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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Duo Group Rings
Yuanlin Li
Brock University
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Contents
1 Introduction and Preliminaries
2 Characteristic Zero
3 Characteristic 2
4 General Case
5 References
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Some Notations1. R – Associative ring with 1.2. kG – Group ring over a commutative ring k.3. Mn(R) – n × n matrix ring over R.Definition 1Let R be an associative ring with iden-tity.Call R left (right) duo if every left (right) ideal is anideal, and call R duo if it is both left and right duo.DefineR to be reversible if αβ = 0 implies βα = 0, and sym-metric if αβγ = 0 implies αγβ = 0 for all α, β, γ ∈R.Finally, say that R has the “SI” property if αβ = 0implies αRβ = 0 for all α, β ∈ R.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Remark Let k be a commutative ring with identity andG be any group. Using the standard involution ∗ on thegroup ring kG, defined by (
∑aigi)
∗ =∑
aig−1i for all
ai ∈ k and gi ∈ G, we see that kG is left duo if and onlyif it is right duo.Marks [3] has clarified the relationships among duo,reversible and symmetric rings. Moreover, he provedthe following result [3, Proposition 6].Proposition 1.1 Let k be a commutative ring withidentity, and let G be a finite group. Then the groupring kG is reversible if and only if kG has the “SI” prop-erty.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Remark Let k be a commutative ring with identity andG be any group. Using the standard involution ∗ on thegroup ring kG, defined by (
∑aigi)
∗ =∑
aig−1i for all
ai ∈ k and gi ∈ G, we see that kG is left duo if and onlyif it is right duo.Marks [3] has clarified the relationships among duo,reversible and symmetric rings. Moreover, he provedthe following result [3, Proposition 6].Proposition 1.1 Let k be a commutative ring withidentity, and let G be a finite group. Then the groupring kG is reversible if and only if kG has the “SI” prop-erty.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Remark Let k be a commutative ring with identity andG be any group. Using the standard involution ∗ on thegroup ring kG, defined by (
∑aigi)
∗ =∑
aig−1i for all
ai ∈ k and gi ∈ G, we see that kG is left duo if and onlyif it is right duo.Marks [3] has clarified the relationships among duo,reversible and symmetric rings. Moreover, he provedthe following result [3, Proposition 6].Proposition 1.1 Let k be a commutative ring withidentity, and let G be a finite group. Then the groupring kG is reversible if and only if kG has the “SI” prop-erty.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Remark Let k be a commutative ring with identity andG be any group. Using the standard involution ∗ on thegroup ring kG, defined by (
∑aigi)
∗ =∑
aig−1i for all
ai ∈ k and gi ∈ G, we see that kG is left duo if and onlyif it is right duo.Marks [3] has clarified the relationships among duo,reversible and symmetric rings. Moreover, he provedthe following result [3, Proposition 6].Proposition 1.1 Let k be a commutative ring withidentity, and let G be a finite group. Then the groupring kG is reversible if and only if kG has the “SI” prop-erty.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Remark Let k be a commutative ring with identity andG be any group. Using the standard involution ∗ on thegroup ring kG, defined by (
∑aigi)
∗ =∑
aig−1i for all
ai ∈ k and gi ∈ G, we see that kG is left duo if and onlyif it is right duo.Marks [3] has clarified the relationships among duo,reversible and symmetric rings. Moreover, he provedthe following result [3, Proposition 6].Proposition 1.1 Let k be a commutative ring withidentity, and let G be a finite group. Then the groupring kG is reversible if and only if kG has the “SI” prop-erty.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Remark Let k be a commutative ring with identity andG be any group. Using the standard involution ∗ on thegroup ring kG, defined by (
∑aigi)
∗ =∑
aig−1i for all
ai ∈ k and gi ∈ G, we see that kG is left duo if and onlyif it is right duo.Marks [3] has clarified the relationships among duo,reversible and symmetric rings. Moreover, he provedthe following result [3, Proposition 6].Proposition 1.1 Let k be a commutative ring withidentity, and let G be a finite group. Then the groupring kG is reversible if and only if kG has the “SI” prop-erty.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
Remark Let k be a commutative ring with identity andG be any group. Using the standard involution ∗ on thegroup ring kG, defined by (
∑aigi)
∗ =∑
aig−1i for all
ai ∈ k and gi ∈ G, we see that kG is left duo if and onlyif it is right duo.Marks [3] has clarified the relationships among duo,reversible and symmetric rings. Moreover, he provedthe following result [3, Proposition 6].Proposition 1.1 Let k be a commutative ring withidentity, and let G be a finite group. Then the groupring kG is reversible if and only if kG has the “SI” prop-erty.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
