Minimal polynomial are same for $\lambda$ and $\lambda ^2$

Question

Let $L/K$ be an algebraic field extension. Let $\lambda$ and $\lambda ^2$ have the same minimal polynomial over $K$ where $\lambda \in L$ and $\lambda$ is nonzero Prove that $\lambda$ is a root of unity.

My attempt: It is clear that $K(\lambda)$=$K(\lambda^2)$ after that how I conclude? Any help/hint in this regards would be highly appreciated. Thanks in advance!

Answer 1

One way to think about it is this: In the Galois closure of $K(\lambda)$, there is an automorphism $\tau$ taking $\lambda$ to $\lambda^2$. Since $\tau$ is finite order, for some $k$, $\tau^k$ is trivial, so $\lambda^{2^k} = \lambda$, and $\lambda$ is a root of unity.

Answer 2

This means that $\lambda^2$ is a Galois conjugate of $\lambda$ over $K$. Therefore $\lambda^4$ is a Galois conjugate of $\lambda^2$, and so also a Galois conjugate of $\lambda$. So all of $\lambda,\lambda^2,\lambda^4,\lambda^8,\ldots$ are Galois conjugates. But $\lambda$ only has a finite number of Galois conjugates....