# Automorphisms of the splitting field of the m-th cyclotomic polynomial.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Let $p(x)=x^m-1$ be a polynomial over $\mathbb Q$ and $E$ be the splitting field for $p$ over $\mathbb Q$. We know that $p$ has $\phi(m)$ primitive roots in $E$, where $\phi$ is the Euler's totient function. Let $\omega$ be a primitive root of $p$.

Define $\theta_k:E\to E$ as $\theta_k(q(\omega))=q(\omega^{k})$ for all $q[x]\in \mathbb Q[x]$.

I want to show that $\theta_k$ is an isomorphism for each $k$ satisfying $\gcd(k,m)=1$.

The only problem here is to show that $\theta_k$ is well defined. This is equivalent to showing that $q(\omega)=0 \Rightarrow q(\theta_k(\omega))=0$, for each $q(x)\in \mathbb Q[x]$.

I don't want to use that $\Phi_m$, the $m$-th cyclotomic polynomial, is irreducible over $\mathbb Q$. The reason for this is that in Herstein's book Problem $13$ and $14$ of section $5.6$ show the irreducibility of $\Phi_m$ using the fact that $\theta_k$ is an automorphism for each $k$ with $\gcd(k,m)=1$.

Note that it is clear that the set of all $\mathbb Q$ automorphisms of $E$ is a subset of $\{\theta_k:\gcd(k,m)=1\}$.

#### Deveno

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MHB Math Scholar
This seems rather obvious.

Suppose the minimal polynomial of $$\displaystyle \omega$$ is $$\displaystyle r(x) \in \Bbb Q[x]$$. Since any automorphism of $$\displaystyle E$$ that fixes $$\displaystyle \Bbb Q$$ must send $$\displaystyle \omega$$ to another root of $$\displaystyle r$$, if:

$$\displaystyle q(\omega) = 0$$ then $$\displaystyle q(x) = r(x)s(x)$$ for some $$\displaystyle s(x) \in \Bbb Q[x]$$ so:

$$\displaystyle q(\theta_k(\omega)) = r(\theta_k(\omega))s(\theta_k(\omega)) = 0$$.

The basic idea behind all of this is that the roots of $$\displaystyle r(x)$$ are all *primitive* $$\displaystyle m$$-th roots of unity...which are the primitive $$\displaystyle m$$-th roots of unity? Precisely:

$$\displaystyle \{\omega^k : \text{gcd}(k,m) = 1\}$$.

The special case $$\displaystyle m = 6$$ should prove illuminating.

Something else to consider is this: the $$\displaystyle m$$-th roots of unity form a cyclic (multiplicative) subgroup of $$\displaystyle E^{\ast}$$. Any (field) automorphism of $$\displaystyle E$$ fixing $$\displaystyle \Bbb Q$$ *must* induce a (group) automorphism of this subgroup of $$\displaystyle E^{\ast}$$, in particular, it must send a generator to another generator.

That is, what you are really doing is establishing a bijection between:

$$\displaystyle \text{Gal}(E/\Bbb Q) \leftrightarrow \text{Aut}(\langle \omega \rangle)$$

#### caffeinemachine

##### Well-known member
MHB Math Scholar
This seems rather obvious.

Suppose the minimal polynomial of $$\displaystyle \omega$$ is $$\displaystyle r(x) \in \Bbb Q[x]$$. Since any automorphism of $$\displaystyle E$$ that fixes $$\displaystyle \Bbb Q$$ must send $$\displaystyle \omega$$ to another root of $$\displaystyle r$$, if:

$$\displaystyle q(\omega) = 0$$ then $$\displaystyle q(x) = r(x)s(x)$$ for some $$\displaystyle s(x) \in \Bbb Q[x]$$ so:

$$\displaystyle q(\theta_k(\omega)) = r(\theta_k(\omega))s(\theta_k(\omega)) = 0$$.
Hello Deveno.

I have some doubts. Please consider them.

We have $r(x)$ as the minimal polynomial of $\omega$ over $\mathbb Q$. Let $k$ be an integer with $\gcd(k,m)=1$. Let $q(x)\in\mathbb Q[x]$ such that $q(\omega)=0$. We want to show that $q(\theta_k(\omega))=0$.

We have, from $q(\omega)=0$, that $q(x)=r(x)s(x)$, for some $s(x)\in\mathbb Q[x]$.

Thus $q(\theta_k(\omega))=q(\omega^k)=r(\omega^k)\cdot s(\omega^k)$.

To show that $q(\theta_k(\omega))=0$, we must argue that $r(\omega^k)=0$ or $s(\omega^k)=0$. I think you claim that $r(\omega^k)=0$ (as we know hardly anything about $s(x)$). Am I right? If yes, then can you please give me some details.

