How do we find the minimal polynomial of $\alpha = a + b \sqrt{d}$ over $\mathbb{Q}$? Let $\alpha = a + b \sqrt{d} \in \mathbb{Q} \left(\sqrt{d} \right) = \{a+b \sqrt{d}:a,b \in \mathbb{Q} \}.$
The minimal polynomial $m(x)$ of an algebraic number $\alpha \in \mathbb{C}$ is the monic polynomial of smallest degree, with coefficients in $\mathbb{Q}$ such that $m(\alpha) = 0.$ How do we find the minimal polynomial of $\alpha = a + b \sqrt{d}$ over $\mathbb{Q}$?
 A: $$\alpha-a=b\sqrt{d}$$
$$(\alpha-a)^2=b^2d$$
$$\alpha^2-2\alpha a+a^2=b^2d$$
$$f(x)=x^2-2x a+a^2-b^2d$$
A: Hint: If $\alpha \in K$ is a root of an irreducible polynomial $f \in F[x]$, then $\sigma(\alpha)$ is a root of $f$ for all field automorphisms $\sigma \in \text{Aut}(K/F)$.  Can you think of an automorphism $\sigma$ of $\mathbb{Q}(\sqrt{d})$ that fixes $\mathbb{Q}$?  What are the possibilities for $\sigma(\sqrt{d})$?
A: Another way to determine the minimal polynomial, which is nice in the quadratic case, is to compute the trace and norm of the element. 
$${\rm Tr}(a-b\sqrt{d})=(a-b\sqrt{d})+(a+b\sqrt{d})=2a$$
$${\rm N}(a-b\sqrt{d})=(a-b\sqrt{d})(a+b\sqrt{d})=a^2-b^2d$$
Then the coefficient of $x^{n-1}$, in our case $x$, is $-{\rm Tr}(a-b\sqrt{d})$ and the constant coefficient if $(-1)^{n}{\rm N}(a-b\sqrt{d})$.
(For reference, see Robert Ash's book on Algebraic Number Theory.)
- There are many other references in fact, but this is one that I have found helpful.
A: Let $x= a+ b\sqrt{d}$.  Then $x- a= b\sqrt{d}$ so $(x- a)^2= b^2 d$.  Those are now all integers and since the number has a square root, this is the minimal polynomial.
(Of course, that is the same as $x^2- 2ax+ a^2- b^2d$.)
