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Maximal and Prime Ideals
If 𝑅 is a ring and 𝑁 is an ideal in 𝑅 then 𝑅/𝑁 is also a ring (a factor ring). The
question is under what conditions on 𝑅 and 𝑁 will 𝑅/𝑁 have special features
(for example, be an integral domain or a field)?
Ex. If 𝑅 = ℤ, an integral domain, and 𝑁 = 𝑝ℤ, for a prime 𝑝, then
the factor ring ℤ/𝑝ℤ ≅ ℤ𝑝 which is a field.
Ex. The ring ℤ × ℤ is not an integral domain because if 𝑎, 𝑏 ∈ ℤ, and nonzero
(0, 𝑎), (𝑏, 0) ∈ ℤ × ℤ but (0, 𝑎)(𝑏, 0) = (0, 0),
However, let 𝑁 = {(𝑛, 0)| 𝑛 ∈ ℤ}. 𝑁 is an ideal in ℤ × ℤ because
for any (𝑎, 𝑏) ∈ ℤ × ℤ
(𝑎, 𝑏)(𝑛, 0) = (𝑎𝑛, 0) ∈ 𝑁
and (𝑛, 0)(𝑎, 𝑏) = (𝑛𝑎, 0) ∈ 𝑁.
Then (ℤ × ℤ)/𝑁 is isomorphic to ℤ under the map:
(0, 𝑘) + 𝑁 → 𝑘, 𝑘 ∈ ℤ.
Thus the factor ring of a ring can be an integral domain even if the
original ring is not.
Ex. 𝑁 = {0, 5} ⊆ ℤ10 is an ideal of ℤ10 , and ℤ10 /𝑁 has 5 elements:
0 + 𝑁, 1 + 𝑁, 2 + 𝑁, 3 + 𝑁, 4 + 𝑁.
ℤ10 /𝑁 ≅ ℤ5 under the map:
𝑘 + 𝑁 → 𝑘.
Thus if 𝑅 is not even an integral domain it’s still possible for 𝑅/𝑁 to be a
field.
, 2
Ex. ℤ is an integral domain but ℤ/8ℤ ≅ ℤ8 is not.
Thus, a factor ring may have a stronger structure than the original ring
(like the example ℤ10 /𝑁 ≅ ℤ5 ) or a weaker structure than the original
ring (like ℤ/8ℤ ≅ ℤ8 ).
Def. Every non-zero ring 𝑅 has at least two ideals. The entire ring 𝑅 is an ideal,
called the improper ideal of 𝑅. And {0} is an ideal of 𝑅 called the trivial
ideal of 𝑅. A proper, nontrivial ideal of a ring 𝑅 is an ideal 𝑁 of 𝑅 such
that 𝑁 ≠ 𝑅 and 𝑁 ≠ {0}.
Theorem: If 𝑅 is a ring with unity and 𝑁 is an ideal of 𝑅 containing a unit, then
𝑁 = 𝑅.
Proof: Let 𝑁 be an ideal of 𝑅, and suppose that 𝑢 ∈ 𝑁 a unit in 𝑅.
Thus the condition 𝑎𝑁 ⊆ 𝑁 for all 𝑎 ∈ 𝑅 implies that 𝑢 −1 𝑁 ⊆ 𝑁.
Since 𝑢 ∈ 𝑁 ⟹ 𝑢 −1 (𝑢) = 1 ∈ 𝑁.
But then 𝑎𝑁 ⊆ 𝑁 ⟹ 𝑎(1) ⊆ 𝑁 for all 𝑎 ∈ 𝑅.
Thus 𝑁 = 𝑅.
Corollary: A field contains no proper nontrivial ideals.
Proof: Since every non-zero element of a field is a unit, any nontrivial
ideal of a field contains a unit and must equal the field.
Maximal and Prime Ideals
If 𝑅 is a ring and 𝑁 is an ideal in 𝑅 then 𝑅/𝑁 is also a ring (a factor ring). The
question is under what conditions on 𝑅 and 𝑁 will 𝑅/𝑁 have special features
(for example, be an integral domain or a field)?
Ex. If 𝑅 = ℤ, an integral domain, and 𝑁 = 𝑝ℤ, for a prime 𝑝, then
the factor ring ℤ/𝑝ℤ ≅ ℤ𝑝 which is a field.
Ex. The ring ℤ × ℤ is not an integral domain because if 𝑎, 𝑏 ∈ ℤ, and nonzero
(0, 𝑎), (𝑏, 0) ∈ ℤ × ℤ but (0, 𝑎)(𝑏, 0) = (0, 0),
However, let 𝑁 = {(𝑛, 0)| 𝑛 ∈ ℤ}. 𝑁 is an ideal in ℤ × ℤ because
for any (𝑎, 𝑏) ∈ ℤ × ℤ
(𝑎, 𝑏)(𝑛, 0) = (𝑎𝑛, 0) ∈ 𝑁
and (𝑛, 0)(𝑎, 𝑏) = (𝑛𝑎, 0) ∈ 𝑁.
Then (ℤ × ℤ)/𝑁 is isomorphic to ℤ under the map:
(0, 𝑘) + 𝑁 → 𝑘, 𝑘 ∈ ℤ.
Thus the factor ring of a ring can be an integral domain even if the
original ring is not.
Ex. 𝑁 = {0, 5} ⊆ ℤ10 is an ideal of ℤ10 , and ℤ10 /𝑁 has 5 elements:
0 + 𝑁, 1 + 𝑁, 2 + 𝑁, 3 + 𝑁, 4 + 𝑁.
ℤ10 /𝑁 ≅ ℤ5 under the map:
𝑘 + 𝑁 → 𝑘.
Thus if 𝑅 is not even an integral domain it’s still possible for 𝑅/𝑁 to be a
field.
, 2
Ex. ℤ is an integral domain but ℤ/8ℤ ≅ ℤ8 is not.
Thus, a factor ring may have a stronger structure than the original ring
(like the example ℤ10 /𝑁 ≅ ℤ5 ) or a weaker structure than the original
ring (like ℤ/8ℤ ≅ ℤ8 ).
Def. Every non-zero ring 𝑅 has at least two ideals. The entire ring 𝑅 is an ideal,
called the improper ideal of 𝑅. And {0} is an ideal of 𝑅 called the trivial
ideal of 𝑅. A proper, nontrivial ideal of a ring 𝑅 is an ideal 𝑁 of 𝑅 such
that 𝑁 ≠ 𝑅 and 𝑁 ≠ {0}.
Theorem: If 𝑅 is a ring with unity and 𝑁 is an ideal of 𝑅 containing a unit, then
𝑁 = 𝑅.
Proof: Let 𝑁 be an ideal of 𝑅, and suppose that 𝑢 ∈ 𝑁 a unit in 𝑅.
Thus the condition 𝑎𝑁 ⊆ 𝑁 for all 𝑎 ∈ 𝑅 implies that 𝑢 −1 𝑁 ⊆ 𝑁.
Since 𝑢 ∈ 𝑁 ⟹ 𝑢 −1 (𝑢) = 1 ∈ 𝑁.
But then 𝑎𝑁 ⊆ 𝑁 ⟹ 𝑎(1) ⊆ 𝑁 for all 𝑎 ∈ 𝑅.
Thus 𝑁 = 𝑅.
Corollary: A field contains no proper nontrivial ideals.
Proof: Since every non-zero element of a field is a unit, any nontrivial
ideal of a field contains a unit and must equal the field.
