I have this exercise in my worksheet in the discrete mathematics course.I don't understand the part that deals with prime numbers in integer-divisibility.
"Show that for a prime number $p$, if a $p\mid a^n$ then $p\mid a$"
Can somebody show how ?
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Thomas Russell
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MathDisease
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2Usual question: what have you tried so far? Any idea? – Umberto Jan 13 '14 at 19:21
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2The answer below gives one way to go about it. An easier way, imo, would simply utilise the fundamental theorem of arithmetic and work contrapositively. If you can show that for prime p, $p \nmid a \implies p \nmid a^n$, you're done. – G. H. Faust Jan 13 '14 at 19:32
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See also: If $p$ is prime and $p \mid a^k$, then $p \mid a$, and hence $p^k \mid a^k$. and $p$ prime, $p\mid a^k \Rightarrow p^k\mid a^k$ – Martin Sleziak Dec 28 '19 at 15:03
2 Answers
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Hint
By Euclid's lemma we have: if $p$ is prime then $$p|ab\Rightarrow p|a\lor p|b$$
Now write $a^n=a\times a^{n-1}$ and use a descending induction.
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So we're working in $\mathbb{Z}$, though mainly $\mathbb{Z}^+$, right? And $p, a, n \in \mathbb{Z}^+$, with $p$ prime, $a > 1$, $n > 1$, right?
We accept without proof that $\mathbb{Z}$ is a unique factorization domain (though if you want proof, this can be found easily enough).
We're given the fact that $p|a^n$, but we don't want to just jump to the conclusion that $p|a$. Could it be possible that $p|a^2$ is true but $p|a$ is false? That would mean we can solve $a^2 = pq = rs$ in positive integers such that $p$ and $q$ are distinct from $r$ and $s$, $\gcd(p, q) = \gcd(p, r) = \gcd(p, s) = 1$ and $r \neq 1$, $s \neq 1$. But this contradicts the fact that $\mathbb{Z}$ is a unique factorization domain. Therefore we must have $a = pq$ so that $a^2 = p^2 q^2$ does not lead to a different distinct factorization. After this, it is a simple matter to extend this result to $n = 3$ and higher.
For the sake of learning these things with greater confidence, it is important to work out some concrete examples. Here's one: $2|10^2$ and $2|10$; similarly, $5|10^2$ and $5|10$.
Now I'd like to make a little foray into algebraic number theory. If we're working in a non-UFD, $p|a^n$ does not necessarily mean $p|a$, though we might want to call $p$ "irreducible" rather than "prime" (in $\mathbb{Z}$ all primes are irreducible, so we generally don't bother to make that distinction). For example, in $\mathbb{Z}[\sqrt{-6}]$ we have $2|(\sqrt{-6})^4$, but $2\nmid\sqrt{-6}$.
Robert Soupe
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