Showing $(2,x)$ is maximal in $\Bbb Z[x]$

Question

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 $\ker\phi =(2,x)$.

Answer 1

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 $P(X) = 2a + XQ(X)\in (2,X)$.

Answer 2

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 $\Bbb Z/2$.