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Let $K = \mathbb{Q}(\alpha_1,\alpha_2,...\alpha_n)$, where the $\alpha_i$ are the roots of some irreducible polynomial (and hence they are pairwaise distinct since the polynomial is separable). Then $K/\mathbb{Q}$ is a finite extension. By the primitive element theorem there exists a $\alpha$ such that $\mathbb{Q}(\alpha) = K$. Galois ("Sur les conditions de resolubilite des equations par radicaux", Lemme II; see here) was able (without proof) to choose $\alpha = u_1 \alpha_1 + \cdots + u_n \alpha_n$ with $u_i \in \mathbb{Q}$ such that all the elements $\sigma(\alpha) := u_1 \alpha_{\sigma(1)} + \cdots + u_n \alpha_{\sigma(n)}$ are distinct for every permutation $\sigma$ of the symmetric group. Distinct in this sense means that $\sigma(\alpha) \neq \tau(\alpha)$ for different $\sigma, \tau \in S_n$. Is this always true and if so, does somebody have a reference for this?

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  • $\begingroup$ Do you mean every permutation $\sigma \in Gal(\mathbb{Q}[\alpha]/K)$? $\endgroup$ Jan 2, 2016 at 12:01
  • $\begingroup$ No, I mean every permutation from the symmetric group. $\endgroup$
    – user276611
    Jan 2, 2016 at 12:03
  • $\begingroup$ Maybe I am misunderstanding your question, but this can't work for every permutation in $S_n$ because of the identity. $\endgroup$ Jan 2, 2016 at 12:05
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    $\begingroup$ I edited the question to reflect of what is meant by "distinct". $\endgroup$
    – user276611
    Jan 2, 2016 at 12:07
  • $\begingroup$ What are the $\alpha_i $ ? From your notation they could be any elements generating $K $, in which case the answer is clearly negative (as some of them could be equal). $\endgroup$ Jan 2, 2016 at 12:31

2 Answers 2

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Yes, it is true. The following more general fact is true:

Theorem 1. Let $\mathbf{k}$ be an infinite field. Let $V$ be a $\mathbf{k}$-vector space. Let $v_{1},v_{2},\ldots,v_{n}$ be finitely many distinct elements of $V$. Then, there exists some $\left( a_{1},a_{2} ,\ldots,a_{n}\right) \in\mathbf{k}^{n}$ such that the $n!$ elements $a_{\sigma\left( 1\right) }v_{1}+a_{\sigma\left( 2\right) }v_{2} +\cdots+a_{\sigma\left( n\right) }v_{n}$, for $\sigma$ ranging over the symmetric group $S_{n}$, are pairwise distinct.

[Notice that my notations are different from yours. My $\mathbf{k}$, $V$, $v_{i}$ and $a_{i}$ correspond to your $\mathbb{Q}$, $K$, $\alpha_{i}$ and $u_{i}$, respectively (but of course, my setting is more general).]

The main tool for proving Theorem 1 is the following theorem, which doubles as a well-known exercise:

Theorem 2. Let $\mathbf{k}$ be an infinite field. Let $V$ be a finite-dimensional $\mathbf{k}$-vector space. Then, $V$ cannot be written as a union of finitely many proper subspaces of $V$. (A proper subspace of $V$ means a $\mathbf{k}$-vector subspace of $V$ distinct from $V$.)

Theorem 2 is proven in many places; for example, see A finite-dimensional vector space cannot be covered by finitely many proper subspaces? or (for a stronger statement) If a field $F$ is such that $\left|F\right|>n-1$ why is $V$ a vector space over $F$ not equal to the union of $n$ proper subspaces of $V$ or (also for a stronger statement) A vector space over $R$ is not a countable union of proper subspaces or https://mathoverflow.net/q/26/ .

Proof of Theorem 1. Let $G$ be the set $\left\{ \left( \sigma,\tau\right) \in S_{n}\times S_{n}\ \mid\ \sigma\neq\tau\right\} $. Clearly, the set $G$ is finite (since $S_{n}$ is finite).

