Is the inverse of $(x+y, zw+z+w, y^2)\subset Q[x,y,z,w]$ under inclusion $g: Z[x,y,z,w]\to Q[x,y,zw]$ in $Z[x,y,z,w]$ still be $(x+y, zw+z+w, y^2)$? In general, suppose
$g: Z[x,y,z,w] \to Q[x,y,z,w]$
is the canonical inclusion, and $I=(f_1,...,f_n)\subset Q[x,y,z,w]$ be an ideal such that for each $f_i$, its coefficients are 1, and for each two generators $f_i, f_j$, their terms are different with each other.
My questions is whether the inverse image of $I$ under the inclusion $g$, for which is the ideal of $Z[x,y,z,w]$, i.e. $Z[x,y,z,w]\cap I$, still has the form $I=(f_1,...,f_n)$?
If the coefficients are not 1 and two polynomials have common terms, there has counterexample, for example
$I=(x-y, x+y)\subset Q[x,y,z,w]$, it is clear that $x\in I$, however, $x$ is not in$Z[x,y,z,w]\cap I=(x-y, x+y)\subset Z[x,y,z,w]$.
Especially I am interested in the following two ideals:
$I_1:=(x+y,zw+z+w,y^2)$
$I_2:=(x+y+z+w+xy+xz+xw+yz+yw+zw, xyz+xyw+yzw+xzw+xyzw)$
I want to ask if the inverse image of these two ideals in $Z[x,y,z,w]$ still keep the form?
 A: Let $I=(f_1,\dots,f_n)$ be an ideal of $\mathbb{Z}[x_1,\dots,x_n]$ and $J$ the corresponding ideal in $\mathbb{Q}[x_1,\dots,x_n]$.  In other words, $J=I\otimes_{\mathbb{Z}}\mathbb{Q}$.
Let $G$ be a (strong and reduced) Groebner basis for $I$.  Note that $G$ is a Groebner basis for $J$ as well: Let $g\in J$, then $g=\sum h_i f_i$ where $h_i\in\mathbb{Q}[x_1,\dots,x_n]$.  For a large enough $N\in\mathbb{Z}$, $Ng = \sum (Nh_i)f_i$ is in $I$ since $Nh_i\in \mathbb{Z}[x_1,\dots,x_n]$.  Thus, the leading monomials of both $I$ and $J$ are the same.
If any $f\in G$ has nontrivial content (that is, the $\gcd$ of the leading coefficients), then $J\cap\mathbb{Z}[x_1,\dots,x_n]\not=I$, by dividing through by the content.  On the other hand, if all the leading coefficients of elements of $G$ are $1$'s, then $J\cap\mathbb{Z}[x_1,\dots,x_n]=I$.  In particular, the division algorithm will never produce a fractional coefficient and the algorithm will be identical in both $I$ and $J$.
In your examples, graded-reverse-lex Groebner bases (as computed by Macaulay2) have Groebner bases with every leading coefficient $1$, so the intersection is an equality.
Note: In the case where the Groebner basis elements have leading coefficient not equal to $1$, but are content free, the situation is more complicated: $(2x+y)$ has the desired property, but $(2x+y,3y+2z)$ does not have the desired property.  There are content-free Groebner bases for both ideals, but the leading coefficients are not $1$ in this case.
