I am new in field theory and i began studying it recently.
I have a question about a finite field extension
The basis of the extension $\mathbb{Q}$$(\sqrt{1+\sqrt{3}})$/$\mathbb{Q}$ is the set $\{1,\sqrt{1+\sqrt{3}},\sqrt{1+\sqrt{3}}^2,\sqrt{1+\sqrt{3}}^3\}$ becase of the monic polynomial $(x^2-1)^2-3=0$ ?
Or am i missing something?
Thank you in advance.
On
Yes, the extension is biquadratic, i.e., has degree $4$, because of the arguments given in the answer to this duplicate, for $a=1$ and $b=3$. Indeed, the polynomial $(x^2-1)^2-3$ is irreducible, because neither $b=3$ is a square nor $a^2-b=-2$ is a square.
Since $\bigl(\sqrt{1+\sqrt{3}}\,\bigr)^2=1+\sqrt{3}$, we see that $\mathbb{Q}(\sqrt{3})\subseteq\mathbb{Q}(\sqrt{1+\sqrt{3}}\,)$.
Thus if the given extension is not of degree $4$, it has degree $2$. On the other hand, the decomposition of $(x^2-1)^2-3=x^4-2x^2-2$ over the reals is $$ (x-\sqrt{1+\sqrt{3}}\,)(x+\sqrt{1+\sqrt{3}}\,)(x^2-1+\sqrt{3}\,) $$ and the last factor has nonreal roots. Since the product of the two linear factors is $x^2-1-\sqrt{3}$, we conclude that the polynomial has no factor of degree $2$ with rational coefficients.
Hence $x^4-2x^2-2$, being irreducible, is indeed the minimal polynomial of $\sqrt{1+\sqrt{3}}$.