In modern geometry, given an equilateral triangle, one can't construct a square with the same area with the use of Hilbert tools. Why is this? The claim seems untrue to me, so there must be something wrong with my understanding.
First, given an equilateral triangle of side length $s$, the area of the triangle is $\frac{s^2\sqrt{3}}{4}$, so it seems to construct a square would require that $\frac{s\sqrt[4]{3}}{2}$ be a constructible number. Obviously $2$ is constructible, and so is $s$ since it is a given side. Isn't $\sqrt[4]{3}$ also constructible, since we have $ \mathbb{Q}\subseteq\mathbb{Q}(\sqrt{3})\subseteq\mathbb{Q}(\sqrt[4]{3}) $ so $\deg(\mathbb{Q}(\sqrt[4]{3})/\mathbb{Q})=4=2^2$? What am I missing?