Let $\pi:Y\to X$ be a dominant, finite morphism of nonsingular varieties over an algebraically closed field $\Bbbk$. Assume furthermore that for all $Q\in Y$, with $P=\pi(Q)$, we have 
$$\mathcal O\_{Y,Q}=\mathcal O\_{X,P}[T\_1,\ldots,T\_k]/(T\_1^n-x\_1,\ldots,T\_k^n-x\_k)$$
for certain $x\_i\in\mathcal O\_{X,P}$. In other words, we have adjoined certain $n$-th roots. 


Now, let $R\subseteq Y$ be the ramification divisor and $H=\pi(R)$ the branch locus (both now with the reduced subscheme structure). Let $\mathcal I(H)$ be the ideal sheaf of $H$ and $\mathcal I(H)\_P$ the stalk at $P$. Can I conclude that this ideal $\mathcal I(H)\_P\subseteq \mathcal O\_{X,P}$ is generated by the product $x\_1\cdots x\_k$? Or, correspondingly, can I conclude that $\mathcal I(R)\_Q$ is generated by the product $y\_1\cdots y\_k$?


**Edit**. As pointed out in the answer by Dmitry Vaintrob, I want to assume $n$ not divisible by $\mathrm{char}(\Bbbk)$. Furthermore, we assume that the $x\_i$ are reduced and coprime. 



Question about local description of the branch locus
There are two issues. You presumably mean that $x\_i$ are themselves reduced and have no common divisors, or else the answer would be trivially no: just take $x\_1=x\_2$ and then the reduced branching ideal is generated by $x\_1,$ not $x\_1x\_2.$ You also need to be careful what you mean in characteristic $p$, as then $T^p-x$ is everywhere ramified, no matter what $x$ is. Otherwise (in characteristic $0$ or relatively prime to $n$ and with $x\_i$ reduced and having no common divisors) I don't see any reason why this would be false, just by writing out equations for the ramification and branching loci: ramification divisors under fiber products satisfy $R\_{Y\_1\times\_X Y\_2\to X}=R\_{Y\_1\to X}\times\_X Y\_2\cup Y\_1\times R\_{Y\_2\to X}$, so it's enough to consider a single $T\_1$ and you have $T\_1^n-x\_1=0, nT\_1^{n-1}=0,$ which turns into $T\_1=0, T\_1^{n-1}=x\_1$ (and the projection of this to $X$ is again $x\_1=0$). 


Is this what you were asking or am I missing something? If this is what you need, I don't deserve the bounty as I didn't say anything deep.


 **Edit:** 
PS - The fact that your ring is a fiber product follows from the following general statement. Suppose $U=S[x\_1,\dots, x\_n]/(R\_1,\dots, R\_k)$ and $V = S[x\_1',\dots, x\_m']/(R\_1',\dots, R\_l')$ are commutative algebras over $S$ given by generators and relations, where $R\_i$ are relations on the $x$'s and $R\_j'$ are relations on the $y$'s. Then the tensor product can be written in generators and relations as follows.
$$U\otimes\_S V=S[x\_1,\dots, x\_n, x\_1',\dots, x\_m']/(R\_1,\dots, R\_k, R\_1',\dots, R\_l')$$ 
You can prove this in one line by using universal properties in the category of commutative $S$-algebras. 
Namely, a map of $S$-algebras from the tensor product $U\otimes\_S V$ to some $S$-algebra $W$ is equivalent to a pair of maps $f\_1:U\to W, f\_2:V\to W$ both over $S$, and a map from $S[x\_1,\dots, x\_n]/(R\_1,\dots, R\_k)$ to $W$ is equivalent to choosing a set of elements $w\_1,\dots, w\_n$ of $W$ satisfying relations $R\_1,\dots, R\_k$. Putting this together, $U\otimes\_S V$ and the ring defined by the two sets of generators and relations as above satisfy the exact same universal property.


