Rings, Maximal Ideals and Fields

Maximal Ideals and Fields

Let R be a commutative ring with kernel K and quotient ring S. Let K be a maximal ideal. If x is not in K, consider the ideal generated by x and K. Characterize the ideal as p(x)+K, where p is a polynomial with no constant term and coefficients in R. Verify that these expressions are closed under addition and multiplication, and generated by x and K. Let x map to y in S, and remember that K maps to 0. Since x and K span 1, the ideal generated by y also includes 1. Some polynomial in y, without a constant term, = 1. This means y is invertible. The choice of x was arbitrary, so everything in S is invertible, and S is a field.

Conversely, if K is not maximal, Let H properly contain K, and note that the image of H in S does not include 1. The nonzero elements in this image are not invertible. The quotient of a commutative ring is a field iff the kernel is maximal.

If R is noncommutative p(x), and hence p(y), could be a polynomial with coefficients on either side. Thus p(y) may not be divisible by y on the left or the right, and the quotient need not be a division ring.

If the kernel K is a prime ideal in R, R being commutative again, there are no elements x and y outside of K with xy in K. Moving to S, there are no zero divisors, hence S is an integral domain. Conversely, if K is not prime, and xy lies in K, then x and y map to zero divisors and S is not an integral domain.

As a corollary, a maximal ideal in a commutative ring is also prime. This because a field has no zero divisors. In contrast, Z cross 0 is a prime ideal in Z cross Z, but it is not maximal.

Noncommutative Example

consider the n×n matrices over a field. This ring has only two ideals, 0 and itself. This is called a simple ring.

Now let R be the direct product of two such rings. A maximal ideal in the product is all of one component cross a maximal ideal in the other. In this case the first ring is crossed with 0 in the second. This ideal is maximal and prime, yet the quotient, the second ring, admits zero divisors. It is not a division ring - not even a domain.