Show only elements fixed by Galois group of cyclotomic field are those elements in $\mathbb{Q}$

Question

Currently working through a book that uses a bit of Galois Theory in some chapter. If a theorem is stated I do prefer to at least have read proofs/proved results at some point in the past before proceeding (even though I'm sure the author isn't lying to me about the theorem being true). Unfortunately I have note gone over Galois Theory in the past and its quite a distraction to do so now for one theorem.

The theorem in the book states this:

Let $ \omega_n = e^{ 2\pi i/n }$, the primitive $n$-th root of unity. Put $F := \mathbb{Q} (\omega_n )$, the smallest field subfield of $\mathbb{C}$ containing $ \omega_n $. Let $ \Gamma $ be the automorphisms of $F$ that fix $\mathbb{Q}$. Let $\alpha \in F $. If $ \sigma ( \alpha ) = \alpha $ for all $ \sigma \in \Gamma $ then $ \alpha \in \mathbb{Q} $.

I want to prove this using only facts I know, which include the following (thanks to contributors in this thread Looking for shorter proof that order of Galois group of cyclotomic field is finite );

Facts I Know, Which Can Be Used Without Proving:

• $ \mathbb{Q} ( \omega_{n} ) $ has finite dimension as a vector space over $ \mathbb{Q} $,

• $ \omega_{n} $ is algebraic over $ \mathbb{Q} $,

• the definition of a minimal polynomial in $ \mathbb{Q} [x] $ for an element $ a \in \mathbb{Q} ( \omega_{n} ) $

• Let $\sigma$ be an automorphism of $\mathbb{Q}(\omega_n)$ fixing $\mathbb{Q}$. For any polynomial $p(X)$ over $\mathbb{Q}$ and $\alpha \in \mathbb{Q}(\omega_n)$ we have $\sigma(p(\alpha)) = p(\sigma(\alpha))$.

• Any element $\alpha \in \mathbb{Q}(\omega_n)$ can be written in the form $\alpha = \frac{p(\omega_n)}{q(\omega_n)}$ for some polynomials $p, q$ over $\mathbb{Q}$ such that $q(\omega_n) \neq 0$.

• Let $\sigma, \tau$ be automorphisms of $\mathbb{Q}(\omega_n)$ that fix $\mathbb{Q}$. Then $\sigma(\omega_n) = \tau(\omega_n) \iff \sigma = \tau$.

• Let $p$ be any polynomial over $\mathbb{Q}$ such that $p(\omega_n) = 0$ (not necessarily the minimal polynomial). Then for any automorphism $\sigma$ of $\mathbb{Q}(\omega_n)$ fixing $\mathbb{Q}$, we have $p(\sigma(\omega_n)) = 0$.

Thing of Which I Have NOT Read Proofs, So Can't Be Used Unless Proved (Happy for Them To Be Proved, Then Used Though):

• $ \mathbb{Q} ( \omega_{n} ) $ has dimension $ \phi (n) $, where $ \phi (n) $ is the number of positive integers less than $ n $ that are coprime to $ n $,

• The set $ \{ \omega_{n} , \omega_{n}^{2} , \dots , \omega_{n}^{n-1} \} $ spans $ \mathbb{Q} ( \omega_{n} ) $ as a $\mathbb{Q}$-vector space

• The Fundamental Theorem of Galois Theory

• The minimal polynomial of some $ \beta \in F $ has no repeated roots

• If $ m $ is the minimal polynomial of some $ \beta \in F $ then $\Gamma $ acts transitively on the roots of $m$

My attempt so far:

Let $\alpha \in F $. Suppose $ \sigma ( \alpha ) = \alpha $ for all $ \sigma \in \Gamma $.

It is simple to verify that there exists an automorphism $ \gamma \in \Gamma $ that sends elements of $F$ to their complex conjugates. If $\alpha = a + bi$ with $a,b \in \mathbb{R}$ and $ b \neq 0$, then $\gamma ( \alpha ) \neq \alpha $. So we can assume $ \alpha \in \mathbb{R}$.

I'm not sure how to proceed from here.

Answer 1

What is the book you are reading and do you really need to have your question answered for all $n$ or only some special case like prime $n$ or prime power $n$?

Your fifth bullet point can be simplified: each element of $\mathbf Q(\omega_n)$ is $f(\omega_n)$ for some polynomial $f(x)$ in $\mathbf Q[x]$. This is because your $1/q(\omega_n)$ can be rewritten as a polynomial in $\omega_n$. That is because it is a standard result in field theory that when $K$ is a field and $\alpha$ is algebraic over $K$, the field $K(\alpha)$ is the same as the ring $K[\alpha]$ of polynomials in $\alpha$ with coefficients in $K$. This then makes the second fact you have not yet proved easy to show.

