I want to prove the primitive element theorem for finite Galois extensions of $\mathbb Q$ along the following lines. Suppose $E/\mathbb Q$ is a finite Galois extension with basis $x_1, \ldots, x_n$. By the Steinitz exchange theorem we can assume $x_1 = 1 \in \mathbb Q$. Set $\alpha = 1 + x_2 + \ldots x_n$ and consider the equation $$ \sum_{k=0}^m c_k \alpha^k = 0. $$ If we substitute the sum for $\alpha$, multiply out, write cross products of the $x_i$'s in terms of sums of the $x_i$, and rearrange we find an equation in the $x_i$ where the coefficients came from this rearrangement. Hence we have $n$ equalities (the terms before the $x_i$) and $m$ indeterminates (the $c_k$'s). So for $m = n$ we can solve for the $c_k$ non-trivially. This gives us a polynomial $f = c_n X^n + \ldots + c_1 X + c_0$ with $f(\alpha) = 0$, to show that its root generates $E$ over $\mathbb Q$ we have to show that $c_n \ne 0$ and $c_n^{-1} \cdot f$ is the minimal polynomial of $\alpha$, but here I do not know how to proceed?
So is it possible to show that the minimal polynomial of $\alpha$ has degree $n$, or to alter the above proof so as to give this result?
The element $\alpha$ is not necessarily primitive. The first counterexample that occured to me is with $E=\Bbb{Q}(\sqrt2,\sqrt3)$. Let $$ \mathcal{B}=\{x_1=1,x_2=\sqrt2+\sqrt6,x_3=\sqrt3, x_4=-\sqrt2-\sqrt3\}. $$ The coordinates of elements of $\mathcal{B}$ with respect to the "natural" basis $\{1,\sqrt2,\sqrt3,\sqrt6\}$ form the matrix $$ A=\pmatrix{1&0&0&0\cr0&1&0&-1\cr0&0&1&-1\cr0&1&0&0\cr}. $$ We easily see that $\det A=1$, so $\mathcal{B}$ is a basis. But $$ \alpha=1+x_2+x_3+x_4=1+\sqrt6 $$ is clearly not primitive.
If there are any intermediate fields $K$ between $E$ and $\Bbb{Q}$ it should be possible to select the basis in such a way that $\alpha\in K$. The trick of including $1$ in the basis only serves to prove that $\alpha\notin\Bbb{Q}$.