Affine space and common zero sets turn equations into geometric objects
Objective: Why does V(xy)=V(x)∪V(y) need the assumption that k is a field or at least has no zero divisors?
Algebraic geometry begins with a deliberately simple question: which coordinate points make all the named polynomial equations equal to zero? The answer is a set of points. It may look like a curve or surface over the real numbers, but the definition works over any field and does not require distance, angle, or a drawing. Before calculating, name the base field, ambient space, coordinate variables, point set, defining ideal, and every map being used. The exact result to understand is: If (S) denotes the ideal generated by S, then V(S)=V((S)). Moreover V(S∪T)=V(S)∩V(T), while V(ST)=V(S)∪V(T), where ST={fg:f∈S,g∈T}. Thus adding equations intersects solution sets, whereas multiplying equations takes a union.
The new object means: Let k be a field. Affine n-space over k, written A^n_k and read “affine n-space over k,” is the set k^n of ordered n-tuples a=(a_1,…,a_n). For S⊆k[x_1,…,x_n], define V(S)={a∈A^n_k:f(a)=0 for every f∈S}; V(S) is read “the common zero set of S.” An affine algebraic set is any subset of A^n_k equal to V(S) for some S. Read each displayed formula as a sentence about points, functions, or neighborhoods; do not manipulate an unnamed symbol merely because it resembles earlier algebra.
Rebuild the worked problem “In A^2_k with coordinates x and y, determine V(xy), prove it is the union of the two coordinate axes, and compare it with V(x,y).” from the definitions. Check both directions of every set equality, identify the corresponding ring map, and keep this failure boundary visible: A^n_k is a set of k-valued coordinate tuples, not Euclidean n-space unless k=R and one deliberately adds Euclidean structure. Over a finite field it has finitely many points, and over C a single equation can describe four real dimensions rather than an ordinary plotted surface.
If (S) denotes the ideal generated by S, then V(S)=V((S)). Moreover V(S∪T)=V(S)∩V(T), while V(ST)=V(S)∪V(T), where ST={fg:f∈S,g∈T}. Thus adding equations intersects solution sets, whereas multiplying equations takes a union.
Every element h of (S) is a finite sum h=Σr_if_i with f_i∈S. If a makes every f_i zero, then h(a)=Σr_i(a)f_i(a)=0; hence V(S)⊆V((S)). The reverse inclusion holds because S⊆(S).
A point lies in V(S∪T) exactly when it kills every polynomial in S and every polynomial in T, which says exactly that it lies in both V(S) and V(T).
If every product fg vanishes at a, then either all f∈S vanish at a or all g∈T vanish at a: otherwise choose f(a)≠0 and g(a)≠0, whose product is nonzero in the field k. This proves V(ST)=V(S)∪V(T).
In A^2_k with coordinates x and y, determine V(xy), prove it is the union of the two coordinate axes, and compare it with V(x,y).
- A point (a,b) lies in V(xy) exactly when the evaluated product ab equals 0 in k. Because a field has no nonzero zero divisors, ab=0 means a=0 or b=0.
- The condition a=0 describes V(x), the y-axis, and b=0 describes V(y), the x-axis. Therefore V(xy)=V(x)∪V(y).
- By contrast, V(x,y)=V(x)∩V(y) requires a=0 and b=0 simultaneously, so it is only the origin {(0,0)}.
Result: V(xy)=V(x)∪V(y) is the union of the two coordinate axes, whereas V(x,y)={(0,0)} is their intersection.