We note that this result remains valid for anarbitrary group G.We also note that the “SI” property is simplythe statement that left annihilators and rightannihilators are ideals, hence it is obvious thatduo rings have the “SI” property. It now fol-lows from Proposition 4 that if kG is a duoring, then it is reversible.However, the converse is not true, as the fol-lowing example shows.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
We note that this result remains valid for anarbitrary group G.We also note that the “SI” property is simplythe statement that left annihilators and rightannihilators are ideals, hence it is obvious thatduo rings have the “SI” property. It now fol-lows from Proposition 4 that if kG is a duoring, then it is reversible.However, the converse is not true, as the fol-lowing example shows.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
We note that this result remains valid for anarbitrary group G.We also note that the “SI” property is simplythe statement that left annihilators and rightannihilators are ideals, hence it is obvious thatduo rings have the “SI” property. It now fol-lows from Proposition 4 that if kG is a duoring, then it is reversible.However, the converse is not true, as the fol-lowing example shows.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
We note that this result remains valid for anarbitrary group G.We also note that the “SI” property is simplythe statement that left annihilators and rightannihilators are ideals, hence it is obvious thatduo rings have the “SI” property. It now fol-lows from Proposition 4 that if kG is a duoring, then it is reversible.However, the converse is not true, as the fol-lowing example shows.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
We note that this result remains valid for anarbitrary group G.We also note that the “SI” property is simplythe statement that left annihilators and rightannihilators are ideals, hence it is obvious thatduo rings have the “SI” property. It now fol-lows from Proposition 4 that if kG is a duoring, then it is reversible.However, the converse is not true, as the fol-lowing example shows.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
We note that this result remains valid for anarbitrary group G.We also note that the “SI” property is simplythe statement that left annihilators and rightannihilators are ideals, hence it is obvious thatduo rings have the “SI” property. It now fol-lows from Proposition 4 that if kG is a duoring, then it is reversible.However, the converse is not true, as the fol-lowing example shows.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
We note that this result remains valid for anarbitrary group G.We also note that the “SI” property is simplythe statement that left annihilators and rightannihilators are ideals, hence it is obvious thatduo rings have the “SI” property. It now fol-lows from Proposition 4 that if kG is a duoring, then it is reversible.However, the converse is not true, as the fol-lowing example shows.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
We note that this result remains valid for anarbitrary group G.We also note that the “SI” property is simplythe statement that left annihilators and rightannihilators are ideals, hence it is obvious thatduo rings have the “SI” property. It now fol-lows from Proposition 4 that if kG is a duoring, then it is reversible.However, the converse is not true, as the fol-lowing example shows.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
example 1.1 Let Q8 = 〈a, b|a4 = 1, a2 =b2, ab = a−1〉 be the quaternion group of order8. The integral group ring ZQ8 is a reversiblering, but not a duo ring.Proof. It follows from Theorem 3.1 in [1] thatthe rational group algebra QQ8 is reversible.As a subring of QQ8, ZQ8 is clearly reversible.R = ZQ8 is not a duo ring since the left idealR(a + 2b) generated by a + 2b is not a rightideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
example 1.1 Let Q8 = 〈a, b|a4 = 1, a2 =b2, ab = a−1〉 be the quaternion group of order8. The integral group ring ZQ8 is a reversiblering, but not a duo ring.Proof. It follows from Theorem 3.1 in [1] thatthe rational group algebra QQ8 is reversible.As a subring of QQ8, ZQ8 is clearly reversible.R = ZQ8 is not a duo ring since the left idealR(a + 2b) generated by a + 2b is not a rightideal.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
example 1.1 Let Q8 = 〈a, b|a4 = 1, a2 =b2, ab = a−1〉 be the quaternion group of order8. The integral group ring ZQ8 is a reversiblering, but not a duo ring.Proof. It follows from Theorem 3.1 in [1] thatthe rational group algebra QQ8 is reversible.As a subring of QQ8, ZQ8 is clearly reversible.R = ZQ8 is not a duo ring since the left idealR(a + 2b) generated by a + 2b is not a rightideal.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
example 1.1 Let Q8 = 〈a, b|a4 = 1, a2 =b2, ab = a−1〉 be the quaternion group of order8. The integral group ring ZQ8 is a reversiblering, but not a duo ring.Proof. It follows from Theorem 3.1 in [1] thatthe rational group algebra QQ8 is reversible.As a subring of QQ8, ZQ8 is clearly reversible.R = ZQ8 is not a duo ring since the left idealR(a + 2b) generated by a + 2b is not a rightideal.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