#### Deveno

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Ok, explicitly, suppose:

$$\displaystyle r(x) = a_0 + a_1x + \cdots + a_{n-1}x^{n-1} + x^n$$, where the $$\displaystyle a_j \in \Bbb Q$$.

Then:

$$\displaystyle r(\theta_k(\omega)) = a_0 + a_1\theta_k(\omega) + \cdots + a_{n-1}(\theta_k(\omega))^{n-1} + (\theta_k(\omega))^n$$

$$\displaystyle = \theta_k(a_0) + \theta_k(a_1)\theta_k(\omega) + \cdots + \theta_k(a_{n-1})(\theta_k(\omega))^{n-1} + (\theta_k(\omega))^n$$

(since $$\displaystyle \theta_k$$ fixes $$\displaystyle \Bbb Q$$)

$$\displaystyle = \theta_k(a_0 + a_1\omega + \cdots + a_{n-1}\omega^{n-1} + \omega^n)$$

(since $$\displaystyle \theta_k$$ is a ring-homomorphism)

$$\displaystyle = \theta_k(r(\omega)) = \theta_k(0) = 0$$.

For the special case $$\displaystyle m = 4$$, this is the usual proof that if a complex number (in some subfield of $$\displaystyle \Bbb C$$) satisfies a rational polynomial, so does its complex-conjugate because the complex-automorphism:

$$\displaystyle a + bi \to a + bi^3 = a - bi$$ is just complex conjugation.

I want to stress that the key idea here is to simplify the "field" situation (we are considering the extension field $$\displaystyle E$$ of $$\displaystyle \Bbb Q$$) by considering a special group associated with this extension field, its Galois group (being a splitting field, it's a normal extension). In this case, the Galois group has a particularly "nice" structure, it is cyclic. We can say a lot rather easily about cyclic groups, using divisibility relations of INTEGERS.

It turns out these divisibility relationships extend rather nicely to the cyclotomic polynomials, because of this tight association between the subfield lattice of $$\displaystyle E$$ and the subgroup lattice of $$\displaystyle \text{Gal}(E/\Bbb Q)$$:

$$\displaystyle \phi(m) = \prod_{d|m} \phi(d) \leftrightarrow \Phi_m(x) = \frac{x^m - 1}{\prod_{d|m} \Phi_d(x)}$$ (where d is kept less than m on both sides).

Another key observation: not only is $$\displaystyle \omega$$ a primitive $$\displaystyle m$$-th root of unity, it is also a primitive ELEMENT of $$\displaystyle E$$ in that $$\displaystyle E = \Bbb Q(\omega)$$ (adjoining $$\displaystyle \omega$$ to $$\displaystyle \Bbb Q$$ gives us ALL the roots of $$\displaystyle x^m - 1$$).

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Ok, explicitly, suppose:

$$\displaystyle r(x) = a_0 + a_1x + \cdots + a_{n-1}x^{n-1} + x^n$$, where the $$\displaystyle a_j \in \Bbb Q$$.

Then:

$$\displaystyle r(\theta_k(\omega)) = a_0 + a_1\theta_k(\omega) + \cdots + a_{n-1}(\theta_k(\omega))^{n-1} + (\theta_k(\omega))^n$$

$$\displaystyle = \theta_k(a_0) + \theta_k(a_1)\theta_k(\omega) + \cdots + \theta_k(a_{n-1})(\theta_k(\omega))^{n-1} + (\theta_k(\omega))^n$$

(since $$\displaystyle \theta_k$$ fixes $$\displaystyle \Bbb Q$$)

$$\displaystyle = \theta_k(a_0 + a_1\omega + \cdots + a_{n-1}\omega^{n-1} + \omega^n)$$ (since $$\displaystyle \theta_k$$ is a ring-homomorphism) $\leftarrow$ Problematic Step.

$$\displaystyle = \theta_k(r(\omega)) = \theta_k(0) = 0$$.
Forgive me if I am being a blockhead. But I am still not convinced . Referring to the thing marked as 'Problematic Step':

You say that 'since $\theta_k$ is a ring homomorphism ...'. Weren't we trying to show that $\theta_k$ is well defined? Once we show that it is well defined only then we can say that it is a homomorphism. Moreover, $\theta_k$ is a map form a field to a field. So either it is a trivial homomorphism or a monomorphism.

Also, I don't see where have you used the fact that $\gcd(k,m)=1$.
This fact is crucial since $\theta_k$ is certainly not an automorphism of $E$ if $\gcd(k,m)\neq 1$. (In fact I think $\theta_k$ won't even be well defined if $\gcd(k,m)\neq 1$. I can come up with a counterexample in some time assuming I am not wrong about this.)

