Showing $(2,x)$ is maximal in $\Bbb Z[x]$ I need to show $(2,x)$ is maximal in $\Bbb Z[x]$ by using a function that goes from $$\Bbb Z[x]\stackrel{\phi}{\longrightarrow} \Bbb Z_{2}$$ I defined a function and showed onto homomorphism and $$(2,x) \subset \ker\phi$$ but can't show the other direction to get $\ker\phi =(2,x)$.  
 A: Suppose that $P$ is a polynomial so that when evaluated at $0$ and reduced modulo $2$, we obtain zero. This simply means that $P(0)$ is even. But then certainly $P(X) = 2a + XQ(X)\in (2,X)$. 
A: Write it as a composite $\Bbb Z[x]\to\Bbb Z\to\Bbb Z/2$. The first map is $p(x)\mapsto p(0)$ and the second is $n\mapsto n\mod 2$. You know by the fourth isomorphism theorem that ideals of $\Bbb Z$ are exactly those from $\Bbb Z[x]$ containing the kernel of the map $\Bbb Z[x]\to\Bbb Z$, in particular $(2,x)\to (2)\subseteq\Bbb Z$. Then the composite kernel is exactly $(2,x)$ and the isomorphism theorem gives the equality of $\Bbb Z[x]/(2,x)$ and $\Bbb Z/2$.
A: So first you define $\phi:\Bbb{Z}[x]\to\Bbb{Z}$ by
$\qquad \qquad$$\phi(p(x))=p(0)\; \forall\;\; p(x)\in\Bbb{Z}[x]$ 
This is clearly a homomorphism.  
To show that this is also onto, let $n\in \Bbb{Z}$. We define  $p(x)=n$ for all $x$.
Then $\phi(p(x))=p(0)=n$. Thus $\phi$ is onto.  
Now \begin{align} 
\text{ker} \phi &=\{p(x)\in \Bbb{Z}[x]:\phi(p(x))=0\}\\ &=\{ p(x)\in \Bbb{Z}[x]:p(0)=0\}\\\end{align}
Now $p(0)=0\implies p(x) \;\text{is constant free}\implies \text{ker}\phi=\langle x \rangle$
So by First Isomorphism theorem, $\Bbb{Z}[x]/\langle x \rangle \cong\Bbb{Z} $. But $\Bbb{Z}$ is not a field.  
We now define $\psi:\Bbb{Z}\to\Bbb{Z}_2$ by
$\qquad \qquad$ $\psi(x)=[x]_2$ 
Again $\psi$ is an onto homomorphism.  
So finally consider $(\phi\circ\psi):\Bbb{Z}[x]\to\Bbb{Z}_2$. So  $(\phi\circ\psi)$ is onto homomorphism.
Also \begin{align} \text{ker}(\phi\circ\psi)&=\{p(x)\in\Bbb{Z}[x]: (\phi\circ\psi)(p(x))\in [0]_2\}\\ &=\langle 2,x \rangle\\ 
\end{align}
Therefore by First Isomorphism theorem, $\Bbb{Z}[x]/\langle 2,x \rangle\cong \Bbb{Z}_2$. Therefore $\langle 2,x \rangle$ is a maximal ideal.