example 1.1 Let Q8 = 〈a, b|a4 = 1, a2 =b2, ab = a−1〉 be the quaternion group of order8. The integral group ring ZQ8 is a reversiblering, but not a duo ring.Proof. It follows from Theorem 3.1 in [1] thatthe rational group algebra QQ8 is reversible.As a subring of QQ8, ZQ8 is clearly reversible.R = ZQ8 is not a duo ring since the left idealR(a + 2b) generated by a + 2b is not a rightideal.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
example 1.1 Let Q8 = 〈a, b|a4 = 1, a2 =b2, ab = a−1〉 be the quaternion group of order8. The integral group ring ZQ8 is a reversiblering, but not a duo ring.Proof. It follows from Theorem 3.1 in [1] thatthe rational group algebra QQ8 is reversible.As a subring of QQ8, ZQ8 is clearly reversible.R = ZQ8 is not a duo ring since the left idealR(a + 2b) generated by a + 2b is not a rightideal.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
example 1.1 Let Q8 = 〈a, b|a4 = 1, a2 =b2, ab = a−1〉 be the quaternion group of order8. The integral group ring ZQ8 is a reversiblering, but not a duo ring.Proof. It follows from Theorem 3.1 in [1] thatthe rational group algebra QQ8 is reversible.As a subring of QQ8, ZQ8 is clearly reversible.R = ZQ8 is not a duo ring since the left idealR(a + 2b) generated by a + 2b is not a rightideal.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 In this talk, we shall study when the groupalgebras KG of torsion groups over fields areduo rings.
2 Our main result is that the group algebra KGis a duo ring if and only if KG is reversible.
3 It was shown in [1] that if the group algebraKG of a torsion group G is reversible, theneither G is an abelian group, or G is a Hamil-tonian group and the characteristic of K is 0or 2. If G is abelian, then clearly RG is bothreversible and duo.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 In this talk, we shall study when the groupalgebras KG of torsion groups over fields areduo rings.
2 Our main result is that the group algebra KGis a duo ring if and only if KG is reversible.
3 It was shown in [1] that if the group algebraKG of a torsion group G is reversible, theneither G is an abelian group, or G is a Hamil-tonian group and the characteristic of K is 0or 2. If G is abelian, then clearly RG is bothreversible and duo.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 In this talk, we shall study when the groupalgebras KG of torsion groups over fields areduo rings.
2 Our main result is that the group algebra KGis a duo ring if and only if KG is reversible.
3 It was shown in [1] that if the group algebraKG of a torsion group G is reversible, theneither G is an abelian group, or G is a Hamil-tonian group and the characteristic of K is 0or 2. If G is abelian, then clearly RG is bothreversible and duo.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 In this talk, we shall study when the groupalgebras KG of torsion groups over fields areduo rings.
2 Our main result is that the group algebra KGis a duo ring if and only if KG is reversible.
3 It was shown in [1] that if the group algebraKG of a torsion group G is reversible, theneither G is an abelian group, or G is a Hamil-tonian group and the characteristic of K is 0or 2. If G is abelian, then clearly RG is bothreversible and duo.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 In this talk, we shall study when the groupalgebras KG of torsion groups over fields areduo rings.
2 Our main result is that the group algebra KGis a duo ring if and only if KG is reversible.
3 It was shown in [1] that if the group algebraKG of a torsion group G is reversible, theneither G is an abelian group, or G is a Hamil-tonian group and the characteristic of K is 0or 2. If G is abelian, then clearly RG is bothreversible and duo.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 In this talk, we shall study when the groupalgebras KG of torsion groups over fields areduo rings.
2 Our main result is that the group algebra KGis a duo ring if and only if KG is reversible.
3 It was shown in [1] that if the group algebraKG of a torsion group G is reversible, theneither G is an abelian group, or G is a Hamil-tonian group and the characteristic of K is 0or 2. If G is abelian, then clearly RG is bothreversible and duo.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 In this talk, we shall study when the groupalgebras KG of torsion groups over fields areduo rings.
2 Our main result is that the group algebra KGis a duo ring if and only if KG is reversible.
3 It was shown in [1] that if the group algebraKG of a torsion group G is reversible, theneither G is an abelian group, or G is a Hamil-tonian group and the characteristic of K is 0or 2. If G is abelian, then clearly RG is bothreversible and duo.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 We may always assume that K has charac-teristic 0 or 2 and G is a Hamiltonian group,i.e. G = Q8 × E2 × E′
2, where Q8 is the quater-nion group of order 8, E2 is an elementaryabelian 2-group, and E′
2 is an abelian groupall of whose elements are of odd order.
