Let $n$ be an integer greater than $1$ and let $\sqrt[n]{2}\in \mathbb{R}$ be the unique positive $n$-th root of $2$. Show that the real numbers $1, (\sqrt[n]{2})^2,\cdot\cdot\cdot,(\sqrt[n]{2})^{n-1}$ are linearly independent over $\mathbb{Q}$.

Using the definition of linear independence, the problem is equivalent to showing that $$c_1+c_2 (\sqrt[n]{2})^2+\cdot\cdot\cdot+c_{n-1}(\sqrt[n]{2})^{n-1}=0$$ has only the trivial solution $c_1=c_2=\cdot\cdot\cdot=c_{n-1}=0$

I believe there is something special about each $\sqrt[n]{2}$ being a unique positive $n$-th root of 2 that allows us to state linear independence. In particular, this implies that each $(\sqrt[n]{2})^i$ is linearly independent where $i\in \{0,2,3,\cdot\cdot\cdot,n-1\}$.

I don't know if my intuition is correct and how to prove this linear independence rigorously.

  • $\begingroup$ Hint: What is the minimal polynomial of $\sqrt[n]{2}$? $\endgroup$ – Slade May 13 '16 at 3:44

Hint: if there were a linear dependence relation between these elements, then $\sqrt[n]{2}$ would satisfy a (monic) polynomial of degree $\leqslant n-1$ (after dividing by the highest nonzero coefficient). On the other hand, $\sqrt[n]{2}$ is a root of $X^{n} - 2$, which is irreducible over $\mathbb{Q}$ by Eisenstein at $2$. Why is this first bit in contradiction with the second fact?


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