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Ring Theory (part 3): Homomorphisms, Ideals, and Quotients (by Evan Dummit, 2018, v. 1.01)
Contents
3 Homomorphisms, Ideals, and Quotients 1
3.1 Ring Isomorphisms and Homomorphisms................................. 1 3.1.1 Ring Isomorphisms.......................................... 1 3.1.2 Ring Homomorphisms........................................ 4 3.2 Ideals and Quotient Rings.......................................... 7 3.2.1 Ideals................................................. 8 3.2.2 Quotient Rings............................................ 9 3.2.3 Homomorphisms and Quotient Rings................................ 11 3.3 Properties of Ideals.............................................. 13 3.3.1 The Isomorphism Theorems..................................... 13 3.3.2 Generation of Ideals......................................... 14 3.3.3 Maximal and Prime Ideals...................................... 17 3.3.4 The Chinese Remainder Theorem.................................. 20 3.4 Rings of Fractions.............................................. 21
3 Homomorphisms, Ideals, and Quotients
In this chapter, we will examine some more intricate properties of general rings. We begin with a discussion of isomorphisms, which provide a way of identifying two rings whose structures are identical, and then examine the broader class of ring homomorphisms, which are the structure-preserving functions from one ring to another. Next, we study ideals and quotient rings, which provide the most general version of modular arithmetic in a ring, and which are fundamentally connected with ring homomorphisms. We close with a detailed study of the structure of ideals and quotients in commutative rings with 1.
3.1 Ring Isomorphisms and Homomorphisms
- We begin our study with a discussion of structure-preserving maps between rings.
3.1.1 Ring Isomorphisms
- We have encountered several examples of rings with very similar structures.
- For example, consider the two rings R = Z/ 6 Z and S = (Z/ 2 Z) × (Z/ 3 Z).
◦ Here are the addition and multiplication tables in R:
- 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4
◦ Now compare those tables to the tables in S:
- (0, 0) (1, 1) (0, 2) (1, 0) (0, 1) (1, 2) (0, 0) (0, 0) (1, 1) (0, 2) (1, 0) (0, 1) (1, 2) (1, 1) (1, 1) (0, 2) (1, 0) (0, 1) (1, 2) (0, 0) (0, 2) (0, 2) (1, 0) (0, 1) (1, 2) (0, 0) (1, 1) (1, 0) (1, 0) (0, 1) (1, 2) (0, 0) (1, 1) (0, 2) (0, 1) (0, 1) (1, 2) (0, 0) (1, 1) (0, 2) (1, 0) (1, 2) (1, 2) (0, 0) (1, 1) (0, 2) (1, 0) (0, 1)
◦ Notice that these tables look quite similar (especially given the artful reordering of the labels of the elements in S). ◦ Indeed, if we relabel each entry n in the rst set of tables with the ordered pair corresponding to its reduction modulo 2 and 3 (so that 1 becomes (1, 1), 2 becomes (0, 2), and so forth) we will obtain the second set of tables!
- For another example, consider the rings R = (Z/ 2 Z) × (Z/ 2 Z) and S = F 2 [x]/(x^2 + x).
◦ Here are the addition and multiplication tables in R:
- (0, 0) (1, 1) (1, 0) (0, 1) (0, 0) (0, 0) (1, 1) (1, 0) (0, 1) (1, 1) (1, 1) (0, 0) (0, 1) (1, 0) (1, 0) (1, 0) (0, 1) (0, 0) (1, 1) (0, 1) (0, 1) (1, 0) (1, 1) (0, 0)
◦ Now compare those tables to the tables in S:
- 0 1 x x + 1 0 0 1 x x + 1 1 1 0 x + 1 x x x x + 1 0 1 x + 1 x + 1 x 1 0
· 0 1 x x + 1 0 0 0 0 0 1 0 1 x x + 1 x 0 x x 0 x + 1 0 x + 1 0 x + 1 ◦ Here, if we relabel (0, 0) as 0, (1, 1) as 1 , (1, 0) as x, and (0, 1) as x + 1, the rst pair of tables becomes the second set of tables.
- As a third example, consider the rings R = C = {a+bi : a, b ∈ R} and S =
{[
a b −b a
]
∈ M 2 × 2 (R) : a, b ∈ R
◦ Notice that S is a subring of M 2 × 2 (R): we have
[
a b −b a
]
[
c d −d c
]
[
a − c b − d −(b − d) a − c
]
and [ a b −b a
]
[
c d −d c
]
[
ac − bd ad + bc −(ad + bc) ac − bd
]
◦ Compare these to the addition and multiplication operations in C: (a + bi) − (c + di) = (a − c) + (b − d)i and (a + bi) · (c + di) = (ac − bd) + (ad + bc)i.
◦ Upon identifying the complex number a + bi with the matrix
[
a b −b a
]
, we see that the ring structure of S is precisely the same as the ring structure of C.
- Let us formalize the central idea in the examples above: in each case, we see that there is a way to relabel the elements of R using the elements of S in a way that preserves the ring structure.
◦ The desired relabeling is a function ϕ : R → S with the property that ϕ is a bijection (so that each element of R is labeled with a unique element of S) and that ϕ respects the ring operations. ◦ Explicitly, we require ϕ(r 1 + r 2 ) = ϕ(r 1 ) + ϕ(r 2 ) and ϕ(r 1 · r 2 ) = ϕ(r 1 ) · ϕ(r 2 ) for all elements r 1 and r 2 in R.
- De nition: Let R and S be rings. A ring isomorphism ϕ from R to S is a bijective^1 function ϕ : R → S such that ϕ(r 1 + r 2 ) = ϕ(r 1 ) + ϕ(r 2 ) and ϕ(r 1 · r 2 ) = ϕ(r 1 ) · ϕ(r 2 ) for all elements r 1 and r 2 in R. (^1) Recall that a function ϕ : R → S is injective (one-to-one) if ϕ(x) = ϕ(y) implies x = y, and ϕ is surjective (onto) if for every s ∈ S there exists an r ∈ R with ϕ(r) = s. A bijective function is one that is both injective and surjective. Equivalently, ϕ is a bijection if it possesses a two-sided inverse function ϕ−^1 : S → R with ϕ(ϕ−^1 (s)) = s and ϕ−^1 (ϕ(r)) = r for every r ∈ R and s ∈ S.