2 We first deal with the special case when G =Q8 in the next two sections.
3 The general case will be handled in the lastsection. As mentioned earlier, Q8 = 〈a, b|a4 =1, a2 = b2, bab−1 = a−1〉.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 We may always assume that K has charac-teristic 0 or 2 and G is a Hamiltonian group,i.e. G = Q8 × E2 × E′
2, where Q8 is the quater-nion group of order 8, E2 is an elementaryabelian 2-group, and E′
2 is an abelian groupall of whose elements are of odd order.
2 We first deal with the special case when G =Q8 in the next two sections.
3 The general case will be handled in the lastsection. As mentioned earlier, Q8 = 〈a, b|a4 =1, a2 = b2, bab−1 = a−1〉.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 We may always assume that K has charac-teristic 0 or 2 and G is a Hamiltonian group,i.e. G = Q8 × E2 × E′
2, where Q8 is the quater-nion group of order 8, E2 is an elementaryabelian 2-group, and E′
2 is an abelian groupall of whose elements are of odd order.
2 We first deal with the special case when G =Q8 in the next two sections.
3 The general case will be handled in the lastsection. As mentioned earlier, Q8 = 〈a, b|a4 =1, a2 = b2, bab−1 = a−1〉.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 We may always assume that K has charac-teristic 0 or 2 and G is a Hamiltonian group,i.e. G = Q8 × E2 × E′
2, where Q8 is the quater-nion group of order 8, E2 is an elementaryabelian 2-group, and E′
2 is an abelian groupall of whose elements are of odd order.
2 We first deal with the special case when G =Q8 in the next two sections.
3 The general case will be handled in the lastsection. As mentioned earlier, Q8 = 〈a, b|a4 =1, a2 = b2, bab−1 = a−1〉.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 We may always assume that K has charac-teristic 0 or 2 and G is a Hamiltonian group,i.e. G = Q8 × E2 × E′
2, where Q8 is the quater-nion group of order 8, E2 is an elementaryabelian 2-group, and E′
2 is an abelian groupall of whose elements are of odd order.
2 We first deal with the special case when G =Q8 in the next two sections.
3 The general case will be handled in the lastsection. As mentioned earlier, Q8 = 〈a, b|a4 =1, a2 = b2, bab−1 = a−1〉.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 We may always assume that K has charac-teristic 0 or 2 and G is a Hamiltonian group,i.e. G = Q8 × E2 × E′
2, where Q8 is the quater-nion group of order 8, E2 is an elementaryabelian 2-group, and E′
2 is an abelian groupall of whose elements are of odd order.
2 We first deal with the special case when G =Q8 in the next two sections.
3 The general case will be handled in the lastsection. As mentioned earlier, Q8 = 〈a, b|a4 =1, a2 = b2, bab−1 = a−1〉.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Introduction and Preliminaries
1 We may always assume that K has charac-teristic 0 or 2 and G is a Hamiltonian group,i.e. G = Q8 × E2 × E′
2, where Q8 is the quater-nion group of order 8, E2 is an elementaryabelian 2-group, and E′
2 is an abelian groupall of whose elements are of odd order.
2 We first deal with the special case when G =Q8 in the next two sections.
3 The general case will be handled in the lastsection. As mentioned earlier, Q8 = 〈a, b|a4 =1, a2 = b2, bab−1 = a−1〉.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic Zero
Let G = Q8 and K be a field of characteristic 0. The fol-lowing result can be derived from some known results ingroup rings along with the Wedderburn decomposition ofKQ8.Theorem 2.1 The following statements are equivalent:
(1) R = KQ8 is duo.(2) R = KQ8 is reversible.(3) The equation 1 + x2 + y2 = 0 has no solutions in K.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic Zero
Let G = Q8 and K be a field of characteristic 0. The fol-lowing result can be derived from some known results ingroup rings along with the Wedderburn decomposition ofKQ8.Theorem 2.1 The following statements are equivalent:
(1) R = KQ8 is duo.(2) R = KQ8 is reversible.(3) The equation 1 + x2 + y2 = 0 has no solutions in K.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic Zero
1 Proof: It follows from Lemma 7.4.9 in [4] that KQ8
has the Wedderburn decomposition KQ8∼= K ⊕ K ⊕
K ⊕ K ⊕ H(K),where H(K) is the quaternion algebraover K.
