Ideals
In our study of group theory, we learned that we cannot simply form a factor group using
any subgroup; we require a specific kind of structure called a normal subgroup.
Similarly, in ring theory, to construct a new structure by "dividing" a ring, we need a special
kind of subring called an Ideal.
Recall that a normal subgroup \(H\) of a group \(G\) is characterized by the property that
\(gHg^{-1} = H\) for all \(g \in G\), which ensures that left and right cosets coincide.
For rings, the analogous condition must account for both operations. The key insight is that
an ideal must "absorb" multiplication from the entire ring, not just from itself.
Definition: Ideal
A subring \(A\) of a ring \(R\) is called a (two-sided) ideal of \(R\) if for every
\(r \in R\) and every \(a \in A\), both \(ra\) and \(ar\) are in \(A\).
The condition "\(ra \in A\) and \(ar \in A\)" is the "absorption" property. Unlike a mere subring,
which only needs to be closed under multiplication within itself, an ideal must remain closed
when multiplied by any element of the parent ring. This stronger condition is precisely
what guarantees that the coset multiplication in the factor ring will be well-defined.
To verify that a subset is an ideal, we use a streamlined criterion analogous to the subgroup test.
Theorem: Ideal Test
A nonempty subset \(A\) of a ring \(R\) is an ideal of \(R\) if
- \(a - b \in A\) whenever \(a, b \in A\).
- \(ra\) and \(ar\) are in \(A\) whenever \(a \in A\) and \(r \in R\).
Condition (1) ensures that \(A\) is a subgroup under addition (using the one-step subgroup test),
while condition (2) is the absorption property. Together, they guarantee that \(A\) is a subring
with the additional absorption requirement.
Example: Ideals in \(\mathbb{Z}\)
Consider the set \(n\mathbb{Z} = \{nk \mid k \in \mathbb{Z}\}\) for a fixed integer \(n\).
This is the set of all multiples of \(n\).
Verification using the Ideal Test:
-
For any \(nk_1, nk_2 \in n\mathbb{Z}\), we have
\(nk_1 - nk_2 = n(k_1 - k_2) \in n\mathbb{Z}\).
-
For any \(nk \in n\mathbb{Z}\) and any \(r \in \mathbb{Z}\), we have
\(r(nk) = n(rk) \in n\mathbb{Z}\). Since \(\mathbb{Z}\) is commutative,
\((nk)r = r(nk) \in n\mathbb{Z}\) as well.
Thus \(n\mathbb{Z}\) is an ideal of \(\mathbb{Z}\).
Such ideals, generated by a single element, are called principal ideals
and denoted \(\langle n \rangle\).
Definition: Principal Ideal
Let \(R\) be a commutative ring with unity and let \(a \in R\). The set
\[
\langle a \rangle = \{ra \mid r \in R\}
\]
is an ideal of \(R\) called the principal ideal generated by \(a\).
Every ideal of \(\mathbb{Z}\) is principal - a fact that makes \(\mathbb{Z}\) particularly well-behaved.
However, not all rings share this property. For instance, let \(\mathbb{Z}[x]\) denote the ring of all
polynomials with integer coefficients. (This is called a polynomial ring.) The ideal
\[
I = \langle x, 2 \rangle = \{2f(x) + xg(x) \mid f(x), g(x) \in \mathbb{Z}[x]\}
\]
consists of all polynomials with even constant term and cannot be generated by a single element.
Factor Rings
The concept of a factor ring (or quotient ring) is the ring-theoretic generalization of modular arithmetic.
In basic arithmetic, we construct \(\mathbb{Z}_n\) by taking the integers \(\mathbb{Z}\) and identifying
all multiples of \(n\) with zero. In the language of rings, we say we "mod out" by the principal ideal
\(\langle n \rangle = n\mathbb{Z}\).
We can extend this idea to any ring \(R\) and any ideal \(A\). By treating the elements of \(A\) as "zero,"
we collapse the ring into a simplified structure \(R/A\). The elements of this new ring are cosets -
translates of the ideal that partition the original ring. This construction is one of the most powerful tools in algebra,
allowing us to manufacture new fields (like Galois Fields for cryptography) by carefully choosing the ring \(R\)
and the ideal \(A\).
