Commutative algebra - Department of Mathematical Sciences - old ...

88 7. MODULES OF FINITE LENGTH
Proof. Let M be artinian. A chain in N gives a chain in M, so N is artinian. A
chain in L gives a chain in M, which becomes stationary, so also the original chain,
so L is artinian. Conversely if N and L are artinian and Mi a chain in M, then
the induced chains in N and L become stationary. By the snake lemma 3.2.4 the
original chain is stationary.
7.3.5. Corollary. Let f : M → N be a homomorphism.
(1) M is artinian if and only if Ker f, Im f are artinian.
(2) N is artinian if and only if Im f, Cok f are artinian
Proof. Use the sequences 3.1.8.
7.3.6. Proposition. Let f : M → M be a homomorphism on an artinian module.
Then the following are equivalent
(1) f is injective
(2) f is an isomorphism
Proof. There is a number n such that Im f ◦n = Im f ◦2n . For x ∈ M there is y
such that f ◦n (x) = f ◦2n (y). Then f ◦n (x − f ◦n (y)) = 0 so x = f ◦n (y) since f is
injective. It follows that f ◦n is surjective and so is f.
7.3.7. Proposition. Given submodules N, L ⊂ M. Then the following are equivalent.
(1) M/N, M/L are artinian
(2) M/N ∩ L is artinian
Proof. use the sequences 3.2.8.
7.3.8. Corollary. Given ideals I, J ⊂ R. Then the following are equivalent.
(1) R/I, R/J are artinian
(2) R/I ∩ J is artinian
7.3.9. Corollary. Let R be an artinian ring and I an ideal. Then the factor ring
R/I is artinian.
The product of two artinian rings is artinian.
7.3.10. Proposition. An artinian domain is a field.
Proof. By 7.3.6 scalar multiplication with a nonzero element is an isomorphism.
7.3.11. Proposition. Let R be an artinian ring. Then all prime ideals are maximal
and there are only finitely many such.
Proof. By 7.3.10 primes are maximal. If Pi are different maximal ideals, then
P1 · · · Pn+1 ⊂ P1 · · · Pn is a strictly decreasing chain. So there are only finitely
many maximal ideals.
7.3.12. Proposition. Let R be an artinian ring.
(1) The factor ring R/ √ 0 is a finite product of fields.
(2) The nilradical √ 0 is nilpotent, √ 0 k = 0 for some k.