Integral Extensions, Going-Up, and Noether Normalization

Integral = algebraic, with monic coefficients in the base

For field extensions, "algebraic" means satisfying a polynomial over the base. For rings we need the polynomial to be monic — that single word is everything. Given R ⊆ S, an element s ∈ S is integral over R if s^n + r_{n−1}s^{n−1} + … + r_0 = 0 for some r_i ∈ R, with leading coefficient 1. Over a field monic is free, so this exactly recovers algebraic; over Z it forces s to be an algebraic *integer*, not just any algebraic number.

The workhorse lemma — the determinant trick — says s is integral over R iff R[s] is a finitely generated R-module. From it you get that the set of all elements integral over R, the integral closure, is a subring, and that integrality is transitive.

Determinant trick (why integral <=> finite module):
Suppose M = R[s] is generated as an R-module by m_1,...,m_n, with s*M c M:
      s * m_i = sum_j a_ij m_j ,   a_ij in R.
In matrix form: (s*I - A) * [m_1; ...; m_n] = 0, where A = (a_ij).
Multiply by the adjugate:  det(s*I - A) * m_i = 0 for every i.
Since 1 is an R-combination of the m_i, det(s*I - A) * 1 = 0.
But det(s*I - A) = s^n + (lower) is a MONIC polynomial in s with R-coeffs.
So s satisfies a monic equation over R  =>  s is integral.  QED

Example: sqrt(2) is integral over Z: x^2 - 2 = 0 (monic). It IS an alg. integer.
         (1/2) is NOT integral over Z: any monic x^n + ... + c_0 = 0 with
         x = 1/2 clears to 1 + 2(...) = 0, forcing 2 | 1.  Impossible.

Going-up: primes lift along integral maps

If R ⊆ S is integral, the induced map Spec S → Spec R is geometrically a finite, surjective map — a branched cover. The going-up theorem makes the surjectivity and more precise: every prime of R is hit, and chains of primes downstairs lift to chains upstairs.

1. Lying over. For every prime P of R there is a prime Q of S with Q ∩ R = P. So Spec S → Spec R is surjective.
2. Going up. Given P_1 ⊆ P_2 in R and a prime Q_1 of S over P_1, there is Q_2 ⊇ Q_1 over P_2. Chains lift.
3. Incomparability. Two distinct primes of S over the same P are never nested. So fibers are discrete.

Noether normalization: every variety covers affine space

Now the structural climax. Noether normalization: if A is a finitely generated algebra over a field k, there exist elements y_1,…,y_d ∈ A, algebraically independent over k, such that A is a finitely generated *module* — i.e. integral — over the polynomial subring k[y_1,…,y_d]. Geometrically: every affine variety admits a finite surjection onto affine space A^d, and d is its dimension.

Toy case: A = k[x,y]/(xy - 1)  (the hyperbola, dimension 1).
  x and y are not independent (xy = 1). Is k[x] a normalizing subring?
  No: y = 1/x is NOT integral over k[x]
  (it would satisfy a monic poly over k[x], impossible as above).

  Fix by a linear change: set t = x + y. Claim A is integral over k[t].
  From xy = 1 and x + y = t: x, y are the two roots of
        T^2 - t*T + 1 = 0     (monic over k[t]).
  So x and y are each integral over k[t]. Hence A = k[t][x,y] is
  a finite module over k[t]. Here d = 1: the hyperbola finitely covers A^1.

(The generic linear change of coordinates works whenever k is infinite;
 over finite fields use a slightly twisted substitution.)