Definition: Factor Ring (Quotient Ring)
Let \(R\) be a ring and \(A\) an ideal of \(R\). The factor ring (or quotient ring)
\(R/A\) is defined as:
\[
R/A = \{r + A \mid r \in R\}
\]
where \(r + A = \{r + a \mid a \in A\}\) is the coset of \(A\) containing \(r\). The ring operations are defined by:
\[
(s + A) + (t + A) = (s + t) + A
\]
and
\[
(s + A)(t + A) = st + A.
\]
For these operations to be valid, they must be well-defined: the result must not depend on
which representative we choose from each coset. This is precisely where the ideal property becomes essential.
Theorem: Existence of Factor Rings
Let \(R\) be a ring and let \(A\) be a subring of \(R\). The set of cosets \(\{r + A \mid r \in R\}\) is
a ring under the operations
\[
(s + A) + (t + A) = (s + t) + A
\]
and
\[
(s + A)(t + A) = st + A
\]
if and only if \(A\) is an ideal of \(R\).
Proof:
First, suppose \(A\) is an ideal and let \(s + A = s' + A\) and \(t + A = t' + A\).
By definition, \(s = s' + a\) and \(t = t' + b\) for some \(a, b \in A\).
We need to show \(st + A = s't' + A\). Then
\[
st = (s' + a)(t' + b) = s't' + at' + s'b + ab
\]
and since \(A\) absorbs \(at' + s'b + ab\),
\[
st + A = s't' + at' + s'b + ab + A = s't' + A.
\]
Thus, multiplication is well-defined when \(A\) is an ideal.
We prove the second direction by contrapositive: if \(A\) is not an ideal, then the cosets
do not form a ring under the given operations.
Suppose \(A\) is a subring of \(R\) but not an ideal. Then there exist \(a \in A\) and
\(r \in R\) such that \(ar \notin A\) or \(ra \notin A\). For convenience, say \(ar \notin A\).
Since \(a \in A\), we have \(a + A = 0 + A\). Now consider the products:
\[
(a + A)(r + A) = ar + A \neq A \quad \text{(since } ar \notin A \text{)}
\]
\[
(0 + A)(r + A) = 0 \cdot r + A = A
\]
The same cosets \(a + A = 0 + A\) and \(r + A\) yield different products.
Thus multiplication is not well-defined, and the set of cosets is not a ring.
Example: \(\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n\)
Consider the factor ring \(\mathbb{Z}/n\mathbb{Z}\) where \(n\mathbb{Z} = \langle n \rangle\).
The cosets are:
\[
0 + n\mathbb{Z}, \quad 1 + n\mathbb{Z}, \quad 2 + n\mathbb{Z}, \quad \ldots, \quad (n-1) + n\mathbb{Z}
\]
These are exactly the residue classes modulo \(n\). The operations in \(\mathbb{Z}/n\mathbb{Z}\)
coincide with modular arithmetic in \(\mathbb{Z}_n\), confirming that they are isomorphic as rings.
Prime & Maximal Ideals
Not all factor rings are created equal. If we mod out \(\mathbb{Z}\) by \(\langle 6 \rangle\), we get \(\mathbb{Z}_6\),
which has zero-divisors (since \(2 \cdot 3 = 6 \equiv 0\)). However, if we mod out by \(\langle 5 \rangle\), we get
\(\mathbb{Z}_5\), which is a perfect field. What is the fundamental difference between the ideal
\(\langle 6 \rangle\) and the ideal \(\langle 5 \rangle\)?
The answer lies in the classification of ideals. Just as integers can be composite or prime, ideals can
be classified based on their internal structure. This section introduces prime ideals and
maximal ideals. These concepts provide a precise algebraic criterion for characterizing
the structure of factor rings. Prime ideals yield integral domains, while maximal ideals yield fields. This
correspondence is the engine behind constructing the finite fields used in AES encryption and Elliptic Curve Cryptography.
Definition: Prime Ideal
A prime ideal \(A\) of a commutative ring \(R\) is a proper ideal of \(R\) such that
for all \(a, b \in R\), if \(ab \in A\), then \(a \in A\) or \(b \in A\).
The definition mirrors the property of prime numbers: if a prime \(p\) divides a product \(ab\),
then \(p\) must divide \(a\) or \(b\). Indeed, in \(\mathbb{Z}\), an ideal \(\langle n \rangle\) is prime
if and only if \(n\) is a prime number (or \(n = 0\)).
Definition: Maximal Ideal
A maximal ideal \(A\) of a commutative ring \(R\) is a proper ideal of \(R\) such that,
whenever \(B\) is an ideal of \(R\) and \(A \subseteq B \subseteq R\), then \(B = A\) or
\(B = R\).
