Proving that $ P[x] + \langle x\rangle$ is a prime ideal of $R[x]$

Question

This question is from a previous year exam of masters and I was unable to prove it.

Let $R$ be a commutative ring with $1$ and $P$ be a prime ideal of $R$. Consider the polynomial ring $R[x]$ and let $P[x]$ be the ideal of $R[x]$ consisting of polynomials whose coefficients all belong to $P$. Show that the ideal $$P[x] + \langle x\rangle=\left\{ f(x) + xg(x) : f(x) \in P[x] , g(x) \in R[x]\right\}$$ is a prime ideal of $R[x]$.

Let $A(x)$, $B(x) \in R[x]$ lies in $P[x] +\langle x\rangle$, then $A(x) B(x) = f(x) + x g(x)$, $f(x) \in P[x] $, $g(x) \in R[x]$.

Now, I can get that $A(x) B(x) - g(x)$ belongs to $P[x]$ which is a prime ideal but unfortunately this will not deduce it.

Background: I have done a masters level course on algebra.

Can you please tell what argument should I use?

Answer 1

So let $A(x)B(x)=f(x)+xg(x)$ and $f(x)=f_0+xf_1(x)$. Then $A(x)B(x)=f_0+x(g(x)+f_1(x))$, where $f_0\in P$. Let $A(x)=a+xA_1(x)$ and $B(x)=b+xB_1(x)$. Then $ab=f_0$. Since $P$ is a prime ideal, then $a\in P$ or $b\in P$. If say $a\in P$, then $A(x)\in P[x]+\langle x\rangle$.