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Let $D$ and $E$ be integral domains. Show that if $\operatorname{char}(D)=p$ and $n \cdot 1_D = 0_D$ for some $n \in \Bbb N$, then $p \mid n$.

Since $\operatorname{char}(D)=p$ we have that $p \cdot 1_D =1_D + \dots + 1_D = 0$. Also $p$ is the smallest such number. We get that $$n \cdot 1_D=0_D=p\cdot 1_D$$ and because $p$ is the smallest such number we must have that either $n =p$ or $n$ is a multiple of $p$ and in any case $p$ divides $n$.

Can I draw the conclusion that either $n=p$ or $n=kp$ for some $k \in \Bbb Z$ from $n \cdot 1 _D =p \cdot 1_D$?

Robert Shore
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Walker
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    How is $E$ involved? – anomaly Aug 23 '22 at 18:33
  • Use the Euclidean algorithm to express $n$ as $ap+r$ with $0 \leq r \lt p$. Now what do you know? – Robert Shore Aug 23 '22 at 18:33
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    Hint: $, K := { n\in \Bbb Z\ :\ n\cdot 1_D = 0},$ is an ideal in the $\rm\color{#c00}{PID}$ $:!\Bbb Z.,$ $,{\rm char}(D) = p,$ means $,K = p\Bbb Z,$ so $,n\in K=p\Bbb Z\iff p\mid n,$ in $\Bbb Z.\ \ $ – Bill Dubuque Aug 23 '22 at 18:36
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    This was a question with two parts, but I figured the second part out which involved $E$. @anomaly – Walker Aug 23 '22 at 18:42
  • From here: Recall an R-algebra is a ring A containing a central image of the ring R. This image is $,\cong$ R/I so it is characterized by the kernel I. For example, if R = $\mathbb Z$ then an R-algebra is simply a ring A, and the kernel $\rm\ I = (n)\ $ characterizes the canonical image of $\mathbb Z$ in A, via $\rm 1\mapsto 1_A.,$ Therefore we say that A has characteristic n because n characterizes the canonical image of $:\mathbb Z:$ in A. – Bill Dubuque Aug 23 '22 at 18:45
  • @BillDubuque: Do you want to turn your first comment on the main post into an answer? I don't think there's anything more to say about the question than what you wrote there. – anomaly Aug 23 '22 at 19:48
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    @anomaly I didn't answer since it is likely a dupe, but I haven't had a chance yet to locate good targets – Bill Dubuque Aug 23 '22 at 19:51

2 Answers2

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By the Euclidean algorithm, $n = ap+r$ where $a, r \in \Bbb Z, 0 \leq r \lt p$. But then $0=n \cdot 1_D = (ap+r) \cdot 1_D = ap \cdot 1_D + r \cdot 1_D = a \cdot (p \cdot 1_D) + r \cdot 1_D = a \cdot 0 + r \cdot 1_D = r \cdot 1_D.$

By the definition of characteristic, since $r \lt p$, this forces $r=0$; in other words, $n=ap$ and $p \mid n$.

Robert Shore
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    There's no need to repeat the proof that (Euclidean) $\Bbb Z$ is a PID - rather simply invoke it and exploit the powerful (ideal) abstraction that was designed exactly for such purposes. – Bill Dubuque Aug 23 '22 at 18:40
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Hint: Consider $D$ as an additive group.

  1. $\operatorname{char}(D)=p$ iff the additive order of $1$ is $p$

  2. $n \cdot 1_D = 0_D$ implies $p\mid n$

Note: Let $(G, \cdot) $ be a group and $|a|=n$.Then $a^m=e$ implies $n\mid m$.

Sourav Ghosh
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