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Homomorphisms and Factor/Quotient Rings
Just as we discussed group homomorphisms and factor/quotient groups, there
are analogous notions for rings.
Recall that:
Def. A map 𝜙 of a ring 𝑅 into a ring 𝑅′ is a (ring) homomorphism if:
𝜙(𝑎 + 𝑏) = 𝜙(𝑎) + 𝜙(𝑏) and
𝜙(𝑎𝑏) = 𝜙(𝑎)𝜙(𝑏) for all 𝑎, 𝑏 ∈ 𝑅.
We saw earlier that 𝜙: ℤ → ℤ𝑛 by 𝜙(𝑚) = 𝑚 (𝑚𝑜𝑑 𝑛) is a ring
homomorphism.
Ex. Projection homomorphism: Let 𝑅1 , 𝑅2 , … , 𝑅𝑛 be rings for each 𝑖, the map:
𝜋𝑖 : 𝑅1 × 𝑅2 × … × 𝑅𝑛 → 𝑅𝑖
𝜋𝑖 (𝑟1 , 𝑟2 , … , 𝑟𝑛 ) = 𝑟𝑖
is a homomorphism. This homomorphism projects an element in
𝑅1 × 𝑅2 × … × 𝑅𝑛 on to its 𝑖𝑡ℎ component.
𝜋𝑖 is a homomorphism because addition and multiplication in
𝑅1 × 𝑅2 × … × 𝑅𝑛 are defined componentwise. For example:
𝜋𝑖 ((𝑟1 , 𝑟2 , … , 𝑟𝑛 ) + (𝑠1 , 𝑠2 , … , 𝑠𝑛 )) = 𝜋𝑖 (𝑟1 + 𝑠1 , 𝑟2 + 𝑠2 , … , 𝑟𝑛 + 𝑠𝑛 )
= 𝑟𝑖 + 𝑠𝑖
= 𝜋𝑖 (𝑟1 , 𝑟2 , … , 𝑟𝑛 ) + 𝜋𝑖 (𝑠1 , 𝑠2 , … , 𝑠𝑛 ),
for any(𝑟1 , … , 𝑟𝑛 ), (𝑠1 , … , 𝑠𝑛 ) ∈ 𝑅1 × … × 𝑅𝑛 .
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Theorem: Let 𝜙 be a homomorphism of a ring 𝑅 into a ring 𝑅′.
1. If 0 is the additive identity in 𝑅, then 𝜙(0) = 0′ is the additive
identity in 𝑅′.
2. If 𝑎 ∈ 𝑅, then 𝜙(−𝑎) = −𝜙(𝑎).
3. If 𝑆 is a subring of 𝑅, then 𝜙[𝑆] is a subring of 𝑅′.
4. If 𝑆′ is a subring of 𝑅′ then 𝜙 −1 [𝑆 ′ ] is a subring of 𝑅.
5. If 𝑅 has unity 1, then 𝜙(1) is unity for 𝜙[𝑅].
Proof: 1. and 2. follow from the theorem on pages 5-6 of the section called
Group Homomorphisms (which I‘ll refer to as the Group Homomorphism
theorem), since 𝜙 is a group homomorphism on (𝑅, +).
For 3. and 4., by the Group Homomorphism theorem, 𝜙[𝑆, +] is a subgroup of
𝑅′ and 𝜙 −1 [𝑆 ′ , +′] is a subgroup of 𝑅. Thus we only need to show that 𝜙[𝑆]
and 𝜙 −1 [𝑆 ′ ] are closed under multiplication.
3. If 𝜙(𝑥1 ), 𝜙(𝑥2 ) ∈ 𝜙[𝑆] then 𝜙(𝑥1 )𝜙(𝑥2 ) = 𝜙(𝑥1 𝑥2 ) ∈ 𝜙[𝑆]
4. If 𝑥1 , 𝑥2 ∈ 𝜙 −1 [𝑆′] then 𝜙(𝑥1 𝑥2 ) = 𝜙(𝑥1 )𝜙(𝑥2 ) ∈ 𝑆′
so 𝑥1 𝑥2 ∈ 𝜙 −1 [𝑆 ′ ].
For 5. Notice that :
𝜙(𝑥 ) = 𝜙(1𝑥 ) = 𝜙(1)𝜙(𝑥 )
𝜙(𝑥 ) = 𝜙(𝑥1) = 𝜙(𝑥 )𝜙(1)
So 𝜙(1) is unity for 𝜙[𝑅].