A maximal ideal is "as large as possible" without being the entire ring. There is no ideal
strictly between a maximal ideal and the ring itself. In \(\mathbb{Z}\), the maximal ideals are
exactly the ideals \(\langle p \rangle\) where \(p\) is prime - a fact that reflects the
intimate connection between primality and maximality in principal ideal domains.
The following two theorems reveal the power of these definitions: they provide an algebraic criterion
for determining the structure of the factor ring.
Theorem: \(R/A\) is an Integral Domain \(\iff\) \(A\) is Prime
Let \(R\) be a commutative ring with unity and let \(A\) be an ideal of \(R\). Then \(R/A\) is
an integral domain if and only if \(A\) is a prime ideal.
Proof:
First, suppose \(R/A\) is an integral domain. Let \(ab \in A\) for some \(a, b \in R\).
Then in \(R/A\):
\[
(a + A)(b + A) = ab + A = A \quad \text{(the zero element of } R/A\text{)}
\]
Since \(R/A\) has no zero-divisors, either \(a + A = A\) or \(b + A = A\).
This means \(a \in A\) or \(b \in A\), so \(A\) is prime.
Next, suppose \(A\) is prime and \((a + A)(b + A) = 0 + A = A\) in \(R/A\).
Then \(ab + A = A\), which means \(ab \in A\). Since \(A\) is prime, \(a \in A\) or \(b \in A\).
Thus, either \(a + A = A\) or \(b + A = A\) is the zero coset in \(R/A\), so \(R/A\) has no zero-divisors.
Theorem: \(R/A\) is a Field \(\iff\) \(A\) is Maximal
Let \(R\) be a commutative ring with unity and let \(A\) be an ideal of \(R\). Then \(R/A\) is
a field if and only if \(A\) is a maximal ideal.
Proof:
First, suppose \(R/A\) is a field and \(B\) is an ideal of \(R\) that properly contains
\(A\) (i.e., \(A \subsetneq B \subseteq R\)). Choose \(b \in B\) but \(b \notin A\).
Then \(b + A \neq A\) is a nonzero element of the field \(R/A\),
so it has a multiplicative inverse, say \(c + A\), such that \((b + A)(c + A) = 1 + A\).
This means \(1 - bc \in A \subset B\). Since \(b \in B\) and \(B\) is an ideal, \(bc \in B\).
Therefore \(1 = (1 - bc) + bc \in B\). Since \(1 \in B\), we have \(B = R\). Thus, \(A\)
is maximal.
Next, suppose \(A\) is maximal and let \(b \in R\) but \(b \notin A\). We must show that
\(b + A\) has a multiplicative inverse.
Consider \(B = \{br + a \mid r \in R, a \in A\}\). This is an ideal of \(R\) that properly contains
\(A\). Since \(A\) is maximal, we must have \(B = R\). Thus \(1 \in B\), say \(1 = bc + a'\), where
\(a' \in A\). Then
\[
1 + A = bc + a' + A = bc + A = (b + A)(c + A).
\]
This shows that \(c + A\) is the multiplicative inverse of \(b + A\). Since every nonzero element
of \(R/A\) has a multiplicative inverse, \(R/A\) is a field.
Corollary: Maximal Ideals are Prime
Every maximal ideal in a commutative ring with unity is a prime ideal.
This follows immediately: if \(A\) is maximal, then \(R/A\) is a field, which is certainly an integral domain,
so \(A\) is prime. The converse is false in general; for example, \(\langle 0 \rangle\) is prime in \(\mathbb{Z}\)
(since \(\mathbb{Z}\) is an integral domain) but not maximal.
Ring Homomorphisms
In Structural Group Theory, we established that
homomorphisms are the "structure-preserving maps" that allow us to compare different groups. We saw that the
kernel of a group homomorphism is not just any subgroup, but specifically a normal subgroup.
We now extend this concept to rings. Because a ring involves two operations - addition and multiplication - a
ring homomorphism must preserve the structure of both operations simultaneously.
Definition: Ring Homomorphism & Isomorphism
A ring homomorphism \(\phi\) from a ring \(R\) to a ring \(S\) is a mapping from \(R\) to \(S\) that
preserves the two ring operations. That is, for all \(a, b \in R\):
- \(\phi(a + b) = \phi(a) + \phi(b)\)
- \(\phi(ab) = \phi(a)\phi(b)\)
A ring homomorphism that is both one-to-one and onto is called a
ring isomorphism.