I will read the deeper insights you have been trying to give in both of your posts once I understand this. I will comment on them too. Thanks.

#### Deveno

##### Well-known member
MHB Math Scholar
I understand your concern: you think it could be that because the ring-homomorphism

$$\displaystyle \Bbb Q[x] \to \Bbb Q(\omega)$$ given by:

$$\displaystyle q(x) \mapsto q(\omega)$$ is surjective, it could be that we might have:

$$\displaystyle q(\omega) = 0$$ while $$\displaystyle q(\omega^k) \neq 0$$.

However, consider this:

$$\displaystyle \Bbb Q(\omega) = \Bbb Q(\omega^k)$$.

Since $$\displaystyle \omega^k \in \Bbb Q(\omega)$$ it is clear that:

$$\displaystyle \Bbb Q(\omega^k) \subseteq \Bbb Q(\omega)$$.

But since gcd(k,m) = 1, there exists integers a,b with:

ak + bm = 1, so that:

$$\displaystyle (\omega^k)^a = (\omega^k)^a(1^b) = (\omega^k)^a(\omega^m)^b = \omega^{ak+bm} = \omega$$

which shows that $$\displaystyle \omega \in \Bbb Q(\omega^k)$$.

Now think about what $$\displaystyle q(\omega)$$ actually looks like: it is a polynomial of degree < the degree of the minimal polynomial for $$\displaystyle \omega$$ (since only the remainder upon division by the minimal polynomial remains).

Since $$\displaystyle \Bbb Q(\omega^k)$$ has the same degree over $$\displaystyle \Bbb Q$$ (that is, the same dimension as a $$\displaystyle \Bbb Q$$-vector space), the only non-zero elements $$\displaystyle q(\omega)$$ get mapped to non-zero elements $$\displaystyle q(\omega^k)$$. Thus we have a bijective mapping at least.

As I indicated before, the special case $$\displaystyle m = 6$$ is instructive:

Let $$\displaystyle \omega$$ be a primitive sixth root of unity. We can factor like so:

$$\displaystyle x^6 - 1 = (x^3 + 1)(x^3 - 1)$$.

The primitive root must be a root of $$\displaystyle x^3 + 1$$, since the roots of $$\displaystyle x^3 - 1$$ are third roots of unity (and therefore not PRIMITIVE sixth roots of unity). We can further factor:

$$\displaystyle x^3 + 1 = (x + 1)(x^2 - x + 1)$$, and since -1 is a square root of unity, the primitive root(s) must be roots of $$\displaystyle x^2 - x + 1$$.

The only possible values for k are 1 and 5. We calculate:

$$\displaystyle (\omega^5)^2 - \omega^5 + 1 = \omega^4 - \omega^5 + 1 = \omega^3(\omega - \omega^2) + 1$$

$$\displaystyle = \omega^3 + 1 = -1 + 1 = 0$$

(since $$\displaystyle \omega^2 - \omega = -1$$).

In this case, there are exactly two automorphisms of $$\displaystyle \Bbb Q(\omega)$$:

$$\displaystyle \text{id}:\Bbb Q(\omega) \to \Bbb Q(\omega)$$

which sends $$\displaystyle a + b\omega \to a + b\omega$$, and:

$$\displaystyle \sigma:\Bbb Q(\omega) \to \Bbb Q(\omega)$$

which sends $$\displaystyle a + b\omega \to a + b\omega^5$$.

(note that if we write $$\displaystyle \omega = c + di$$ that since $$\displaystyle \omega^5 = \overline{\omega}$$ we have:

$$\displaystyle \sigma(a + b\omega) = \sigma(a + b(c + di)) = a + b(c - di) = \overline{a + b\omega}$$

that is, $$\displaystyle \sigma$$ is just complex conjugation).

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#### caffeinemachine

##### Well-known member
MHB Math Scholar
I understand your concern: you think it could be that because the ring-homomorphism

$$\displaystyle \Bbb Q[x] \to \Bbb Q(\omega)$$ given by:

$$\displaystyle q(x) \mapsto q(\omega)$$ is surjective, it could be that we might have:

$$\displaystyle q(\omega) = 0$$ while $$\displaystyle q(\omega^k) \neq 0$$.
Yes! That's exactly what I am worried about.

However, consider this:

$$\displaystyle \Bbb Q(\omega) = \Bbb Q(\omega^k)$$.

Since $$\displaystyle \omega^k \in \Bbb Q(\omega)$$ it is clear that:

$$\displaystyle \Bbb Q(\omega^k) \subseteq \Bbb Q(\omega)$$.

