Minimal polynomial of $\sqrt{3+2\sqrt{3}}$

Question

Let $a=\sqrt{3+2\sqrt{3}}$. Then \begin{align*} &a=\sqrt{3+2\sqrt{3}}\\ &\implies a^2=3+2\sqrt{3}\\ &\implies a^2-3 = 2\sqrt{3}\\ &\implies (a^2-3)^2 = 4\cdot 3=12\\ &\implies (a^2-3)^2-12=0. \end{align*}

So, $a$ a root of $$(x^2-3)^2-12=x^4-6x^2-3.$$

By Eisenstein's criterion, $x^4-6x^2-3$ is irreducible over $\mathbb Q$. So, $a$ is algebraic of degree $4$.

---

However, I've seen that the degree of $\mathbb Q(\sqrt{3+2\sqrt{3}})$ over $\mathbb Q$ is $2$.

What is wrong with my attempt?

Answer 1

You are unquestionably right. A much more advanced viewpoint:

The ring $R=\Bbb Z[\sqrt3\,]$ is the integer-ring of the quadratic number field $K=\Bbb Q(\sqrt3\,)$. It’s “well known” (and easy to prove) that $R$ is a Principal Ideal Domain (class number is $1$).

We have the factorization $z=3+2\sqrt3=\sqrt3(2+\sqrt3\,)$, in which the two factors are respectively, an indecomposable element, and a generator of the (free part of the) unit group. In particular, neither is a square, and they are independent modulo squares. Therefore, the square root $\sqrt z$ generates a quadratic extension of $K$. That does it.

Answer 2

I think you are right.

Another way: $$\sqrt{3+2\sqrt3}=\sqrt[4]3\sqrt{2+\sqrt3}=\frac{\sqrt[4]3}{\sqrt2}\sqrt{4+2\sqrt3}=$$ $$=\frac{\sqrt[4]3}{\sqrt2}\sqrt{(1+\sqrt3)^2}=\frac{\sqrt[4]3}{\sqrt2}(1+\sqrt3),$$ which gives the same answer.