Show that $\mathbb{Q}(\sqrt{5})$ is real closed

Question

Problem: Show that $\mathbb{Q}(\sqrt{5})$ is real closed.

There is a useful criterion to solve this problem that is Artin-Schreier criterion. How to apply this to this problem?

Answer 1

$\mathbb{Q}(\sqrt{5})$ is formally-real but not real-closed. Recall:

Formally-real: $-1$ is not a sum of squares.

Real-closed: formally real and does not admit any algebraic, real extension.

So $\mathbb{Q}(\sqrt{5})$ is not real-closed because it has formally-real extensions such as $\mathbb{Q}(\sqrt[4]5)$ or $\mathbb{Q}(\sqrt2,\sqrt5)$.

To see that it is formally-real you can do one of two things:

1. Show that $\mathbb{Q}(\sqrt5)$ embeds in $\mathbb{R}$ (which you know is formally-real). There are two embeddings depending on whether you want $\sqrt{5}$ to be the positive or negative root in $\mathbb{R}$.

2. Prove that $\mathbb{Q}(\sqrt{5})$ is formally-real from first principles.

For example, we can try to define an ordering on $\mathbb{Q}(\sqrt{5})$. Let's say we want $\sqrt{5} > 0$ and let's say $a + b\sqrt{5} > 0$ if $a > -b\sqrt{5}$ then—depending on the signs of $a$ and $b$—we check that either $a^2 > 5b^2$ or $a^2 < 5b^2$.

Or, we can assume that $-1 = \sum (a_i + b_i\sqrt{5})^2$ and try to reach a contradiction. Possibly we might need to use the above ordering to reach this contradiction.

Answer 2

This is very far from true. The algebraic closure of a real closed field is a finite extension.