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 co-efficients all belong to $P.$ Show that the ideal $$P[x] + \langle x \rangle : = \left \{f(x) + x g(x)\ \big |\ f(x) \in P[x],g(x) \in R[x] \right \}$$ is a prime ideal of $R[x].$
My attempt $:$ What I first observe is that $$f(x) \in P[x] + \langle x \rangle \iff f(0) \in P.$$ Because if $f$ is a zero polynomial then we are through. Otherwise we can write $f(x) = \sum\limits_{k=1}^{n} a_k x^k,$ where $a_1,a_2, \cdots, a_n \in R$ and $a_n \neq 0.$ But then $$f(x) = a_0 + x \sum\limits_{k=1}^{n} a_k x^{k-1} = f(0) + x \sum\limits_{k=1}^{n} a_k x^{k-1}.$$ So $f(x) \in P[x] + \langle x \rangle \iff f(0) \in P[x] \iff f(0) \in P,$ which proves our claim.
Now let $A(x), B(x) \in R[x]$ such that $A(x) B(x) \in P[x] + \langle x \rangle.$ Then there exist $f(x) \in P[x]$ and $g(x) \in R[x]$ such that $A(x) B(x) = f(x) + x g(x).$ But then $A(0) B(0) = f(0).$ Since $f(x) \in P[x]$ we have $f(0) \in P.$ Hence $A(0) B(0) \in P.$ Since $P$ was a prime ideal of $R$ it follows that either $A(0) \in P$ or $B(0) \in P.$ But then by our previous observation we conclude that either $A(x) \in P[x] + \langle x \rangle$ or $B(x) \in P[x] + \langle x \rangle.$ This shows that $P[x] + \langle x \rangle$ is a prime ideal of $R[x].$
QED
Does my proof hold good? Please verify it.
Thanks in advance.