Let $\varphi:(A,\mathfrak{m})\to(B,\mathfrak{n})$ be a local morphism of regular local $\mathbb{C}$-algebras (**of the same dimension**) which makes $B$ integral over $A$. 


Let $\mathfrak{m}=(x\_1,\ldots,x\_d)$ be a (*fixed*) regular system of parameters of $A$ and assume that there exist principal prime ideals $\mathfrak{q}\_i\subset B$ such that $\mathfrak{q}\_i\cap A=(x\_i)$. By going up, $\sum\_i\mathfrak{q}\_i=\mathfrak{n}$. Is it possible to choose a regular system of parameters $\mathfrak{n}=(y\_1,\ldots,y\_d)$ for $B$ such that $\varphi(x\_i)=y\_i^{n\_i}$ for certain integers $n\_i$ and such that $y\_i$ generats $\mathfrak{q}\_i$? If not, can I add any conditions that will give me such a result?



Local coordinate system under finite integral extension
Are you assuming that the $x\_i$ are fixed? In that case, if one of the $x\_i$ happens to be sent to something such that $B/x\_i$ is reduced but singular, this seems to be a problem. 


For example, consider $\mathbb{C}[[s,t]] \to \mathbb{C}[[x,y]]$ where $s$ is sent to $x^2 - y^3$ and $t$ is sent to $y$. 


The only choice for the $n\_i$ are to be $1$. But $B / (x^2 - y^3)$ isn't regular, so $(x^2 - y^3, y)$ can't generate the maximal ideal (as is obvious anyways). 


In this case, if you are allowed to change $x\_i$, then things are better, but I'm not sure if you want that or not.


