Let z1 through zn be the images of x1 through xn in C. Note that xi-zi is a perfectly good polynomial, and an element of R. Apply f, and get f(xi)-f(zi), or 0. Thus the ideal H generated by the n elements xi-zi is contained in M.

Is H all of M? Mod out by H, and each generator turns into an element of C. The quotient ring is equal to C, H is maximal, and H = M. All maximal ideals have been characterized.

Since C is integral over K, the ring C[x1…xn] is integral over K[x1…xn]. Push H up to a maximal ideal in the smaller ring, then raise this up to a maximal ideal in the larger ring. (There are always primes over primes in an integral extension.) Thus we can embed H in a maximal ideal M in C[x1…xn]. This maximal ideal is generated by xi-zi, so that M vanishes on the point z1,z2,z3…zn in Cn. Since H vanishes on the same point, the algebraic set H′ is nonempty.

Note that the image space has to be Cn; the point z1,z2,…zn may not lie in Kn.

Again, C must be closed, as shown by x7-x mod 7.

Let H be a proper ideal in the polynomial ring, hence H′ is a nonempty region in Cn. What about H′′? It certainly includes H. In fact, H′′ = rad(H).

Assume pn evaluates to 0 for every z in H′. Rewrite pn(x) as p(x)n. Now p(z) raised to the nth is 0 in a field, hence p(z) = 0, and p is in H′′.

Conversely, let p lie in H′′, and assume p is nonzero. Adjoin the indeterminant y to produce the larger ring K[x1,x2,…xn,y]. Let H and y*p-1 generate the ideal J. What does J′ look like?

Suppose z1,z2,…zn,w is in J′. The vector z vanishes on h, and on h′′. Since z1 through zn satisfy p, w*p-1 can never equal 0. Therefore J′ is the empty set. Apply weak nullstellensatz, and J is the entire ring. In particular, J includes 1.

If each coefficient ci is a polynomial in the extended ring, write the following:

1 = c0*(yp-1) + c1h1 + c2h2 + c3h3 + … cnhn

The polynomials h1 h2 etc come from the set H, which is not necessarily an ideal in the extended ring, but it contributes to J.

Let a ring homomorphism map the extended ring into the fraction field K(x1,x2,…xn). Map xi to xi/1, and map y to 1/p. (This is why p has to be nonzero.) Apply this to our representation of 1. The first term on the right hand side drops to 0, since y*p-1 becomes 0.

Look at the image of a coefficient ci. After obtaining a common denominator, ci becomes a polynomial in x1 through xn, divided by a power of p. Multiply through by a sufficiently high power of p, and pn is generated by polynomials drawn from H. Thus p lies in rad(H), and H′′ = rad(H).

In any ring, the zariski topology equates every radical ideal with a closed set. In this sense, the two zariski topologies are the same.

for example, let x2-yz generate a quadratic surface in 3 space. This polynomial is irreducible in the ufd K[x,y,z], hence it is prime. It generates a prime ideal, which is a radical ideal. In the same way, the polynomial y generates a radical ideal that corresponds to a plane in 3 space. The intersection is a line, with x and y set to 0. This is demonstrated by an ideal H, generated by y and x2-yz. This ideal contains x2, but not x, hence it is not a radical ideal.

Union is a simpler operation. The ideal that vanishes on the union of two algebraic sets is the radical of the product of the two ideals. This is equal to the radical of the intersection. This in turn is equal to the intersection of the radicals. The union of two algebraic sets is realized by taking the intersection of the corresponding radical ideals.