Björk , Conditions that subrings of artinian rings are artinian

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Example 2.1. Firstly we put К = Q(xj^, x^, .. .) so that К is the field of rational functions in infinitely many variables. We also put Kq= Qix^, x^, . , .) so that Kq is a subfield of K, The map / which sends x^ -> х^,^^ defines an isomorphism between К and Kq, Next we define a ring R as follows: R = eK + zKisdi 2-dimensional right Z-space, e is the unit of R, z^ = 0 and kz == zf(k). Then R is right but not left artinian. Next we put M = JJ KqxI, so that M is a ÜTo-subspace of K, We also put L = 2J Rzx^, so that Z/ is a left ideal of R. Now we consider the left i?-module R/L and put

E = Hom^(/?/L, R/L).

We claim that E is not semiprimary. For E ^ U/L, where U = [r ^ R: Lr <: L], We see that U = T + zK where T -= {k^ K\ Mk<=z M holds}. If f//L is semiprimary it is easily seen that T must be a field. But T is not a field here, for we can show that T = M holds.

Part 2

Let i? be a ring and let G{R) be its associated Lie ring. We define the sets D" ductively as follows: D« = G{R) and i)"+^ = {\xy\:x,y^ i)"}. Then G(R) is solvable if D" = 0 for some n. We also define the sets C„ :

Co = G(R) and C,^i ^{\xy\:xe G(R) and y € Q.

Then G{R) is nilpotent if C^ = 0 for some n. Now we state the main results of this section.

Theorem 1. 2. Let R be a left coperfect ring such that G(R) is solvable. Then R is right coperfect.

Theorem 2. 2. Let R be a left artinian ring such that G{R) is nilpotent. Then R is right artinian.

Proof of Theorem 1. 2. Since R is left coperfect it follows that RIJ(R) is artinian. Using the Theorem of H. Bass it is then sufficient to prove that J(R) is left T-nilpotent, so let us assume the contrary and derive a contradiction.

We say that an element x in J{R) has an infinite chain if there exists an infinite sequence (x^) in J{R) for which xx^ • • • :г:„ Ф 0 for all n. So by our assumption there is an element which has an infinite chain. Using the fact that R is left coperfect we can choose Xi in J(R) such that ж^ has an infinite chain while the left principal ideal Rx^^ is minimal under this condition. This means in particular that iî x(: J(R), then xx^ has no infinite chain. Inductively we can choose x^ in J(R) such that ж^ • • • ж„ have infinite chains, while each Rx^ is minimal under the condition that (x^- - - Xn_-^)Xn has an finite chain.

Now {x„) is the sequence above. For each aï > 3 we put

^l ' ' ' Xy^ = I X1X2 I ^3 * ' * ^n » (^2*^1/*^3 * * ' »^n*

Since X2X^ has no infinite chain it follows that x^- - - Xn= \ x^x^ I ^3 • * • ^„ for all n > N^ say.

If n > N1 we put Xj^' - ' x^= \ x^x^ 1 I x^x^ I ^5 * * * ^» + I ^i^'^^2 I {^4^3)^5 * * * ^и- The condition about x^ implies that x^x^x^x^ has no infinite chain and x^x-^x^x^ has no infinite chain because x^x^ has no infinite chain. We conclude that

0^1 • • • Ж„ = I X-^X2 I I ^3^4 I ^5 * * ' *^n

holds for all гг > iVa say.