So basically the question is the one of the title of the post, but let me show you the context:
Let $\mathbb K$ be an algebraic closed field and let $f\in \mathbb K[X,Y]$ be a polynomial such that is monic over $Y$ and $(0,0)\in V_{\mathbb K}(f)$. You can also use that $V_{\mathbb K}(X)$ does not belong to the tangent cone of $V_{\mathbb K}(f)$ in $(0,0)$.
I already know that $f$ factorises in $\mathbb K\{\{X\}\}[Y]$ where $\mathbb K\{\{X\}\}$ is the Puiseux's series ring. As a consequence: $$ f=\prod_i(Y-q_i) $$ where $q_i\in\mathbb K\{\{X\}\}$. If we put toghether the conjugate roots we can write $$ f=f_1\cdot...\cdot f_t $$ where $f_i=\prod_{j=1}^{r_i}(Y-q_{ij})\in\mathbb K[[X]]$ and $q_{ij}\in\mathbb K[[X^{1/r_i}]]$ are the conjugate roots. Notice that $\mathbb K[[X]]$ is the formal power series ring. Here is the diagram with everything I know so far: $$ \begin{array}[c] a& & \mathbb K[[X^{1/r_i}]] &\cong &\mathbb K[[X]]^{r_i} \\ & & \qquad\cup \\ \frac{\mathbb K[[X]][Y]}{\left<f_i\right>}&\cong & \mathbb K[[X]][q_i]&\cong &\mathbb K[[X]]^{r_i} \\ & & \qquad\cup \\ & & \;\;\;\mathbb K[[X]] \end{array} $$
In the diagram $q_i$ is one of the roots $q_{ij}$. My question is:
Can we prove that $\mathbb K[[X^{1/r_i}]]=\mathbb K[[X]][q_i]$ under all the given assumptions?
Let me know if you need more information or if something I wrote is wrong. Thanks in advance!