Let $X$ and $Y$ be projective varieties and $φ\colon X \to Y$ a regular birational map. If $q\in Y$ and $φ^{-1}(q)$ is finite and $φ(p) = q$, is it true that $\mathcal O_{X,p}$ is an integral extension of $φ^*(\mathcal O_{Y,q})$?
By $\mathcal O_{X,p}$ I mean the set of rational functions which are regular on $p$. I know that $\mathcal O_{X,p}$ is a local ring, with maximal ideal $m_p$, the set of rational functions regular and vanishing at $p$. $k(X)$ is the set of rational functions on $X$. $φ^*\colon k(Y) \to k(X)$ is the $k$-algebra homomorphism sending rational function $f\in k(Y)$ to $f \circ φ$. I know that $φ^*$ is an isomorphism, since $φ$ is birational. I also know that $φ$ is surjective.
What I have to show is that, if $f \in \mathcal O_{X,p}$, then there exist $n >0$ and $g_0,\dots,g_{n-1} \in \mathcal O_{Y,q}$ such that $$f^n+ g_{n-1}f^{n-1} +\cdots +g_0 = 0. $$ But I have no clue how to proceed.
I'm new to algebraic geometry, so please don't use too advanced arguments, as I won't be able to understand. Or rather, you can, but if you do, please give some reference.
Edit: I know that there is an open set $V⊆Y$ such that for all $q\in V$, $ φ^*(\mathcal O_{Y,q}) = \mathcal O_{X,p}$, but this is not what I want. That said, I'm not sure that the answer to my question is affirmative.
Edit: maybe we need the additional assumption that $Y$ is normal, in which case $φ^*(\mathcal O_{Y,q}) = \mathcal O_{X,p}$. Yet, I don't know how to prove this one either.