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Question: Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$

Thoughts: the "standard" method of starting by squaring (twice) to get rid of the square roots, because I don't have a nice way of showing the resulting polynomial is irreducible. So..

Attempt: It would be great if I could get our $\alpha$ in the form $$(a+b)^2=a^2+2ab+b^2=\sqrt{3+2\sqrt{2}}.$$ So, $$3+2\sqrt{2}=(\sqrt{2})^2+2\sqrt{2}+1=(\sqrt{2}+1)^2.$$So, $$\alpha=\sqrt{3+2\sqrt{2}}=\sqrt{2}+1.$$So, $$(\alpha-1)^2=2\\ \alpha^2-2\alpha+1=2\\ \alpha^2-2\alpha-1=0.$$So let $f(x)=x^2-2x-1$. Since $f(x)$ is irreducible over $\mathbb{Q}$ by the Rational Roots Test (since it has degree $2$), $f(x)$ is monic, and $f(\alpha)=0$, we conclude that $f(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$.

Does this look okay?

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    $\begingroup$ What you have done is good. If you were to try squaring twice to remove the square roots you would get a quartic which would be reducible and would have the minimal polynomial as an irreducible factor. $\endgroup$
    – Ben
    Jul 16, 2021 at 22:10
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    $\begingroup$ @Ben Yeah, I was stuck there for a bit when good ol' Eisenstein's didn't work. $\endgroup$
    – User7238
    Jul 16, 2021 at 22:12

2 Answers 2

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OP's proof is good, and there is not much room left to simplify. The following are just alternatives.

  • Let $\beta = \sqrt{3 - 2\sqrt{2}}$ and note that $\alpha\beta=1$ and $\alpha^2+\beta^2=6$. Given that $0 \lt \beta \lt \alpha$ it follows that $\alpha-\beta=\sqrt{(\alpha-\beta)^2}=\sqrt{\alpha^2+\beta^2-2\alpha\beta}=\sqrt{6-2}=2$.

    Then $\alpha(-\beta)=-1$ and $\alpha+(-\beta)=2$, so $\alpha, -\beta$ are the roots of $x^2 - 2 x - 1\,$.

  • The "standard" method can also work. Suppose "starting by squaring (twice) to get rid of the square roots", then $x^4 -6x^2+1=0=(x^2-1)^2-4x^2=(x^2-2x-1)(x^2+2x-1)$. Only the first quadratic factor has a root larger than $2$, which must be $\alpha$.

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  • $\begingroup$ Why would having a "root larger than $2$" justify that it MUST be $\alpha$. I intuitively understand that, it is just one of those things that as soon as I think it is true, I can see myself finding some trivial counterexample. $\endgroup$
    – User7238
    Jul 16, 2021 at 23:36
  • $\begingroup$ @User7238 Because $\alpha$ must be the root of one of the factors, and we know $\alpha \gt 2$ since $\alpha^2 = 3+2\sqrt{2}\gt 4$. (Actually, $\alpha \gt 1$ would suffice, since the other factor has both roots smaller than $1$.) $\endgroup$
    – dxiv
    Jul 16, 2021 at 23:38
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Since the comment has answered the OP's stated question, it is open season.

Alternative approach

If $a$ is rational and $b$ is irrational, then $(a+b)$ is irrational.

Further, if $a$ is irrational, then $\sqrt{a}$ is also irrational.

The proof to the 2nd assertion above is that if $\sqrt{a}$ can be rationally expressed as $\frac{P}{Q}$, this implies that $a$ has the rational expression $\frac{P^2}{Q^2}$, which contradicts the premise that $a$ is irrational.

Using the two assertions, you have that since $\sqrt{2}$ is known to be irrational, so is $\left[3 + 2\sqrt{2}\right]$, and therefore, so is $\sqrt{3 + 2\sqrt{2}}.$

This implies that there can not be any polynomial equation of degree 1, with form $x + b = 0,$ with $(b)$ rational, whose root is $\sqrt{3 + 2\sqrt{2}}.$

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