Over an algebraically closed field k, which ideals of k[x, y] are maximal?
AExactly the ideals (x - a, y - b) for a, b in k
BExactly the principal ideals (f) with f irreducible
CExactly the prime ideals
DOnly the ideal (x, y)
This is the weak Nullstellensatz: maximal ideals correspond to points of k². An irreducible f generates a prime ideal that is not maximal, since (f) is contained in (x - a, y - b) for any point on the curve f = 0. The zero ideal is prime but not maximal.
Question 2 True / False
In k[x] with k algebraically closed, let J = (x²). The polynomial x vanishes at every point of V(J) but does not belong to J.
TTrue
FFalse
Answer: True
V(J) = {0}, and x vanishes there. But x is not a multiple of x², so x ∉ J. Its square x² is in J, which matches the strong Nullstellensatz: I(V(J)) = √J = (x).
Question 3 Short Answer
Explain why the Nullstellensatz fails over ℝ using the ideal (x² + y² + 1) in ℝ[x, y].
Think about your answer, then reveal below.
Model answer: x² + y² + 1 has no real zeros, so its zero set in ℝ² is empty, yet the ideal is proper, not the unit ideal.
Over an algebraically closed field, an empty zero set forces the ideal to contain 1. Over ℝ, a proper ideal can have no zeros, so the link between ideals and zero sets breaks down.
Question 4 Short Answer
What is the role of the extra variable in the Rabinowitsch trick?
Think about your answer, then reveal below.
Model answer: If f vanishes on V(J), then J together with 1 - yf has no common zero in kⁿ⁺¹, so by the weak form 1 lies in J + (1 - yf); substituting y = 1/f and clearing denominators shows some power of f lies in J.
The new variable converts 'f vanishes wherever J does' into 'a system has no solutions,' which is what the weak Nullstellensatz handles. Clearing the denominators 1/fᵐ is what produces the power of f.