Find the generator of the kernel of an evaluation homomorphism 
$\phi:\mathbb{R}[x] \to \mathbb{R}$ defined by $f(x)\to f(1+\sqrt {2}$)

I noticed that in the solution, we say that $1+\sqrt{2}$ is a root for the polynomials we consider. Can someone explain why that's the case and show me the steps for this problem briefly?
 A: Maybe you could try to solve some "easier" problems before the one you were given, for instance:
1)
What is the kernel of
$$\phi_1 : \Bbb R[X] \to \Bbb R, \qquad f \mapsto f(1)$$
i.e. what is the set of all the polynomials $f \in \Bbb R[X]$ such that $f(1)=0$ ? Try first to find an example of such a polynomial. For instance... 

$f(X)=X-1$.

Can you find another one?

 $f(X)=2X-2$, or $f(X)=(X^{2017}+\pi)(X-1)$.

Now you should see that $(X-1) \subset \ker(\phi_1)$. Conversely, if $f(1)=0$, can you say that $f$ is a multiple of $X-1$ in $\Bbb R[X]$? I let you think about it.

2)
What is the kernel of
$$\phi_2 : \Bbb R[X] \to \Bbb R, \qquad f \mapsto f(\sqrt 2)$$
As before, first try to give me an example of polynomial $f \in \Bbb R[X]$ such that $f(\sqrt 2)=0$, except $f=0$ of course ;-)

 $f(X)=X^2-2$.

You can probably see that $(X^2-2) \subseteq \ker(\phi_2)$. Conversely, if $f(\sqrt 2)=0$, then try to perform an euclidean divison by $X^2-2$:
$$f(X)=(X^2-2)q(X)+r(X), \qquad \deg(r)<\deg(X^2-2)=2$$
Now evaluate at $x=\sqrt 2$ in order to get
$$f(\sqrt 2)=0 = 0\cdot q(\sqrt2)+r(\sqrt2)=a+b\sqrt 2,$$
where $r(X)=a+bX$ has degree $1$. It must follow that $a=b=0$ since $\sqrt 2$ is irrational (it implies that $1,\sqrt 2$ are linearly independent over $\Bbb Q$).
You just got $r(X)=0$, which yields $f(X)=(X^2-2)q(X)$, and that's it! You proved that $f(X) \in (X^2-2)$, and therefore $\ker(\phi_2) = (X^2-2)$.

What about $1+\sqrt 2$ now? There are two ways:


*

*Assume that you computed the kernel of $f \mapsto f(a)$, and you proved that it is equal to $(P(X))$ (true since $\Bbb R[X]$ is Euclidean, hence a PID). Then the kernel of $f \mapsto f(1+a)$ is simply $(P(X-1))$. Here, in part 2) above, we found what is the kernel of $f \mapsto f(\sqrt 2)$, so you immediately get the kernel of $f \mapsto f(1 +\sqrt 2)$.

*Another way is to "imitate" what we did in 1) and 2). First find a non-zero polynomial $f$ such that $f(1+\sqrt 2)$, and with minimal degree. It was $X-1$ for $a=1$ in 1) and it was $X^2-2$ for $a=\sqrt 2$ in 2). I let you find it for $a=1+\sqrt 2$. Then apply some Euclidean division as we did, and you'll find the kernel of $\phi$.
Here is the answer:

 $$\ker(\phi) = \{g(X)(X^2-2X-1) \mid g \in \Bbb R[X]\} = ((X-1)^2-2) = (X^2-2X-1),$$ so a generator is $X^2-2X-1$ (but $3X^2-6X-3$ is also a generator).

