What is Centre of Polynomial ring? Recently this question come to me, what it the centre of a polynomial ring $R[x]$ where $R$ is a ring ( we do not know if its commutative or even until) . intuitionally I can see $Z(R[x])=Z(R)[x]$. so is I am right? 
My attempt:-
Since $p(x)q(x)=\sum_{i=0}^{n} a_i x^i \sum_{j=0}^{m} b_j x^j= \sum_{i=0}^{n+m}c_i x^i$. where $c_i$ is a product of some coefficient of in $p(x),q(x)$. so if those coefficient commute ( that is they are in $Z(R)$ since we take them arbitrary) later we can recover $q(x)$ and $p(x)$ after we commute the cofficent. 
 A: You are correct in your intuition.
As you noticed: if $f\left(x\right)=a_{0}+\cdots+a_{n}x^{n}\in Z\left(R\right)\left[x\right]$
and $g\left(x\right)=b_{0}+\cdots+b_{m}x^{m}\in Z\left(R\right)\left[x\right]$
then it is straightforward to verify $f\left(x\right)g\left(x\right)=g\left(x\right)f\left(x\right)$.
If conversely $h\left(x\right)=c_{0}+\cdots+c_{k}x^{k}\in Z\left(R\left[x\right]\right)$
then for each $r\in R$ we have: $$rc_{0}+\cdots+rc_{k}x^{k}=rh\left(x\right)=h\left(x\right)r=c_{0}r+\cdots+c_{k}rx^{k}$$
The multiplications after first and second $=$ sign takes place in the polynomial ring where $r$ stands for a constant polynomial.
This justifies the conclusion that $c_{i}\in Z\left(R\right)$ for each $i$.
A: Evidently that $Z(R)[x]\subset Z(R[x])$. Let us show the reverse inclusion.
Assume the contraty. 
Let $A(x) = \sum\limits_{j\geq 0}a_jx^j\in Z(R[x])\setminus Z(R)[x]$. Then there exists $j_0\geq 0$ and $b\in R\subset R[x]$ such that $a_{j_0}b\neq ba_{j_0}$. 
So, we obtain
$$
A(x)b \neq bA(x).
$$
This is contradiction with $A(x)\in Z(R[x])$.