2 By Theorem 7.4.6 in [4], H(K) is either a division ringD or the 2 × 2 full matrix ring M2(K) over K,and thelatter possibility arises if and only if x2 + y2 + 1 = 0 hassolutions in K.
3 Thus KQ8 is duo (or reversible) iff H(K) is duo (or re-versible) iff H(K) is a division ring iff x2 + y2 +1 = 0 hasno solutions in K.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic Zero
1 Proof: It follows from Lemma 7.4.9 in [4] that KQ8
has the Wedderburn decomposition KQ8∼= K ⊕ K ⊕
K ⊕ K ⊕ H(K),where H(K) is the quaternion algebraover K.
2 By Theorem 7.4.6 in [4], H(K) is either a division ringD or the 2 × 2 full matrix ring M2(K) over K,and thelatter possibility arises if and only if x2 + y2 + 1 = 0 hassolutions in K.
3 Thus KQ8 is duo (or reversible) iff H(K) is duo (or re-versible) iff H(K) is a division ring iff x2 + y2 +1 = 0 hasno solutions in K.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic Zero
1 Proof: It follows from Lemma 7.4.9 in [4] that KQ8
has the Wedderburn decomposition KQ8∼= K ⊕ K ⊕
K ⊕ K ⊕ H(K),where H(K) is the quaternion algebraover K.
2 By Theorem 7.4.6 in [4], H(K) is either a division ringD or the 2 × 2 full matrix ring M2(K) over K,and thelatter possibility arises if and only if x2 + y2 + 1 = 0 hassolutions in K.
3 Thus KQ8 is duo (or reversible) iff H(K) is duo (or re-versible) iff H(K) is a division ring iff x2 + y2 +1 = 0 hasno solutions in K.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic Zero
1 Proof: It follows from Lemma 7.4.9 in [4] that KQ8
has the Wedderburn decomposition KQ8∼= K ⊕ K ⊕
K ⊕ K ⊕ H(K),where H(K) is the quaternion algebraover K.
2 By Theorem 7.4.6 in [4], H(K) is either a division ringD or the 2 × 2 full matrix ring M2(K) over K,and thelatter possibility arises if and only if x2 + y2 + 1 = 0 hassolutions in K.
3 Thus KQ8 is duo (or reversible) iff H(K) is duo (or re-versible) iff H(K) is a division ring iff x2 + y2 +1 = 0 hasno solutions in K.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic Zero
1 Proof: It follows from Lemma 7.4.9 in [4] that KQ8
has the Wedderburn decomposition KQ8∼= K ⊕ K ⊕
K ⊕ K ⊕ H(K),where H(K) is the quaternion algebraover K.
2 By Theorem 7.4.6 in [4], H(K) is either a division ringD or the 2 × 2 full matrix ring M2(K) over K,and thelatter possibility arises if and only if x2 + y2 + 1 = 0 hassolutions in K.
3 Thus KQ8 is duo (or reversible) iff H(K) is duo (or re-versible) iff H(K) is a division ring iff x2 + y2 +1 = 0 hasno solutions in K.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic Zero
1 Proof: It follows from Lemma 7.4.9 in [4] that KQ8
has the Wedderburn decomposition KQ8∼= K ⊕ K ⊕
K ⊕ K ⊕ H(K),where H(K) is the quaternion algebraover K.
2 By Theorem 7.4.6 in [4], H(K) is either a division ringD or the 2 × 2 full matrix ring M2(K) over K,and thelatter possibility arises if and only if x2 + y2 + 1 = 0 hassolutions in K.
3 Thus KQ8 is duo (or reversible) iff H(K) is duo (or re-versible) iff H(K) is a division ring iff x2 + y2 +1 = 0 hasno solutions in K.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
Theorem 3.1 The following statements are equivalent:
(1) KQ8 is duo.(2) KQ8 is reversible.(3) The equation 1 + x + x2 = 0 has no solutions in K.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Proof: Note that (1) ⇒ (2) ⇒ (3) is clear.We shallprove (3) ⇒ (1) by using the following two lemmas.
2 For any v ∈ R = KQ8, we shall prove that Rv is anideal, and so KQ8 is left duo and thus duo.Write v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b, where each ai, aj ∈ K.If theaugmentation ε(v) of v is not zero, then v is a unit, andso clearly Rv is an ideal.
3 Next we assume that ε(v) = 0, and we shall verifythat both va and vb are in Rv. The crucial step is toshow that va + av = βv for some β =
∑3i=0 a′
iai +
(∑7
j=4 a′ja
j−4)b ∈ KG.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 From now on, we always assume that 1 + x + x2 = 0has no solutions in K.
