Quantitative linear independence of algebraic numbers - a poor man's version.

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The celebrated Subspace Theorem of Schmidt has the following implication.

Theorem. Let $1,\alpha_1,\dots,\alpha_n \in \mathbb{R}$ be algebraic numbers linearly independent over $\mathbb{Q}$, and $\varepsilon > 0$. Then, there is a constant $c > 0$ such that for all $k \in \mathbb{Z}^n$ we have $$ \left| \sum_{i=1}^n k_i \alpha_i \bmod{1} \right| > c \lVert k \rVert_2^{-n-\varepsilon}. $$

This is a very elegant and powerful result, and it is optimal in the sense that the exponent at $\lVert k \rVert_2$ cannot be made smaller than $-n$. At the same time, for some applications it is not terribly important what the exponent is.

There is a rather trivial way to obtain the bound $\left| \sum_{i=1}^n k_i \alpha_i \bmod{1} \right| > c \lVert k \rVert_2^{-C},$ where $C$ depends on $\alpha_1,\dots,\alpha_n$: just multiply all the Galois conjugates of $\sum_{i=1}^n k_i \alpha_i - l$, where $l$ is the integer part.

I am curious about something in between. Namely, I don't care how large the constant gets, but I would like it to only depend on $n$. Of course, one can just quote Schmidt, but that feels like killing a butterfly with a cannon. More precisely, I would like to have the following:

Proposition. For any $n$, there exists $C > 0$, such that the following holds. Let $1,\alpha_1,\dots,\alpha_n \in \mathbb{R}$ be algebraic numbers linearly independent over $\mathbb{Q}$. Then, there is a constant $c > 0$ such that for all $k \in \mathbb{Z}^n$ we have $$ \left| \sum_{i=1}^n k_i \alpha_i \bmod{1} \right| > c \lVert k \rVert_2^{-C}. $$