You do not need the fundamental theorem of Galois theory to prove what you want, as what you want to prove is needed to prove the fundamental theorem of Galois theory. That $\Gamma$ acts transitively on the roots of the minimal polynomial of each $\beta$ in $F$ is very closely related to the fact that you are trying to show. I think you being unrealistic in hoping for a shortcut that bypasses proving any of the results you say you have not yet proved.

Answer 2

I think the easiest way to do this without explicitly using Galois theory is to show that $1, \omega, \omega^2, ... , \omega^{\varphi(n)-1}$ is a basis for $F$ over $\mathbb Q$, where $\omega = e^{2\pi i /n}$ is the standard primitive $n$th root of unity. From here, if

$$x = \sum\limits_{j=0}^{\varphi(n)-1} c_j \omega^j \tag{$c_j \in \mathbb Q$}$$

is fixed by every element of $\Gamma$, it's not difficult to see that $c_1 = c_2 = \cdots = c_{\varphi(n)-1} = 0$, so $x = c_0 \in \mathbb Q$.

How to see this is a basis? The minimal polynomial $\mu(X)$ of $\omega$ divides

$$X^n-1 = (X-1)(X - \omega)(X- \omega^2) \cdots (X - \omega^{n-1})$$

and in fact

$$\mu(X) = \prod\limits_{ \textrm{GCD}(k,n)=1} (X - \omega^k)$$

This is because the roots of $\mu(X)$ are precisely the primitive $n$th roots of unity (there are some gaps to fill in here, where you will have to make some ad hoc "Galois theory"-like arguments). From here, you may deduc that $F$ has degree $\varphi(n)$ over $\mathbb Q$, and that the aforementioned list is actually a basis.

As others have mentioned, you can't get something for nothing here, and will need to either invoke standard field theory results or fill in the gaps yourself in an ad hoc way.

Answer 3

Firstly, it is immediate that $\mathbb{Q}(\omega_n) =\mathbb{Q}[\omega_n]$ (see KCd's comment, here you can use the extended euclidean algorithm to solve for $r(x)$ such that $m_{\omega_n}q + r = 1$, this is a consequence of irreducibility of minimal polynomials. Here we know such a minimal polynomial exists since $\omega_n^n - 1 = 0$ so we can choose the polynomial of least degree as below). It then follows that any element can be written in the form $\sum_0^{n-1} a_i\omega_n^i$, this proves the second fact, from this we can choose a basis $\{1,\omega_n^{k_1},...,\omega_n^{k_r}\}$ for $F/\mathbb{Q}$. Alternatively you can show $\mathbb{Q}[\omega_n]$ is a field by noticing $\omega_n^{n-1} = \omega_n^{-1}$ and take inverses by partial fraction decomposition.

Now we can represent any element $\alpha \in F$ as a matrix, with the $j$-th column being the image of the $j$-th basis element under multiplication by $\alpha$. By the [Cayley-Hamilton Theorem] we have that $\alpha$ satisfies its characteristic polynomial. Thus for any element the set of polynomials which it satisfies is non-empty and we can pick an element of minimal degree. This minimal degree polynomial must be seperable (i.e. distinct roots) since otherwise $(f,f') \neq 1$, and since $\deg f' = \deg f - 1$ this contradicts minimality since $\alpha$ must then satisfy one of $(f,f')$ or $\frac{f}{(f,f')}$ of strictly smaller degree. Now if $m_\alpha$ is the minimal polynomial of $\alpha$, and $\beta$ is any other root of $m_\alpha$ we have the isomorphism $\alpha \mapsto \beta$ via $\mathbb{Q}[\alpha] \cong \mathbb{Q}[x]/m_\alpha \cong \mathbb{Q}[\beta]$ (universal property of polynomial rings + first isomorphism theorem). This can be extended to an isomorphism $F \to F$ via the [Isomorphism Extension Theorem] (or if you insist on checking the details yourself note that you can easily do the extension "one element at a time" using the above isomorphism for simple extensions).

Finally note that $\deg m_\alpha = 1$ if and only if $\alpha \in \mathbb{Q}$, so for any $\alpha \not \in \mathbb{Q}$ we can choose some $\beta \neq \alpha$ satisfying $m_\alpha$ and thus we have an automorphism $F \overset{\sigma}{\to} F$ with $\sigma: \alpha \mapsto \beta$.

