Commutative algebra - Department of Mathematical Sciences - old ...
Commutative algebra - Department of Mathematical Sciences - old ...
Commutative algebra - Department of Mathematical Sciences - old ...
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7.3. ARTINIAN RINGS 89<br />
Pro<strong>of</strong>. (1) Let P1, . . . , Pn be the prime and maximal ideals 7.3.11. The nilradical<br />
√ 0 = P1 ∩ · · · ∩ Pn by 5.1.7. Conclude by Chinese remainders 1.4.3. (2) √ 0 k =<br />
√ 0 k+1 for some k. If √ 0 k = 0 let (a) be minimal among ideals I such that<br />
I √ 0 k = 0. By minimality (a) = (a) √ 0 k , so a = ab for some b ∈ √ 0 k . But b is<br />
nilpotent 1.3.8, so a = 0 gives a contradiction. It follows, that √ 0 k = 0.<br />
7.3.13. Proposition. A ring R is artinian if and only if it has finite length.<br />
Pro<strong>of</strong>. The factor module √ 0 i / √ 0 i+1 is an artinian module over R/ √ 0 which is<br />
a product <strong>of</strong> fields, so it has finite length.<br />
7.3.14. Corollary. Let R be an artinian ring and M a module. The following are<br />
equivalent.<br />
(1) M is finite.<br />
(2) M has finite length.<br />
(3) M is finite presented.<br />
7.3.15. Corollary. Let R be artinian and R → S a finite ring homomorphism.<br />
(1) S is artinian.<br />
(2) A finite length S-module N is by restriction <strong>of</strong> scalars a finite length Rmodule.<br />
(3) A finite length R-module M gives by change <strong>of</strong> rings M ⊗R S as finite length<br />
S-module.<br />
7.3.16. Proposition. Let M be a R-module <strong>of</strong> finite length.<br />
(1) The ring R/ Ann(M) is artinian.<br />
(2) There are only finitely many prime ideals Ann(M) ⊂ P .<br />
(3) Any prime ideal Ann(M) ⊂ P is maximal.<br />
(4) M is finite presented.<br />
Pro<strong>of</strong>. (1) Let x ∈ M, then R/ Ann(x) Rx is artinian. If x1, . . . , xn generate<br />
M, then Ann(M) = Ann(x1)∩· · ·∩Ann(Xn). By 7.3.8 R/ Ann(M) is artinian.<br />
7.3.17. Example. Let K be a field. The ring R = <br />
artinian.<br />
(1) The maximal ideals in R are<br />
(2) The simple types are<br />
N<br />
Pi = {a : N → K|a(i) = 0}<br />
R/Pi Ji = {a : N → K|a(j) = 0, i = j}<br />
(3) Ji are the only simple ideals and the sum is an ideal<br />
<br />
Ji = J = R<br />
(4) J is an ideal which has no complement in R.<br />
i<br />
K = {a : N → K} is not<br />
7.3.18. Exercise. (1) Show that a vector space is artinian if and only if it is finite dimensional.<br />
(2) Show that Z is not artinian.<br />
(3) Show that R[X] is not artinian for a nonzero ring R.<br />
(4) Show that Q[X]/(X 2 − X) is artinian.