The $\mathbf{k}$-vector space $\mathbf{k}^n$ is finite-dimensional. Hence, Theorem 2 (applied to $\mathbf{k}^n$ instead of $V$) shows that $\mathbf{k}^n$ cannot be written as a union of finitely many proper subspaces of $\mathbf{k}^n$. In other words, any union of finitely many proper subspaces of $\mathbf{k}^n$ must be a proper subset of $\mathbf{k}^n$.

For every $\sigma\in S_{n}$, we define a map $v_{\sigma}:\mathbf{k} ^{n}\rightarrow V$ as follows: For any $\left( a_{1},a_{2},\ldots ,a_{n}\right) \in\mathbf{k}^{n}$, we set $v_{\sigma}\left( a_{1} ,a_{2},\ldots,a_{n}\right) =\sum\limits_{i=1}^{n}a_{\sigma\left( i\right) }v_{i}$. This map $v_{\sigma}$ is $\mathbf{k}$-linear.

Now, let $\left( \sigma,\tau\right) \in G$. We shall show that $\operatorname*{Ker}\left( v_{\sigma}-v_{\tau}\right) $ is a proper subspace of $\mathbf{k}^n$.

Indeed, the map $v_{\sigma}-v_{\tau}$ is $\mathbf{k}$-linear (since $v_{\sigma}$ and $v_{\tau}$ are $\mathbf{k}$-linear), and thus $\operatorname*{Ker}\left( v_{\sigma}-v_{\tau}\right) $ is a $\mathbf{k} $-vector subspace of $\mathbf{k}^{n}$.

Moreover, $\sigma\neq\tau$ (since $\left( \sigma,\tau\right) \in G$). Assume (for the sake of contradiction) that $\operatorname*{Ker}\left( v_{\sigma }-v_{\tau}\right) =\mathbf{k}^{n}$. Thus, $v_{\sigma}-v_{\tau}=0$, so that $v_{\sigma}=v_{\tau}$.

Let $g\in\left\{ 1,2,\ldots,n\right\} $.

We shall use the notation $\delta_{u,v}$ for the element $ \begin{cases} 1, & \text{if }u=v;\\ 0, & \text{if }u\neq v \end{cases} \in\mathbf{k}$ whenever $u$ and $v$ are two objects. For every permutation $\pi\in S_{n}$, we have

$v_{\pi}\left( \delta_{1,g},\delta_{2,g},\ldots,\delta_{n,g}\right) =\sum\limits_{i=1}^{n}\underbrace{\delta_{\pi\left( i\right) ,g}} _{=\delta_{i,\pi^{-1}\left( g\right) }}v_{i}$ (by the definition of $v_{\pi }$)

(1) $=\sum\limits_{i=1}^{n}\delta_{i,\pi^{-1}\left( g\right) } v_{i}=v_{\pi^{-1}\left( g\right) }$.

Applying (1) to $\pi=\sigma$, we obtain

(2) $v_{\sigma}\left( \delta_{1,g},\delta_{2,g},\ldots,\delta _{n,g}\right) =v_{\sigma^{-1}\left( g\right) }$.

Applying (1) to $\pi=\tau$, we obtain $v_{\tau}\left( \delta_{1,g} ,\delta_{2,g},\ldots,\delta_{n,g}\right) =v_{\tau^{-1}\left( g\right) }$. Since $v_{\sigma}=v_{\tau}$, this rewrites as $v_{\sigma}\left( \delta _{1,g},\delta_{2,g},\ldots,\delta_{n,g}\right) =v_{\tau^{-1}\left( g\right) }$. Comparing this with (2), we obtain $v_{\sigma^{-1}\left( g\right) }=v_{\tau^{-1}\left( g\right) }$. Since $v_{1},v_{2},\ldots,v_{n}$ are distinct, this shows that $\sigma^{-1}\left( g\right) =\tau^{-1}\left( g\right) $.

