Corrigendum

This article refers to:

Goldie Extending Rings

Article title: Goldie Extending Rings

Authors: Evrim Akalan, Gary F. Birkenmeier, and Adnan Tercan

Journal: Communications in Algebra

Bibliometrics: Volume 40, Issue 2, pages 423–428

DOI: 10.1080/00927872.2010.529096

Evrim Akalan ¹ , Gary F. Birkenmeier ² , and Adnan Tercan ¹

¹ Department of Mathematics, Hacettepe University, Beytepe Campus, Ankara, Turkey

² Department of Mathematics, University of Louisiana at Lafayette, Lafayette, LA, USA

This corrigendum is written to correct the statements and proofs of Theorem 4, Theorem 9, and Corollary 10 of Akalan et al. [2].

Key Words: Goldie extending rings; Split-null extension; Trivial extension.

2010 Mathematics Subject Classification: Primary: 16D10, 16D50; Secondary: 16D80.

In the following revised and corrected version of Theorem 4, S(R, M) denotes the split-null extension (also called the trivial extension) of M by R, where M is a (R, R) −bimodule. Also a ring R is Abelian if every idempotent is central.

Theorem 4.

Let R be a ring, M be an ideal of R and S denotes S(R, M).

(i) If S_S is (𝒢 −)extending, then R_R is (𝒢 −)extending,

(ii) If _RM is faithful and R_R is 𝒢 −extending, then S_S is 𝒢 −extending,

(iii) If _RM is faithful, R_R is extending, and R is Abelian, then S_S is extending and Abelian.

Proof First observe that for any (R, R) −bimodule M and (e, m) = (e, m)² ∈ S, then e = e ² ∈ R, em + me = m and eme = 0. So (e, me) = (e, me)², (e, m)(e, me) = (e, me) and (e, me)(e, m) = (e, m). Hence (e, m)S = (e, me)S.

(i) Assume that S _S is extending. Let X _R  ≤ R _R . Then (X, X ∩ M) _S  ≤ S _S . There exists (e, m) = (e, m)² ∈ S such that (X, X ∩ M) _S  ≤ ^ess (e, m)S _S . Since (x, 0) = (e, m)(x, 0), for all x ∈ X, then ex = x for all x ∈ X and mX = 0. So (X, X ∩ M) _S  ≤ (e, 0)S _S .

Let 0 ≠ er ∈ eR. We need to find t ∈ R such that 0 ≠ ert ∈ X. Since (X, X ∩ M) _S  ≤ ^ess (e, m)S _S , there exists (v, k) ∈ S such that 0 ≠ (e, me)(r, 0)(v, k) = (erv, erk + merv) ∈ (X, X ∩ M). So (erv, erk + merv) = (e, 0)(erv, erk + merv) = (erv, erk). Hence 0 ≠ erv ∈ X or 0 ≠ erk ∈ X ∩ M ⊆ X. Therefore R _R is extending.

Next assume that S _S is 𝒢 −extending. Let X _R  ≤ R _R . Then there exists (e, m) = (e, m)² ∈ S such that [(X, X ∩ M) ∩ (e, m)S] _S  ≤ ^ess (e, m)S _S and [(X, X ∩ M) ∩ (e, m)S] _S  ≤ ^ess (X, X ∩ M) _S . To show that R _R is 𝒢 −extending, let 0 ≠ er ∈ eR and 0 ≠ y ∈ X. We need to find t, v ∈ R such that 0 ≠ ert ∈ X ∩ eR and 0 ≠ yv ∈ X ∩ eR. Since [(X, X ∩ M) ∩ (e, m)S] _S  ≤ ^ess (e, m)S _S there exists (t, k) ∈ S such that 0 ≠ (e, me)(r, 0)(t, k) = (ert, erk + mert) ∈ (X, X ∩ M) ∩ (e, m)S. If 0 ≠ ert, then 0 ≠ ert ∈ X ∩ eR. Otherwise ert = mert = 0 and 0 ≠ erk ∈ X ∩ eR. Therefore (X ∩ eR) _R  ≤ ^ess eR. Since [(X, X ∩ M) ∩ (e, m)S] _S  ≤ ^ess (X, X ∩ M) _S , there exists (a, h) ∈ S such that 0 ≠ (y, 0)(a, h) = (ya, yh) ∈ (X, X ∩ M) ∩ (e, m)S. Then 0 ≠ (ya, yh) = (e, me)(ya, yh) = (eya, eyh + meya). If 0 ≠ eya, then 0 ≠ eya ∈ X ∩ eR. Otherwise eya = meya = 0 and 0 ≠ eyh ∈ X ∩ eR. Therefore (X ∩ eR) _R  ≤ ^ess X _R . Consequently, R _R is 𝒢 −extending.

