$(B/m)[x] = B[x]/M$? Assume $K$ is a field and $B$ is a subring of $K$ and $x \in K$. Let $m$ be a maximal ideal of $B$. Let $m^e$ denote the extension of $m$ in $B[x]$. Let $M$ be a maximal ideal in $B[x]$ containing $m^e$. 
Can someone explain to me why $(B/m)[\bar{x}] = B[x]/M$ where $\bar{x}$ is the image of $x$ in $B[x]/M$? Thank you.
 A: I'm afraid your question doesn't really make sense because $\mathfrak m^e$ might be equal to $B[x]$, and then of course no maximal ideal  $M\supset \mathfrak m^e $ exists. Here is an example of such failure:  
Let $K=\mathbb Q, B=\mathbb Z, x=\frac{1}{2}$ and $\mathfrak m=(2)\subset B=\mathbb Z$.
Then $\mathfrak m^e=(1)=B[ x]=\mathbb Z[\frac{1}{2}]$ because $2\cdot  \frac{1}{2} =1\in \mathfrak m^e$ .
Thus no maximal $M$ with  $\mathfrak m^e\subset M\subset B$ can exist.  
Edit
However if $M$ exists then $M\cap B=\mathfrak m$ (by maximality of $M$) and you have an injective  morphism $$B/\mathfrak m\to B[x]/M \quad (INJ)$$  An element of $B[x]/M$ is the class $\overline {\Sigma b_jx^j}$ of an element $\Sigma b_jx^j\in B[x]$  and thus can be written $ \Sigma \overline {b_j}\cdot \overline {x} ^j\in (B/\mathfrak m) [\bar x]$ as you wish, once you notice that $(INJ)$ allows you to see $B/\mathfrak m$ as a subring of $B[x]/M$.
 With this identification  you have indeed $ B[x]/M=(B/m)[\bar{x}]$
