Degree of the extension $\mathbb{Q}(\sqrt{3 + 2\sqrt{2}})$.

The following question is from Abstract Algebra by Dummit and Foote.

Determine the degree of the extension $$\mathbb{Q}(\sqrt{3 + 2\sqrt{2}})$$ over $$\Bbb Q$$.

This question has been answered here and the correct answer is $$2$$. But I got the answer as $$4$$ and couldn't find my mistake.

My attempt : I started out by constructing a polynomial $$f (x)\in \Bbb{Q}[x]$$ such that $$\sqrt{3 + 2\sqrt{2}}$$ is a root of $$f(x)$$.

$$x=\sqrt{3 +2\sqrt{2}}\;\implies x^2=3 +2\sqrt{2}.$$ Taking $$3$$ to the LHS and squaring gives$$x^4-6x^2+9=8.$$ Hence $$\sqrt{3 +2\sqrt{2}}$$ is a root of $$f (x)=x^4-6x^2+1.$$

By Rational Root Theorem, the only possible rational roots are $$\pm1$$ and neither of these satisfy $$f(x)=0$$. Hence $$f(x)$$ is a monic irreducible polynomial which has $$\sqrt{3 +2\sqrt{2}}$$ as a root. By the uniqueness property of minimal polynomial, $$f(x)$$ is the minimal polynomial for $$\sqrt{3 +2\sqrt{2}}$$. Since the degree of the extension is same as the degree of minimal polynomial, we have $$[\mathbb{Q}(\sqrt{3 +2\sqrt{2}}):\Bbb Q]=4.$$

Thank you.

• Just because a quartic has no roots doesn't mean it's irreducible. Consider $(x^2 + 1)(x^2 + 3)$ over $\mathbb{Q}$, for instance. – André 3000 Jan 23 at 13:42
• @André3000 Thanks a lot. That cleared everything. – Thomas Shelby Jan 23 at 13:55

Because $$3+2\sqrt2=(1+\sqrt2)^2$$ and $$x^4-6x^2+1=x^4-2x^2+1-4x^2=(x^2-2x-1)(x^2+2x-1).$$
• I have alreday seen this one and I understand the proof. But I don't understand how I get $4$ as answer. Can you go through my "proof"? – Thomas Shelby Jan 23 at 13:25
• Okay. So lack of rational roots only means that they cannot be factored into linear factors in $\Bbb Q$. The polynomial can still be factored as a product other irreducible polynomials. Am I right? – Thomas Shelby Jan 23 at 13:37
• The polynomial, where $\sqrt{3+2\sqrt2}$ is a root must be irreducible above $\mathbb Q$. The polynomial $x^4-6x^2+1$ is not irreducible avove $\mathbb Q$, which says $\mathbb Q\left(\sqrt{3+2\sqrt2}\right):\mathbb Q<4$. – Michael Rozenberg Jan 23 at 13:40