Pythagorean closure

Im reading the book "Galois Theory" by Ian Stewart $$(4$$th Edition$$)$$. Here the author defines the Pythagorean closure as follows:

Definition:The Pythagorean closure $$\mathbb{Q}^{PY}$$ of $$\mathbb{Q}$$ is the smallest subfield $$K \subseteq \mathbb{C}$$ with the property $$z \in K \Rightarrow \pm\sqrt{z} \in K$$.

A few pages later he says without further explanation:

Suppose $$\alpha \in \mathbb{Q}^{PY}$$. Then by definition there is a tower: $$\mathbb{Q} =L_0 \subseteq L_1 \subseteq ... \subseteq L_n \supseteq \mathbb{Q}(\alpha)$$ such that $$[L_{j+1} : L_j] = 2$$ for all $$j$$.

I understand that each quadratic adjunction has degree $$2$$. But for me it is not obvious why the existence of such tower follows from the definition. How can it be constructed?

• The field consisting of the union of all such towers is closed under taking square roots. – Lord Shark the Unknown Feb 10 at 20:43
• What means the union of towers? Why is this a field? – Philipp Feb 10 at 22:11

1 Answer

Note: appending a square root to a field $$K$$, as in $$K(\sqrt{a})$$ for $$a \in K$$, makes a Galois extension of degree $$2$$.

Theorem The pythagorean closure $$\mathbb{Q}^{PY}$$ of $$\mathbb{Q}$$ is equal to the union of all subfields $$L \subset \mathbb{C}$$ for which there is a filtration $$L_0 \subset L_1 \subset \cdots \subset L_n = L$$ where each consecutive extension of degree $$2$$.

On the one hand, it must include these extensions, being closed under square roots and field operations. On the other hand, this construction forms a field. For two $$L_0 \subset L_1 \subset \cdots \subset L_n = L$$ and $$M_0 \subset M_1 \subset \cdots \subset M_n = M$$, we can make $$L_0 \subset L_1 \subset \cdots \subset L_n \subset L_n M_0 \subset L_n M_1 \subset \cdots \subset L_n M_n$$, and each will have degree $$1$$ or $$2$$.

Now an element $$\alpha$$ in the pythagorean closure is going to be contained in one of these $$L$$ such that there is a filtration $$L_0 \subset L_1 \subset \cdots \subset L_n = L$$. Then $$\mathbb{Q}(\alpha)$$ is also contained in $$L$$.