- If ϕ : R → S is an isomorphism, then ϕ(0R) = 0S , and if R has a 1, then so does S, and ϕ(1R) = 1S. ◦ Proof: For any r ∈ R, we have ϕ(r) = ϕ(r + 0R) = ϕ(r) + ϕ(0R): thus, by additive cancellation in S we see ϕ(0R) = 0S. ◦ Likewise, if R has a 1, then let s ∈ S be arbitrary and r = ϕ−^1 (s). Then s · ϕ(1R) = ϕ(r)ϕ(1R) = ϕ(r · (^1) R) = ϕ(r) = s, and likewise ϕ(1R) · s = s, so ϕ(1R) is a multiplicative identity in S.
- From the proposition, we immediately see that being isomorphic is an equivalence relation on any collection of rings. In general, it is not easy to determine whether two given rings are isomorphic, and even if two rings are isomorphic, there is no general method for constructing an isomorphism between them.
◦ Two isomorphic rings have the same additive and multiplicative structures. Thus, any statement that only depends on the ring operations must be identical in two isomorphic rings. ◦ Thus, for example, if ϕ : R → S is an isomorphism, then R is commutative if and only if S is commutative, and R has a 1 if and only if S has a 1. Likewise, R has zero divisors if and only if S has zero divisors, and the cardinalities of any two isomorphic rings (along with their sets of units) must be equal. ◦ So, for example, we see that M 2 × 2 (R) is not isomorphic to the ring of real quaternions H, since the former has zero divisors and the latter does not. ◦ Likewise, we see that none of the rings Z/mZ for m > 1 are isomorphic to one another, since they all have dierent cardinalities. ◦ In a similar way, the ring R is not isomorphic to C since the polynomial equation x^2 + 1 = 0 has no solutions in R, but does have solutions in C. ◦ As a nal example, the rings Z/ 4 Z and (Z/ 2 Z) × (Z/ 2 Z) are not isomorphic: there are two solutions to x^2 = 0 in the rst ring (namely, 0 and 2) while there is only one solution to x^2 = 0 in the second ring (namely, (0, 0)). Alternatively, the rst ring has 2 units, while the second ring has only 1.
- We also remark that there can exist nontrivial isomorphisms of a ring with itself. Such maps are known as automorphisms.
◦ Remark (for those who like group theory): The set of automorphisms of a ring forms a group under function composition.
- Example: Show that the complex conjugation map ϕ(a + bi) = a − bi is an isomorphism from C to C.
◦ It is easy to see that ϕ is a bijection, since it is its own inverse function. ◦ Furthermore, it is a straightforward calculation that ϕ(z + w) = ϕ(z) + ϕ(w) and ϕ(zw) = ϕ(z)ϕ(w) for any complex numbers z and w, so ϕ is an isomorphism.
3.1.2 Ring Homomorphisms
- We now study maps that respect the structure of ring operations without the requirement that they be bijections.
- De nition: A function ϕ : R → S is a ring homomorphism if ϕ(r 1 + r 2 ) = ϕ(r 1 ) + ϕ(r 2 ) and ϕ(r 1 · r 2 ) = ϕ(r 1 ) · ϕ(r 2 ) for all elements r 1 and r 2 in R.
◦ Note of course that any isomorphism is a homomorphism, but the reverse is not typically true.
- Example: If m > 1 , show that the map ϕ : Z → Z/mZ de ned by ϕ(a) = a, so that ϕ maps the integer a to its associated residue class a modulo m, is a ring homomorphism.
◦ From our results on residue classes, we see ϕ(a + b) = a + b = a + b = ϕ(a) + ϕ(b), and likewise ϕ(a · b) = a · b = a · b = ϕ(a) · ϕ(b). Thus, ϕ is a homomorphism. ◦ Notice that this map is surjective but not injective (since for example ϕ(0) = ϕ(m)), so it is not an isomorphism.
- In essentially the same way, we see that the reduction modulo p map inside F [x] is also a homomorphism:
- Example: Let F be a eld with R = F [x] and let p(x) ∈ R be nonzero. Then the map ϕ : R → R/pR given by ϕ(a) = a, mapping the polynomial a to its associated residue class a modulo p, is a ring homomorphism.
◦ From our results on residue classes, we see ϕ(a + b) = a + b = a + b = ϕ(a) + ϕ(b), and likewise ϕ(a · b) = a · b = a · b = ϕ(a) · ϕ(b). Thus, ϕ is a homomorphism. ◦ In the next section, we will generalize the ideas in these two examples and describe a general procedure for constructing a quotient ring.
- Example: Let R be a commutative ring and a ∈ R. Show that the evaluation at a map ϕa : R[x] → R de ned by ϕa(p) = p(a) is a ring homomorphism.
◦ We have ϕa(p + q) = (p + q)(a) = p(a) + q(a) = ϕa(p) + ϕa(q) by the de nition of polynomial addition. ◦ Likewise, we have ϕa(rbxb^ · rcxc) = rbrcab+c^ = (rbab)(rcac) = ϕa(rbxb)ϕa(rcxc) because R is commuta- tive. ◦ Then for any polynomials p and q we see ϕa(pq) = ϕa(p)ϕa(q) by applying distributivity and the fact that ϕa respects multiplication of individual terms and addition.
- Example: Let R and S be any rings. The zero map Z : R → S given by Z(r) = 0S for every r ∈ R is a ring homomorphism.
- Example: If S is a subring of R, the map ι : S → R given by ι(s) = s is a ring homomorphism. This map is called the inclusion map (since it simply re ects the set inclusion of S inside R).
- There exist numerous examples of maps that satisfy only one of the two requirements for being a homomor- phism.
◦ Non-Example: The function f : Z → Z given by f (n) = 2n is not a homomorphism. Explicitly, although it satis es f (m + n) = 2(m + n) = f (m) + f (n), it is not multiplicative since f (1 · 1) = 2 while f (1) · f (1) = 4. ◦ Non-Example: The function f : R → R given by f (x) = x^2 is not a homomorphism. Explicitly, although it satis es f (xy) = (xy)^2 = f (x)f (y), it is not additive since f (1 + 1) = 4 while f (1) + f (1) = 2.
- Here are a few more examples (and non-examples) of homomorphisms:
- Example: Determine whether the map ϕ : M 2 × 2 (R) → R given by ϕ
([
a b c d
])
= b is a ring homomorphism.
◦ We see that ϕ
([
a 1 b 1 c 1 d 1
]
[
a 2 b 2 c 2 d 2
])
= b 1 + b 2 = ϕ
([
a 1 b 1 c 1 d 1
])
([
a 2 b 2 c 2 d 2
])
◦ However, ϕ
([
a 1 b 1 c 1 d 1
]
[
a 2 b 2 c 2 d 2
])
= a 1 b 2 + b 1 d 2 while ϕ
([
a 1 b 1 c 1 d 1
])
· ϕ
([
a 2 b 2 c 2 d 2
])
= b 1 b 2 ,
and these expressions are not equal in general. Thus, ϕ is is not a homomorphism.
