proof verification $\frac{3+2\sqrt{6}}{1-\sqrt{6}}$ is an algebraic integer

Question

Is $$\frac{3+2\sqrt{6}}{1-\sqrt{6}}$$ an algebraic integer?

An algebraic integer means an algebraic number in some algebraic number field $K\supset \Bbb Q$ that is the root of a monic polynomial $f\in \Bbb Z[x]$. Here I guess we are in $\Bbb Q({\sqrt{6}})=\Bbb{Q}(\alpha)/(\alpha^2-6)$, and with that:

$$\beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies \beta= \frac{(3+2\sqrt{6})(1+\sqrt6)}{-5}\implies -5\beta =15+5\sqrt{6}\implies (-5\beta-15)^2=25(6)\implies 5^2\beta^2+2(15)(5)\beta+15^2-5^2(6)=0,$$ and thus $5^2x^2+6(5)^2x+5^2(3)\in\Bbb{Z}[x]$ is the desired polynomial.

Is this correct?

Answer 1

It is mostly correct, but there is a small problem at the end. You got that $\beta$ is a root of $5^2x^2+6\times5^2x+3\times5^2\in\mathbb{Z}[x]$ and you deduced that $\beta$ is an algebraic integer. How? An algebraic integer is a root of a monic polynomial with integer coefficients. That's easy, though: $\beta$ is also a root of $x^2+6x+3$.

Answer 2

$$ \beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies $$

$$\beta= \frac{(3+2\sqrt{6})(1+\sqrt 6)}{-5}\implies $$

$$-5\beta =15+5\sqrt{6}\implies $$

$$ \beta =-3-\sqrt 6$$

$$ \beta ^2 + 6\beta +3=0 $$

Answer 3

Your approach - finding the minimal polynomial and observing that it is a monic polynomial over $\Bbb Z$ - is correct, but the following is probably easier:

$$ \frac{3 + \sqrt6}{1 - \sqrt6} = \frac{3 + \sqrt6}{1 - \sqrt6} \frac{1 + \sqrt6}{1 + \sqrt6} = \frac{15 + 5 \sqrt6}{-5} = -3 + \sqrt6. $$ Since $\sqrt 6$ is an algebraic integerer, so is $-3 + \sqrt6$ and its minimal polynomial is $(x + 3)^2 - 6 = x^2 + 6x + 3$.