If $a+\sqrt{b}$ is a root of a polynomial equation with integer coefficients, so is $a-\sqrt{b}$ I tried to use the Briot-Ruffini method but it didn't work.
The question I need help is:
"Prove that, if a polynomial equation with integer coefficients has the irrational number $a+\sqrt{b}$ as a root, with $a,b \in \mathbb{Z} $, $b$ a prime number, so is $a-\sqrt{b}$."
 A: Hint: $a+\sqrt b$ is a root of $(x-(a+\sqrt b))(x - (a-\sqrt b)) = x^2 -2ax + (a^2-b)$.
(This is the minimal polynomial of $a + \sqrt b$ over $\mathbb Z$, and hence, it divides any other polynomial that has $a + \sqrt b$ as a root)
A: If $p(x)$ is a polynomial with integer coefficients, and $a$ is an integer, then $q(x)=p(a+x)$ is also a polynomial with integer coefficients.  If $a+\sqrt b$ is a root of $p(x)$, then $q(\sqrt b)=p(a+\sqrt b)=0$.  Now any polynomial with integer coefficients can be written in the form $r(x^2)+xs(x^2)$, where $r$ and $s$ also have integer coefficients.  If we write $q$ this way, then we have
$$0=q(\sqrt b)=r(b)+s(b)\sqrt b$$
which implies $r(b)=s(b)=0$ if $b$ is not a perfect square.  But this implies
$$p(a-\sqrt b)=q(-\sqrt b)=r(b)-s(b)\sqrt b=0$$
hence $a-\sqrt b$ is also a root of $p(x)$.
A: Hint $ $ The set $I$ of $\,f\in\Bbb Q[x]\,$ with root $\,w = a+\sqrt b\,$ is an ideal, i.e. is closed under subtraction & multiplication by any $\,g\in \Bbb Q[x].\,$ So by the Euclidean algorithm $\,I = (g)$ is principal, generated by any element of minimal degree (else $\,0\ne f\ {\rm mod}\ g = f - q\:\! g \in I\,$ and has degree smaller than $\,g).$
By $\,\sqrt b\not\in\Bbb Q,\,$ a min degree $\,g \in I\,$ has degree $\,2,\,$ e.g. $\,g = (x-w)(x-w'),$ $\, w' = a-\sqrt b = $ conjugate of $\,w.\,$  Hence $\ f(w) = 0\iff f\in (g)\iff g\mid f\iff w,w'\,$ are roots of $\,f.$
Remark $\ $ The same idea works generally to show that ideals in Euclidean domains (i.e. domains enjoying division with "smaller" remainder) are principal, generated by any element of minimal Euclidean size (above the size is the polynomial degree)
A: Conjugation $\mathbb{Q}(\sqrt{b}) \to \mathbb{Q}(\sqrt{b}):z= x+y\sqrt{b} \to \bar{z}=x-y\sqrt{b}$ is a field automorphism that fixes $\mathbb{Q}.$ So if $z\in \mathbb{Q}(\sqrt{b})$ is the root of a polynomial with rational coefficients i.e. $\exists \ a_i\in \mathbb{Q}$ such that  $$a_n z^n + a_{n-1}z^{n-1} + \ldots + a_1 z +a_0=0,$$
then conjugating both sides of the equation shows that $\bar{z}$ is a root of the same polynomial. 

A similar result is the Conjugate Root Theorem. More generally, if $K\subseteq L$ is a field extension and $\varphi:L\to L$ is an automorphism of $L$ that fixes $K,$ then $$ l \in L \text{ is a root of } p\in K[X] \implies \varphi(l) \text{ is a root of } p\in K[X]. $$
This is used frequently when you study Field/Galois theory. 
A: Another way, similar to Barry Cipra's method: 
Assume that $P$ is a polynomial with rational coefficients and let $a$ and $b$ be rationals such that $\sqrt{b}$ is irrational such that $P(a+\sqrt{b})=0$. 
By the binomial theorem, $P(a+\sqrt{b})+P(a-\sqrt{b})$ is rational. To see why, write $P(x)=\sum_{k=0}^nc_kx^k$, and expand $P(a \pm \sqrt{b})$; when you sum $P(a+\sqrt{b})$ and $P(a-\sqrt{b})$, the terms containing $\sqrt{b}$ cancel. 
By similar reasoning $P(a+\sqrt{b})-P(a-\sqrt{b})$ is some rational multiple of $\sqrt{b}$.
But $P(a+\sqrt{b})=0$, thus $P(a-\sqrt{b})$ is both rational and some rational multiple of $\sqrt{b}$. Hence $P(a-\sqrt{b})=0$.
