Contraction of an ideal 
Let $f: \mathbb{Z}[X] \longrightarrow \mathbb{Z}[\sqrt{2}]$ be a ring homomorphism sending $X$ to $\sqrt{2}$.

I am asked to compute a few contractions, and I am wondering if I could get some help in understanding the reasoning.  I also was told that I need to compute the kernel of $f$, which I found to be $(X^2 - 2)$.  
I was also told that the kernel of $f$ is contained in every contraction.  Is this true or did I possible misunderstand?  If it is true, why is it so?
Here are a couple problems that I tried, and I am wondering if they are correct.
$i)$  $(0)^c = f^{-1}(0) = \ker(f) = (X^2 - 2)$
$ii)$ $(\sqrt{2})^c = f^{-1}(\sqrt{2}) = (X,X^2 - 2) = (X,2)$
(How does $(X, X^2 - 2)$ simplify to be $(X,2)$?)
$iii)$ $(2)^c = f^{-1}(2) = (X^2, X^2 - 2)$ (How does this ideal "simplify"?)
Thank you for your help!!!

Note that for any ring homomorphism $f \colon R \to S$, and any ideal $J \subseteq S$, we have - due to $0 \in J$ - 
$$ J^c = f^{-1}[J] \supseteq f^{-1}[0] = \ker f $$
so the kernel is contained in every contraction.
In the case here, we have $\ker f = (X^2 - 2)$, as you already found. For the three problems, we have
(i) $0^c = \ker f = (X^2 - 2)$ 
(ii) $(\sqrt 2)^c = f^{-1}[(\sqrt 2)] = (X,X^2-2) = (X,2)$

As $f(X) = \sqrt 2$, $f(X^2 -2) = 0$, we have $(X^2 - 2, X) \subseteq (\sqrt 2)^c$. 
Now let $p \in (\sqrt 2)^c$, that is $f(p) \in (\sqrt 2)$. Divide $p$ by $X^2 - 2$, giving 
  $$ p(X) = (X^2 - 2)q(X) + r(X) $$
  where $\deg r < 2$, dividing $r$ by $X$, we get 
  $$ p(X) = (X^2 - 2)q(X) + q_2(X)X + r_2(X) $$
  where $\deg r_2 < 1$. Under $f$ we have 
  $$ f(p) = 0 \cdot f(q) + f(q_2) \cdot \sqrt 2 + f(r_2) $$
  hence $f(r_2) \in (\sqrt 2)$. As $\sqrt 2$ is irrational and $r_2$ is constant, we must have $r_2 = 0$. That is 
  $$ p(X) = (X^2 - 2)q(X) + q_2(X)X  \in (X^2 - 2,  X).$$
  For the simplification:
  Note that $X \in (X^2 - 2, X)$ and $2 = X\cdot X - (X^2 -2) \in (X^2 - 2, X)$, hence $(X,2) \subseteq (X^2-2,X)$, and on the other hand $X^2 - 2 = X \cdot X + (-1) \cdot 2 \in (X,2)$, hence $(X^2 -2, X) \subseteq (X,2)$.

(iii) $(2)^c = f^{-1}[(2)] = (X^2, X^2 - 2) = (X^2, 2)$

Along the same line as above, we have: $(X^2, X^2 - 2) \subseteq (2)^c$. Now let $p \in (2)^c$, 
  divide $p$ by $X^2 - 2$, giving 
  $$ p(X) = (X^2 - 2)q(X) + r(X) $$
  where $\deg r < 2$. Under $f$ we have 
  $$ f(p) = f(r) $$
  hence $f(r) \in (2)$. Now $r(X) = aX + b$ for some $a,b \in \mathbf Z$, as $f(r) = a\sqrt 2 + b \in (2)$, both $a$ and $b$ are even, say $a = 2a_1$, $b = 2b_1$. That is 
  $$ p(X) = (X^2 - 2)q(X) + 2 a_1X + 2 b_1  
        = (X^2 - 2)\bigl(q(X) - a_1X - b_1\bigr)
               + X^2(a_1X + b_1X) \in (X^2 , X^2- 2). $$
For the simplification we get that $X^2 \in (X^2 - 2, X^2)$ and $2 = 1\cdot X^2 - (X^2 -2) \in (X^2 - 2, X^2)$, hence $(X^2,2) \subseteq (X^2-2,X^2)$, and on the other hand $X^2 - 2 = 1 \cdot X^2 + (-1) \cdot 2 \in (X^2,2)$, hence $(X^2 -2, X) \subseteq (X^2,2)$. 

