endgroup$ – Matt Samuel May 27 '15 at 0:05
  • $\begingroup$ Thanks for the correction, the correct equality would be $pa=s(ta)$. $\endgroup$ – user156441 May 27 '15 at 0:21
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First prove a lemma that $(sa)b = a(sb)$ by induction on natural numbers $s$.

Now it is easy to see that if $s$ "kills" one nonzero element of $D$, it kills every nonzero element of $D$. For if $sa = 0$, then $a(sb) = 0$, and if $a,b\neq 0$, then this implies $sb = 0$ by $D$ being a domain.

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  • $\begingroup$ Doesn't $(sa)b=a(sb)$ follow from the commutativity of $D$ and the distributive law of $D$? I mean, $(sa)b=(a+...+a)b=ab+...+ab=a(b+...+b)=a(sb)$. Btw, thanks for the answer! $\endgroup$ – user156441 May 27 '15 at 0:35
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    $\begingroup$ Yes, it follows from the distributivity of multiplication over addition in $D$, but since we have repeated addition (aka "multiplication" by $s$), there is an element of induction in proving this. $\endgroup$ – hardmath May 27 '15 at 0:46
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Hint: consider the quotient field $F$ of $D$; then $F$ has the same characteristic as $D$ and it has an identity.

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