Now, let us forget that we fixed $g$. We thus have shown that $\sigma ^{-1}\left( g\right) =\tau^{-1}\left( g\right) $ for every $g\in\left\{ 1,2,\ldots,n\right\} $. In other words, $\sigma^{-1}=\tau^{-1}$. In other words, $\sigma=\tau$. This contradicts $\sigma\neq\tau$. This contradiction proves that our assumption (that $\operatorname*{Ker}\left( v_{\sigma }-v_{\tau}\right) =\mathbf{k}^{n}$) was wrong. Hence, $\operatorname*{Ker} \left( v_{\sigma}-v_{\tau}\right) \neq\mathbf{k}^{n}$. Since $\operatorname*{Ker}\left( v_{\sigma}-v_{\tau}\right) $ is a $\mathbf{k} $-vector subspace of $\mathbf{k}^{n}$, this yields that $\operatorname*{Ker} \left( v_{\sigma}-v_{\tau}\right) $ is a proper subspace of $\mathbf{k}^{n}$.

Now, let us forget that we fixed $\left( \sigma,\tau\right) $. Thus, for every $\left( \sigma,\tau\right) \in G$, the set $\operatorname*{Ker}\left( v_{\sigma}-v_{\tau}\right) $ is a proper subspace of $\mathbf{k}^{n}$. Hence, $\bigcup_{\left( \sigma,\tau\right) \in G}\operatorname*{Ker}\left( v_{\sigma}-v_{\tau}\right) $ is a union of finitely many proper subspaces of $\mathbf{k}^n$ (since $G$ is finite). Therefore, $\bigcup_{\left( \sigma,\tau\right) \in G}\operatorname*{Ker}\left( v_{\sigma}-v_{\tau}\right) $ must be a proper subset of $\mathbf{k}^n$ (since any union of finitely many proper subspaces of $\mathbf{k}^n$ must be a proper subset of $\mathbf{k}^n$). In other words, there exists some $a\in \mathbf{k}^n$ such that $a\notin\bigcup_{\left( \sigma,\tau\right) \in G} \operatorname*{Ker}\left( v_{\sigma}-v_{\tau}\right) $. Consider this $a$.

We have $a\notin\bigcup_{\left( \sigma,\tau\right) \in G}\operatorname*{Ker} \left( v_{\sigma}-v_{\tau}\right) $. In other words,

(3) $a\notin\operatorname*{Ker}\left( v_{\sigma}-v_{\tau}\right) $ for every $\left( \sigma,\tau\right) \in G$.

Now, let $\sigma$ and $\tau$ be two distinct elements of $S_{n}$. Thus, $\left( \sigma,\tau\right) \in S_{n}\times S_{n}$ and $\sigma\neq\tau$. In other words, $\left( \sigma,\tau\right) \in G$. Hence, (3) shows that $a\notin\operatorname*{Ker}\left( v_{\sigma}-v_{\tau}\right) $. In other words, $\left( v_{\sigma}-v_{\tau}\right) \left( a\right) \neq0$. Hence, $0\neq\left( v_{\sigma}-v_{\tau}\right) \left( a\right) =v_{\sigma}\left( a\right) -v_{\tau}\left( a\right) $, so that $v_{\sigma}\left( a\right) \neq v_{\tau}\left( a\right) $.

Let us forget that we fixed $\sigma$ and $\tau$. We thus have shown that $v_{\sigma}\left( a\right) \neq v_{\tau}\left( a\right) $ for any two distinct elements $\sigma$ and $\tau$ of $S_{n}$. In other words,

(4) the $n!$ elements $v_{\sigma}\left( a\right) $, for $\sigma$ ranging over the symmetric group $S_{n}$, are pairwise distinct.

