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?
Answer
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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