Homomorphisms and Factor/Quotient Rings
Just as we discussed group homomorphisms and factor/quotient groups, there
are analogous notions for rings.
Recall that:
Def. A map 𝜙 of a ring 𝑅 into a ring 𝑅′ is a (ring) homomorphism if:
𝜙(𝑎 + 𝑏) = 𝜙(𝑎) + 𝜙(𝑏) and
𝜙(𝑎𝑏) = 𝜙(𝑎)𝜙(𝑏) for all 𝑎, 𝑏 ∈ 𝑅.
We saw earlier that 𝜙: ℤ → ℤ𝑛 by 𝜙(𝑚) = 𝑚 (𝑚𝑜𝑑 𝑛) is a ring
homomorphism.
Ex. Projection homomorphism: Let 𝑅1 , 𝑅2 , … , 𝑅𝑛 be rings for each 𝑖, the map:
𝜋𝑖 : 𝑅1 × 𝑅2 × … × 𝑅𝑛 → 𝑅𝑖
𝜋𝑖 (𝑟1 , 𝑟2 , … , 𝑟𝑛 ) = 𝑟𝑖
is a homomorphism. This homomorphism projects an element in
𝑅1 × 𝑅2 × … × 𝑅𝑛 on to its 𝑖𝑡ℎ component.
𝜋𝑖 is a homomorphism because addition and multiplication in
𝑅1 × 𝑅2 × … × 𝑅𝑛 are defined componentwise. For example:
𝜋𝑖 ((𝑟1 , 𝑟2 , … , 𝑟𝑛 ) + (𝑠1 , 𝑠2 , … , 𝑠𝑛 )) = 𝜋𝑖 (𝑟1 + 𝑠1 , 𝑟2 + 𝑠2 , … , 𝑟𝑛 + 𝑠𝑛 )
= 𝑟𝑖 + 𝑠𝑖
= 𝜋𝑖 (𝑟1 , 𝑟2 , … , 𝑟𝑛 ) + 𝜋𝑖 (𝑠1 , 𝑠2 , … , 𝑠𝑛 ),
for any(𝑟1 , … , 𝑟𝑛 ), (𝑠1 , … , 𝑠𝑛 ) ∈ 𝑅1 × … × 𝑅𝑛 .
, 2
Theorem: Let 𝜙 be a homomorphism of a ring 𝑅 into a ring 𝑅′.
1. If 0 is the additive identity in 𝑅, then 𝜙(0) = 0′ is the additive
identity in 𝑅′.
2. If 𝑎 ∈ 𝑅, then 𝜙(−𝑎) = −𝜙(𝑎).
3. If 𝑆 is a subring of 𝑅, then 𝜙[𝑆] is a subring of 𝑅′.
4. If 𝑆′ is a subring of 𝑅′ then 𝜙 −1 [𝑆 ′ ] is a subring of 𝑅.
5. If 𝑅 has unity 1, then 𝜙(1) is unity for 𝜙[𝑅].
Proof: 1. and 2. follow from the theorem on pages 5-6 of the section called
Group Homomorphisms (which I‘ll refer to as the Group Homomorphism
theorem), since 𝜙 is a group homomorphism on (𝑅, +).
For 3. and 4., by the Group Homomorphism theorem, 𝜙[𝑆, +] is a subgroup of
𝑅′ and 𝜙 −1 [𝑆 ′ , +′] is a subgroup of 𝑅. Thus we only need to show that 𝜙[𝑆]
and 𝜙 −1 [𝑆 ′ ] are closed under multiplication.
3. If 𝜙(𝑥1 ), 𝜙(𝑥2 ) ∈ 𝜙[𝑆] then 𝜙(𝑥1 )𝜙(𝑥2 ) = 𝜙(𝑥1 𝑥2 ) ∈ 𝜙[𝑆]
4. If 𝑥1 , 𝑥2 ∈ 𝜙 −1 [𝑆′] then 𝜙(𝑥1 𝑥2 ) = 𝜙(𝑥1 )𝜙(𝑥2 ) ∈ 𝑆′
so 𝑥1 𝑥2 ∈ 𝜙 −1 [𝑆 ′ ].
For 5. Notice that :
𝜙(𝑥 ) = 𝜙(1𝑥 ) = 𝜙(1)𝜙(𝑥 )
𝜙(𝑥 ) = 𝜙(𝑥1) = 𝜙(𝑥 )𝜙(1)
So 𝜙(1) is unity for 𝜙[𝑅].