- Example: Determine whether the map ϕ : (Z/ 15 Z) → (Z/ 15 Z) given by ϕ(a) = 10a is a ring homomorphism.
◦ We have ϕ(a + b) = 10(a + b) = 10a + 10b = ϕ(a) + ϕ(b). ◦ Likewise, ϕ(ab) = 10ab = 100ab = (10a)(10b) = ϕ(a)ϕ(b), since 10 ≡ 100 (mod 15). ◦ Therefore, ϕ is a homomorphism.
- Example: Let R be the ring of in nitely dierentiable real-valued functions on R. Determine whether the derivative map D : R → R given by D(f ) = f ′^ is a ring homomorphism.
◦ We have D(f + g) = (f + g)′^ = f ′^ + g′^ = D(f ) + D(g), so D is additive. ◦ However, D does not respect ring multiplication, since for example D(x · x^2 ) = 3x^2 while D(x) · D(x^2 ) = 2 x. Therefore, ϕ is not a homomorphism.
- The kernel ker ϕ is a subring of R. In fact, if x ∈ ker ϕ, then rx and xr are in ker ϕ for any r ∈ R: in other words, ker ϕ is closed under multiplication by arbitrary elements of R. ◦ Proof: Since ϕ(0R) = 0S , the kernel contains 0. Furthermore, if r 1 and r 2 are in ker ϕ then ϕ(r 1 − r 2 ) = ϕ(r 1 ) − ϕ(r 2 ) = 0 and ϕ(r 1 r 2 ) = ϕ(r 1 )ϕ(r 2 ) = 0 · 0 = 0 ◦ Thus, kerϕ contains 0 and is closed under subtraction and multiplication, so it is a subring. ◦ Moreover, if x ∈ ker ϕ then ϕ(rx) = ϕ(r)ϕ(x) = ϕ(r)0 = 0 and likewise ϕ(xr) = ϕ(x)ϕ(r) = 0ϕ(r) = 0.
- The kernel is zero (i.e., ker ϕ = { 0 }) if and only if ϕ is injective. In particular, ϕ is an isomorphism if and only if ker ϕ = { 0 } and im ϕ = S. ◦ Proof: If ϕ(a) = ϕ(b), then ϕ(a − b) = ϕ(a) − ϕ(b) = 0, so a − b ∈ ker ϕ. Thus, if the only element in ker ϕ is 0, then we must have a − b = 0 so that a = b. ◦ Conversely, if x ∈ ker ϕ and ϕ is injective, then ϕ(x) = 0 = ϕ(0) implies x = 0. ◦ The second statement follows from the facts that ker ϕ = { 0 } is equivalent to ϕ being injective and im ϕ = S is equivalent to ϕ being surjective.
3.2 Ideals and Quotient Rings
- Our next task is to generalize the idea of modular arithmetic into general rings.
◦ To motivate our discussion, recall the ideas behind the construction of Z/mZ and R/pR where R = F [x]: we rst de ned modular modular congruences and studied their properties, and then we constructed residue classes and showed that the collection of all residue classes had a ring structure.
- In both Z and F [x], we de ned modular congruences using divisibility, but let us take a broader approach: if I is a subset of R (whose properties we intend to characterize in a moment) let us say that two elements a, b ∈ R are congruent modulo I if a − b ∈ I.
◦ This is a generalization of both types of congruence we have described thus far: for Z/mZ, the set I consists of the multiples of m, while for R/pR, the set I consists of the multiples of p. ◦ We would like congruence modulo I to be an equivalence relation: this requires a ≡ a (mod I), a ≡ b (mod I) implies b ≡ a (mod I), and a ≡ b (mod I) and b ≡ c (mod I) implies a ≡ c (mod I). ◦ It is easy to see that these three conditions require 0 ∈ I, that I be closed under additive inverses, and that I be closed under addition. (Thus, I is in fact closed under subtraction.) ◦ Furthermore, we would like the congruences to respect addition and multiplication: if a ≡ b (mod I) and c ≡ d (mod I), then we want a + c ≡ b + d (mod I) and ac ≡ bd (mod I). ◦ In terms of ring elements, this is equivalent to the following: if b = a + r and d = c + s for some r, s ∈ I, then we want (b+d)−(a+c) = r +s to be in I, and we also want bd−ac = (a+r)(c+s)−ac = as+rc+rs to be in I. ◦ The rst condition clearly follows from the requirement that I is closed under addition. It is a bit less obvious how to handle the second condition, but one immediate implication follows by setting a = c = 0: namely, that rs ∈ I. ◦ Thus, I must be closed under multiplication, so it is in fact a subring of R. ◦ But the well-de nedness of multiplication actually requires more: since 0 ∈ I, we can set r = 0 to see that as ∈ I, and we can also set s = 0 to see that rc ∈ I. ◦ So in fact, I must be closed under (left and right) multiplication by arbitrary elements of R, in addition to being a subring. It is then easy to see that this condition is also su cient to ensure that a ≡ b (mod I) and c ≡ d (mod I) imply a + c ≡ b + d (mod I) and ac ≡ bd (mod I). ◦ Our last task is to de ne residue classes and then the ring operations: we de ne the residue class a (modulo I) to be the set of ring elements b congruent to a modulo I, which is to say, a = {a + r : r ∈ I}. ◦ Then we take the operations on residue classes to be a + b = a + b and a · b = a · b: then from our properties of congruences, we can verify that these operations are well-de ned and that the collection of residue classes forms a ring.
3.2.1 Ideals
- Now that we have established the basic properties of the classes of the sets I we can use to construct congru- ences, we can run through the discussion more formally.
- De nition: A subring I of a ring R is called a left ideal of R if it is closed under arbitrary left multiplication by elements of R, and it is called a right ideal if it is closed under arbitrary right multiplication by elements of R.
◦ Explicitly, I is a left ideal if I contains 0 and for any x, y ∈ I and any r ∈ R, the elements x − y and rx are in I, while I is a right ideal if I contains 0 and for any x, y ∈ I and any r ∈ R, the elements x − y and xr are in I.
- De nition: A subset I of a ring R that is both a left and a right ideal is called an ideal of R (or, for emphasis, a two-sided ideal).
◦ Explicitly, I is an ideal if I contains 0 and for any x, y ∈ I and any r ∈ R, the elements x − y, rx, and xr are all in I. ◦ If R is commutative, then left ideals, right ideals, and two-sided ideals are the same. (As we will mention below, when R is not commutative, there may exist left ideals that are not right ideals and vice versa.)
