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I managed to think of a proof of fundamental theorem of arithmetic using roots of unity however not too sure about it. Finding one factorization is easy, we just focus on uniqueness or on Euclid's lemma.

Euclid's lemma. Let $p$ be a prime and $a,b \in \mathbb{Z}$. If $p \mid ab$ then $p \mid a$ or $p \mid b$.

Proof: Assume the contrary. We might also assume $a<p,b<p$.

Let $\omega$ be a primitive $p^{th}$ root of unity. [proving existence doesn't use fundamental theorem of arithmetic hopefully I hope]

as $p \mid ab$ we know that $\omega^{ab}=1$

then $(\omega^a)^{b}=1$

But then $p|b$ . [as we know $\omega^a$ is also a primtive root ].

However as like fundamental theorom of arethemtic is really fundamental i am not tooo sure if i am creating a circular reasoning somewhere .

I thought for a second I am assuming existence of primtive root.

However when you say let $t$ be the smallest positive integer such that $(\omega)^t=1$ then $t \mid p$ then $t=1,p$ by that you easily come to the result.

And for why $\omega^a$ is also a primitive root say $t$ is the smallest integer such that $(\omega^a)^{t}=1$. Assume it's not primitive then $t<p$ and $t \mid p$ so $t=1$.

So $\omega^a=1$, however that forces $p(x)=x^{p}-1$ to have a double root however checking its derivative we get an easy contradiction.

Anne Bauval
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Lucid
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  • "we know $\omega^a$ is also a primtive root": I don't. "Assume its not primtive then $t<p$ and $t|p$": why $t\mid p$? "however that forces $p(x)=x^{p}-1$ has a double root": how so? – Anne Bauval May 02 '24 at 11:49
  • Assume the contrary then let $r$ be the remainder when $p$ is divided by $t$ then $\omega^r=1$ for r will be something of form $p-tk$ and $0<r<t$ @AnneBauval – Lucid May 02 '24 at 11:50
  • And as for the double root if $\omega^a=1$ and p did not divide $a$ we know 1 is a double root – Lucid May 02 '24 at 11:51
  • Thank you, the first comment helped me. I still don't understand the second one. – Anne Bauval May 02 '24 at 11:58
  • Does this answer your question? If a prime $p\mid ab$, then $p\mid a$ or $p\mid b$ (Euclid's Lemma) Your proof is essentially correct. Bernard's answer in this link is not an alternative one. It is yours (up to exchanging $a$ and $b$), just simplified, thus made shorter than most of those on this site. – Anne Bauval May 02 '24 at 15:05
  • @Anne That "short" direct proof of Euclid's Lemma is the most common proof given in textbooks, and almost surely the most common proof on this site. The slightly longer proofs in some answers in the linked dupe (using ideas from the theory of ideals or cyclic groups) are pedagogically far superior if one goes on to study advanced number theory or abstract algebra, so one should not denigrate them. – Bill Dubuque May 02 '24 at 17:51
  • I already knew you are conviced of this "far superiority", and I agree to disagree, as previously shortly explained. – Anne Bauval May 02 '24 at 18:07
  • Wait can u please explain a bit on what u said @BillDubuque i didnt quite understand – Lucid May 02 '24 at 18:14
  • @Lucid What is your knowledge level? Which of these topics do your know: congruences / modular arithmetic, groups, rings. – Bill Dubuque May 02 '24 at 18:29
  • @Anne Please do tell what you think is "scary" about the first few paragraphs here, That can be understood by any high school student, and once they learn that fundamental Lemma they will be able to reapply it in many other common proofs (e.g. the gcd Bezout identity, the Order Theorem, etc). It will further serve as good intuition once they advance to the study of (cyclic) groups & (principal) ideals. – Bill Dubuque May 02 '24 at 18:38
  • @BillDubuque i guess ik all of them , but group rings not at too advanced level – Lucid May 02 '24 at 18:50

1 Answers1

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Your proof is actually correct (if rephrased a little bit) and is very nice! I cleaned it up a little bit.

Let $p$ be a prime and let $a,b$ be natural numbers s.t. $p\vert ab$. Then $ab=kp$ for some natural number $k$. Let $C_p$ be the cyclic group of order $p$. We have that any $\omega\in C_p-\{e\}$ has order $p$ by Lagrange's theorem.

Now on the one hand we have for any $\omega\in C_p-\{e\}$ that $$\omega^{ab}=\omega^{kp}=(\omega^p)^k=e^k=e.$$

On the other hand if we assume that $p$ doesn't divide $a$ and $b$ we have $$w^{ab}=(w^a)^b=\gamma^b=\phi\neq e$$ for some $\gamma,\phi\in C_p-\{e\}$ which would lead to a contradiction. To show the last chain of equalities we use division with remainder.

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    This proof relies in an essential way on knowing that all nontrivial elements in a cyclic group of prime order are generators of the group, and this seems like a disguised way of saying that if $a \not\equiv 0 \bmod p$ and $b \not\equiv 0 \bmod p$, then $ab \not\equiv 0 \bmod p$, which is what the OP wants to show in the first place. – KCd May 02 '24 at 12:49
  • @KCd You can definitely proof the wished for statement more efficiently using Lagrange's theorem. But the (most direct) proof of Lagrange's theorem does not use anything which makes the argument circular. The point is that the proof is not circular and works (which I believe answers the question asked). –  May 02 '24 at 12:52
  • @KCd If you so want this is just one way (not the shortest) to lift the disguised form (which you get for free by Lagrange's theorem) in question to the wished for form. –  May 02 '24 at 13:07