Note on why $\mathbf{\beta \in F}$: We have that since $\mathbb{Q}[x]$ is a PID, $m_{\omega_n} \vert x^n - 1$. Since our $F$ contains all roots of the latter polynomial, the same is true for the former. From the fourth fact you know, it follows that field isomorphisms must permute the roots of $m_{\omega_n}$, thus $F = Q(\omega_n) \subset \sigma(\mathbb{Q}(\omega_n))$ for any field isomorphism (in particular the one sending $\alpha \mapsto \beta$). Now since $\omega_n$ generates $F/\mathbb{Q}$, the same must be true of $\sigma(\omega_n)$ for $\sigma(F)/\mathbb{Q}$, i.e. $\sigma(\mathbb{Q}(\omega_n)) \subset \mathbb{Q}(\sigma(\omega_n)) \subset F$. Thus $\beta \in F$.

Answer 4

Proved this in the end by piecing together bits of information from answers, comments and links provided in this post. Let me know of any places where there are mistakes. These are the steps of the proof I put together:

As in the original post, continue using the notation $F:= \mathbb{Q} (\omega_n)$ and $\Gamma$ for the set of automorphisms of $F$.

Assume $n \geq 3$ because the main theorem that we want to prove is trivial for $n=1$ and $n=2$. Then $\omega _{n} \notin \mathbb{Q}$.

The following two results are some that I have read a proof of in the past. Probably should have put it in the "results I know" section of the original post, but did not know they would be used.

(1) State result on constructing field using irreducible polynomial of $K[x]$ where $K$ is a field

Let $K$ be a field. Let $p(x) \in K[x]$ be a polynomial of degree $d$ that is irreducible over $K$. Let $\langle p(x) \rangle$ be the ideal generated by $p(x)$. Then $K[x] / \langle p(x) \rangle$ is a field. Furthermore, if $\alpha = x + \langle p(x) \rangle $ then every element of $K[x] / \langle p(x) \rangle$ can be expressed uniquely in the form $ k_{d-1} \alpha ^{d-1} + \dots + k_{1} \alpha ^{1} + k_{0} $, where $k_{i} \in K$ for each $i \in \{ 0,1, \dots ,d-1 \}$.

(2) State result on isomorphism between an algebraic field extension of $K$ and a quotient of $K [x]$

Suppose $K$ is a subfield of $\mathbb{C}$. Let $p(x) \in K [x]$ be an irreducible polynomial. Let $\rho \in \mathbb{C}$ be a root of $p$. Then there exists a field isomorphism $\psi : K ( \rho ) \rightarrow K [x] / \langle p(x) \rangle $ with $ \psi ( \rho ) = x + \langle p(x) \rangle $.

(Could generalise this to fields other than $\mathbb{C}$ but this result is enough for what we need.)

(3) Show $\{ 1, \omega_{n} , \dots , \omega ^{d-1} \}$ is basis for $\mathbb{Q} (\omega _{n} )$, where $d$ degree of minimal polynomial of $\omega _{n}$ over $\mathbb{Q}$

Let $m(x) \in \mathbb{Q}$ be minimal polynomial of $\omega _{n}$ over $\mathbb{Q}$. By (2), there exists an isomorphism $ \psi_{1} : \mathbb{Q} ( \omega _{n} ) \rightarrow \mathbb{Q} [x] / \langle m(x) \rangle $ with $ \psi_{1} ( \omega _{n} ) = x + \langle m(x) \rangle $.

Using (1) and the isomorphism $ \psi_{1}^{-1} $, we can see that $\{ 1, \omega_{n} , \dots , \omega ^{d-1} \}$ is basis for $\mathbb{Q} (\omega _{n} )$, where $d$ degree of minimal polynomial of $\omega _{n}$ over $\mathbb{Q}$.

(4) Find all of the roots of $m(x)$ and show degree of $m(x)$ is $ \phi (n) $

Using proof from this post,

prove cyclotomic polynomial is a minimal polynomial

we know that the minimal polynomial is \[ m(x) = \prod _{\substack{ k \in \{ 1, \dots , n\}, \\ gcd(k,n) = 1 } } (x- \omega _{n}^{k} ) . \] We can see this has $ \phi (n) $ distinct roots, where $ \phi(n) $ is the number of positive numbers less than $n$ that are coprime to $n$. Can also see $m(x)$ has no repeated roots. So $ \phi (n) $ is the degree of $m(x)$. Another fact we can see is each root generates the group $ \langle \omega _{n} \rangle $.

(5) Prove $\Gamma$ acts transitively on roots of $m(x)$

Let $ \psi _{1} $ be the isomorphism defined in (3). Let $ \omega_{n} ^{\prime} $ be a root of the minimal polynomial $m(x)$. Then by (2), there exists an isomorphism $ \psi_{2} : \mathbb{Q} ( \omega _{n}^{\prime} ) \rightarrow \mathbb{Q} [x] / \langle m(x) \rangle $ with $ \psi_{2} ( \omega _{n}^{\prime} ) = x + \langle m(x) \rangle $.

