Algebraic Geometry, Noether Normalization on an Integral Domain

Noether Normalization on an Integral Domain

Before I state the theorem, here are a few comments on notation. In earlier sections I have used a subscript t, as in R_t, to indicate a purely transcendental ring inside R. In this section I will use the notation R^t for a transcendental extension above R. This is not R to the t power; it is R adjoin finitely many indeterminants.

The notation R/f means the fraction ring of R by f, where f is an element of R. Elements of R are in the numerator and powers of f are in the denominator. Since R is an integral domain (for this theorem), the powers of f never reach 0, thus R/f is well defined.

Let S be a finitely generated R algebra, where R and S are integral domains, and R embeds in S. There is a transcendental extension R^t between R and S, and an element f in R, with S/f integral over R^t/f.

In traditional noether normalization the base ring R is a field, and S becomes a transcendental extension followed by an integral extension. In this case R is not a field, so we need to employ the element f to achieve a similar result.

Let V be the nonzero elements of R. Let these act as denominators, so that S/V is a finitely generated R/V algebra.

Of course R/V is a field; call it K. Thus S/V is a finitely generated K algebra. Apply noether normalization to obtain an intermediate transcendental extension K^t inside S/V.

If z_j is the j^th generator of S, z_j/1 is integral over K^t. Write a monic polynomial p_j with coefficients in K^t that is satisfied by z_j/1, and do this for each j. This is not a conceptually simple polynomial. Each of the coefficients is itself a polynomial in x₁ x₂ x₃ etc, which are the indeterminants that build K^t. And the coefficients on these polynomials lie in K. A sample polynomial might look like this.

z³ + (4x₁x₃/7-8x₅³/9) z² + (x₄⁶x₅³/8-4x₂²) z + 7x₃+1

Let f be the product of the denominators of all the coefficients in these polynomials. For example, the above polynomial contributes 7×8×9 = 504. This becomes the value of f that satisfies the theorem.

If x_i is one of the indeterminants that builds K^t, multiply it by f to get y_i. The variables x_i are algebraically independent with respect to K, and they are certainly algebraically independent when polynomials are restricted to R. Scaled instances of these variables will also be independent. Therefore, adjoining the variables y_i to R produces a purely transcendental extension that I will call R^t.

Focus on z_j, one of the generators of S. The monic polynomial p_j is satisfied by z_j/1. Look at a coefficient on a term in p_j. This is a polynomial in the variables x₁ x₂ x₃ etc, with coefficients in K. Look at a term in this polynomial. Multiply and divide by a power of f, so that each x_i turns into a y_i. Then multiply top and bottom by some portion of f, so that the denominator becomes a perfect power of f. Each coefficient of p_j is now something in R^t/f.

Each z_j is integral over R^t/f, and S is integral over R^t/f. Since f belongs to R, the entire ring S/f is integral over R^t/f, and that completes the proof.

Mapping into a Closed Field

Let a ring homomorphism h map R into an algebraically closed field C. Assume h(f) is nonzero. Map each of the indeterminants y_i to anything in C, thus extending h to R^t. Then extend h to R^t/f. This is unique; h(1/f) has to equal 1/h(f).

Next extend h to S/f, which is integral over R^t/f. Finally restrict h to S. the ring homomorphism on R has been extended to all of S.

If R embeds in C then h(f) is automatically nonzero, and the embedding can be extended to a homomorphism on S.