- Here are a few basic examples of ideals:
◦ Example: The subrings nZ are ideals of Z, since they are clearly closed under arbitrary multiplication by elements of Z. ◦ Example: If R = F [x] and p is any polynomial, the subring pR of multiples of p is an ideal of F [x], since it is closed under arbitrary multiplication by polynomials in F [x]. ◦ Non-example: The subring Z of Q is not an ideal of Q, since it is not closed under arbitrary multiplication by elements of Q, since for example if we take r =
∈ Q and x = 4 ∈ Z, the element rx =
is not in Z.
◦ Example: For any ring R, the subrings { 0 } and R are ideals of R. We refer to { 0 } as the trivial ideal (or the zero ideal) and refer to any ideal I 6 = R as a proper ideal (since it is a proper subset of R).
- Here are a few more examples (and non-examples) of ideals.
- Example: In the polynomial ring Z[x], determine whether the set S of polynomials with even constant term (i.e., the polynomials of the form 2 a 0 + a 1 x + a 2 x^2 + · · · + anxn^ for integers ai) forms an ideal.
◦ It is easy to see that 0 ∈ S and that S is closed under subtraction. ◦ Furthermore, if q(x) is any other polynomial, and p(x) ∈ S, then p(x)q(x) also has even constant term, so it is also in S. ◦ Thus, S is closed under multiplication by elements of Z[x], so it is an ideal.
- Example: Determine whether the set S of upper-triangular 2 × 2 matrices is a left ideal or a right ideal of M 2 × 2 (R).
◦ The upper-triangular matrices form a subring, so we need only determine whether they are closed under multiplication by arbitrary 2 × 2 matrices on the left and the right.
◦ We can see that if r =
[
]
and x =
[
]
then x is upper-triangular but rx =
[
]
. Thus,
S is not a left ideal.
◦ Indeed, with the same choices, we have xr =
[
]
, so S also is not a right ideal.
- Example: Determine whether the set S = { 0 , 2 , 4 , 6 } of even residue classes is an ideal of Z/ 8 Z.
◦ We have 0 ∈ S, and it is a straightforward calculation to see that S is closed under subtraction, since the sum of two even residue classes modulo 8 will still be even.
◦ Proof: We are given a−b ∈ I and b−c ∈ I, so since I is closed under addition, we see (a−b)+(b−c) = a − c ∈ I.
- If a ≡ b (mod I) and c ≡ d (mod I), then a + c ≡ b + d (mod I). ◦ Proof: We are given a−b ∈ I and c−d ∈ I, so since I is closed under addition, we see (a−b)+(c−d) = (a + c) − (b + d) ∈ I.
- If a ≡ b (mod I) and c ≡ d (mod I), then ac ≡ bd (mod I). ◦ Proof: We are given a − b ∈ I and c − d ∈ I. Then since I is closed under arbitrary left and right multiplication, we see that (a − b)c and b(c − d) are also in I. Hence ac − bd = (a − b)c + b(c − d) is also in I since I is closed under addition.
- Now we can de ne residue classes:
- De nition: If I is an ideal of the ring R, then for any a ∈ R we de ne the residue class of a modulo I to be the set a = a + I = {a + x : x ∈ I}. This set is also called the coset of I represented by a.
◦ We will use the notation a and a+I interchangeably. (The latter is intended to evoke the idea of adding a to the set I.) ◦ We observe, as with our previous examples of residue classes, that any two residue classes are either disjoint or identical and that they partition R: speci cally, a = b if and only if a ≡ b (mod I) if and only if a − b ∈ I.
- All that remains is to verify that the residue classes form a ring, in the same way as in Z and F [x]:
- Theorem (Quotient Rings): Let I be an ideal of the ring R. Then the collection of residue classes modulo I forms a ring, denoted R/I (read as R mod I), under the operations a + b = a + b and a · b = ab. (This ring is called the quotient ring of R by I.) If R is commutative then so is R/I, and likewise if R has a 1 then so does R/I.
◦ Remark: The notation R/I is intended to emphasize the idea that I represents a single element (namely, 0 ) in the quotient ring R/I, and the other elements in R/I are translates of I. In this way, R/I is the ring obtained from R by collapsing or dividing out by I, whence the name quotient ring. ◦ The proof of this fact is exactly the same as in the cases of Z and F [x], and only requires showing that the operations are well-de ned. ◦ Proof: First we must show that the addition and multiplication operations are well-de ned: that is, if we choose dierent elements a′^ ∈ ¯a and b′^ ∈ ¯b, the residue class of a′^ + b′^ is the same as that of a + b, and similarly for the product. ◦ To see this, if a′^ ∈ ¯a then a′^ ≡ a (mod I), and similarly if b′^ ∈ b then b′^ ≡ b (mod I). ◦ Then a′^ + b′^ ≡ a + b (mod I), so a′^ + b′^ = a + b. Likewise, a′b′^ ≡ ab (mod I), so a′b′^ = ab. ◦ Thus, the operations are well-de ned. ◦ For the ring axioms [R1]-[R6], we observe that associativity, commutativity, and the distributive laws follow immediately from the corresponding properties in R: the additive identity in R/I is ¯ 0 and the additive inverse of a is −a. ◦ Finally, if R is commutative then so will be the multiplication of the residue classes, and if R has a 1 then the residue class 1 is easily seen to be a multiplicative identity in R/I.
- This general description of quotient rings generalizes the two examples we have previously discussed: Z/mZ and R/pR where R = F [x].
◦ To be explicit, Z/mZ is the quotient of Z by the ideal mZ, while F [x]/p is the quotient of the polynomial ring F [x] by the principal ideal (p) consisting of all multiples of p. ◦ It is not hard to see that the integer congruence a ≡ b (mod m), which we originally de ned as being equivalent to the statement m|(b − a), is the same as the congruence a ≡ b (mod I) where I is the ideal mZ, since b − a ∈ mZ precisely when b − a is a multiple of m.
- Here are some additional examples of quotient rings:
- Example: If R is any ring, the quotient ring of R by the zero ideal, namely R/ 0 , is (isomorphic to) R itself, while the quotient ring of R by itself, namely R/R, is (isomorphic to) the trivial ring { 0 }.
- Example: In R = Z[x], with I consisting of all multiples of x^2 + 1, describe the structure of the quotient ring R/I.
