Algebraic Geometry, Maximal Ideals of Integer Polynomials

Maximal Ideals of Integer Polynomials

Let M be a maximal ideal in Z[x1,x2,…xn]. Contract M into the subring Z to get a prime ideal P. If P is nonzero, it consists of the multiples of some prime p, and we can proceed with our proof. If P = 0, then M is not a maximal ideal, but that's not easy to prove. Perhaps m contains 1+px, and the introduction of p would span 1.

Suppose M is maximal in the ring of n indeterminants, yet M contracts back to 0 in Z. Let M1, M2, M3, … be the contraction of M into the subrings spanned by one indeterminant, then two indeterminants, then three, and so on. Thus Mn = M, and M0 = 0. Let Q0 be any prime ideal in Z, such as the multiples of p. Then let Q1 be a prime ideal in Z[x1], lying over Q0, and containing M1. Why can we do this? It is the going up property in a rather complicated theorem about chains of prime ideals in polynomial rings over a ufd. If you accept this result, then do it again, creating Q2 lying over Q1 and containing M2. This continues all the way up to Mn, and creates a larger ideal,which is a contradiction. Therefore the Maximal ideal M contains the multiples of some prime p in Z.

Mod out by the ideal generated by p, and M becomes a maximal ideal in Zp[x1,x2,…xn]. Let's try to generate the maximal ideal of the polynomial ring over Zp. We can always include p to produce the maximal ideal in the integer polynomials.

I'll do this for x and y, you can carry on from there. The contraction of M back into Zp[x] yields a prime ideal, and Zp[x] is a pid, so we're talking about 0, which is impossible since the contraction of M back to Zp is all of Zp, or the ideal generated by some irreducible polynomial g(x). So our finite field of order p is expanded to a larger finite field according to the degree of g.

You can find such a polynomial for each indeterminant, by contracting to that particular subring, but this doesn't necessarily span a maximal ideal. Consider x2+1 and y2+4 mod 7. Mod out by the first and get the finite field of order 49; but then the second splits into y+2x times y-2x. Seen another way, the ideal spanned by x2+1 and y2+4 can be enhanced by y-2x, and still does not span 1. So bring in y-2x instead, and then y2+4 is implied.

After x, y satisfies some irreducible polynomial h(y) over our extended finite field. If h has degree > 1, h extends the field further. This continues through all the indeterminants.

Finitely Generated Ring

Let R be a finitely generated ring. Thus R is the quotient of Z[x1,x2,…xn]. Every maximal ideal in R lifts to a maximal ideal in the polynomial ring, which has finite index. In fact one can use the generators of the maximal ideal upstairs to span the maximal ideal downstairs. Build an ideal based on p and the irreducible polynomials, and map this into R.