I am considering function $f(s,t,x,y,z)$ in $\mathbb{R}^5 = {[s,t,x,y,z]}$ satisfing
$f \equiv s \pmod{ (x-s,y-t,z-s-t)}$
$f \equiv s+2t \pmod{ (x-s,y+t,z-s-t)}$
$f \equiv -s \pmod{ (x-s,y+t,z+s+t)}$
$f \equiv -s \pmod{ (x+s,y-t,z-s-t)}$
$f \equiv s+2t \pmod{ (x+s,y-t,z+s+t)}$
$f \equiv s \pmod{ (x+s,y+t,z+s+t)}$.
I found such $f$ in fraction form $f= (z-y)x/s = -(z-x)y/t + (s+t) = (x-y)z/(s+t) + t$, but I cannot find $f$ in polynomial form. Does such polynomial exist? If not, why?
Geometrically, I am considering the ring $A=\mathbb{R}[s,t,x,y,z]/I$, where $I = (x^2-s^2,y^2-t^2,z^2-(s+t)^2, (s-x)(t-y)(s+t+z), (s+x)(t+y)(s+t-z))$ has vanishing set corresponding to the union of 6 planes defined by the 6 ideals described above.
Since $f$ on any two planes agree on their intersection, I expected $f$ can extend to whole function on $A$, but it seems I cannot. How should I understand it?
Why could I expect $f$ can extend?
I found counterexample in lower dimension.
There is no polynomial function in $k^2$ satisfying
$f(x,y)=y$ on $x=0$, $f(x,y)=0$ on $y=0$, and $f(x,y)=0$ on $y=x$
(because $f$ must divisible by $y(y-x)$)
If rational function is allowed, I have for example $y(y-x)/(y+x)$ (excluding the origin)