SOME CONDITIONS ON FIXED RINGS

This paper deals with a question about the ascending and descending chain conditions on two-sided ideals. Using the ideas of the skew group ring,


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E.E.REITER then we have (Soc R) [ RGc qoc RG.Thus for a semiprime ring with no IGI- torsion, Soc R If R G Soc RG.
We also answer a question about the ascending and descending chain conditions on two-sided ideals.It is well known that if IGI-I R (i.e., if IGI is invertible in R) and if R satisfies the ascending or descending chain condition on left (right) ideals, then R G also satisfies the same condition.
However, the techniques used for one-sided ideals have not been applied to two- sided ideals.In this paper, we prove the analogous result for two-sided ideals by using the skew group ring of R and G,'R,G @ Y Ru Addition in R*G is componentwise-multiplication is given by the relations (ru)(su h) (rs g )Ugoh g extended linearly.The inverse question, posed as follows, remains open: If R is semiprime and IGI -I R, and if R G satisfies the ascending (descending) chain condition on two-sided ideals, then must R satisfy the ascending (descending) chain condition on two-sided ideals?The analsous question is true for one- sided ideals.
For a left and right Noetherian ring, the Artin radical has been defined as the sum of all Artinian left ideals of the ring.If R is semiprime and R, then A(R) is defined if and only if A(RG) is defined.In the final section of this paper, we prove that under these conditions, A(R) R G 2. THE SOCLES OF R AND RG.
First, consider the following theorem relating R and RGymodules.The proof is a variation of the proof of Maschke's theorem.Note that all R,G- modules are certainly R-modules; on the other hand, the left R*G-submodules of R are the G-invariant left ideals of R (I R such that I g c I J g G).
THEOREM 2.1 Let R be a ring and G a finite group of automorphisms of R, and let @ N / M / N' / 0 be a short exact sequence of left (right) R'G- modules such that the map m / IGlm is a bijection on M. Then if the sequence iii splits over R, it also splits over R,G.
PROOF.See Fisher-Osterburg [7,Theorem 1.3].They assume IGI -I R, but the proof can be carried out easily without this as long as for any m M, N, there exist m' and n' in M and N respectively such that IGlm' m and IGln' n.
From this, we obtain: COROLLARY 2.2 Let M be a left (or right) R'G-module with IGI a blJection on M. If M is semisimple as a left (or right) R-module, then M is R,G-semisimple.

PROOF. Immediate.
Now we can apply this to solve the socle question.THEOREM 2.3 Let R be a ring with no IGl-torsion, and let Soc R denote the left (or right) socle of R. Then Soc R R G c Soc RG.
PROOF.We give the proof for the left socle only.Note that Soc R is a G- inarlant left ideal of R; thus it is a left R,G-module.The simple components = m.ft s of R. Th =no b =h= y GI o GI .a biJection on Soc R. We can apply Corollary 2.2: Soc R is R,G-semisimple, i.e., n Soc R =i--@K i' where each K i is a minimal G-invariant left ideal of R.
We now claim that each K i I'1 R G is a minimal left ideal of RG.For, suppose that X is a left ideal of R c such that 0 X_c K.I RG" Then 0 # RX__C R(K i RG) ___c Ki; this implies RX K i.However, the fact that GIRX RX forces RX R G X, and the claim is proved.
, i= l,...n, since the sum is direct and each i is G-invariant.These equations show that each x i E RG.
REMARK.This theorem was subsequently proved by R. Diop [6], and, with the extra assumption that GI -I R, by M. Lorenz [9].
E.E.REITER We note the following corollary.
COROLLARY 2.4 Let R be a semiprime ring with no Gl-torsion.Then (Soc R) G= Soc RG.
PROOF.By a theorem of Fisher and Osterburg [7], if R is semlprlme with no Gl-torsion, then Soc R G c__ Soc R.
3. CHAIN CONDITIONS ON TWO-SIDED IDEALS.
To deal with two-slded ideals in R and RG, we again consider the skew group ring.In particular, we will use the following lemma.
LEMMA 3.1 Let R be any finite group.If R has the ascending (descending) chain condition on two-slded ideals, then R,G also satlfles the ascending (descending) chain condition on two-slded ideals.
PROOF.The proof follows the method of the Hilbert Basis Theorem.For G {gl gn }, I R'G, we define I k {r R: for some rl,...rk_ I R, ,n.For a chain of ideals gk-i gk be the k th associated ideal of I. as defined above.
I I, 12 The n chains of ideals I I, 12,,... lj,,..., i, n must each terminate; this forces the chain I. in R,G to terminate.
It may be mentioned that the above lemma is true for any ring T and over- ring S such that there exists a finite subset of S, {Sl, Sn } such that n S Ts and s.T Ts.. i=l i z It is well known (and easily proved) that for any ring S that has the ascendiug (descending) chain condition on two-sided ideals, any subring of the 2 form eSe where e e must also satisfy the same condition.We use this fact, together with Lemma 3.1, to prove the following theorem.THEOREM 3.2 If IGI -I R and R satlfies the ascending (descending) chain conditions on two-sided ideals, then R G satisfies the ascending (descending) chain condition on two-sided ideals.If IGI -I R and R has a composition series of two-sided ideals, then R G has such a composition series also.
PROOF.Consider the element e (I/IGI) 7 u. 6R,G.It is easy to show G 2 that e e Also, we claim that R G e(R,G)e.To prove this, use the ring isomorphism r / (rUgl)e where gl is the identity of G to obtain R G RGe e(R*G) e.Then, since R,G has the ascending (descending) chain condition on two-sided ideals, so does RG.
In the preceeding Theorem 3.2, we may weaken the assumption that R has no GI -I R to the assumption that R has no Gl-torsion for the cases of the descending chain condition or the composition series property.This is true since IGI is in the center of R and RIGI k, k 1,2, is a descending chain of two-sided ideals which must terminate, forcing IGI to be invertible in R.
We state this as the following corollary.
COROLLARY 3.3.If a ring R has no IGl-torsion, then if R satisfies the descending chain condition (has a composition series of two-sided ideals), R G must also satisfy the descending chain condition (have a composition series of two-sided ideals).
PROOF.Evident from Theorem 3.2 and the paragraph above.
We conclude this section by noting an example due to Chuang and Lee [3].
This is a commutative Noetherian ring R with involution, such that R has no 2-torsion and the subring generated by the symmetric elements is not Noetherian.
In a commutative ring, an involution is an automorphism of order 2, and the subring generated by the symmetric elements is the fixed ring.This example, then, shows that for the ascending chain condition, the hypotheses of Theorem 3o 2 cannot be weakened to assume merely no 4. THE ARTIN RADICALS OF R AND RG.
In a left and right Noetherian ring R, the Artin radical A(R) is defined as the sum of all right ideals of R that are Artinian as right R-modules.This radical turns out to be a two-sided ideal of R; it is also equal to the sum of the left ideals of R that are left R-Artinian.

