# 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