Help finding $p(i+\sqrt{2}) = 0$, $p(x) \in \mathbb{Q}$ I am doing a proof and I need to use that $\mathbb{Q}(i, \sqrt{2})=\mathbb{Q}(i+\sqrt{2})$. I know this because I have calculated the degree of the extension and it is $1$. Since $\mathbb{Q}(i, \sqrt{2})$ over $\mathbb{Q}$ is finite then $\mathbb{Q}(i+\sqrt{2})$ over $\mathbb{Q}$ is also finite. That means $i+\sqrt{2}$ is algebraic over $\mathbb{Q}$. However, I can't find any polynomial $p(x)\in \mathbb{Q}$ which meets the condition $p(i+\sqrt{2}) = 0$. Can someone give me an example of a polynomial in $\mathbb{Q}$ that meets this condition?
 A: We may set
$\alpha = i + \sqrt 2; \tag 1$
then
$\alpha^2 = i^2 + 2i\sqrt 2 + (\sqrt 2)^2 = -1 + 2 + 2i\sqrt 2; = 1 + 2i\sqrt 2, \tag 2$
whence
$\alpha^2 - 1 = 2i\sqrt 2, \tag 3$
or
$(\alpha^2 - 1)^2 = -8, \tag 4$
whence
$\alpha^4 - 2\alpha^2 + 1 = -8; \tag 5$
finally,
$\alpha^4 - 2\alpha^2 + 9 = 0. \tag 7$
Thus $i + \sqrt 2$ satisfies the quartic polynomial
$x^4 - 2x^2 + 9 \in \Bbb Z[x] \subset \Bbb Q[x]. \tag 8$
A: Try to compute $(x - (i + \sqrt 2))(x - (-i + \sqrt 2))(x - (i - \sqrt 2))(x - (-i - \sqrt 2))$. You should get a polynomial of degree $4$ which has all coefficients in $\Bbb Q$.
A: Here’s yet another method, which is what I often use:
The minimal polynomial for $i$ is (of course) $g(X)=X^2+1$. Then a polynomial for $i+\sqrt2$ will be $g(X-\sqrt2)=(X-\sqrt2\,)^2+1=X^2-2\sqrt2+3$, which of course is not a $\Bbb Q$-polynomial. But if you multiply it by its conjugate namely $\overline g(X)=X^2+2\sqrt2+3$, you’ll get the same final result as everybody else,
namely $g\overline g(X)=X^4-2X^2+9$ .
