Prove that an ideal is not maximal 
Ring $\mathbb Z[x],$ ideal is $(x)$. How to prove that this is NOT a maximal ideal? 

I can't imagine ideal, part of which would be $(x)$.
 A: If it was maximal then $\mathbb{Z}[x] / (x)$ would be a field. So every non zero element has to have multiplicative inverse. $2 + (x)$ is not the zero element so there should be $f + (x)$ such that $(2 + (x))(f + (x)) = 1 + (x)$. So $2f - 1 \in (x)$, which means that if the constant term of $f$ is $a$, then $2a - 1 = 0 \Rightarrow a = \frac{1}{2}$. And this is a contradicton since $f \in \mathbb{Z}[x]$.
A: Consider the ideal $(x, 2)$ of $\Bbb Z[x]$. $(x, 2)$ is a proper ideal of $\Bbb Z[x]$ since $3 \notin (x, 2)$. Hence $(x) \subset (x, 2) \subset \Bbb Z[x]$, implying $(x)$ is not maximal.
A: If $R$ is a commutative ring, then the universal property of the polynomial ring is as follows:

for any ring homomorphism $f\colon R\to S$ and any element $s\in S$, there exists one and only one ring homomorphism $f_s\colon R[x]\to S$ such that $f_s(r)=f(r)$, for all $r\in R$, and $f_s(x)=s$.

Specialize this to $R=S$, $s=0$, $f$ the identity. Then
$$
\ker f_0=(x)
$$
because $f_0(x)=0$ by definition, so $(x)\subseteq\ker f_0$; if $p\in R[x]$, write $p(x)=a+xq(x)$, where $a\in R$. Then $f_0(p)=a$, so $p\in\ker f_0$ implies $a=0$ and so $p\in(x)$.
Thus $\ker f_0=(x)$ and so, by the homomorphism theorem,
$$
R[x]/(x)\cong R
$$
Therefore $(x)$ is maximal in $R[x]$ if and only if $R$ is a field.