◦ It is easy to see that I is an ideal of R, since it is a subring that is closed under arbitrary multiplication by elements of R. ◦ From our discussion of polynomial rings, we know that the residue classes in R/I are represented uniquely by residue classes of the form a + bx where a, b ∈ Z. Note that in this quotient ring, we have x^2 + 1 = 0, which is to say, x^2 = − 1. ◦ The addition in this quotient ring is given by a + bx+c + dx = (a + c) + (b + d)x while the multiplication is given by a + bx · c + dx = (ac − bd) + (ad + bc)x, which follows from the distributive law and the fact that x^2 = − 1. ◦ In this case, the quotient ring is isomorphic to the ring of Gaussian integers Z[i], with the isomorphism ϕ : R/I → Z[i] given by ϕ(a + bx) = a + bi.
- Example: In R = Z/ 8 Z, with I = { 0 , 4 }, describe the structure of the quotient ring R/I.
◦ It is easy to see that I is an ideal of R, since it is a subring that is closed under arbitrary multiplication by elements of R. (Indeed, it is the principal ideal generated by 4.) ◦ Since each residue class contains 2 elements, and R has 8 elements in total, there are four residue classes. With this observation in hand, it is not hard to give a list: 0 = I = { 0 , 4 }, 1 = 1 + I = { 1 , 5 }, 2 = 2 + I = { 2 , 6 }, and 3 = 3 + I = { 3 , 7 }. ◦ Notice, for example, that in the quotient ring R/I, we have 1 + 3 = 0, 2 · 2 = 0, and 2 · 3 = 2: indeed, we can see that the structure of R/I is exactly the same as Z/ 4 Z (the labelings of the elements are even the same).
- Example: In the polynomial ring R = Z[x], with I consisting of the polynomials with even constant term (i.e., the polynomials of the form 2 a 0 + a 1 x + a 2 x^2 + · · · + anxn^ for integers ai), describe the structure of the quotient ring R/I.
◦ We observe that there are only two residue classes, namely 0 and 1 : to see this observe that p(x) ∈ 0 when the constant term of p is even, and p(x) ∈ 1 when the constant term of p is odd. ◦ Then it is fairly easy to see that the structure of this quotient ring is the same as Z/ 2 Z (or more formally, it is isomorphic to Z/ 2 Z), since 1 + 1 = 0.
3.2.3 Homomorphisms and Quotient Rings
- Although homomorphisms and quotient rings may not immediately appear to be connected, in fact they are quite deeply related.
◦ To begin, observe that if ϕ : R → S is a ring homomorphism, then the kernel of ϕ is an ideal of R. ◦ In fact, we proved this fact earlier when we introduced the kernel, but let us remark again: if x ∈ ker ϕ and r ∈ R, then ϕ(rx) = ϕ(r)ϕ(x) = ϕ(r)0 = 0 and likewise ϕ(xr) = ϕ(x)ϕ(r) = 0ϕ(r) = 0. ◦ Thus, we can use homomorphisms to construct new ideals. ◦ Equally importantly, we can also do the reverse: we can use ideals to construct homomorphisms. ◦ The key observation in this direction is that the map ϕ : R → R/I associating a ring element to its residue class (i.e., with ϕ(a) = a) is a ring homomorphism. ◦ Indeed, the two parts of the de nition of homomorphism were precisely the properties we arranged for the residue classes modulo I to possess: ϕ(a + b) = a + b = a + b = ϕ(a) + ϕ(b) and ϕ(a · b) = a · b = a · b = ϕ(a) · ϕ(b).
- Example: If R is any commutative ring, show that R[x]/(x) is isomorphic to R.
◦ Let ϕ : R[x] → R be the evaluation at 0 homomorphism ϕ(p) = p(0). This map is clearly surjective since for any r ∈ R we have ϕ(r) = r. ◦ Furthermore, the kernel of this homomorphism is precisely the collection of polynomials p(x) = a 0 + a 1 x + · · · + anxn^ with p(0) = 0, which is easily seen to be the ideal I = (x) consisting of polynomials divisible by x. ◦ Thus, by the rst isomorphism theorem, for I = (x) we have R[x]/I ∼= R.
- Example: Show that Z/ 12 Z is isomorphic to (Z/ 3 Z) × (Z/ 4 Z).
◦ We seek a surjective homomorphism ϕ : Z → (Z/ 3 Z) × (Z/ 4 Z) whose kernel is 12 Z. ◦ Once this idea is suggested, it is not hard to come up with a candidate, namely, ϕ(a) = (a mod 3 , a mod 4). ◦ It is easy to verify that map is a homomorphism (since the individual maps of reduction mod 3 and reduction mod 4 are homomorphisms) and it is likewise fairly easy to see that the map is surjective by checking that the images of 0, 1, ... , 11 represent all of the elements in (Z/ 3 Z) × (Z/ 4 Z). ◦ Finally, the kernel of the map consists of all integers a with ϕ(a) = (0, 0), which is equivalent to saying a ≡ 0 (mod 3) and a ≡ 0 (mod 4), so that 3 |a and 4 |a: thus, the kernel is precisely 12 Z. ◦ Therefore, by the rst isomorphism theorem applied to this map ϕ, we conclude that Z/ 12 Z is isomorphic to (Z/ 3 Z) × (Z/ 4 Z). ◦ Remark: In fact, we could have avoided checking surjectivity explicitly by instead observing that the rst isomorphism theorem yields an injective homomorphism ψ : Z/ 12 Z → (Z/ 3 Z) × (Z/ 4 Z), which must therefore also be surjective since there are 12 elements in both sets.
3.3 Properties of Ideals
- Now that we have established basic properties of ideals, homomorphisms, and quotient rings, we embark on a deeper study of these topics.
3.3.1 The Isomorphism Theorems
- We begin by discussing several fundamental theorems about rings, subrings, and ideals that are collectively known as the isomorphism theorems. We have already proven the rst one:
- Theorem (First Isomorphism Theorem): If ϕ : R → S is a homomorphism of rings, then R/ ker ϕ is isomorphic to im ϕ.
- Theorem (Second Isomorphism Theorem): If A is a subring of R and B is an ideal of R, then A + B = {a + b : a ∈ A, b ∈ B} is a subring of A, A ∩ B is an ideal of A, and (A + B)/B is isomorphic to A/(A ∩ B).
