Commutative feebly invo-clean group rings.pdf 1. Introduction and Conventions Throughout the current paper, we will assume that all groups G are multiplicative abelian and all rings R with Jacobson radical J (R) are associative, containing the identity element 1 which differs from the zero element 0. The standard terminology and notation are mainly in agreement with [9 and 10], whereas the specific notion and notation shall be explained explicitly below. As usual, both objects R and G form the group ring R[G] of G over R . The next concepts appeared in [1, 2, and 3], respectively. Definition 1.1. A ring R is said to be invo-clean if, for each r∈ R, there exist an involution v and an idempotent e such that r=v+e. If r=v+e or r=v-e, the ring is called weakly invo-clean. The next necessary and sufficient condition for a commutative ring R to be invo-clean was established in [1, 2], namely: A ring R is invo-clean if, and only if, R = R1 × R2, where R1 is a nil-clean ring with z= 2z for all z ∈ J(R1), and R2 is a ring of characteristic 3 whose elements satisfy the equation x3 = x. Moreover, it was proved in [6] that a ring R is weakly invo-clean ⇔ either R is invo-clean or R can be decomposed as R = K × Z5, where K = {0} or K is invo-clean. The above two notions could be expanded as follows: Definition 1.2. A ring R is said to be feebly invo-clean if, for each r∈ R, there exist an involution v and idempotents e, f such that r=v+e-f. We will give up in the sequel an useful criterion for a commutative ring to be feebly invo-clean in order to be successfully applied to commutative group rings (compare with Proposition 2.2). P.V. Danchev It was asked in [6] to find a suitable criterion only in terms of the commutative unital ring R and the abelian group G when the group ring R[G] is feebly invo-clean. So, the goal of this short article is to address that question in the affirmative. Some related results in this area can also be found in [4 and 7]. 2. The Characterization Result We begin here with the following key formula from [8] which will be freely used below without concrete citation: Suppose that R is a commutative ring and G is an abelian group. Then J(R[G]) = J(R)[G] г(g - 1)|g ∈ Gp, pr ∈ J(R)}, where Gp designates the p -primary component of G. The next two technicalities are crucial for our further considerations. Lemma 2.1. LeZ K be a commutative ring of characteristic 5. Then K is feebly invo-clean ⇔ = X holds for any x ∈ K . Proof. The "left-to-right" implication is almost trivial as writing x=v+e-f with v2 =1, e2 =e and f2 = f , we have that X5=(v+e-f)5=v5+e5-f5=v+e-f=X, as asserted. As for the "right-to-left" implication, we process like this: Given an arbitrary nonidentity element X in K . Then the subring, S , generated by 1 and X will have the same property, namely its characteristic is again 5 and y5 = y for all y∈ S. So, with no harm of generality, we may replace K by this subring S , and thus it needs to prove the wanted representation property in S only. To that purpose, we claim that S is isomorphic to a quotient of the factor-ring Z5[X]/(X- X) ≡ Z5 × Z5 × Z5 × Z5 × Z5 of the polynomial ring Z5[X] over Z5. In fact, we just consider the map Z5 [X]→S, defined by mapping X → X, which is elementary checked to be a surjective homomorphism with kernel which contains the ideal generated by X5 - X , and henceforth the classical Homomorphism Theorem works to get the desired claim. Working now in the direct product of five copies of the five-element field Z 5 = {0,1,2,3,4|5 = 0}, a plain technical argument gives our wanted initial assertion that S and hence K are both feebly invo-clean. This is subsumed by the presentations 0 =1+0-1, 1=1+0-0, 2 = 1 +1 - 0, 3 = 4 + 0 -1 and 4 = 4 + 0 - 0, where 42 = 1, 12 = 1 and 02 = 0. □ Proposition 2.2. A commutative ring R is feebly invo-clean ⇔ R=P×K for two rings P, K, where P = {0} or P is invo-clean, and K = {0} or K possesses characteristic 5 such that X5 = X, ∀X∈K. Proof. "⇒ ". It follows from the corresponding characterization method used in [3, Theorem 2.6]. " ". Firstly, it needs to show that K is feebly invo-clean. This, however, follows di rectly from Lemma 2.1. Furthermore, one suffices to observe again with [3, Theorem 2.6] at hand that the direct product of such a ring K with an invo-clean ring remains a feebly invo-clean ring, thus getting resultantly that R is feebly invo-clean, as expected. □ We are now ready to proceed by proving the following preliminary statement (see [5] as well). Commutative feebly invo-clean group rings Proposition 2.3. Suppose R is a non-zero commutative ring and G is an abelian group. Then R[G] is invo-clean if, and only if, R is invo-clean having the decomposition R=R1×R2 such that precisely one of the next three items holds: (0) G ={1} or (1) |G |>2 , G2 ={1}, R1 ={0} or R1 is a ring of char(R1)= 2, and R2 ={0}, or R2 is a ring of char (R2 )= 3 or (2) |G |=2, 2r12 = 2r1 for all r1∈ R1 (in addition 4= 0 in R1), and R2 = {0} or R2 is a ring of char (R2 )= 3. Proof. If G is the trivial i.e., the identity group, there is nothing to do, so we shall assume hereafter that G is non-identity. "Necessity." Since there is an epimorphism R[G]→ R, and an epimorphic image of an invo-clean ring is obviously an invo-clean ring (see, e.g., [1]), it follows at once that R is again an invo-clean ring. According to the criterion for invo-cleanness alluded to above, one writes that R=R1×R2, where R1 is a nil-clean ring with a2 = 2a for all a∈ J(R1) and R2 is a ring whose elements satisfy the equation x3 = x. Therefore, it must be that R[G ] ≡ R1 [G] × R2 [G], where it is not too hard to verify by [1] that both R1[G] and R2[G] are invo-clean rings. First, we shall deal with the second direct factor R2[G] being invo-clean. Since char(R2 ) = 3, it follows immediately that char(R2 [G]) = 3 too. Thus an application of an assemble of facts from [1, 2] allows us to deduce that all elements in R2[G] also satisfy the equation y3 = y. So, given g∈G⊆R[G], it follows that g3 = g, that is, g2 = 1. Next, we shall treat the invo-cleanness of the group ring R1[G]. Since char(R1) is a power of 2 (see [1]), it follows the same for R1[G]. Consequently, utilizing once again an assortment of results from [1, 2], we infer that R1[G] should be nil-clean, so that z2 = 2z for all z∈J(R1[G]). That is why, invoking the criterion from [7], we have that G is a 2 -group. We claim that even G2 =1. In fact, for an arbitrary g∈G, we derive with the aid of the aforementioned formula from [8] that 1-g∈J(R1[G]), because 2∈J(R1). Hence (1-g)2 =2(1-g) which forces that 1-2g+g2 =2-2g and that g2 =1, as desired. We now assert that char(R1 )= 2 whenever |G |>2. To that purpose, there are two nonidentity elements g≠ h in G with g2=h2=1. Furthermore, again appealing to the formula from [8], the element 1-g+1-h=2-g-h lies in J(R1[G]), because 2∈J(R1). Thus (2-g-h)2 =2(2-g-h) which yields that 2-2g-2h+2gh=0. Since gh ≠1as for otherwise g=h-1 =h, a contradiction, this record is in canonical form. This assures that 2 = 0, as wanted. P.V. Danchev However, in the case when | G ∣= 2, i.e. when G = {1, g|g2 = 1} = , we can con clude that 2r2 = 2r for any r∈R1. Indeed, in view of the already cited formula from [8], the element r(1- g) will always lie in J(R1[G]), because 2∈ J(R1). We therefore may write [r(1-g)]2 =2r(1-g) which ensures that 2r2-2r2g=2r-2rg is canonically written on both sides. But this means that 2r2 = 2r, as pursued. Substituting r = 2, one obtains that 4 = 0. Notice also that 2r2 = 