But since gcd(k,m) = 1, there exists integers a,b with:

ak + bm = 1, so that:

$$\displaystyle (\omega^k)^a = (\omega^k)^a(1^b) = (\omega^k)^a(\omega^m)^b = \omega^{ak+bm} = \omega$$

which shows that $$\displaystyle \omega \in \Bbb Q(\omega^k)$$.
Yes. I understand that $\mathbb Q(\omega)=\mathbb Q(\omega^k)\iff \gcd(k,m)=1$.

Now think about what $$\displaystyle q(\omega)$$ actually looks like: it is a polynomial of degree < the degree of the minimal polynomial for $$\displaystyle \omega$$ (since only the remainder upon division by the minimal polynomial remains).
In other words, given an element $q(\omega)\in\mathbb Q(\omega)$, we can find a polynomial $p(x)\in\mathbb Q[x]$ such that $\deg p(x)<\deg r(x)$ (where $r(x)$ is the minimal polynomial of $\omega$ over $\mathbb Q$) such that $q(\omega)=p(\omega)$.

Since $$\displaystyle \Bbb Q(\omega^k)$$ has the same degree over $$\displaystyle \Bbb Q$$ (that is, the same dimension as a $$\displaystyle \Bbb Q$$-vector space), the only non-zero elements $$\displaystyle q(\omega)$$ get mapped to non-zero elements $$\displaystyle q(\omega^k)$$. Thus we have a bijective mapping at least.
I don't follow this. Sorry . Here's what troubles me:

Let $\gcd(k,m)=1$. Say we have $q(\omega)\neq 0$ for some $q(x)\in\mathbb Q[x]$. We want to argue that $q(\omega^k)\neq 0$. Isn't it? I think your argument is this:

Say the minimal polynomial of $\omega$ over $\mathbb Q$ is $r(x)$ and the minimal polynomial of $\omega^k$ over $\mathbb Q$ is $m(x)$. We don't know for sure that $r(x)=m(x)$ but we know that $\deg r(x)=\deg m(x)$.

Now,
Find a non-zero polynomial $p(x)\in\mathbb Q[x]$ such that $\deg p(x)<\deg r(x)$ and $p(\omega)=q(\omega)$. We know that $p(\omega^k)\neq 0$ since $\deg p(x)<\deg m(x)$. Since $p(\omega^k)=q(\omega^k)$ , we conclude that $q(\omega^k)\neq 0$.

The thing marked in red is something I cannot prove. This red thing clearly would follow if $\theta_k$ were known to be well defined.

If I am right about that being your argument then please clarify my doubt. If not then.. well.. can you please give some more details .

#### Deveno

##### Well-known member
MHB Math Scholar
Well, it works like this:

In $$\displaystyle E = \Bbb Q(\omega), x^m - 1$$ splits completely, as:

$$\displaystyle x^m - 1 = (x - 1)(x - \omega)(x - \omega^2)\dots(x - \omega^{m-1})$$

(these are m roots, and since $$\displaystyle x^m - 1$$ is of degree m, this must be ALL of them).

The minimal polynomial for $$\displaystyle \omega$$ is some sub-product of some of these roots. Which ones? Well, clearly not 1, nor (for even m) -1. In fact, if n|m, we have:

$$\displaystyle x^n - 1|x^m - 1$$.

Well, what if gcd(n,m) = d > 1? Then we have a factor $$\displaystyle x^{m/d} - 1$$ of $$\displaystyle x^m - 1$$ whose roots are all non-primitive (since they have order m/d at most). I'll illustrate with m = 6, with n = 4. Then gcd(4,6) = 2.

Now 6/2 = 3, so $$\displaystyle \omega^4$$ is a root of $$\displaystyle x^3 - 1$$ (we are using facts about the CYCLIC group generated by $$\displaystyle \omega$$ here).

In other words, you can keep discounting powers of $$\displaystyle \omega$$ whose exponent is NOT co-prime to m as roots of the minimal polynomial of $$\displaystyle \omega$$. They are roots of some lower degree polynomial of the form $$\displaystyle x^n - 1$$. This leaves us with $$\displaystyle \phi(m)$$ roots of a polynomial of degree $$\displaystyle \phi(m)$$....ALL primitive roots of $$\displaystyle x^m - 1$$ have the same minimal polynomial. In the m = 6 case, that polynomial is:

$$\displaystyle (x - \omega)(x - \omega^5) = x^2 - (\omega + \omega^5)x + 1$$
$$\displaystyle = x^2 - \omega(1 + \omega^4)x + 1 = x^2 - \omega(-\omega^2)x + 1$$

(since $$\displaystyle x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1)$$, $$\displaystyle \omega^4 + \omega^2 + 1 = 0$$, so $$\displaystyle \omega^4 + 1 = -\omega^2$$)

$$\displaystyle = x^2 + \omega^3x + 1 = x^2 - x + 1$$.