◦ Proof: Clearly A + B contains 0 and (a + b) − (a′^ + b′) = (a − a′) + (b − b′) so it is also closed under subtraction. For multiplication, we observe (a + b)(a′^ + b′) = aa′^ + ba′^ + ab′^ + bb′: the rst term is in A since A is a subring, while the other three terms are in B (hence so is their sum) since B is an ideal. ◦ For the last statement, consider the map ϕ : A → (A + B)/B de ned by ϕ(a) = a + B. This map is well-de ned and a homomorphism by the basic properties of quotient rings, and it is surjective since for any class r + B in (A + B)/B for some r = a + b ∈ A + B, we have ϕ(a) = a + B = r + B. ◦ The kernel of the map ϕ consists of all a ∈ A with a + B = 0 + B, which is (by de nition) equivalent to saying a ∈ B: thus, ker ϕ = A∩B. In particular, A∩B is an ideal since it is a kernel of a homomorphism. ◦ Thus, by applying the rst isomorphism theorem to ϕ, we see that the rings A/(A ∩ B) and (A + B)/B are isomorphic, as claimed.
- Theorem (Third Isomorphism Theorem): If I and J are ideals of R with I ⊆ J, then J/I is an ideal of R/I and (R/I)/(J/I) is isomorphic to R/J.
◦ Proof: De ne the map ϕ : R/I → R/J given by setting ϕ(r + I) = r + J. This map is well-de ned because if r′^ + I = r + I, then since J contains I, we also have r′^ + J = r + J, and it is also surjective since for any class r + J in R/J, we clearly have ϕ(r + I) = r + J. ◦ Furthermore, ϕ is a homomorphism by the basic properties of quotient rings, since for example ϕ((r 1 + r 2 ) + I) = (r 1 + r 2 ) + J = (r 1 + J) + (r 2 + J) = ϕ(r 1 + I) + ϕ(r 2 + I), which shows that ϕ is additive because (r 1 + I) + (r 2 + I) = (r 1 + r 2 ) + I. ◦ Likewise, since (r 1 + I)(r 2 + I) = r 1 r 2 + I, we see that ϕ(r 1 r 2 + I) = r 1 r 2 + J = (r 1 + J)(r 2 + J) = ϕ(r 1 + I)ϕ(r 2 + I) and so ϕ is multiplicative. ◦ The kernel of the map ϕ consists of all r + I in R/I with the property that r + J = 0 + J, which is equivalent to saying r ∈ J: thus, ker ϕ consists of the classes of the form r + I for r ∈ J; this is simply another way of saying that ker ϕ = J/I. ◦ Finally, by applying the rst isomorphism theorem to ϕ, we see that the rings (R/I)/(J/I) and R/J are isomorphic, as claimed.
- Example: Inside R = Z[x], let I be the ideal of all polynomials with zero constant term and J be the ideal of all polynomials with even constant term.
◦ As we have already mentioned, both I and J are ideals of R, and clearly I ⊆ J. ◦ Furthermore, R/I is isomorphic to Z (per the division algorithm), and J/I is isomorphic to 2 Z (the residue classes are represented by the even integers). Also, R/J is isomorphic to Z/ 2 Z (since the residue classes are 0 and 1 ). ◦ Then indeed (R/I)/(J/I) ∼= Z/ 2 Z ∼= R/J, as claimed.
- Theorem (Fourth/Lattice Isomorphism Theorem): If I is an ideal of R, then there is an inclusion-preserving bijection between subrings A of R containing I and the subrings A = A/I of R/I. Furthermore, a subring A of R containing I is an ideal of R if and only if A/I is an ideal of R/I.
◦ Proof: We showed during the proof of the second isomorphism theorem that if A contains I then I is an ideal of A, so the association of A with A = A/I is well-de ned. Conversely, if S is a subring of R/I, then the set A = {r ∈ R : r + I ∈ S} is the unique subring of R containing I with the property that A/I = S. ◦ Furthermore, if B is a subring containing A, then B = B + I contains A = A + I, so the association preserves containment. ◦ For the statements about ideals, we showed during the proof of the third isomorphism theorem that if J is an ideal containing I then J/I is an ideal of R/I. Conversely, if J/I is an ideal of R/I, then for any r ∈ R and x ∈ J we have r(x + I) ∈ J/I, and this is equivalent to saying that rx ∈ J: thus, J is an ideal of R (since it is already a subring, per the above).
- Example: For R = Z and I = 10Z, identify the ideals of R containing I and verify that they all yield ideals of R/I.
◦ The ideals of R containing I are Z, 2 Z, 5 Z, and 10 Z. ◦ The corresponding ideals of R/I = Z/ 10 Z are Z/ 10 Z, 2 Z/ 10 Z = { 0 , 2 , 4 , 6 , 8 }, 5 Z/ 10 Z = { 0 , 5 }, and 10 Z/ 10 Z = { 0 }. ◦ As claimed, each of these is indeed an ideal of Z/ 10 Z.
3.3.2 Generation of Ideals
- In order to study the structure of ideals, we would like a simpler way to describe them. A convenient way is to describe ideals as being generated by subsets of a ring:
◦ If R is a ring with 1 and A is a subset of R, we would like to de ne the ideal generated by A to be the smallest ideal containing A.
- Example: In Z, for any integer n we have (n) = nZ. Since every ideal of Z is of the form nZ, we see that every ideal of Z is principal.
◦ Also, we remark that the notation nZ we have already used is consistent with the de nition above. (The same is true for the notation pR for R = F [x].) ◦ We also remark that if a and b are integers with greatest common divisor d, then (a, b) = (d): this follows from the pair of observations that a and b are both contained in (d) so that (a, b) ⊆ (d), and that d = xa + yb for some integers x and y by the Euclidean algorithm, so that d is contained in (a, b). ◦ Indeed, as a re ection of this fact, many authors write (a, b) to denote the greatest common divisor of a and b.
- Since principal ideals are the easiest to describe, it is often useful to try to determine whether a particular ideal is principal (though this task is not always so easy!):
- Example: Show that the ideal I = (2, x) in Z[x] is not principal.
◦ Note that I = { 2 p(x) + xq(x) : p, q ∈ Z[x]} is the collection of polynomials in Z[x] with even constant term. ◦ If I were principal and generated by some polynomial r(x), then every polynomial in I would be divisible by r(x). Hence, in particular, r(x) would divide 2, so since 2 is a constant polynomial and a prime number, r(x) would have to be one of {± 1 , ± 2 }. ◦ However, since r(x) must also divide x, the only possibility is that r(x) would be either 1 or − 1. But it is easy to see that the ideal generated by 1 (or − 1 ) is all of Z[x], so r(x) cannot be 1 or − 1 , since I 6 = Z[x]. ◦ Thus, there is no possible choice for r, so I is not principal. (Of course, it is still nitely generated!)
- Example: Determine whether or not the ideal I = (2, 1 +
−5) in Z[
−5] is principal.
◦ Suppose this ideal were principal with generator r = a + b
− 5 in Z[
−5].
