Algebraic Geometry, Noether Normalization

Noether Normalization

Being a K algebra, the aforementioned homomorphism must fix K. We only need map the indeterminants into A.

Before diving into this theorem, you might want to review a related theorem from field theory. The ring R can be described as a transcendental extension of K, followed by an algebraic extension. There may be many ways to do this, but in each case the transcendent degree of R, i.e. the number of indeterminants, is always n.

Noether normalization asserts the existence of a transcendental extension R_t inside R, whose image A_t is a transcendent space inside A, with R integral over R_t.

Notice that R is R_t adjoin x₁ through x_n. If each of these elements is integral over R_t, then R is a finite integral extension of R_t, and a finitely generated R_t module.

If x₁ is integral over R_t, apply the ring homomorphism and the image of x₁ is integral over A_t. The same holds for x₂ through x_n, hence A is integral over A_t, and A is a finitely generated A_t module.

Consider all the ways one can build R as a transcendental extension followed by an integral extension. There is at least one way to do this - set R_t = R. Field theory fixes the transcendent degree of R/K, hence each representation of R adjoins n indeterminants, then brings in the integral elements like the icing on the cake. Select a representation y₁ through y_n so that the maximum number of indeterminants map to 0 in A. Here R_t is K adjoin y₁ through y_n, and R is integral over R_t.

Assume the first l indeterminants, y₁ through y_l, do not lie in the kernel, and let them have images z₁ through z_l in A. These images produce the desired transcendental extension A_t.

Suppose z₁ through z_l are not algebraically independent, so that some polynomial p(z₁,z₂,z₃…z_l) = 0. Pull this back to find a polynomial p in y₁ through y_l that produces some element in R_t, which I will call w₁. Even though y₁ through y_l do not map to 0 in A, w₁ does.

We're going to select new indeterminants, starting with w₁, that will take the place of y₁ through y_l. With an additional indeterminant mapping to 0, namely w₁, we will have our contradiction, whence z₁ through z_l are algebraically independent. But how do we select w₂ through w_l?

Let m be an integer larger than any of the exponents on any of the variables in p. Build the new indeterminants as follows. As i runs from 2 to l, let w_i = y_i - y₁^mi-1. Equivalently, y_i = w_i + y₁^mi-1. Using the binomial theorem, substitute for each y_i in p, giving an expression in y₁ and w₁ through w_l. This is equal to w₁, so subtract w₁ from both sides and set the resulting polynomial to 0.

Consider the resulting free powers of y₁, without any adjunct factors of w₁ through w_l. for instance, if p includes 23y₁⁷y₂y₃⁵, The first factor contributes y₁ to the seventh, the second factor brings in y₁ to the m, and the third factor brings in y₁ to the 5m². The exponent on y₁, for this particular term in p, is 7+m+5m². And of course the coefficient is 23. Read the exponent in base m and resurrect the original term in p. The polynomial p is encoded in the powers of y₁.

One of the powers of y₁ has a higher exponent than all the others. This comes fromthe term of p with the highest power of y^l, and if there is a tie then the term that has the highest power of y^l-1, and so on. Divide through by its lead coefficient, which we can do because K is a field, and y₁ becomes integral over w₁ through w_l.

All the powers of y₁ are integral over K[w₁…w_l]. Add w_i to an appropriate power of y₁ and find another integral element, thus y₁ through y_l are integral over w₁ through w_l. With R integral over K[y₁…y_n], R is integral over K[w₁…w_l,y_l+1…y_n]. Since the transcendent degree has to equal n, the adjoined elements are algebraically independent. This is a new transcendental extension within R, and one more indeterminant maps to 0 in A. This is a contradiction, therefore z₁ through z_l are algebraically independent, building the transcendental extension A_t in A. We may as well say z_i = y_i for i ≤ l, and the remaining y's drop to 0 in A.

In summary, every finitely generated K algebra is a purely transcendental extension followed by an integral extension.

Mapping an Algebra into the Closure of K