And what all THIS means is that if we write a polynomial in $$\displaystyle \Bbb Q[x]$$ as:

$$\displaystyle p(x) = q(x)r(x) + t(x)$$ then:

$$\displaystyle p(\omega) = q(\omega)r(\omega) + t(\omega) = t(\omega)$$ and:

$$\displaystyle p(\omega^k) = q(\omega^k)r(\omega^k) + t(\omega^k) = t(\omega^k)$$.

Which is what you should EXPECT, an automorphism of $$\displaystyle E$$ just permutes any roots of factors of $$\displaystyle x^m - 1$$ to other roots within the same factor.

If you feel more adventurous, try explicitly factoring $$\displaystyle x^{12} - 1$$ over $$\displaystyle \Bbb Q$$. First step: factor out all of the roots of $$\displaystyle x^6 - 1$$ (which powers of a twelfth primitive root are these going to be?).

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Well, it works like this:

In $$\displaystyle E = \Bbb Q(\omega), x^m - 1$$ splits completely, as:

$$\displaystyle x^m - 1 = (x - 1)(x - \omega)(x - \omega^2)\dots(x - \omega^{m-1})$$

(these are m roots, and since $$\displaystyle x^m - 1$$ is of degree m, this must be ALL of them).

The minimal polynomial for $$\displaystyle \omega$$ is some sub-product of some of these roots. Which ones? Well, clearly not 1, nor (for even m) -1. In fact, if n|m, we have:

$$\displaystyle x^n - 1|x^m - 1$$.

Well, what if gcd(n,m) = d > 1? Then we have a factor $$\displaystyle x^{m/d} - 1$$ of $$\displaystyle x^m - 1$$ whose roots are all non-primitive (since they have order m/d at most). I'll illustrate with m = 6, with n = 4. Then gcd(4,6) = 2.

Now 6/2 = 3, so $$\displaystyle \omega^4$$ is a root of $$\displaystyle x^3 - 1$$ (we are using facts about the CYCLIC group generated by $$\displaystyle \omega$$ here).

In other words, you can keep discounting powers of $$\displaystyle \omega$$ whose exponent is NOT co-prime to m as roots of the minimal polynomial of $$\displaystyle \omega$$. They are roots of some lower degree polynomial of the form $$\displaystyle x^n - 1$$. This leaves us with $$\displaystyle \phi(m)$$ roots of a polynomial of degree $$\displaystyle \phi(m)$$....ALL primitive roots of $$\displaystyle x^m - 1$$ have the same minimal polynomial.
The last paragraph does suggest that $\omega$ and $\omega^k$ both have the same minimal polynomial over $\mathbb Q$. But it doesn't prove that fact. I think you'll agree.

If you feel more adventurous, try explicitly factoring $$\displaystyle x^{12} - 1$$ over $$\displaystyle \Bbb Q$$. First step: factor out all of the roots of $$\displaystyle x^6 - 1$$ (which powers of a twelfth primitive root are these going to be?).
$x^{12}-1=(x−1)(x+1)(x^2+x+1)(x^2+1)(x^2−x+1)(x^4−x^2+1)$.

#### PaulRS

##### Member
Let $p(x)=x^m-1$ be a polynomial over $\mathbb Q$ and $E$ be the splitting field for $p$ over $\mathbb Q$. We know that $p$ has $\phi(m)$ primitive roots in $E$, where $\phi$ is the Euler's totient function. Let $\omega$ be a primitive root of $p$.

Define $\theta_k:E\to E$ as $\theta_k(q(\omega))=q(\omega^{k})$ for all $q[x]\in \mathbb Q[x]$.

I want to show that $\theta_k$ is an isomorphism for each $k$ satisfying $\gcd(k,m)=1$.

The only problem here is to show that $\theta_k$ is well defined. This is equivalent to showing that $q(\omega)=0 \Rightarrow q(\theta_k(\omega))=0$, for each $q(x)\in \mathbb Q[x]$.

I don't want to use that $\Phi_m$, the $m$-th cyclotomic polynomial, is irreducible over $\mathbb Q$. The reason for this is that in Herstein's book Problem $13$ and $14$ of section $5.6$ show the irreducibility of $\Phi_m$ using the fact that $\theta_k$ is an automorphism for each $k$ with $\gcd(k,m)=1$.
That's strange indeed.

If all of these were well-defined, then it follows trivially that $\Phi_m$ is irreducible (and viceversa). So one would expect this proof to be at least as hard as the proof that $\Phi_m$ is irreducible.

