Theorem 1. Let I be an ideal of R. Then R/I is a ring under the operations
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(r + I) + (s + I) = (r + s) + I
(r + I)(s + I) = (rs) + I
.
Proof. If the operations are well-defined, then the ring axioms are simple to check, since they’re
just inherited from the properties of R as a ring. Definition: A subring I of R with the property that ri ∈ I and ir ∈ I for any r ∈ R and i ∈ I
is called an ideal.
We also sometimes talk about left and right ideals. A left ideal is one that absorbs multiplication
from the left (ri ∈ I for any r ∈ R), while a right ideal absorbs multiplication from the right (ir ∈ I
for any r ∈ R.) (Note: R/I is not a ring for ideals that are only one-sided, or for subrings I that
are not ideals.)
The analogy here is that ideals are to rings as normal subgroups are to groups. Just like normal
subgroups are our favorite subgroups, ideals are our favorite type of subring, because it makes sense
to take the quotient by them.
Example: Consider the ring Z. Then 6Z is an ideal of Z. The quotient Z/6Z is arithmetic mod
6.
Example: Let Z[x] be the polynomial ring over Z in one variable. We can consider Z to be
a subring of Z[x] (the constant polynomials). However, Z is not an ideal. Hence, we cannot put
sensical operations on the cosets of Z in Z[x]
However, consider the set
I = {a1 x + a2 x2 + . . . + an xn | n ∈ N, ai ∈ Z}.
These are the polynomials with zero constant term. These do form an ideal in Z[x]. Can we figure
out what the quotient ring is? Let’s think about what R/I looks like - two polynomials f and g
are in the same coset if f − g has zero constant term, i.e. if f and g have the same constant term.
So the equivalence classes in R/I are the sets of polynomials that all have the same constant term.
Thus the cosets of I in R are:
{n + I | n ∈ Z}
. It’s not hard to see that R/I is isomorphic to Z.
Now, if ideals are so much like normal subgroups, what about the isomorphism theorems?
Short answer: Yes.
All the isomorphism theorems for groups have analogues in the world of rings as well.
Theorem 2. Let φ : R → S be a homomorphism. Then
=(φ) ∼
= R/ ker(φ).
Theorem 3. Let A be a subring of R and let B be an ideal of R. Then
A + B = {a + b | a ∈ A, b ∈ B}
is a subring of R, A ∩ B is an ideal of A, and
A/(A ∩ B) ∼
= (A + B)/B.
Theorem 4. Let R be a ring and let I and J be ideals of R with I ⊆ J. Then
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(R/I)/(J/I) ∼
= R/J.
Theorem 5. Let I be an ideal of R. Then there is a bijection between the subrings of R containing
I and the subrings of R/I given by
A ↔ A/I.
This bijection respects inclusion and A is an ideal of R if and only if A/I is an ideal of R/I.
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Ideals
Let’s take a closer look at these ideal thingies.
Lemma 6. Let R be a ring and let I and J be ideals of R.
• I + J = {i + j |i ∈ I, j ∈ J} is an ideal of R.
• IJ = {i1 j1 + . . . in jn | ik ∈ I, jk ∈ J} is an ideal of R
• I ∩ J is an ideal of R, which contains IJ.
Proof. Let’s check IJ as an example. If I add two things of this form, I get another one, just
longer. Negatives of elements that are in IJ are also in IJ, since −ij = (−i)j and −i ∈ I. If I
multiply on the left by any element of R:
n
n
n
X
X
X
r(
ik jk ) =
(rik )jk =
i0k jk .
k=1
k=1
k=1
Multiplying on the right is the same. The other two cases are similar. Examples.
In Z, our favorite ideals are nZ. (In fact, these are the only ideals.)
What is 2Z + 3Z? We get Z. In general, since GCD(m, n) can be written as am + bn for some
a, b, mZ + nZ = GCD(m, n)Z.
What is (2Z)(3Z)? We get 6Z. In general, (mZ)(nZ) = mnZ.
(mZ) ∩ (nZ) = LCM(m, n)Z.
How do we make ideals? In groups, we could take a single element and look at all its powers to
make a subgroup. Let’s see how this works in rings.
For now, assume that R has an identity.
Definition: The ideal generated by a, denoted (a), is defined to be the smallest ideal containing
the element a.
How can we think about building such a thing? Assume R is a ring with identity. We know
I should be closed under multiplication, so powers of a should be in there. But more is true - I
should actually absorb multiplication from both sides, so for any r, s ∈ R, we should have ras ∈ I.
So, is {ras | r ∈ R, s ∈ S} an ideal? Unfortunately, not quite. It is likely not closed under
addition. So how about
I = {r1 as1 + r2 as2 + . . . rn asn | ri , sj ∈ R}?
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Luckily, yes, this is an ideal. If I add up two things in here, I just get something longer in there.
And if I multiply on the left or right by anything in r, I’ll just have different ri or si ’s, but they’ll
still be elements of R. So that is the ideal generated by a.