2r for all r∈R1 and a2 = 2a for all a∈J(R1) will imply that a2 = 0. "Sufficiency." Foremost, assume that (1) is true. Since R1 has characteristic 2, whence it is nil-clean, and G is a 2-group, an appeal to [7] allows us to get that R1[G] is nil-clean as well. Since z2 =2z=0 for every z∈ J(R1), it is routinely checked that δ2 =2δ=0 for each δ∈J(R1[G]), exploiting the formula from [8] for J(R1[G]) and the fact that R1[G] is a modular group algebra of characteristic 2. That is why, by a consultation with [1], one concludes that R1[G] is invo-clean, as expected. Further, by a new usage of [1], we derive that R2[G] is an invo-clean ring of characteristic 3. To see that, given x∈R2[G], we write x= ∑g∈Grgg with rg ∈R2 satisfying rg3 = rg. Since G2 =1 will easily imply that g3 = g, one obtains that x3=(∑g∈Grgg)3=∑g∈Grg3g3=∑g∈Grgg=x, as needed. We finally conclude with the help of [1] that Λ[G] ≡ R1 [G] × R2[G] is invo-clean, as expected. Let us now point (2) be fulfilled. Since G2 = 1, similarly to (1), R2 being invo-clean of characteristic 3 implies that R2[G] is invo-clean, too. In order to prove that R1[G] is invo-clean, we observe that R1 is nil-clean with 2∈ J(R1). According to [7], the group ring R1[G] is also nil-clean. What remains to show is that for any element δ of J (R1[G]) the equality δ2 =2δ is valid. Since in conjunction with the explicit formula quoted above for the Jacobson radical, an arbitrary element in J (R1[G]) has the form j + j'g + r(1 - g), where j, j'∈ J(R1) and r ∈ R1, we have that [j + j'g + r(1 - g)]2 ∈ (J(R1)2 + 2J(R1))[G] + r2(1 - g)2. However, using the given conditions, z2=2z=2z2 and thus z2 =2z=0 for any z∈ J(R1). Consequently, one checks that [j + j’g + r(1 - g)]2 = r2(1 - g)2 = 2r2(1 - g) = 2r(1 - g) = 2[j + j'2, G2 ={1}, P≠{0} with R1 ={0} or R1 is a ring of char(R1)=2 and R2 ={0} or R2 is a ring of char (R2) = 3 or (2.3) |G|=2, P≠{0} with 2r12 =2r1 for all r1∈R1 (in addition 4=0 in R1) and R2 ={0} or R2 is a ring of char (R2) = 3. Proof. If G is trivial, there is nothing to prove because of the validity of the isomorphism R[G]= R, so let us assume hereafter that G is non-trivial. "Necessity." As the feebly invo-cleanness of the group ring R[G] implies the same property for R , utilizing Proposition 2.2 we come to the fact that R[G]=P[G]×K[G] will imply feebly invo-cleanness of both group rings P[G] and K[G] whence P[G] is necessarily invo-clean whereas K[G] is either zero or a subdirect product of a family of copies of the field Z5. After that, under the presence of P∣G | ≠ {0}, we just need apply Proposition 2.3 to deduce the described above things in points (2), (2.2) and (2.3). Letting now P[G] = {0}, we shall deal only with K[G]. To that goal, what we now assert is that the group ring K∣G| having the property x^ = x for all x ∈ K[G] with char (K[G]) = 5 yields that K has the property y^ = y for all y ∈ K with char (K) = 5 and G4 = {1} . Indeed, since K ⊆ K[G] and G ⊆ K[G], this can be extracted elementarily thus substantiating our initial statement after all. "Sufficiency." Item (2) ensures that R∣G | = P[G ] × K [G ] and so it is simple verified that the feebly invo-cleanness of both P[G] and K[G] will assure feebly invo-cleanness of R[G] as well. That is why, we will be concentrated separately on these two group rings. Firstly, the stated above conditions are a guarantor with the aid of Proposition 2.3 that P[G] is invo-clean. Secondly, it is pretty easily seen that as y5 = y and g5 = g for all y∈ K and g∈ G, because K is a subdirect product of copies of the field Z5 possessing characteristic 5 and G4 = {1}, we may conclude that x^ = x holds in K[G] too, as required. This substantiates our former assertion after all. □

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