(ii) Assume _R M is faithful and R _R is 𝒢 −extending. Recall from [3, Lemma 4(iv)] that ({0} × M) _S  ≤ ^ess S _S . Let 0 ≠ Y _S  ≤ S _S . Observe that [Y ∩ ({0} × M)] _S  ≤ ^ess Y _S . Consider the set X = {m ∈ M|(0, m) ∈ Y}. Then X _R  ≤ R _R . Since R _R is 𝒢 −extending, there exists e = e ² ∈ R such that (X ∩ eR) _R  ≤ ^ess X _R and (X ∩ eR) _R  ≤ ^ess eR _R . Let V = {(0, x) ∈ S|x ∈ X ∩ eR}. Note that V _S  ≤ Y _S . We will show that V _S  ≤ ^ess Y _S and V _S  ≤ ^ess (e, 0)S. Let 0 ≠ (0, y) ∈ Y ∩ ({0} × M). Hence y ∈ X. So there exists a ∈ R such that 0 ≠ ya ∈ X ∩ eR. Then 0 ≠ (0, y)(a, 0) = (0, ya) ∈ V. Thus V _S  ≤ ^ess [Y ∩ ({0} × M)] _S , so V _S  ≤ ^ess Y _S .

Now let 0 ≠ (e, 0)(r, k) = (er, ek) ∈ (e, 0)S. If 0 ≠ er, there exists b ∈ R such that 0 ≠ erb ∈ X ∩ eR. Since _R M is faithful, there exists m ₁ ∈ M such that 0 ≠ erbm ₁ ∈ X ∩ eR. Then 0 ≠ (er, ek)(0, bm ₁) = (0, erbm ₁) ∈ V. So assume 0 = er. Then 0 ≠ ek. So there exists d ∈ R such that 0 ≠ ekd ∈ X ∩ eR. Hence 0 ≠ (0, ek)(d, 0) = (0, ekd) ∈ V. So V _S  ≤ ^ess (e, 0)S _S . By [1, Proposition 1.5], S _S is 𝒢 −extending.

(iii) If f = f ² ∈ S, then f = (e, m) where e = e ² ∈ R and m = em + me. Since R is Abelian and eme = 0, then m = 0. Hence f = (e, 0). So f is central in S. Therefore S is Abelian. By part (ii) and [1, Proposition 1.8(iii)], S _S is extending.

In the last sentence of the statement of Theorem 9, after “unity” insert “and ”.

In the second sentence of the last paragraph of the proof of Theorem 9, replace “” with “”.

In the last paragraph of the proof of Theorem 9, replace sentences 6 and 7 with the following: “Then f = (b, 0) where b = b ² ∈ T, or f = (−e, 1) where e = e ² ∈ T. But (b, 0)(a, x) ≠ (a, x) and (−e, 1)(a, x) ≠ (a, x).”

In the statement of Corollary 10 parts (i) and (ii) after “Then” insert “T is a ℤ-algebra and”.

In the second last sentence of the first paragraph of the proof of Corollary 10(i), replace “(e, 0)T _T ” with “”.

In sentence 8 of the proof of Claim 1 in the proof of Corollary 10(ii), replace “ ∈ ” with “ ⊆ ”.

REFERENCES

[1] Akalan, E., Birkenmeier, G. F., Tercan, A. (2009). Goldie extending modules. Comm. Algebra 37:663–683.

[2] Akalan, E., Birkenmeier, G. F., Tercan, A. (2012). Goldie extending rings. Comm. Algebra 40:423–428.

[3] Birkenmeier, G. F. (1989). Split-null extensions of strongly right bounded rings. Publicaciones Matematiques 33:37–44.