Question

Let f₁​=x₃−z₄,f₂=xy−z₂−z₃. Let J₁=⟨f₁​,f₂​⟩ and J₂​=⟨f³₁,f³₂⟩ be ideals of C[x,y,z]. Show that V(J₁)=V(J₂), but J₁​=J₂ (Hint: Show that f₁​∈/J₂ by considering the degrees of its terms.)

Answer 1

The varieties V(J₁) and V(J₂) are equal because points that zero f₁ and f₂ in J₁ also zero their cubes, hence zeroing f₁³ and f₂³ in J₂. However, J₁ and J₂ are not equal because f₁ cannot be composed from the cubed functions in J₂ due to the degree of the terms.

Step-by-step explanation:

To show that V(J₁) equals V(J₂) where J₁ is the ideal ⟨f₁, f₂⟩ and J₂ is the ideal ⟨f₁³, f₂³⟩ in C[x,y,z], we need to understand that the variety of an ideal, denoted V(J), is the set of all points that make the functions in the ideal zero. For any point in V(J₁), both f₁ and f₂ are zero, and thus their cubes f₁³ and f₂³ are also zero. Hence, any point in V(J₁) is in V(J₂), and vice versa, as zeroing the original functions results in zeroing their cubes, so V(J₁) = V(J₂).

However, J₁ is not equal to J₂ because the degree of a polynomials' cubed term is by cube law thrice the original, and since f₁ is a polynomial in J₁ but not J₂, considering the degrees of its terms as suggested by the hint validates that f₁ cannot result from the cubes in J₂. Thus, we conclude that J₁ ≠ J₂.