For all the homomorphisms to be well-defined we must have that $g(\omega) = 0$ implies $0=\theta_k (g(\omega)) = g(\omega^k)$ for all $k$ coprime to $m$, but then this means that the minimal polynomial $g(X)\in {\mathbb Q}[X]$ for $\omega$ has $\omega^k$ as roots for all $k$ coprime to $m$ (in $E$)... thus $(X-\omega^k) | g(X)$ for each $k\in\{1,\ldots,m\}$ coprime to $m$, and so $\Phi_m( X) | g(X)$ over $E$. But both are rational polynomials... hence $\Phi_m(X) | g( X)$ over $\mathbb Q$ too (uniqueness of quotient and remainder over $E$). Thus $g(X) = \Phi_m(X)$ is irreducible.

I don't quite see an answer to this, that avoids proving that $\Phi_m$ is irreducible (*). Correct me if I am wrong, what Deveno proved in his last post seems to be that you can "distinguish" (i.e. distinct minimal polynomial) between primitive roots and non-primitive roots when you work over $\mathbb Q$. But not that you can't "distinguish" between primitive roots.

(*) This is not that easy either, there is a beautiful proof in Dummit&Foote Theorem 41 (Third edition).

#### Deveno

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MHB Math Scholar
What I am suggesting is there is a quasi-inductive nature to $$\displaystyle \Phi_m(x)$$.

Clearly, for any m, $$\displaystyle x - 1$$ is a factor of $$\displaystyle x^m - 1$$. So we can always factor out the primitive first root of unity. Now if d < m and d|m, we can factor out $$\displaystyle x^d - 1$$, all of whose roots are NOT primitive m-th roots of unity, leaving some other polynomial (which is still in $$\displaystyle \Bbb Q[x]$$). This is appealing to:

$$\displaystyle \sum_{d|m} \phi(d) = m$$.

For example:

12 = 1 + 1 + 2 + 2 + 2 + 4

as caffeinemachine has shown, this partition of 12 corresponds to the degrees of the polynomial factors of $$\displaystyle x^{12} - 1$$.

Now, given that the degree of the minimal polynomial of a primitive root of unity is $$\displaystyle \phi(m)$$, after we divide out all the minimal polynomials of "lesser degree" primitive roots of unity, what we have left is a polynomial of degree $$\displaystyle \phi(m)$$ containing all $$\displaystyle \phi(m)$$ primitive roots of unity.

The sticking point, as I see it, is the statement I have bolded above. Which is, in my opinion, tantamount to the irreducibility of $$\displaystyle \Phi_m(x)$$. But that was not part of the original problem.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Now, given that the degree of the minimal polynomial of a primitive root of unity is $$\displaystyle \phi(m)$$, after we divide out all the minimal polynomials of "lesser degree" primitive roots of unity, what we have left is a polynomial of degree $$\displaystyle \phi(m)$$ containing all $$\displaystyle \phi(m)$$ primitive roots of unity.
Yes. If we know that the degree of the minimal polynomial of a primitive $m$-th root of unity is $\phi(m)$ then we'd be done. I don't think we have come up with a proof of that yet in this thread.

The sticking point, as I see it, is the statement I have bolded above. Which is, in my opinion, tantamount to the irreducibility of $$\displaystyle \Phi_m(x)$$. But that was not part of the original problem.
It is true that the original problem I posted does not ask to establish the irreducibility of $\Phi_m$ over $\mathbb Q$. But the original problem is equivalent to doing the same.

#### Deveno

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MHB Math Scholar
My bad, I thought (for some reason) that it was stated in the first post that the minimal polynomial of a primitive root had degree $$\displaystyle \phi(m)$$.

And in light of this, I agree, we have a number of equivalent "hard" statements:

a) $$\displaystyle \Phi_m(x)$$ is irreducible

b) $$\displaystyle \theta_k$$ is well-defined

c) $$\displaystyle \Phi_m(x)$$ is the minimal polynomial for a primitive m-th root of unity

My copy of Herstein is an older one, if you could post a pdf file of the exact problem you are working on, I would appreciate it.

Stepping back a bit, it is clear that any automorphism of $$\displaystyle E/\Bbb Q$$ is completely determined by the image of $$\displaystyle \omega$$. So perhaps it is easier to show DIRECTLY that:

$$\displaystyle a_0 + a_1\omega +\cdots+ a_{\text{deg}(r) - 1}\omega^{\text{deg}{r} - 1} \mapsto a_0 + a_1\omega^k +\cdots+ a_{\text{deg}(r) - 1}(\omega^k)^{\text{deg}{r} - 1}$$

is such an automorphism (the $$\displaystyle \Bbb Q$$-linearity is clear, the quotient ring (field) inherits the necessary properties from the ring $$\displaystyle \Bbb Q[x]$$), so only the multiplicative property and injectiveness need be shown, I think we have addressed the injectiveness already).

