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

Question: Find the minimal polynomial of $$alpha=sqrt{3+2sqrt{2}}$$ 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 $$(a+b)^2=a^2+2ab+b^2=sqrt{3+2sqrt{2}}$$. So, $$3-2sqrt{2}=(sqrt{2})^2+2sqrt{2}+1=(sqrt{2}+1)^2$$. So, $$alpha=sqrt{3+2sqrt{2}}=sqrt{2}+1$$.
So, $$(alpha-1)^2=2 implies alpha^2-2alpha+1=2impliesalpha^2-2alpha-1=0$$. 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?