2 Lemma 3.2 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
with ε(v) = 0. If∑3
i=0 ai 6= 0, then va, vb ∈ Rv and thusRv is an ideal.
3 Remark 3.3 We note that since v = (a0 + a4b + a2b2 +a6b3) + (a1 + a7b + a3b2 + a5b3)a, if a0 + a2 + a4 + a6 6= 0,by the symmetry of a and b, Lemma 3.2 shows that Rvis an ideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 From now on, we always assume that 1 + x + x2 = 0has no solutions in K.
2 Lemma 3.2 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
with ε(v) = 0. If∑3
i=0 ai 6= 0, then va, vb ∈ Rv and thusRv is an ideal.
3 Remark 3.3 We note that since v = (a0 + a4b + a2b2 +a6b3) + (a1 + a7b + a3b2 + a5b3)a, if a0 + a2 + a4 + a6 6= 0,by the symmetry of a and b, Lemma 3.2 shows that Rvis an ideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 From now on, we always assume that 1 + x + x2 = 0has no solutions in K.
2 Lemma 3.2 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
with ε(v) = 0. If∑3
i=0 ai 6= 0, then va, vb ∈ Rv and thusRv is an ideal.
3 Remark 3.3 We note that since v = (a0 + a4b + a2b2 +a6b3) + (a1 + a7b + a3b2 + a5b3)a, if a0 + a2 + a4 + a6 6= 0,by the symmetry of a and b, Lemma 3.2 shows that Rvis an ideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 From now on, we always assume that 1 + x + x2 = 0has no solutions in K.
2 Lemma 3.2 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
with ε(v) = 0. If∑3
i=0 ai 6= 0, then va, vb ∈ Rv and thusRv is an ideal.
3 Remark 3.3 We note that since v = (a0 + a4b + a2b2 +a6b3) + (a1 + a7b + a3b2 + a5b3)a, if a0 + a2 + a4 + a6 6= 0,by the symmetry of a and b, Lemma 3.2 shows that Rvis an ideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 From now on, we always assume that 1 + x + x2 = 0has no solutions in K.
2 Lemma 3.2 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
with ε(v) = 0. If∑3
i=0 ai 6= 0, then va, vb ∈ Rv and thusRv is an ideal.
3 Remark 3.3 We note that since v = (a0 + a4b + a2b2 +a6b3) + (a1 + a7b + a3b2 + a5b3)a, if a0 + a2 + a4 + a6 6= 0,by the symmetry of a and b, Lemma 3.2 shows that Rvis an ideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 From now on, we always assume that 1 + x + x2 = 0has no solutions in K.
2 Lemma 3.2 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
with ε(v) = 0. If∑3
i=0 ai 6= 0, then va, vb ∈ Rv and thusRv is an ideal.
3 Remark 3.3 We note that since v = (a0 + a4b + a2b2 +a6b3) + (a1 + a7b + a3b2 + a5b3)a, if a0 + a2 + a4 + a6 6= 0,by the symmetry of a and b, Lemma 3.2 shows that Rvis an ideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 From now on, we always assume that 1 + x + x2 = 0has no solutions in K.
2 Lemma 3.2 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
with ε(v) = 0. If∑3
i=0 ai 6= 0, then va, vb ∈ Rv and thusRv is an ideal.
3 Remark 3.3 We note that since v = (a0 + a4b + a2b2 +a6b3) + (a1 + a7b + a3b2 + a5b3)a, if a0 + a2 + a4 + a6 6= 0,by the symmetry of a and b, Lemma 3.2 shows that Rvis an ideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 From now on, we always assume that 1 + x + x2 = 0has no solutions in K.
2 Lemma 3.2 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
with ε(v) = 0. If∑3
i=0 ai 6= 0, then va, vb ∈ Rv and thusRv is an ideal.
3 Remark 3.3 We note that since v = (a0 + a4b + a2b2 +a6b3) + (a1 + a7b + a3b2 + a5b3)a, if a0 + a2 + a4 + a6 6= 0,by the symmetry of a and b, Lemma 3.2 shows that Rvis an ideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 From now on, we always assume that 1 + x + x2 = 0has no solutions in K.
2 Lemma 3.2 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
with ε(v) = 0. If∑3
i=0 ai 6= 0, then va, vb ∈ Rv and thusRv is an ideal.
