Proof Check for a Theorem on Cyclotomic Polynomials and Connection with Eisenstein's Criterion.

Question

The main help I need is with proof-check of a theorem on divisors of cyclotomic values.

IF: For $\Phi_n(a) \equiv 0 \pmod{p}$ where $p$ is any prime and $n \in \mathbb{N}$ and $a$ is any integer.

To prove: $n \vert p-1$ or $p \vert n$

Proof: If $\Phi_n(a)$ is divisble by $p$ $\implies p \vert a^n-1$ as $\Phi_n(a) \vert a^n-1$. $$a^n \equiv 1 \pmod{p}$$

Case 1: If $n$ is the order of $a$ under moduli prime $p$ or $n = \text{ord}_p(a)$ then we say $n \vert p-1$.

Case 2: If $m$ is the order of $a$ under moduli prime $p$ or $m = \text{ord}_p(a)$ then $$a^m-1 = \prod_{d|m} \Phi_d(a) \equiv 0 \pmod{p}$$ as $m \vert n$ and $\Phi_n(a) \equiv 0 \pmod{p}$

$$a^n-1 \equiv 0 \pmod{p^2}$$

If $l$ is the order of $a$ under moduli prime $p^2$ or $l = \text{ord}_{p^2}(a)$. $$a^l-1 = \prod_{d|l} \Phi_l(a) \equiv 0 \pmod{p^2}$$ and as once again $l|n$ and $\Phi_n(a) \equiv 0 \pmod{p}$

We get $$a^n-1 \equiv 0 \pmod{p^3}$$

The same result is found if $l=m$ by repeating the process.

But if $v_p(a^n-1) = x $,

$$a^n-1 \equiv 0 \pmod{p^x}$$

Is max it can go,

Therefore, at $\pmod{p^x}$ the order should hit $n$ to stop this increment of $p$ power.

Later Addition for Clarification:

This is because: If the order is some other number $t(p^x|a^t-1)$, then $t|n$.So, $a^t-1|a^n-1$ and p also divides $\Phi_n(a)$, this means $a^n-1$ is divisble by $p^{x+1}$ which can not happen.So, $t = n$ to prevent this. $$n|p^{x-1}(p-1), x \geq 2$$

If we prove that $n$ Does not divide $p-1$, We can guarantee that $p|n$.

Now, for the sake of contradiction let us assume $n|p-1$.

By Lifting the Exponent Lemma, this can be easily shown.

$$v_p(a^{p-1}-1) = v_p(f)+v_p(a^m-1),p-1 = f\cdot m$$ $$v_p(f) = 0$$ P-adic valuation of $a^{p-1}-1$ and $a^m-1$ is same.

Let us say that $v_p(a^m-1) = i$, this means that $p^i| a^m-1$,

with $p|\Phi_n(a) $, $p^{i+1}|a^{p-1}-1$ but the p adic valuations were same. This is a contradiction.

Hence, $n$ does not divide p-1.

This means $p$ will divide $n$

Hence, Proved.

Question_1:

My friend said that I proved Eisenstein's criterion's, I cannot make any sense of it, what does it mean? Is there any connection, I did not know about Eisenstein's criterion, did some search and figured what It meant. Still unable to see any connection.

Question_2

A better proof?

Answer 1

I'm going to first have a look on your proof, and write a connection with Eisenstein's criterion, and then write my proof which uses lifting the exponent lemma.

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But if $v_p(a^n-1) = x $, $$a^n-1 \equiv 0 \pmod{p^x}$$ Is max it can go,

Therefore, at $\pmod{p^x}$ the order should hit $n$ to stop this increment of $p$ power. $$n|p^{x-1}(p-1), x \geq 2$$

It seems to me that you think the following claim is true :

Claim 1 : If $v_p(a^n-1) = x $, then $ \text{ord}_{p^x}(a)=n$.

The claim 1 is not true. Take $a=2,n=12,p=3$ for which we have $x=2$ and $\text{ord}_{p^x}(a)=6\not=n$.

Added 1 :

I forgot to consider the condition $\Phi_n(a)\equiv 0\pmod p$.

Your claim is the following :

Claim 2 : If $\Phi_n(a)\equiv 0\pmod p$ and $v_p(a^n-1)=x$, then $\text{ord}_{p^x}(a)=n$.

I can see that claim 2 is true (you might want to add a proof of your claim in your proof).

In conclusion, I have not found any error in your solution, so I think your proof is correct.

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Eisenstein's criterion gives a sufficient condition for a polynomial with integer coefficients to be irreducible over the rational numbers.

