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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?

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

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