ABSTRACT
Proof. Let p be a prime ideal o f A. Then B = A/p is an Artinian integral domain. Let xeB , x Φ 0. By the d.c.c. we have (jcn) = (xn+1) for some n9 hence xn = xn+ly for some y g B. Since B is an integral domain and x Φ 0, it follows that we may cancel xn, hence xy = 1. Hence x has an inverse in B, and therefore B is a field, so that p is a maximal ideal. ■
Corollary 8.2. In an Artin ring the nilradical is equal to the Jacobson radical. ■
Proposition 8.3. An Artin ring has only a finite number o f maximal ideals. Proof. Consider the set o f all finite intersections m1 n - · - n m „ where the m* are maximal ideals. This set has a minimal element, say nij η · · ·π τηη; hence for any maximal ideal m we have m n ntj η · · · n ntn = m1 η · · · n tnn, and therefore η ΐ2 ΐη1η · · · π mn. By (1.11) m 2 m, for some /, hence m = m* since m, is maximal. ■
Proposition 8.4. In an Artin ring the nilradical is nilpotent. Proof. By d.c.c. we have fRk = 9fefc+x = · · · = a say, for some k > 0. Suppose a Φ 0, and let Σ denote the set of all ideals b such that ab φ 0. Then Σ is not empty, since a e Σ. Let c be a minimal element of Σ; then there exists x e c such that xa Φ 0; we have (x) £ c, hence (x) = c by the minimality of c. But (χα)α = χα2 * xa Φ 0, and xa £ (x), hence xa = (x) (again by minimality). Hence x = xy for some y ea , and therefore x = xy = xy2 = · · = xyn = · · ·. But y 6 a = filk 2 9£, hence y is nilpotent and therefore x = xyn = 0. This contradicts the choice of x, therefore a = 0. ■
By a chain of prime ideals of a ring A we mean a finite strictly increasing sequence p0 <=■ pt cr... c £n; the length o f the chain is n. We define the
dimension o f A to be the supremum of the lengths of all chains of prime ideals in A: it is an integer >0, or +00 (assuming A φ 0). A field has dimension 0; the ring Z has dimension 1.
