Let $R$ be a commutative ring with unity and let $P$ be prime ideal of $R$ consider the polymial ring $R[x]$ and let $P[x]$ be the ideal of $R[x]$ consisting of polynomials whose coefficient all belongs to $P$ show that the ideal $$P[x]+<x>$$ is prime ideal.
I know that $P[x] $ is also prime ideal. I don't know how to show that $P[x]+<x>$ is prime ideal.
Please give some hint.
Thank you.
On
Hint: take an element $q$ of $P[X] + \langle X \rangle$. This means that there exists a polynomial $p$ with coefficients in $P$ and a polynomial $r$ with coefficients in $R$ such that
$$ q = p + Xr. $$
In particular, all but the $0$-th degree term of $q$ belong to $R$ but not necessarily to $P$. Conversely, suppose that $q = a_nX^n + \dots + a_1X + a_0 \in R[X]$ is such that $a_0 \in P$. Then, we have
$$ q = a_0 + X(a_nX^{n-1} + \dots + a_1) \in P + \langle X \rangle \subset P[X] + \langle X \rangle. $$
This shows that $P[X] + \langle X \rangle = P \oplus \langle X \rangle$. On the other hand, $R[X]$ can be written as $\langle X \rangle \oplus R$.
Think about what these descriptions tell you about $R[X]/P[X] + \langle X \rangle$.
Concretely, we have that $R[X]/P[X] + \langle X \rangle = \frac{\langle X \rangle \oplus R}{\langle X \rangle \oplus P} \simeq \frac{\langle X \rangle}{\langle X \rangle} \oplus \frac{R}{P} \simeq 0 \oplus R/P \simeq R/P.$ Since $P$ is prime, the former quotient is a domain, and thus $P[X] + \langle X \rangle$ is prime.
Hint. An ideal $I ⊆ A$ of a ring $A$ is prime if and of $A/I$ is an integral domain. For $A = R[X]$ and $I = P[X] + (X)$ in your case, use that you have an isomorphism $$R[X]/P[X] \to (R/P)[X],$$ sending the residue class $[X]$ to $X$. (You might already know both this trick to show something is a prime ideal and this isomorphism from the proof for the fact that $P[X]$ is prime in $R[X]$.)