◦ Then r would necessarily divide 2, meaning that 2 = rs for some s ∈ Z[
−5]. By taking norms, we see that 4 = N (2) = N (r)N (s). ◦ Likewise, since r divides 1 +
− 5 , we would have 1 +
−5 = rt for some t ∈ Z[
−5], so by taking norms we would have 6 = N (1 +
−5) = N (r)N (t). ◦ Since N (r) = a^2 + 5b^2 is a nonnegative integer, we see that N (r) must divide both 4 and 6, hence is either 1 or 2. However, it is easy to see that there are no integer solutions to a^2 + 5b^2 = 2, and the only elements of norm 1 are 1 and − 1. ◦ As in the examples above, the ideal generated by 1 (or − 1 ) is all of Z[
−5], but (2, 1 +
−5) 6 = Z[
−5]
since every element a + b
− 5 in the ideal has a + b even. ◦ Thus, I is not principal.
- Example: Determine whether the ideal I = (x^3 , x + 3) in Q[x] is principal.
◦ In the same way as in the example above, if I were principal and generated by a polynomial r(x), then every polynomial in I would be divisible by r. ◦ Here, since x^3 and x + 3 are relatively prime, we can see that any generator would necessarily divide their gcd, which is 1.
◦ In fact, 1 is a generator of I: via the Euclidean algorithm, we can see that 1 = −
x^3 + (
x + 1 27
x^2 )(x + 3), and so since both x^3 and x + 3 are in I, we see that 1 is also in I.
◦ Then since 1 is in I, so is p(x) · 1 = p(x) for any p(x) ∈ Q[x], meaning that in fact I = Q[x] and I indeed is principal (and generated by 1).
- We can in fact generalize the argument from the last example above:
- Proposition (Ideals of F [x]): If F is a eld, then every ideal in F [x] is principal.
◦ Proof: Let I be an ideal of F [x]. If I is the zero ideal we are done, so assume I contains a nonzero element. ◦ We claim that I = (d), where d is the monic greatest common divisor of all the elements in I. (Equiva- lently, d is the monic polynomial of largest degree dividing all the elements of I: such a polynomial must exist by the well-ordering axiom.) ◦ If d divides every polynomial in I, then clearly I ⊆ (d). ◦ Conversely, since d is the gcd, by the Euclidean algorithm and the well-ordering axiom we can write d = x 1 p 1 + x 2 p 2 + · · · + xnpn for some polynomials xi ∈ F [x] and pi ∈ I: then we see that d ∈ I, and hence (d) ⊆ I. Thus, I = (d) is principal as claimed.
- As we also saw above, when R is a ring with 1, then 1 is a generator of R. We can likewise generalize this statement:
- Proposition (Ideals and Units): If I is an ideal of the ring R with 1, then I = R if and only if I contains a unit.
◦ Proof: If I = R then certainly I contains a unit (namely, 1). ◦ Conversely, if u ∈ I is a unit with ur = 1, then since I is an ideal we have 1 = ur ∈ I, and then for any s ∈ R, the element s = 1s is also in I, and so I = R.
- Since every nonzero element in a eld is a unit, we immediately see that the only nonzero ideal of a eld is the full ring. The converse is also true:
- Corollary (Ideals of Fields): A commutative ring R with 1 is a eld if and only if the only ideals of R are 0 and R.
◦ Proof: If F is a eld and I is any nonzero ideal, then I contains some nonzero element r. Since F is a eld, r is a unit, and so by the proposition above, I = R. ◦ Conversely, if the only ideals of R are 0 and R, let r ∈ R be any nonzero element. Then (r) contains r 6 = 0 so it cannot be the zero ideal, so we must have (r) = R. ◦ By the previous proposition, this means (r) contains 1: then rs = 1 for some s ∈ R, so r is a unit. Hence every nonzero element of R is a unit, so R is a eld as claimed. ◦ Remark: In fact, the proof above shows that the only ideals of a division ring R are 0 and R. However, the converse direction does not hold: there exist noncommutative rings R with zero divisors whose only ideals are 0 and R. (One such ring is M 2 × 2 (R), although this is not completely trivial to prove.)
3.3.3 Maximal and Prime Ideals
- An important class of ideals are those that are maximal under inclusion (i.e., which are not contained in any other ideal except the full ring):
- De nition: If R is a ring, a maximal ideal of R is an ideal M 6 = R with the property that the only ideals of R containing M are M and R.
◦ Example: If F is a eld, then since the only ideals of F are 0 and F , the zero ideal is a maximal ideal of F. ◦ Example: In Z, the ideal mZ is contained in nZ precisely when n divides m. Accordingly, the maximal ideals of Z are precisely the ideals of the form pZ, where p is a prime. ◦ Non-example: The ideal (x) is not a maximal ideal of Z[x] because it is contained in the proper ideal (2, x).
- A general ring need not possess any maximal ideals.
◦ A trivial example is the zero ring, since its only ideal is itself.
◦ Proof: Every ideal of F [x] is principal, and the quotient ring F [x]/(p) is a eld if and only if p is irreducible.
- Example: Determine whether the ideal I = (2, x) is a maximal ideal of R = Z[x].
◦ As we have already shown, the quotient ring R/(2, x) is isomorphic to Z/ 2 Z, which is a eld. Thus, I is a maximal ideal of R.
- Example: Determine whether the ideal I = (2) is maximal in R = Z[
2].
◦ In the quotient ring R/I, the residue class
2 + I is nonzero, but has the property that (
2 + I)^2 =
2 + I = 0 + I is equal to zero. ◦ Thus, the quotient ring R/I has zero divisors hence is not a eld, meaning that I is not a maximal of R.
- In addition to maximal ideals, we have another important class of ideals in commutative rings:
- De nition: If R is a commutative ring, a prime ideal of R is an ideal P 6 = R with the property that for any a, b ∈ R with ab in P , at least one of a and b is in P.
◦ Remark: There is also a de nition of prime ideal in a noncommutative ring, but it is more complicated (ultimately because the de nition above involves products of elements). ◦ As naturally suggested by the name, prime ideals are a generalization of the idea of a prime number in Z: for n > 1 , the ideal nZ is a prime ideal of Z precisely when ab ∈ nZ implies a ∈ nZ or b ∈ nZ. Equivalently (in the language of divisibility) this means n|ab implies n|a or n|b, and this is precisely the condition that n is either a prime number (or zero). ◦ Example: The prime ideals of Z are (0) and the ideals pZ where p is a prime number. ◦ A similar statement holds in R = F [x]: the ideal (p) is prime precisely when p is not a unit and p|ab implies p|a or p|b, and the latter condition is equivalent to saying that p is either irreducible or zero. ◦ Example: The prime ideals of F [x] are (0) and the ideals (p) where p is an irreducible polynomial of positive degree.
