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Irreducibility, Prime Ideals, and Dimension

Which varieties are 'one piece'? The answer is purely algebraic: X is irreducible iff I(X) is prime iff k[X] is a domain. We decompose any variety into irreducible components, then measure size with Krull dimension and transcendence degree.

Irreducible means prime

A nonempty variety X is irreducible if it cannot be written X = X_1 ∪ X_2 as a union of two strictly smaller varieties. The union of the two coordinate axes V(xy) is reducible; a single line is irreducible. The miracle is that this geometric notion has a clean algebraic test. See irreducible variety.

THEOREM.  X irreducible  <=>  I(X) is a PRIME ideal
                          <=>  k[X] is an INTEGRAL DOMAIN.

Why prime <=> irreducible (the key direction):
Suppose I(X) is NOT prime: f*g in I(X) but f,g not in I(X).
Then X = (X cap V(f)) union (X cap V(g)):
  - every point of X kills f*g, so kills f OR kills g;
  - neither piece is all of X, since f,g do not vanish on X.
So X is reducible.  Run the argument backwards for the
converse.  And  k[X] = k[x]/I(X)  is a domain exactly when
I(X) is prime -- the standard ring fact.

Check V(xy):  xy in (xy) but x,y not in (xy), so (xy)
is NOT prime -- matching the visibly reducible axes.
Whereas (xy) ... its prime components are (x) and (y),
the two axes.
The bridge theorem of this guide: geometric irreducibility = algebraic primality.

Every variety splits into irreducible components

Because k[x_1,…,x_n] is Noetherian, the descending chain condition on varieties holds, and a standard argument forces every variety to be a finite union X = X_1 ∪ ⋯ ∪ X_r of irreducible varieties with no X_i contained in another. These X_i are the irreducible components of X, and they are unique. This is the geometric shadow of primary decomposition of the ideal I(X).

Dimension, three ways that agree

How big is a variety? Geometrically, the dimension of X is the largest d for which there is a chain of irreducible closed subsets X_0 ⊊ X_1 ⊊ ⋯ ⊊ X_d ⊆ X. Translated through the dictionary, this is the Krull dimension of k[X]: the length of the longest chain of prime ideals. The same number, two languages. See dimension of a variety.

For an irreducible X there is a third, very computable description: dim X equals the transcendence degree over k of the fraction field of k[X] — the number of coordinates that are 'algebraically free'. Noether normalization makes this concrete: it exhibits k[X] as a finite extension of a polynomial ring k[y_1,…,y_d], so X finitely covers A^d, and d is the dimension.