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 $[\mathbb{Q}(\sqrt{3 +2\sqrt{2}}):\Bbb Q]=4.$

So,where did I go wrong? Please help me find out my mistake.

Thank you.

  • 2
    $\begingroup$ 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. $\endgroup$ – André 3000 Jan 23 at 13:42
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    $\begingroup$ @André3000 Thanks a lot. That cleared everything. $\endgroup$ – Thomas Shelby Jan 23 at 13:55

Because $$3+2\sqrt2=(1+\sqrt2)^2$$ and $$x^4-6x^2+1=x^4-2x^2+1-4x^2=(x^2-2x-1)(x^2+2x-1).$$

  • $\begingroup$ 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"? $\endgroup$ – Thomas Shelby Jan 23 at 13:25
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    $\begingroup$ @Thomas Shelby I added something. See now. $\endgroup$ – Michael Rozenberg Jan 23 at 13:27
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    $\begingroup$ 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? $\endgroup$ – Thomas Shelby Jan 23 at 13:37
  • $\begingroup$ 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 $\mathbb Q\left(\sqrt{3+2\sqrt2}\right):\mathbb Q<4$. $\endgroup$ – Michael Rozenberg Jan 23 at 13:40
  • 1
    $\begingroup$ You are welcome, Thomas! $\endgroup$ – Michael Rozenberg Jan 23 at 13:55

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