- Like with maximal ideals, there is an easy way to test whether an ideal is prime using quotient rings:
- Proposition (Prime Ideals and Quotients): If R is a commutative ring with 1, then the ideal P is prime if and only if R/P is an integral domain.
◦ This proof is essentially just a restatement of the de nition of a prime ideal using residue classes in the quotient ring using the observation that r ∈ P if and only if r = 0 in R/P. ◦ Proof: If R is commutative with 1 and P 6 = R, then R/P is also commutative with 1, so we need only test for zero divisors. ◦ If P is a prime ideal, then ab ∈ P implies a ∈ P or b ∈ P. In the quotient ring, this says that ab = 0 implies a = 0 or b = 0, which is precisely the statement that R/P has no zero divisors. ◦ Conversely, if R/P has no zero divisors, then ab = 0 implies a = 0 or b = 0, which is to say, ab ∈ P implies a ∈ P or b ∈ P. Furthermore, since R/P is not the zero ring (since this possibility is excluded by the de nition of integral domain), we see P 6 = R, and therefore P is a prime ideal of R.
- Corollary: A commutative ring with 1 is an integral domain if and only if 0 is a prime ideal.
◦ Proof: 0 is prime if and only if the quotient R/ 0 ∼= R is an integral domain.
- Corollary: In a commutative ring with 1, every maximal ideal is prime.
◦ Proof: If M is a maximal ideal, then R/M is a eld. Every eld is an integral domain, so M is a prime ideal.
- Example: Determine whether the ideals (x) and (x^2 ) in Z[x] are prime ideals.
◦ Note that (x) is the kernel of the evaluation homomorphism ϕ : Z[x] → Z given by ϕ(p) = p(0), and this homomorphism is surjective. ◦ Thus, by the rst isomorphism theorem, we see that Z[x]/(x) is isomorphic to Z. Since Z is an integral domain, we conclude that (x) is a prime ideal. (Note that it is not maximal, however, since Z is not a eld.) ◦ On the other hand, by the division algorithm, we see that the residue classes in Z[x]/(x^2 ) are of the form a + bx where a, b ∈ Z. Since x · x = 0 but x 6 = 0, we see that Z[x]/(x^2 ) has zero divisors, and so (x^2 ) is not a prime ideal.
3.3.4 The Chinese Remainder Theorem
- We now state an important theorem regarding quotient rings by products of ideals. We rst require a few preliminary de nitions:
- De nition: If R is commutative with 1 and I and J are ideals of R, then the sum I +J = {a+b : a ∈ I, b ∈ J} is de ned to be the set of all sums of elements of I and J, and the product IJ = {a 1 b 1 + · · · + anbn, : ai ∈ I, bi ∈ J} is the set of nite sums of products of an element of I with an element of J.
◦ It is not di cult to verify that I + J and IJ are both ideals of R, and that IJ contains the intersection I ∩ J. ◦ We can also speak of the product I 1 I 2 · · · In of more than two ideals, de ned as the set of nite sums of products of an element from each of I 1 , I 2 ,... , In.
- De nition: If R is commutative with 1, the ideals I and J are comaximal if I + J = R.
◦ Note that aZ + bZ = Z precisely when a and b are relatively prime. (The appropriate notion in general rings is not primality but maximality, so we use the term comaximal rather than coprime.)
- We can now state the theorem:
- Theorem (Chinese Remainder Theorem): Let R be commutative with 1 and I 1 , I 2 ,... , In be ideals of R. Then the map ϕ : R → (R/I 1 ) × (R/I 2 ) × · · · × (R/In) de ned by ϕ(r) = (r + I 1 , r + I 2 ,... , r + In) is a ring homomorphism with kernel I 1 ∩ I 2 ∩ · · · ∩ In. If all of the ideals I 1 , I 2 ,... , In are pairwise comaximal, then ϕ is surjective and I 1 ∩ I 2 ∩ · · · ∩ In = I 1 I 2 · · · In, and thus R/(I 1 I 2 · · · In) ∼= (R/I 1 ) × (R/I 2 ) × · · · × (R/In).
◦ Proof: First, ϕ is a homomorphism since ϕ(a + b) = (a + b + I 1 ,... , a + b + In) = (a + I 1 ,... , a + In) + (b + I 1 ,... , b + In) = ϕ(a) + ϕ(b) and similarly ϕ(ab) = (ab + I 1 ,... , ab + In) = (a + I 1 ,... , a + In) · (b + I 1 ,... , b + In) = ϕ(a)ϕ(b). ◦ The kernel of ϕ is the set of elements r ∈ R such that ϕ(r) = (0 + I 1 ,... , 0 + In), which is equivalent to requiring r ∈ I 1 , r ∈ I 2 , ... , and r ∈ In: thus, ker ϕ = I 1 ∩ I 2 ∩ · · · ∩ In. ◦ For the second statement, we will prove the results for two ideals and then deduce the general statement via induction. ◦ So suppose I and J are ideals of R and ϕ : R → (R/I) × (R/J) has ϕ(r) = (r + I, r + J). We must show that if I + J = R, then I ∩ J = IJ and ϕ is surjective. ◦ If I + J = R then by de nition there exist elements x ∈ I and y ∈ J with x + y = 1. ◦ Then for any r ∈ I ∩ J, we can write r = r(x + y) = rx + yr, and both rx and yr are in IJ: hence I ∩ J ⊆ IJ, and since IJ ⊆ I ∩ J we conclude IJ = I ∩ J. ◦ Furthermore, for any a, b ∈ R we can write ay + bx = a(1 − x) + bx = a + (b − a)x so ay + bx ∈ a + I, and likewise ay + bx = ay + b(1 − y) = b + (a − b)y ∈ b + J. ◦ Then ϕ(ay + bx) = (ay + bx + I, ay + bx + J) = (a + I, b + J), and therefore ϕ is surjective as claimed. ◦ Finally, the statement that R/IJ ∼= (R/I) × (R/J) then follows immediately by the rst isomorphism theorem. This establishes all of the results for two ideals. ◦ For the general statement, we use induction on n: the base case n = 2 was done above, and for the inductive step, it is enough to show that the ideals I 1 and I 2 · · · In are comaximal, since then we may write R/(I 1 I 2 · · · In) ∼= (R/I 1 ) × (R/I 2 · · · In) and apply the induction hypothesis to R/I 2 · · · In.
