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I am considering how to provide an alternative proof of the lemma used in the proof that $\Phi_n$ is irreducible:

Lemma: If $\Phi_n=f_1 f_2\cdots f_r$ is a factorization into monic irreducible polynomials in $\mathbb Q[X]$ then if $\zeta$ is a root of $f_1$ and $p\nmid n$ is a prime number then $\zeta^p$ is a root of $f_1$ as well.

Remark

I will use the following statement that is easily proven:

Consider a field extension of the form $K(\alpha)\supset K$ for some $\alpha$ algebraic over $K$. Then if $f$ is monic and irreducible with $f(\alpha)=0$ it follows that $f=\mbox{Irr}(\alpha,K)$.

My version of the lemma

Lemma: If $\Phi_n=f_1f_2\cdots f_r$ is a factorization into monic irreducible polynomials in $\mathbb Q[X]$ and add to this that $\deg(f_1)\leq\deg(f_2)\leq...\leq\deg(f_r)$, then if $\zeta$ is a root of $f_1$ and $p\nmid n$ is a prime number then $\zeta^p$ is a root of $f_1$ as well.

Analysis: Let $\zeta$ be a root of $f_1$. Then $\mbox{Irr}(\zeta,\mathbb Q)=f_1$. Since $p\nmid n$ we see that both $\zeta$ and $\zeta^p$ are roots of $\Phi_n$. In particular $\zeta^p$ must be a root of some $f_q$. Now since $\zeta^p\in\mathbb Q(\zeta)$ we see that we have a tower of fields $$ \mathbb Q(\zeta)\supset\mathbb Q(\zeta^p)\supset\mathbb Q $$ Since $f_1$ was defined to have minimal degree among the $f_i's$ we see on one hand that $f_q$ must have degree greater than or equal to $f_1$. On the other hand, since $f_q$ is the defining polynomial for $\zeta^p$ over $\mathbb Q$, looking on the tower of fields above it must have degree less than or equal to $f_1$. It follows that $$ \deg(f_q)=\deg(f_1) $$ Thus $\mathbb Q(\zeta)=\mathbb Q(\zeta^p)$.

Now iterating like in the original proof combining different prime powers of $\zeta$ this argument can be repeated to see that each $\zeta^k$ with $\gcd(k,n)=1$ will be a root of some $f_i$ of the same degree as $f_1$. So all the $f_i$'s have the same degree. Furthermore $$ \mathbb Q(\zeta)=\mathbb Q(\zeta^k) $$ for all $k$ relatively prime to $n$.

But then I do not know where this takes us...

String
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    In the next-to-last sentence of the proof, how do you know that $\zeta\mapsto\zeta^p$ is a field automorphism? Specifically, consider where it sends $\zeta^{k+1}$. – Andreas Blass Jan 20 '14 at 13:44
  • @AndreasBlass: My thought is that sending a generator of one basis to a generator of another basis will maintain the entire structure of the field in any respect... – String Jan 20 '14 at 13:55
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    Indeed, your isomorphism works because $\zeta$ and $\zeta^p$ have the same minimal polynomial. But that's what you're trying to prove, so your argument looks circular to me. – Andreas Blass Jan 20 '14 at 16:06
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    Agree with Andreas Blass. The problem is that the claim is equivalent to $\zeta$ and $\zeta^p$ having the same minimal polynomial, which is equivalent to $\zeta\mapsto\zeta^p$ being an automorphism. String, you have not shown that the mapping $\zeta\mapsto\zeta^p$ respects multiplication. More specifically, if $m$ is the degree of the minimal polynomial of $\zeta$, you haven't shown that your mapping satisfies $\phi(\zeta^{m-1}\zeta)=\phi(\zeta^{m-1})\phi(\zeta)$. – Jyrki Lahtonen Jan 20 '14 at 18:35
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    Or in yet other words $\phi(\alpha^i)=\beta^i$ yields a homomorphism of fields only, if $\alpha$ and $\beta$ have the same minimal polynomial, which is exactly what you are trying to prove. So I'm afraid your argument is circular. – Jyrki Lahtonen Jan 20 '14 at 18:37
  • @JyrkiLahtonen: Hmm... I get the point now. It seems to me that different arguments should be used. Perhaps I have been to focused on establishing that $f_q=f_1$. – String Jan 20 '14 at 18:50

1 Answers1

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After modification, your argument succesfully estabishes that all individual primitive $n$-th roots of unity generate the same field over$\def\Q{\Bbb Q}~\Q$, and therefore have minimal polynomials of the same degree. The same is true over larger fields than$~\Q$, where $\Phi_n$ does factor (into equal degree factors), so you have no hope of going from here to the conclusion you seek.

In fact what you have proved is clear from purely group theoretic argments: every primitive $n$-th root of unity generates the group of all $n$-th roots of unity, so for every pair of primitive $n$-th roots of unity, each one can be written as a power of the other, so the two must generate the same field. In fact the proof you gave could be done directly for $k$ relatively prime to $n$ in place of$~p$, as you never use primality. You do of course need that fact (an also that the irredicible factors of$~\Phi_n$ have integer coefficients, in othe words that you are working over$~\Q$) in order to complete the proof by reducing to polynomials over $\Bbb F_p$ and using the Frobenius automorphism, as is done in the classical proof by Dedekind/van der Waerden (not Gauss, who only did the case where $n$ is prime).

Marc van Leeuwen
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  • Yes, thank you! I see this still more clearly each time I look at it. The fact that any primitive $n$-th root generates all $n$-th roots of unity immediately shows that they all generate the same extension. – String Jan 21 '14 at 10:37