Now, let us write $a$ in the form $\left( a_{1},a_{2},\ldots,a_{n}\right) $. Then, for every $\sigma\in S_{n}$, we have

$v_{\sigma}\left( a\right) =v_{\sigma}\left( a_{1},a_{2},\ldots ,a_{n}\right) =\sum\limits_{i=1}^{n}a_{\sigma\left( i\right) } v_{i}=a_{\sigma\left( 1\right) }v_{1}+a_{\sigma\left( 2\right) } v_{2}+\cdots+a_{\sigma\left( n\right) }v_{n}$.

Hence, the statement (4) rewrites as follows: The $n!$ elements $a_{\sigma\left( 1\right) }v_{1}+a_{\sigma\left( 2\right) }v_{2} +\cdots+a_{\sigma\left( n\right) }v_{n}$, for $\sigma$ ranging over the symmetric group $S_{n}$, are pairwise distinct. This proves Theorem 1. $\blacksquare$

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Thanks Darij Grinberg. I found another proof, but am unsure if it is correct or if I have missed some detail:

Let $x_1,\cdots, x_n$ be transzendental over $\mathbb{C}$. For $\sigma,\tau \in S_n$ we have the following equivalent statements (with $u_1,\cdots, u_n \in \mathbb{Q}$):

$ u_1 \alpha_{\sigma(1)}+\cdots+u_n\alpha_{\sigma(n)} = u_1 \alpha_{\tau(1)}+\cdots+u_n\alpha_{\tau(n)}$

$ \sum_{i} \alpha_i ( u_{\sigma^{-1}(i)} - u_{\tau^{-1}(i)} ) = 0$

$(u_1,\cdots,u_n)$ is a zero of the polynomial $\sum_{i} \alpha_i ( x_{\sigma^{-1}(i)} - x_{\tau^{-1}(i)} ) $

With other words: All

$ u_1 \alpha_{\sigma(1)}+\cdots+u_n\alpha_{\sigma(n)}$ are pairwise distinct exactly when $(u_1,\cdots,u_n)$ is not a zero of the polynomial

$\prod_{\sigma,\tau \in S_n, \sigma \neq \tau} \sum_{i} \alpha_i ( x_{\sigma^{-1}(i)} - x_{\tau^{-1}(i)} ) $

The polynomial thus constructed might have coefficients not in $\mathbb{Q}$. But it is not the zero polynomial:

If it was we would find $\sigma,\tau \in S_n$ , $\sigma \neq \tau$ such that:

$x_1 (\alpha_{\sigma(1)}-\alpha_{\tau(1)}) +\cdots+ x_1 (\alpha_{\sigma(n)}-\alpha_{\tau(n)}) = 0$ Because the $x_i$ are transcendental over $\mathbb{C}$ it follows that

$\alpha_{\sigma(i)} = \alpha_{\tau(i)}$ and becaus $\sigma \neq \tau$ we must have $\alpha_i = \alpha_j$ for $i\neq j$ contradicting the fact that the $\alpha_i$ are pairwise distinct.

Now consider the polynomial

$p(x_1,\cdots,x_n) = \prod_{\gamma \in S_n}(\prod_{\sigma,\tau \in S_n, \sigma \neq \tau} \sum_{i} \alpha_i ( x_{\sigma^{-1}(i)} - x_{\tau^{-1}(i)} ))$

It has the polynomial from above as a factor ($\gamma = 1$) and since it is symmetric in the zeros $\alpha_i$ of some other polynomial with rational coefficients, it has also rational coefficients. Suppose there does not exist $(u_1,...,u_n)$ such that the $\alpha_i$ are not pairwise distinct. Then $(u_1,...,u_n)$ must be a zero of $p(x_1,\cdots,x_n)$. Hence the polynomial with rational coefficients $p(x_1,...,x_n)$ has the property that every $(u_1,...,u_n)$ is a zero of this polynomial. But then this polynomial must be the zero polynomial (Cor. 1.7, S.176 Lang, Algebra), contradicting what we have proved above.

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