Restriction of an ideal sheaf to a hypersurface This is from Example 3.82 of Janos Kollar's Lectures on resolution of singularities. Let $I = (x^3+xy+y^2z^4) \subset k[x,y,z]$ (for some field $k$) be an ideal sheaf of $\mathbb{A}^3_k$ and $H = (x+u_1xy^3+u_2y^2+u_3y^2z^2 = 0)$ be a hypersurface in $\mathbb{A}^3_k$, where $u_1,u_2,u_3$ are units. I would like to compute $I|_{H}$.
The book says that by the equation of $H$, we can substitute $x$ by $$x = -y^2(u_2+u_3z^2)(1+u_1y^3)^{-1}$$ which then tells us that $I|_H \subset (y^3,y^2z^4)$.  
I'm confused by why we can take the inverse of $1+u_1y^3$ and how this substitution leads to the result. Thanks!
 A: First, note that $V(1+u_1y^3)$ does not contain $H$ - if we have a point $p=(x,y,z)$ so that $1+u_1y^3=0$, then $p\in H$ iff $u_2y^2+u_3y^2z^2=0$, and since $y\neq 0$, $u_2+u_3z^2=0$. So the intersection of $V(1+u_1y^3)$ with $H$ is at least codimension one in $H$. So the relation $x=-y^2(u_2+u_3z^2)(1+u_1y^3)^{-1}$ is valid on a dense open set $U$ of $H$.
On this dense open set, we can make the substitution $x=-y^2(u_2+u_3z^2)(1+u_1y^3)^{-1}$ and see that on $U$, $$x^3+xy+y^2z^4 = -y^6(u_2+u_3z^2)^3(1+u_1y^3)^{-3}  -y^3(u_2+u_3z^2)(1+u_1y^3)^{-1} + y^2z^4$$ or after factoring, $$x^3+xy+y^2z^4 = -y^3(y^3(u_2+u_3z^2)^3(1+u_1y^3)^{-3}  -(u_2+u_3z^2)(1+u_1y^3)^{-1}) + y^2z^4$$
and so on $U$, we have $I|_H\subset (y^3,y^2z^4)$.
Now we need to reason about the set $H\setminus U$. What we've demonstrated is that $V(y^3,y^2z^4)\cap U \subset V(I|_H)\cap U$ - if $V(y^3,y^2z^4)\cap (H\setminus U)$ is empty, then we're done. This is in fact the case - on closed points, this says that $y=0$, which contradicts $1+u_1y^3=0$.
