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

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?

• 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.
– Ben
Jul 16, 2021 at 22:10
• @Ben Yeah, I was stuck there for a bit when good ol' Eisenstein's didn't work. Jul 16, 2021 at 22:12

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 $$\alpha$$.

• 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. Jul 16, 2021 at 23:36
• @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 $1$.)
– dxiv
Jul 16, 2021 at 23:38

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 $$\sqrt{3 + 2\sqrt{2}}.$$