# Primitive Element theorem, permutations

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?

• Do you mean every permutation $\sigma \in Gal(\mathbb{Q}[\alpha]/K)$? Jan 2, 2016 at 12:01
• No, I mean every permutation from the symmetric group.
– user276611
Jan 2, 2016 at 12:03
• Maybe I am misunderstanding your question, but this can't work for every permutation in $S_n$ because of the identity. Jan 2, 2016 at 12:05
• I edited the question to reflect of what is meant by "distinct".
– user276611
Jan 2, 2016 at 12:07
• 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). Jan 2, 2016 at 12:31

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. $$\blacksquare$$

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.