- Thread starter
- #1

- Feb 29, 2012

- 342

Let $\alpha \in \mathbb{R}$ be a root of $x^3 -2$. Let $K = \mathbb{Q}(\alpha) \subseteq \mathbb{R}$ and $\sigma: K \to K$ a field isomorphism. Prove that $\sigma (x) = x$ for all $x \in K$.

My attempt is as follows: since this is an homomorphism from the field to itself we have that $\sigma (0) = 0$ and $\sigma (1) = 1$. Using this I showed that $\sigma (n) = n$ for all $n \in \mathbb{Z}$ and subsequently that $\sigma (q) = q$ for all $q \in \mathbb{Q}$. Since this is an isomorphism, this means it is surjective and injective, and the only missing piece is $\alpha$, therefore we must have $\sigma (\alpha) = \alpha$. Concluding, we have that $\sigma (x) = x$ for all $x \in K$.

Even if my answer is wrong, do you think this should be awarded zero points at all?

Thank you for all help.

P.S.: There were more passages in my solution, such as $\sigma (0) = \sigma (0+0) = \sigma (0) + \sigma (0)$, thus $\sigma (0) = 0$, and similar ones for $\sigma (1) = 1$, $\sigma(n) = n$ and $\sigma (q) = q$. I don't think it is necessary to repeat them all here, specially since I've just written one of them here. If you find it necessary, though, I will. Thanks.