Why I and I(V(I)) differ: radicals
Consider I = (x^2) in k[x]. Then V(I) = {0}, the single point where x^2 = 0. But I({0}) = (x), strictly bigger than (x^2): the function x vanishes at the origin even though x itself was not in I. The obstruction is exactly powers: f vanishes on V(I) whenever some power f^m lies in I. The set of such f is the radical √I = { f : f^m ∈ I for some m ≥ 1 }.
An ideal is radical if √I = I. Two facts are free: every prime ideal is radical, and √I is always an ideal containing I. Geometrically V cannot see the difference between I and √I, since f and f^m have the same zero set. The natural question is whether radicals are the *only* obstruction.
The two forms of the Nullstellensatz
Hilbert's answer is yes — provided k is algebraically closed. This is the Nullstellensatz ('theorem of zero-loci'), and it comes in two faces.
Let k be ALGEBRAICALLY CLOSED, R = k[x_1, ..., x_n]. WEAK Nullstellensatz. If I is a proper ideal (I != R) then V(I) is NONEMPTY. Contrapositive: the only way to cut out the empty set is with the whole ring. Equivalently, every MAXIMAL ideal of R has the form m_p = (x_1 - a_1, ..., x_n - a_n) for a unique point p = (a_1, ..., a_n) in A^n. STRONG Nullstellensatz. For every ideal I: I( V(I) ) = sqrt(I). How STRONG follows from WEAK -- the 'Rabinowitsch trick': Suppose g vanishes on V(I), with I = (f_1,...,f_r). Add a new variable t and the polynomial 1 - t*g to the list in R[t]. These have NO common zero (where the f_i vanish, g vanishes, so 1 - t*g = 1 != 0). By WEAK, the ideal (f_1,...,f_r, 1 - t*g) = (1). Write 1 as a combo, then substitute t = 1/g and clear denominators: a power g^m lands in I. Hence g in sqrt(I). QED
The perfected dictionary
Now the correspondence is exact. Over an algebraically closed field, V and I are mutually inverse, order-reversing bijections between radical ideals of k[x_1,…,x_n] and affine varieties in A^n. Under it, prime ideals match irreducible varieties (guide 3) and maximal ideals match single points.