Just as with groups, the kernel of a ring homomorphism captures the elements that are "collapsed to zero."
Definition: Kernel of a Ring Homomorphism
Let \(\phi: R \to S\) be a ring homomorphism. The kernel of \(\phi\) is:
\[
\text{Ker } \phi = \{r \in R \mid \phi(r) = 0\}
\]
The following theorem reveals why ideals are the "correct" ring-theoretic analogue of normal subgroups:
the kernel of any ring homomorphism is always an ideal.
Theorem: Kernels are Ideals
Let \(\phi: R \to S\) be a ring homomorphism. Then \(\text{Ker } \phi\) is an ideal of \(R\).
Proof:
Since \(\phi\) is a ring homomorphism, \(\text{Ker } \phi\) is a subgroup of \(R\) under addition,
so \(a - b \in \text{Ker } \phi\) for all \(a, b \in \text{Ker } \phi\).
It remains to verify the absorption property.
Let \(a \in \text{Ker } \phi\) and \(r \in R\). Since \(\phi\) preserves multiplication:
\[
\phi(ra) = \phi(r)\phi(a) = \phi(r) \cdot 0 = 0
\]
Similarly, \(\phi(ar) = 0\). Thus \(ra, ar \in \text{Ker } \phi\), so \(\text{Ker } \phi\) is an ideal of \(R\).
The converse is also true. This is analogous to the group-theoretic result that every normal subgroup is the
kernel of some group homomorphism.
Theorem: Ideals are Kernels
Let \(A\) be an ideal of a ring \(R\). Then \(A\) is the kernel of the mapping \(\gamma: R \to R/A\)
defined by \(\gamma(r) = r + A\).
Proof:
First, we verify that \(\gamma\) is a ring homomorphism. For any \(r, s \in R\):
\[
\gamma(r + s) = (r + s) + A = (r + A) + (s + A) = \gamma(r) + \gamma(s)
\]
and
\[
\gamma(rs) = rs + A = (r + A)(s + A) = \gamma(r)\gamma(s)
\]
For the kernel, observe that \(r \in \text{Ker } \gamma\) if and only if \(\gamma(r) = A\) (the zero element of \(R/A\)),
which occurs if and only if \(r + A = A\), that is, \(r \in A\). Thus \(\text{Ker } \gamma = A\).
The mapping \(\gamma\) is called the natural homomorphism from \(R\) to \(R/A\).
Theorem: First Isomorphism Theorem for Rings (Fundamental Theorem of Ring Homomorphisms)
Let \(\phi: R \to S\) be a ring homomorphism. Then the mapping \(\Phi: R/\text{Ker } \phi \to \phi(R)\)
defined by
\[
\Phi(r + \text{Ker } \phi) = \phi(r)
\]
is a ring isomorphism. In symbols,
\[
R / \text{Ker } \phi \cong \phi(R).
\]
Proof:
Let \(\Phi: R/\text{Ker } \phi \to \phi(R)\) be the mapping defined by \(\Phi(r + \text{Ker } \phi) = \phi(r)\).
From the First Isomorphism Theorem for groups, we already know that \(\Phi\) is a well-defined
isomorphism of additive groups. That is, \(\Phi\) is one-to-one, onto, and preserves addition.
It remains to show that \(\Phi\) preserves multiplication.
For any \(r + \text{Ker } \phi\) and \(s + \text{Ker } \phi\) in \(R/\text{Ker } \phi\):
\[
\begin{align*}
\Phi((r + \text{Ker } \phi)(s + \text{Ker } \phi)) &= \Phi(rs + \text{Ker } \phi) \\\\
&= \phi(rs) \\\\
&= \phi(r)\phi(s) \\\\\
&= \Phi(r + \text{Ker } \phi)\Phi(s + \text{Ker } \phi).
\end{align*}
\]
Thus, \(\Phi\) preserves both addition and multiplication, so it is a ring isomorphism.
The First Isomorphism Theorem provides a powerful principle: every ring homomorphism factors through a quotient.
When we map \(R \to S\) via \(\phi\), we are implicitly:
- Collapsing: identifying all elements of \(\text{Ker } \phi\) with zero.
- Partitioning: grouping elements of \(R\) into cosets of \(\text{Ker } \phi\).
- Relabeling: matching each coset to its image in \(\phi(R)\).
This theorem justifies our entire development: ideals are precisely the substructures that allow us to
form well-defined quotients, and every such quotient arises naturally from a homomorphism.