In other words instead, of defining $$\displaystyle \theta_k$$ on rational polynomials in general, just define it "for polynomials of degree < deg(r)", each of which has a UNIQUE image in the quotient ring (the field $$\displaystyle \Bbb Q(\omega)$$).

Of course, this is harder than it appears, at first glance, because without a *specific* multiplication table for powers of $$\displaystyle \omega$$ it could get a bit messy...

#### caffeinemachine

##### Well-known member
MHB Math Scholar
My bad, I thought (for some reason) that it was stated in the first post that the minimal polynomial of a primitive root had degree $$\displaystyle \phi(m)$$.

And in light of this, I agree, we have a number of equivalent "hard" statements:

a) $$\displaystyle \Phi_m(x)$$ is irreducible

b) $$\displaystyle \theta_k$$ is well-defined

c) $$\displaystyle \Phi_m(x)$$ is the minimal polynomial for a primitive m-th root of unity

My copy of Herstein is an older one, if you could post a pdf file of the exact problem you are working on, I would appreciate it.

Stepping back a bit, it is clear that any automorphism of $$\displaystyle E/\Bbb Q$$ is completely determined by the image of $$\displaystyle \omega$$. So perhaps it is easier to show DIRECTLY that:

$$\displaystyle a_0 + a_1\omega +\cdots+ a_{\text{deg}(r) - 1}\omega^{\text{deg}{r} - 1} \mapsto a_0 + a_1\omega^k +\cdots+ a_{\text{deg}(r) - 1}(\omega^k)^{\text{deg}{r} - 1}$$

is such an automorphism (the $$\displaystyle \Bbb Q$$-linearity is clear, the quotient ring (field) inherits the necessary properties from the ring $$\displaystyle \Bbb Q[x]$$), so only the multiplicative property and injectiveness need be shown, I think we have addressed the injectiveness already).

In other words instead, of defining $$\displaystyle \theta_k$$ on rational polynomials in general, just define it "for polynomials of degree < deg(r)", each of which has a UNIQUE image in the quotient ring (the field $$\displaystyle \Bbb Q(\omega)$$).

Of course, this is harder than it appears, at first glance, because without a *specific* multiplication table for powers of $$\displaystyle \omega$$ it could get a bit messy...
See problem 15 and then see problem 13 and 14.

#### Deveno

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MHB Math Scholar
Ah...a dim memory comes back to me....the problem refers to a PREVIOUS problem where it asks you to establish that:

$$\displaystyle [\Bbb Q(\omega):\Bbb Q] = \phi(m)$$

(problem 8, section 3), which is why I assumed this was already known.

I believe it can be shown that $$\displaystyle \Phi_m(x)$$ has integer coefficients by using complete induction on m, using the division algorithm and Gauss' lemma.

Suppose it can be shown that if $$\displaystyle r(x)$$ is the minimal polynomial of $$\displaystyle \omega$$ that $$\displaystyle r(x) \in \Bbb Z[x]$$ (I *think* this can be done using Gauss' lemma and the fact that $$\displaystyle x^m - 1$$ is monic).

Now, for each value of $$\displaystyle 0 \leq j < m$$, there are UNIQUE polynomials $$\displaystyle q_j, t_j \in \Bbb Z[x]$$ such that:

$$\displaystyle r(x^j) = r(x)q_j(x) + t_j(x)$$ with $$\displaystyle \text{deg}(t_j) < \text{deg}(r)$$ or $$\displaystyle t_j = 0$$.

In particular, for any prime $$\displaystyle p$$, we have:

$$\displaystyle r(\omega^p) = t(\omega)$$ for some $$\displaystyle t(x) \in \{t_0(x),\dots,t_{m-1}(x)\}$$.

Working mod p, we have:

$$\displaystyle r(x^p) = (r(x))^p$$, so $$\displaystyle r(x^p) - (r(x))^p = p(g(x))$$ for some polynomial $$\displaystyle g(x) \in \Bbb Z[x]$$.

Applying the division algorithm to $$\displaystyle g(x)$$, there is a unique polynomial $$\displaystyle h(x) \in \Bbb Z[x]$$ with degree < deg(r) (or h = 0) such that $$\displaystyle h(\omega) = g(\omega)$$. Consequently:

$$\displaystyle t(\omega) = r(\omega^p) = p(g(\omega)) = p(h(\omega))$$

Let $$\displaystyle A$$ be the maximum absolute value of all the coefficients of the $$\displaystyle \{t_j(x)\}$$. If our prime $$\displaystyle p > A$$, clearly we must have:

$$\displaystyle r(\omega^p) = t(\omega) = 0$$, since the only way a prime can divide coefficients of absolute value less than it is if all the coefficients are 0.