3 Remark 3.3 We note that since v = (a0 + a4b + a2b2 +a6b3) + (a1 + a7b + a3b2 + a5b3)a, if a0 + a2 + a4 + a6 6= 0,by the symmetry of a and b, Lemma 3.2 shows that Rvis an ideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Next we may assume that both a0 + a1 + a2 + a3 anda0 + a2 + a4 + a6 are zero. Since ε(v) = 0,the above isequivalent to a0 + a2 = a4 + a6 = a1 + a3 = a5 + a7(∗).
2 Lemma 3.4 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
such that (∗) holds. Then va, vb ∈ Rv and thus Rv is anideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Next we may assume that both a0 + a1 + a2 + a3 anda0 + a2 + a4 + a6 are zero. Since ε(v) = 0,the above isequivalent to a0 + a2 = a4 + a6 = a1 + a3 = a5 + a7(∗).
2 Lemma 3.4 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
such that (∗) holds. Then va, vb ∈ Rv and thus Rv is anideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Next we may assume that both a0 + a1 + a2 + a3 anda0 + a2 + a4 + a6 are zero. Since ε(v) = 0,the above isequivalent to a0 + a2 = a4 + a6 = a1 + a3 = a5 + a7(∗).
2 Lemma 3.4 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
such that (∗) holds. Then va, vb ∈ Rv and thus Rv is anideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Next we may assume that both a0 + a1 + a2 + a3 anda0 + a2 + a4 + a6 are zero. Since ε(v) = 0,the above isequivalent to a0 + a2 = a4 + a6 = a1 + a3 = a5 + a7(∗).
2 Lemma 3.4 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
such that (∗) holds. Then va, vb ∈ Rv and thus Rv is anideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Next we may assume that both a0 + a1 + a2 + a3 anda0 + a2 + a4 + a6 are zero. Since ε(v) = 0,the above isequivalent to a0 + a2 = a4 + a6 = a1 + a3 = a5 + a7(∗).
2 Lemma 3.4 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
such that (∗) holds. Then va, vb ∈ Rv and thus Rv is anideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Characteristic 2
1 Next we may assume that both a0 + a1 + a2 + a3 anda0 + a2 + a4 + a6 are zero. Since ε(v) = 0,the above isequivalent to a0 + a2 = a4 + a6 = a1 + a3 = a5 + a7(∗).
2 Lemma 3.4 Let v =∑3
i=0 aiai + (∑7
j=4 ajaj−4)b ∈ KQ8
such that (∗) holds. Then va, vb ∈ Rv and thus Rv is anideal.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
Theorem 4.1 Let K be a field and let G be a torsion group. ThenKG is a duo ring if and only if one of the following conditionsholds.
(1) G is Abelian.(2) G = Q8 × E2 × E′
2 is Hamiltonian, the characteristic of K is0 and the equation 1 + x2 + y2 = 0 has no solutions in anycyclotomic field K(ξd) for any odd d which is an order of anelement of E′
2.(3) G = Q8 × E′
2, the characteristic of K is 2 and the equation1 + x + x2 = 0 has no solutions in any cyclotomic field K(ξd)for any odd d which is an order of an element of E′
2.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
Theorem 4.1 Let K be a field and let G be a torsion group. ThenKG is a duo ring if and only if one of the following conditionsholds.
(1) G is Abelian.(2) G = Q8 × E2 × E′
2 is Hamiltonian, the characteristic of K is0 and the equation 1 + x2 + y2 = 0 has no solutions in anycyclotomic field K(ξd) for any odd d which is an order of anelement of E′
2.(3) G = Q8 × E′
2, the characteristic of K is 2 and the equation1 + x + x2 = 0 has no solutions in any cyclotomic field K(ξd)for any odd d which is an order of an element of E′
2.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
Theorem 4.1 Let K be a field and let G be a torsion group. ThenKG is a duo ring if and only if one of the following conditionsholds.
(1) G is Abelian.(2) G = Q8 × E2 × E′
2 is Hamiltonian, the characteristic of K is0 and the equation 1 + x2 + y2 = 0 has no solutions in anycyclotomic field K(ξd) for any odd d which is an order of anelement of E′
2.(3) G = Q8 × E′
2, the characteristic of K is 2 and the equation1 + x + x2 = 0 has no solutions in any cyclotomic field K(ξd)for any odd d which is an order of an element of E′
2.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
Theorem 4.1 Let K be a field and let G be a torsion group. ThenKG is a duo ring if and only if one of the following conditionsholds.
