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).
Condition (2) is the absorption property. In particular, taking \(r \in A \subseteq R\) yields that \(A\) is closed
under multiplication, so by the
Subring Test, \(A\) is a subring.
The full absorption requirement (\(r\) ranging over all of \(R\), not just \(A\)) is what distinguishes an ideal
from a mere subring.
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 \(n\), and it is nonempty because it contains \(0 = n \cdot 0\).
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, will be defined formally as principal ideals below.
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 \(a\).
Verification (via the Ideal Test). The set is nonempty since \(a = 1 \cdot a \in \langle a \rangle\)
(using the unity \(1\)). For closure under subtraction, \(r_1 a - r_2 a = (r_1 - r_2) a \in \langle a \rangle\). For
absorption, \(s \cdot (ra) = (sr) a \in \langle a \rangle\). Commutativity gives
\((ra) \cdot s = s \cdot (ra) \in \langle a \rangle\) as well. Hence \(\langle a \rangle\) is an ideal.
Every ideal of \(\mathbb{Z}\) is principal. The zero ideal is \(\langle 0 \rangle\). A nonzero ideal contains a
positive element, since it contains \(-m\) together with each of its members \(m\), and hence a smallest positive
element \(d\). Dividing any member \(m\) by \(d\) gives \(m = qd + t\) with integers \(q, t\) satisfying
\(0 \leq t \lt d\). Since \(t = m - qd\) again lies in the ideal, minimality forces \(t = 0\), so the ideal is
\(\langle d \rangle\). Principality 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.
Suppose, for contradiction, that \(I = \langle p \rangle\). Since \(2 \in I\), we have \(2 = q p\) for some
\(q \in \mathbb{Z}[x]\). Leading coefficients multiply in \(\mathbb{Z}\) without vanishing, so degrees add, and
therefore \(p\) is a constant dividing \(2\), that is, \(p \in \{\pm 1, \pm 2\}\). If \(p = \pm 1\), then
\(1 \in I\), which is impossible because \(1\) has an odd constant term. If \(p = \pm 2\), then
\(x \in \langle 2 \rangle\), contradicting the fact that every element of \(\langle 2 \rangle\) has only even
coefficients while \(x\) has coefficient \(1\).
Integral domains in which every ideal is principal are the
principal ideal domains. The page on
integral domains proves that every Euclidean domain is a principal ideal domain
and every principal ideal domain is a unique factorization domain.
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
\[
(s + A)(t + A) = st + A.
\]
The definition above stipulates the notation and operations, and the next theorem justifies them. The operations are
well-defined (independent of the chosen coset representative) precisely when \(A\) is an ideal.
Absorption is exactly what makes this work.
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:
We first delineate which ring axioms are non-trivial. The additive group structure of \(R/A\) follows directly from
the Factor Groups
construction applied to \((R, +)\) modulo the additive subgroup \(A\), which is normal because \((R, +)\) is
abelian. That structure comprises well-definedness of \((s + A) + (t + A) = (s + t) + A\), associativity, identity,
inverses, and commutativity. The associative law for multiplication and the distributive laws in \(R/A\) descend
from those in \(R\) by a direct coset computation. The only non-trivial check is the well-definedness of
multiplication, which we now establish.
First, suppose \(A\) is an ideal and let \(s + A = s' + A\) and \(t + A = t' + A\). Because these equalities force
\(s \in s' + A\) and \(t \in t' + A\), we have \(s = s' + a\) 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 \(at'\), \(s'b\), \(ab\) all lie in \(A\) by absorption and \(A\) is closed under addition,
\[
st + A = s't' + at' + s'b + ab + A = s't' + A.
\]
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 \(ar \notin A\). The other case runs identically with the
two factors in the opposite order.
Since \(a \in A\), the cosets \(a + A\) and \(0 + A\) coincide (both equal \(A\) itself, by the
coset-equality property
applied to the additive group \((R, +)\)). 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 \(n \geq 1\) and \(n\mathbb{Z} = \langle n \rangle\).
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
\(2 \cdot 3 = 6 \equiv 0 \pmod 6\)). However, if we mod out by \(\langle 5 \rangle\), we get \(\mathbb{Z}_5\),
which is a 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. In a commutative ring with unity, 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 \(\mathbb{Z}\), the ideal \(\langle n \rangle\) is prime exactly when \(|n|\) 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. Primality and maximality are intimately connected in
principal ideal domains, a class of
rings introduced on a later page, where for the polynomial ring \(F[x]\) over a field the ideal
\(\langle p(x) \rangle\) turns out to be maximal exactly when \(p(x)\) is irreducible.
Before stating the two theorems below, we recall that an integral domain and a field satisfy \(1 \neq 0\) by
definition, and we record one small fact. A proper ideal \(A\) of a commutative ring \(R\) with unity never contains
\(1\), because absorption would otherwise force \(r = r \cdot 1 \in A\) for every \(r \in R\), making \(A = R\).
Together, these theorems turn the definitions of prime and maximal ideals into an algebraic criterion for 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. By the
definition of an integral domain,
\(0 + A \neq 1 + A\) in \(R/A\), which means \(1 \notin A\) and hence \(A \neq R\). Thus \(A\) is a proper ideal.
Now 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 \(b + A = A\). Hence \(a \in A\) or \(b \in A\).
Combined with the proper-ideal property established above, this makes \(A\) prime.
Next, suppose \(A\) is prime. In particular, \(A\) is a proper ideal. Since \(R\) is a commutative ring with unity,
so is \(R/A\). Moreover, since \(A\) is proper, \(1 \notin A\), so \(0 + A \neq 1 + A\) in \(R/A\). Our task
therefore reduces to showing \(R/A\) has no zero-divisors.
