The norm of an algebraic number $\alpha$ is the product of its conjugates, $N(\alpha)$.

Suppose that I have an inequality of the form $|x-\alpha*y| > c X^{n-\gamma}$ where $X=max{|x|,|y|}$ and c is some constnat, can this be used to find an upper bound on the norm of $x-\alpha*y$ of the form $|N(x-\alpha*y)| < C *X^{n-\gamma}$ where C is some constant and $\gamma \geq 1$

In the case of $\gamma=1$ it can since: $|N(x-\alpha*y)|=\prod (x-\alpha_i*y)|=|x-\alpha*y|X^{n-1} \prod (x/X-\alpha_i*y/X) \leq |x-\alpha*y| \prod(1|+|\alpha_{i}|)$

It's unclear to me how it might be possible to find a sharper upper bound on the norm of $x-\alpha*y$ by using a sharper exponent in liouiville-type inequalities.

I should add that the constant C should be effective, as it is in the case of $\gamma=1$.

The norm of an algebraic number $\alpha$ is the product of its conjugates, $N(\alpha)$.

Suppose that I have an inequality of the form $|x-\alpha*y| > c X^{n-\gamma}$ where $X=max{|x|,|y|}$ and c is some constnat, can this be used to find an upper bound on the norm of $x-\alpha*y$ of the form $|N(x-\alpha*y)| < C *X^{n-\gamma}$ $|x-\alpha*y|$ where C is some constant and $\gamma \geq 1$

In the case of $\gamma=1$ it can since: $|N(x-\alpha*y)|=\prod (x-\alpha_i*y)|=|x-\alpha*y|X^{n-1} \prod (x/X-\alpha_i*y/X) \leq |x-\alpha*y| \prod(1|+|\alpha_{i}|)X^{n-1}$

It's unclear to me how it might be possible to find a sharper upper bound on the norm of $x-\alpha*y$ by using a sharper exponent in liouiville-type inequalities.

I should add that the constant C should be effective, as it is in the case of $\gamma=1$.

The norm of an algebraic number $\alpha$ is the product of its conjugates, $N(\alpha)$.

Suppose that I have an inequality of the form $|x-\alpha*y| > c X^{n-\gamma}$ where $X=max{|x|,|y|}$ and c is some constnat, can this be used to find an upper bound on the norm of $x-\alpha*y$ of the form $|N(x-\alpha*y)| < C *X^{n-\gamma}$ where C is some constant and $\gamma \geq 1$

In the case of $\gamma=1$ it can since: $|N(x-\alpha*y)|=\prod (x-\alpha_i*y)|=|x-\alpha*y|X^{n-1} \prod (x/X-\alpha_i*y/X) \leq |x-\alpha*y| \prod(1|+|\alpha_{i}|)$

It's unclear to me how it might be possible to find a sharper upper bound on the norm of $x-\alpha*y$ by using a sharper exponent in liouiville-type inequalities.

The norm of an algebraic number $\alpha$ is the product of its conjugates, $N(\alpha)$.

Suppose that I have an inequality of the form $|x-\alpha*y| > c X^{n-\gamma}$ where $X=max{|x|,|y|}$ and c is some constnat, can this be used to find an upper bound on the norm of $x-\alpha*y$ of the form $|N(x-\alpha*y)| < C *X^{n-\gamma}$ where C is some constant and $\gamma \geq 1$

In the case of $\gamma=1$ it can since: $|N(x-\alpha*y)|=\prod (x-\alpha_i*y)|=|x-\alpha*y|X^{n-1} \prod (x/X-\alpha_i*y/X) \leq |x-\alpha*y| \prod(1|+|\alpha_{i}|)$

It's unclear to me how it might be possible to find a sharper upper bound on the norm of $x-\alpha*y$ by using a sharper exponent in liouiville-type inequalities.

I should add that the constant C should be effective, as it is in the case of $\gamma=1$.