So for ANY postive integer $$\displaystyle n$$, if $$\displaystyle p \nmid n$$ for all primes $$\displaystyle p \leq A$$, we have $$\displaystyle r(\omega^n) = 0$$.

Now consider k with gcd(k,m) = 1. Form the number:

$$\displaystyle u = k + m\prod_{q_i \nmid k} q_i$$ where the $$\displaystyle q_i$$ are primes $$\displaystyle \leq A$$.

Let $$\displaystyle p \leq A$$ be a prime. If $$\displaystyle p|k$$ then $$\displaystyle p \nmid m\prod_{q_i \nmid k}q_i$$, since gcd(k,m) = 1. On the other hand if $$\displaystyle p \nmid k$$, then $$\displaystyle p = q_i$$ for some i, so in either case $$\displaystyle p \nmid u$$.

Hence, since $$\displaystyle u \equiv k (\text{mod }m)$$:

$$\displaystyle r(\omega^k) = r(\omega^u) = 0$$.

This shows that any primitive m-th root of unity has the same minimal polynomial as any other (which gets us what we need, I hope? There is one slight hole, where I have indicated a possible approach).

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Suppose it can be shown that if $$\displaystyle r(x)$$ is the minimal polynomial of $$\displaystyle \omega$$ that $$\displaystyle r(x) \in \Bbb Z[x]$$ (I *think* this can be done using Gauss' lemma and the fact that $$\displaystyle x^m - 1$$ is monic).
Knowing that $\Phi_m$ is in $\mathbb Z[x]$ and is a monic, let $r(x)\in\mathbb Z[x]$ be the irreducible (over $\mathbb Z$ and hence over $\mathbb Q$) factor of $\Phi_m$ in $\mathbb Z[x]$ such that $r(\omega)=0$. Thus $r(x)$ is a monic and hence $r(x)$ is the minimal polynomial of $\omega$ over $\mathbb Q$.

So I think the possible hole can be dealt with in this way.

#### Deveno

##### Well-known member
MHB Math Scholar
Just out of curiosity, I wanted to see how this proof played out for m = 6. Of course, we know in this case that:

$$\displaystyle r(x) = x^2 - x + 1$$.

So let's actually compute $$\displaystyle r(x^j)$$ for j = 0,1,2,3,4,5, modulo r(x). Here goes:

$$\displaystyle r(x^0) = r(1) = 1$$, so $$\displaystyle t_0(x) = 1$$<--boring, we know 1 isn't a primitive root.
$$\displaystyle r(x^1) = r(x)$$, so $$\displaystyle t_1(x) = 0$$
$$\displaystyle r(x^2) = x^4 - x^2 + 1 = x^2(x^2 - x + 1) + x^3 - x^2 - x^2 + 1$$
$$\displaystyle = x^2(x^2 - x + 1) + x(x^2 - x + 1) - x^2 - x + 1$$
$$\displaystyle = (x^2 + x)(x^2 - x + 1) - (x^2 - x + 1) - 2x + 2$$
$$\displaystyle = (x^2 + x - 1)(x^2 - x + 1) - 2x + 2$$

so $$\displaystyle t_2(x) = -2x + 2$$

The computations are going to get a bit involved here, so at this point I cheated and used wolfram:

$$\displaystyle r(x^3) = x^6 - x^3 + 1 = (x^4 + x^3 - 2x - 2)(x^2 - x + 1) + 3$$, so $$\displaystyle t_3(x) = 3$$.
$$\displaystyle r(x^4) = x^8 - x^4 + 1 = (x^6 + x^5 - x^3 - 2x^2 - x + 1)(x^2 - x + 1) + 2x$$, so $$\displaystyle t_4(x) = 2x$$

Finally,

$$\displaystyle r(x^5) = x^{10} - x^5 + 1 = (x^8 + x^7 - x^5 - x^4 - x^3 + x + 1)(x^2 - x + 1)$$, so $$\displaystyle t_5(x) = 0$$.

Here, we see that A = 3. So the only primes of interest are p = 2, and p = 3 (the prime divisors of 6, how about that?). What is u in this case? we have only two k to check: k = 1, and k = 5. For k = 1, we have:

u = 1 + 6(2*3) = 37 <--corresponds to our chosen primitive root (37 = 1 mod 6).

For k = 5, we have u = 5 + 6(2*3) = 41 <--corresponds to the root $$\displaystyle \omega^5$$.

This means that $$\displaystyle x^2 - x + 1 = (x - \omega)(x - \omega^5)$$, as expected.

As a bonus, note that indeed the remainder $$\displaystyle t_2(x)$$ is divisible by 2, and the remainder $$\displaystyle t_3(x)$$ is divisible by 3.