Now $ \mathbb{Q} ( \omega _{n} ) = \mathbb{Q} ( \omega _{n}^{\prime} ) $ because we saw in (4) that $ \langle \omega _{n} \rangle = \langle \omega _{n}^{\prime } \rangle $. It follows that $ \psi_{1} \circ \psi_{2} \in \Gamma $ with $ \psi _{1} ( \psi _{2} ( \omega _{n} ) ) = \omega _{n} ^{\prime }$. Clearly this means that $ \Gamma $ acts transitively on the roots of $m(x)$.

(6) Show the set of elements fixed by all automorphisms is a subfield of $\mathbb{Q} (\omega_n)$ containing $\mathbb{Q}$

Let $A = \{ \alpha \in F: \sigma (\alpha ) = \alpha \text{ for all } \sigma \in \Gamma \}$. Let $ \alpha, \beta \in A $. Then, \[ 1 = \sigma (\alpha \alpha^{-1} ) = \sigma (\alpha ) \sigma (\alpha^{-1} ) = \alpha \sigma (\alpha^{-1} ). \] Hence $ \sigma (\alpha^{-1} ) = \alpha^{-1} $. Similarly $\sigma (- \alpha ) = - \alpha$. So $ \sigma (\beta \alpha^{-1} ) = \beta \alpha^{-1} $ and $ \sigma (\beta - \alpha ) = \beta - \alpha $.

This shows $A$ is a subfield of $F$. Clearly $\mathbb{Q} \subseteq A$ by the definition of $\Gamma$.

(7) Find minimal polynomial of $ \omega _n $ over $A$

Let $m^{*} (x) \in A[x]$ be the minimal polynomial of $ \omega _n $ over $A$. Since $\mathbb{Q} \subseteq A$ and $ \omega _n$ is a root of the minimal polynomial $m(x) \in \mathbb{Q}$, it follows $m^{*} (x)$ divides $m(x) $.

Let $ \omega_{n} ^{\prime} $ be a root of $m(x) $ then $\sigma ( \omega _n ) = \omega_{n} ^{\prime} $ for some $ \sigma \in \Gamma$.

By the definition of $A$, the coefficients of $m^{*} (x) $ are fixed by $\sigma$, so $0 = \sigma (0) = \sigma (m^{*} (\omega _n) ) = m^{*} ( \sigma (\omega _n) ) = m^{*} ( \sigma ( \omega_{n} ^{\prime} ) $. Hence $\omega_{n} ^{\prime} $ is a root of $m^{*}$. So $ m(x) $ divides $ m^{*} (x) $. Hence $ m(x) = m^{*} (x) $.

(8) Prove the main result

Notice $ A( \omega _{n} ) = \mathbb{Q} ( \omega _{n} ) $ because $A \subseteq \mathbb{Q} ( \omega _{n} ) $ and $ \mathbb{Q} \subseteq A$. Using (1) and (2), every element of $ \mathbb{Q} ( \omega _{n} ) $ can be expressed uniquely as in the form $ a_{d-1} \omega _{n} ^{d-1} + \dots + a_{1} \omega _{n} ^{1} + a_{0} $, where $d $ is the degree of $m(x)$ and $a_{i} \in A$ for each $i \in \{ 0,1, \dots ,d-1 \}$.

Suppose $A \neq \mathbb{Q}$. By definition, $ \mathbb{Q} \subseteq A$, so there exists $ a \in A \setminus \mathbb{Q}$. Then $a = a \cdot 1$ when expressing $a$ as an element of $\mathbb{Q} ( \omega _{n} ) $ as an $A$-vector space.

But when expressed as an element of $\mathbb{Q} ( \omega _{n} ) $ as an $\mathbb{Q}$-vector space, $a = q_{d-1} \omega _{n} ^{d-1} + \dots + q_{1} \omega _{n} ^{1} + q_{0} $ where $q_{j} \in \mathbb{Q}$ for each $j \in \{0,1, \dots , d-1 \}$ and most importantly $q_{k} \neq 0$ for some $k \neq 0$.

But $ q_{d-1} \omega _{n} ^{d-1} + \dots + q_{1} \omega _{n} ^{1} + q_{0} $ is also a linear combination of basis elements with coefficients in $A$. So we have two distinct linear expressions for $a$ when looking at $\mathbb{Q} ( \omega _{n} ) $ as an $A$-vector space.

This is a contradiction. Thus $A = \mathbb{Q}$.