(1) G is Abelian.(2) G = Q8 × E2 × E′
2 is Hamiltonian, the characteristic of K is0 and the equation 1 + x2 + y2 = 0 has no solutions in anycyclotomic field K(ξd) for any odd d which is an order of anelement of E′
2.(3) G = Q8 × E′
2, the characteristic of K is 2 and the equation1 + x + x2 = 0 has no solutions in any cyclotomic field K(ξd)for any odd d which is an order of an element of E′
2.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
Theorem 4.1 Let K be a field and let G be a torsion group. ThenKG is a duo ring if and only if one of the following conditionsholds.
(1) G is Abelian.(2) G = Q8 × E2 × E′
2 is Hamiltonian, the characteristic of K is0 and the equation 1 + x2 + y2 = 0 has no solutions in anycyclotomic field K(ξd) for any odd d which is an order of anelement of E′
2.(3) G = Q8 × E′
2, the characteristic of K is 2 and the equation1 + x + x2 = 0 has no solutions in any cyclotomic field K(ξd)for any odd d which is an order of an element of E′
2.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
1 Note We need only consider that G is a non-abelian torsiongroup. If KG is duo, then it is reversible. Thus G = Q8×E2×E′
2 and char(K) is 0 or 2.2 If E′
2 is finite, then
KE′2∼= (
∏d| |E′
2|
K(ξd)ad)
where ξd is a primitive dth root of unity over K and ad isequal to the number of elements of order d in E′
2 divided bythe degree of K(ξd) over K, i.e. ad = nd/[K(ξd) : K].
3 Corollary 4.2 KG is duo if and only if KG is reversible.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
1 Note We need only consider that G is a non-abelian torsiongroup. If KG is duo, then it is reversible. Thus G = Q8×E2×E′
2 and char(K) is 0 or 2.2 If E′
2 is finite, then
KE′2∼= (
∏d| |E′
2|
K(ξd)ad)
where ξd is a primitive dth root of unity over K and ad isequal to the number of elements of order d in E′
2 divided bythe degree of K(ξd) over K, i.e. ad = nd/[K(ξd) : K].
3 Corollary 4.2 KG is duo if and only if KG is reversible.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
1 Note We need only consider that G is a non-abelian torsiongroup. If KG is duo, then it is reversible. Thus G = Q8×E2×E′
2 and char(K) is 0 or 2.2 If E′
2 is finite, then
KE′2∼= (
∏d| |E′
2|
K(ξd)ad)
where ξd is a primitive dth root of unity over K and ad isequal to the number of elements of order d in E′
2 divided bythe degree of K(ξd) over K, i.e. ad = nd/[K(ξd) : K].
3 Corollary 4.2 KG is duo if and only if KG is reversible.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
1 Note We need only consider that G is a non-abelian torsiongroup. If KG is duo, then it is reversible. Thus G = Q8×E2×E′
2 and char(K) is 0 or 2.2 If E′
2 is finite, then
KE′2∼= (
∏d| |E′
2|
K(ξd)ad)
where ξd is a primitive dth root of unity over K and ad isequal to the number of elements of order d in E′
2 divided bythe degree of K(ξd) over K, i.e. ad = nd/[K(ξd) : K].
3 Corollary 4.2 KG is duo if and only if KG is reversible.
tu-logo
Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
General Case
1 Note We need only consider that G is a non-abelian torsiongroup. If KG is duo, then it is reversible. Thus G = Q8×E2×E′
2 and char(K) is 0 or 2.2 If E′
2 is finite, then
KE′2∼= (
∏d| |E′
2|
K(ξd)ad)
where ξd is a primitive dth root of unity over K and ad isequal to the number of elements of order d in E′
2 divided bythe degree of K(ξd) over K, i.e. ad = nd/[K(ξd) : K].
3 Corollary 4.2 KG is duo if and only if KG is reversible.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
References
M. Gutan and A. Kisielewicz, Reversible group rings, J. Algebra 279(2004), 280–291.
Y. Li and M.M. Parmenter, Reversible group rings over commutative rings,Submitted.
G. Marks, Reversible and symmetric rings, J. Pure Appl. Algebra 174(2002), 311 - 318.
C. Polcino Milies and S.K. Sehgal, An Introduction to Group Rings,Kluwer Academic Publishers, Dordrecht, 2002.
S.K. Sehgal, Topics in Group Rings, Marcel Dekker, New York, 1978.
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Contents Introduction and Preliminaries Characteristic Zero Characteristic 2 General Case References
Thank you!AUTHOR: Yuanlin LiADDRESS: Department of Mathematics
Brock UniversityCanada
EMAIL: [email protected]