Singular Book 1.8.9 -- radical membership

Recall that an element $f$ is in an ideal $I$ if $1 \in (I, tf-1) \subset R[t]$.

i1 : A = QQ[x,y,z];

i2 : I = ideal"x5,xy3,y7,z3+xyz";

o2 : Ideal of A

i3 : f = x+y+z;

i4 : B = A[t];

i5 : J = substitute(I,B) + ideal(f*t-1)

             5     3   7           3
o5 = ideal (x , x*y , y , x*y*z + z , (x + y + z)t - 1)

o5 : Ideal of B

i6 : 1 % J 

o6 = 0

o6 : B

The polynomial f is in the radical. Let's compute the radical to make sure.

i7 : radical I

o7 = ideal (z, y, x)

o7 : Ideal of A