Suppose \((a + A)(b + A) = 0 + A = A\) 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 \(R/A\). Hence
\(R/A\) is an integral domain.
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. Then \(0 + A \neq 1 + A\), so \(1 \notin A\) and \(A\) is a proper ideal. Let
\(B\) be an ideal of \(R\) that properly contains \(A\) (that is, \(A \subsetneq B \subseteq R\)), and choose
\(b \in B\) with \(b \notin A\). 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 equality means \(1 - bc \in A \subseteq B\). 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. Then \(A\) is proper, so \(1 \notin A\) and \(0 + A \neq 1 + A\) in \(R/A\). It
remains to verify that every nonzero element of \(R/A\) has a multiplicative inverse. The nonzero elements of
\(R/A\) are precisely the cosets \(b + A\) with \(b \in R\), \(b \notin A\) (since \(b + A = 0 + A\) if and only if
\(b \in A\)). Fix such a \(b\). Consider \(B = \{br + a \mid r \in R, a \in A\}\). Note that \(B\) is nonempty,
since it contains \(0 = b \cdot 0 + 0\), and that it satisfies the two conditions of the
Ideal Test:
-
Closure under subtraction:
\((br_1 + a_1) - (br_2 + a_2) = b(r_1 - r_2) + (a_1 - a_2) \in B\).
-
Absorption: for any \(s \in R\),
\(s(br + a) = b(sr) + sa\),
where \(sr \in R\) and \(sa \in A\) (since \(A\) is an ideal). Thus \(s(br + a) \in B\). The right-hand absorption
is identical because \(R\) is commutative.
Furthermore, \(A \subseteq B\) (taking \(r = 0\) yields \(a \in B\) for any \(a \in A\)), and
\(b = b \cdot 1 + 0 \in B\) with \(b \notin A\), so \(A \subsetneq B\).
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.
The corollary follows from the two theorems just proved. If \(A\) is maximal, then \(R/A\) is a field by
R/A is a Field if and only if A is Maximal,
and hence an integral domain by
A Field is an Integral Domain.
Therefore \(A\) is prime by
R/A is an Integral Domain if and only if 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, since \(\langle 0 \rangle \subsetneq \langle 2 \rangle \subsetneq \mathbb{Z}\).
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
(Kernels are Normal).
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:
\[
\operatorname{Ker} \phi = \{r \in R \mid \phi(r) = 0\}.
\]
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 \(\operatorname{Ker} \phi\) is an ideal of \(R\).
Proof:
Since \(\phi\) is a ring homomorphism, it is in particular a homomorphism of the additive groups \((R, +)\) and
\((S, +)\). By the
Properties of Elements under Homomorphism,
\(\operatorname{Ker} \phi\) is a subgroup of \((R, +)\), so \(a - b \in \operatorname{Ker} \phi\) for all
\(a, b \in \operatorname{Ker} \phi\). It remains to verify the absorption property.
Let \(a \in \operatorname{Ker} \phi\) and \(r \in R\). Since \(\phi\) preserves multiplication and, by the
Rules of Multiplication,
any product with \(0\) vanishes in \(S\):
\[
\phi(ra) = \phi(r)\phi(a) = \phi(r) \cdot 0 = 0.
\]
Similarly, \(\phi(ar) = 0\). Thus \(ra, ar \in \operatorname{Ker} \phi\), so \(\operatorname{Ker} \phi\) is an ideal
of \(R\) by the Ideal Test.
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 \(r \in \operatorname{Ker} \gamma\) 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
\(\operatorname{Ker} \gamma = A\).
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
\(\Phi: R/\operatorname{Ker} \phi \to \phi(R)\) defined by
\[
\Phi(r + \operatorname{Ker} \phi) = \phi(r)
\]
is a ring isomorphism. In symbols,
\[
R / \operatorname{Ker} \phi \cong \phi(R).
\]
Proof:
We begin by checking that \(\phi(R) = \{\phi(r) \mid r \in R\}\) is a subring of \(S\). It is nonempty (containing
\(\phi(0) = 0\)), and for any \(\phi(r_1), \phi(r_2) \in \phi(R)\) we have
\(\phi(r_1) - \phi(r_2) = \phi(r_1 - r_2) \in \phi(R)\) and \(\phi(r_1)\phi(r_2) = \phi(r_1 r_2) \in \phi(R)\). By
the Subring Test, \(\phi(R)\) is a
subring of \(S\), so the codomain of \(\Phi\) is a ring in its own right.
By the
First Isomorphism Theorem
applied to the additive groups \((R, +)\) and \((S, +)\), the mapping \(\Phi\) is a well-defined
isomorphism of additive groups, that is, one-to-one, onto, and addition-preserving. It remains to show that
\(\Phi\) preserves multiplication.
For any \(r + \operatorname{Ker} \phi\) and \(s + \operatorname{Ker} \phi\) in \(R/\operatorname{Ker} \phi\):
\[
\begin{align*}
\Phi((r + \operatorname{Ker} \phi)(s + \operatorname{Ker} \phi)) &= \Phi(rs + \operatorname{Ker} \phi) \\\\
&= \phi(rs) \\\\
&= \phi(r)\phi(s) \\\\
&= \Phi(r + \operatorname{Ker} \phi)\Phi(s + \operatorname{Ker} \phi).
\end{align*}
\]
Thus, \(\Phi\) preserves both addition and multiplication, so it is a ring isomorphism.
The First Isomorphism Theorem shows that every ring homomorphism factors through a quotient.
When we map \(R \to S\) via \(\phi\), we are implicitly:
- Collapsing: identifying all elements of \(\operatorname{Ker} \phi\) with zero.
- Partitioning: grouping elements of \(R\) into cosets of \(\operatorname{Ker} \phi\).
- 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.