Download Quotient Rings

Abstract Algebra
Instructor: Mohamed Omar
Lecture - Quotient Rings (Dummit & Foote 7.3)
Nov 17
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Math 171
Quotient Rings
Let R be a ring and suppose I is a subring. Just as we did in the study of groups, we can
define the quotient set R/I and ask ourselves what properties we need of I in order to put
a ring structure on the coset space R/I. First, let us define the quotient set. We define the
equivalence relation ∼ on R by
x ∼ y if and only if x − y ∈ I.
Because the abelian group (I, +) is a subgroup of (R, +), this is indeed an equivalence
relation on R. The equivalence class of x ∈ R is denoted [x]. Observe that
[x] = {y ∈ R | x − y ∈ I}
= {y ∈ R | y − x ∈ I}
= {y ∈ R | y = x + i, i ∈ I}
= x + I.
So in order to have a well-defined ring structure on R/I we would want
(x + I) + (y + I) = (x + y) + I, (x + I)(y + I) = xy + I.
When does this happen? Since (I, +) is a subgroup of (R, +) and both are abelian, (x +
I) + (y + I) = (x + y) + I automatically holds. So, we need to ask ourselves which subgroups
I admit the property on R that (x + I)(y + I) = xy + I for all x, y ∈ R. We will make some
observations that are somewhat weaker (potentially).
• Consider any element in (x + I)(y + I), say (x + i)(y + j) where i, j ∈ I. We have that
(x + i)(y + j) = xy + xj + iy + ij.
Since xj, iy, ij ∈ I, this implies (x + I)(y + I) ⊆ xy + I.
• Let x, y ∈ R be arbitrary. Suppose (x + I)(y + I) = xy + I. Then for any i, j ∈ I we
have xy + xj + iy + ij ∈ xy + I. Hence, for any i, j ∈ I, we must have xj + iy + ij ∈ I,
or equivalently, xj + iy ∈ I. Since this is true for any i, j ∈ I it is particularly true
for when i = 0, which implies xj ∈ I for every j ∈ I. Similarly setting j = 0 and
letting i ∈ I we have iy ∈ I for all i ∈ I. Since x, y were arbitrary this is true for all
x, y ∈ R, so I absorbs products by R.
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The subrings that absorb multiplication by R are given special names.
Definition. Let R be a ring. A subring I of R is called an ideal if for all r ∈ R and i ∈ I
we have ri ∈ I and ir ∈ I.
Theorem 1 If R is a ring and I is an ideal, the quotient set R/I with operations
(x + I) ⊕ (y + I) = (x + y) + I, (x + I) ⊗ (y + I) = xy + I
is a ring.
Proof. Assignment 9.
Theorem 2 Let R be a ring. Then the subring I is an ideal if and only if I = ker(φ) where
φ is a homomorphism whose domain is R.
Proof. (=⇒) If I is an ideal of R, then R/I is a ring. Define the homomorphism φ : R →
R/I by
φ(x) = x + I.
Then ker(φ) is precisely the coset I.
(⇐=) Suppose I = ker(φ) for some homomorphism φ. We want to show I is an ideal. This
was effectively the content of the quiz.
Example.
1. Let n be a positive integer. Then nZ is a subring of Z. Observe that nZ
is an ideal. Indeed, if r ∈ Z and x ∈ nZ then rx = xr is a multiple of n and hence is
in Z. The quotient ring Z/nZ consists of the elements
0 + nZ, 1 + nZ, 2 + nZ, · · · , (n − 1) + nZ
with addition and multiplication mod n.
2. Let R be a ring. Consider the polynomial ring R[x]. The set of polynomials in R[x]
with constant term 0 form an ideal. Indeed if two polynomials p(x), q(x) have constant
term 0, then their sum does. Furthermore, if r(x) ∈ R[x] and p(x) ∈ I, then r(x)p(x)
and p(x)r(x) also has constant term 0. Call this ideal I. We can then form the ring
R[x]/I. As a set,
R[x]/I = {a + I | a ∈ R}.
From this, one can see R[x]/I ∼
= R.
3. The fact that the projection map φ : Z → Z/nZ is a ring homomorphism has important consequence in number theory. Indeed, suppose one wants to determine all
integer solutions to the equation
x2 + y 2 = 3z 2 .
If such a solution exists, then applying φ, we have
(φ(x))2 + (φ(y))2 = 3(φ(z))2 ,
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so we can look at the original equation “mod n” for any n. For instance, if one looks at
the equation mod 4 in this instance, we get a contradiction. In algebra, particularly in
applications to cryptography, this idea is employed to try to factor large polynomials
with integer coefficients, i.e. factor polynomials in Z[x]. The idea here is to use a
natural projection into Z/pZ[x] for various primes p and factor in this new ring.
4. Consider the ring R[x], and let p(x) be any polynomial. Define
I := {f (x)p(x) | f (x) ∈ R[x]}.
Then I is an ideal, so we can form the quotient ring R[x]/I. If deg(p(x)) = n then
the set R[x]/I is
R[x]/I = {a0 + a1 x + · · · + an−1 xn−1 + I | ai ∈ R}.
To see how multiplication works, let’s pick an example. Say p(x) = x2 + 1. Then
((2 + x) + I)((3 − 4x) + I) = (2 + x)(3 − 4x) + I = (6 − 5x − 4x2 ) + I.
But (4x2 ) + I = −4 + I so
(6 − 5x − 4x2 ) + I = (6 − 5x + 4) + I = (10 − 5x) + I
5. Consider the subset of R2 carved out by
V = {(x, y) | x2 = y 3 }.
If we wanted to define real-valued polynomials on V , we could do so by using polynomials in R[X, Y ] by we would want that the polynomial X 2 − Y 3 and all multiples of
it are 0. Hence, in algebraic geometry, we define the polynomials on V to be precisely
the polynomial ring R[X, Y ]/(X 2 − Y 3 ).
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