E.E. REITER
It is easy to prove that if R is left and right Noetherlan and if 'IGI -I R, then R G is left and right Noetherlan.Also, Farkas and Snlder [5] have proved that if R is semlprlme with no Gl-torslon and if R G is left (right) Noetherlan, then R is left (right) Noetherlan.Thus, for R semlprlme with IGI -I R, A(R) is defined if and only if A(RG) is defined.In this section we prove that under these conditions, A(R)i R G A(RG).
To prove the first inclusion, we need several preliminary facts.First, note that if I is any G-invarlant ideal of R, then G (or a homorphlc image of G) acts on R/I by g(r + I) g(r) + I. Furthermore, if IGI -I R, it can be shown (see Fisher and Osterburg [7]) that (R/I)G RG/(I RG).
Also, we require an important theorem due to Bergman and Isaacs [i], which we state below.The trace of an element is defined by tr x [ xg', the trace gG of an ideal L is {tr x: x L}.
THEOREM 4.1 (Bergman-Isaacs) Let G be a finite group acting on a ring R without G -torsion.Then: (i) If R is semlprime, then R G is semlprlme.
(ii) If R is semiprime and L is a G-invariant left ideal of R such that tr L 0, then L 0.
Now we can prove: THEOREM 4.2 If R is semlprlme and GI -I 6 R, then A(R) ] R G _ _ A(RG).
PROOF.First, we prove the following fact about right annihilators: [rR(A(R))]G rRG[A(R) RG].Only one inclusion requires proof.
Let x E rRG [A(R) I RG]; that is, x R G and [A(R) i RG]x O. Now A(R)x is a G-invariant left ideal of R and tr (A(R)x) 0. The Bergman-Isaacs is applicable so we have A(R)x 0, i.e. x [rR(A(R))]G.
To prove the theorem, we show that A(R) R G is an Artlnlan right ideal of RG.By a theorem of Lenagan [8] if I is an ideal of a left and right Noetherian ring such that I is left R-Artinian, then R/rR(1) is an Artinian ring.
Thus, R/rR(A(R)) is an Artinian ring.Since IGI -I R, it follows easily that the fixed ring of this factor ring is Artinian.Using this together with the facts on the preceedlng page, we have the Artlnlan rings: And, from the first part of the proof, we have RG/(rRG[A(R)/ RG] is Artlnlan.
It is true that a right ideal I of a ring S is contained in A(S) if S/r(1) is Artlnlan (see Chatters, Hajarnavls, and Norton [2], the proof of Theorem 1.3).
Applying this, we have A(R) R G c A(RG).
To prove the other inclusion, we need the following theorem.Its proof is variation of a proof by Cohen and Montgomery [4].
-IEORI 4.3 Let R be a semiprime ring, left and right Noetherian with IG1-1 R. If A A(RG) is the Artinian radical of R G then RA is an Artinian left R-module.
PROOF.Since R is left and right Noetherlan, there are left ideals of R, K maximal in RA.And, since A is artlnlan, we may choose a finite family (RG) of these, k I, k n, to minimize K A (K I/ Kn )" We prove first that K 0. For, suppose not.Since G A is Artlnlan, the socle of R G is essential as a left ideal in A. Also, sinc is semiprime, Soc (RG) is a semlslmple Artlnlan ring and has an identity element i S Now, ifK0: c RA is direct, and equality would force Re 0 and hence K 0. Therefore, we can choose a left ideal of R, K0, so that K 0 is maximal in RA and: R(I S -e)__c K0 RA