The contraposition of

Our claim : "If there is an integer $a$ such that $\Phi_n(a)\equiv 0\pmod p$, then $n\mid p-1$ or $p\mid n$"

is

"If $n\not\mid p-1$ and $p\not\mid n$, then there is no integer $a$ such that $\Phi_n(a)\equiv 0\pmod p$"

which gives a sufficient condition for $\Phi_n(x)\pmod p$ to be irreducible.

Maybe this is what your friend's comment meant.

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Finally, I'm going to write my proof which uses lifting the exponent lemma.

Our claim : If there is an integer $a$ such that $\Phi_n(a)\equiv 0\pmod p$, then $n\mid p-1$ or $p\mid n$.

Proof :

For $a=1$, the claim follows since, by induction, we have $$\Phi_n(1)=\begin{cases}0&\text{if $n=1$}\\p&\text{if $n=p^e$ with $e\ge 1$} \\1&\text{otherwise}\end{cases}$$

In the following, $a\ge 2$.

Let $n=p^em$ where $m,e$ are integers such that $\gcd(p,m)=1,m\ge 1,e\ge 0$.

For $p=2$, suppose that $m\gt 1$. Then, we have $\frac{a^{2^em}-1}{a^{2^e}-1}=\sum_{i=0}^{m-1}(a^{2^e})^i\equiv m\equiv 1\pmod 2$, so $2\not\mid \Phi_n(a)$ which is a contradiction. Since we have $m=1$ and $n=2^e$ with $e\ge 0$, the claim follows.

In the following, $p\ge 3$. Let $g:=\text{ord}_p(a)$.

We have $p\mid (a^{n}-1)$, so $g\mid n$. Since $g\mid p-1$, we get $\gcd(p,g)=1$, and so $g\mid m$ which implies $m\ge g$.

By lifting the exponent lemma, $$v_p(a^n-1)=v_p(a^{p^eg}-1)=v_p(n)+v_p(a^g-1)\tag1$$

Suppose that $m\gt g$. From $(1)$, we have $p\not\mid \frac{a^{p^em}-1}{a^{p^eg}-1}$, so $p\not\mid \Phi_n(a)$ which is a contradiction. Since we get $m=g$ and $n=p^eg$ with $e\ge 0$, the claim follows.$\ \blacksquare$

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Added 2 :

Claim 3 : If $\Phi_n(a)\equiv 0\pmod p$ with $a\ge 2$ and $p=2$, then $a$ has to be odd.

Proof : We have $a^n\equiv 1\pmod 2$. Since $a^n$ is odd, we see that $a$ has to be odd.$\ \square$

Claim 4 : If $\Phi_n(a)\equiv 0\pmod p$ with $a\ge 2$ and $p\ge 3$, then $v_p(a^n-1)=v_p(a^{p^eg}-1)=v_p(n)+v_p(a^g-1)$.

Proof : Let $k:=\frac mg$. For $\ell=0,1,\cdots, e-1$, we have $a^{p^{\ell}g}\equiv 1\pmod p$. Let $a^{p^{\ell}g}=Ap+1$. Then, $$\begin{align}\frac{a^{p^{\ell+1}g}-1}{a^{p^{\ell}g}-1}&=\sum_{i=0}^{p-1}(a^{p^{\ell}g})^i \\\\&=\sum_{i=0}^{p-1}(Ap+1)^i \\\\&=\sum_{k=0}^{i}(Ap)^k\sum_{i=0}^{p-1}\binom ik \\\\&=p+\frac{A(p-1)}{2}p^2+(Ap)^2\sum_{i=2}^{p-1}\binom i2+\cdots \\\\&\equiv p\pmod{p^2}\end{align}$$ So, we have $v_p\bigg(\frac{a^{p^{\ell +1}g}-1}{a^{p^{\ell}g}-1}\bigg)=1$ and $v_p\bigg(\frac{a^{p^eg}-1}{a^{g}-1}\bigg)=e$. So, we have $$v_p(a^{p^eg}-1)=v_p(a^g-1)+e\tag2$$ Since $\frac{a^{n}-1}{a^{p^eg}-1}=\frac{a^{p^egk}-1}{a^{p^eg}-1}=\sum_{i=0}^{k-1}(a^{p^eg})^i\equiv k\pmod{a^{p^eg}-1}$, we have $\gcd\bigg(\frac{a^n-1}{a^{p^eg}-1},a^{p^eg}-1\bigg)\mid k$. Since $p\not\mid k$, we have $$v_p(a^n-1)=v_p(a^{p^eg}-1)\tag3$$From $(2)(3)$, the claim 4 follows.$\ \square$