$z(1-x)f=xg+y(1-x)h$ in $k[x,y,z]$. Show $f\in (x,y(1-x))$

Question

Let $k$ be a field, $k[x,y,z]$ a polynomial ring.

If $z(1-x)f=xg+y(1-x)h$ in $k[x,y,z]$ where $f,g,h$ are polynomials in $k[x,y,z]$, show $f\in (x,y(1-x))$, the ideal generated by $x,y(1-x)$.

Attempt) Rewriting the equation,$(1-x)(zf-yh)=xg$. Since $(1-x)$ is an irreducible element in $k[x,y,z]$, $(1-x)$ divides $g$. Dividing $(1-x)$, we get $zf-yh=xg'$ for some polynomial $g'$. $zf=xg'+yh$.

I can't proceed from here. Any help would be appreciated.

Answer 1

Hint: Note that the ideal $(x,y(1-x))=(x,y)$ is prime, so $z(1-x)f\in (x,y(1-x))$ implies $$z\in (x,y(1-x))\qquad\text{ or }\qquad 1-x\in(x,y(1-x))\qquad\text{ or }\qquad f\in(x,y(1-x)).$$