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Let H be a subgroup of G. If for every a element of G, we have aHa-1 = H, then H is called a normal subgroup of G. The set of integers. The set residue classes mod n. commutative group If H is a subgroup of G then aH = {a · h | a in G, h in H} is a left coset Ha = {h · a | a in G, h in H} is a right coset The set of rational numbers. The set of real numbers. Let H G, H . Then H is a subgroup if and only if for a,bH, we have a · b-1 H. Let F be a field, X an indeterminate over F, and let f, g element of F[X], g not equal 0. Then there exist polynomial q and r belonging to F[X] such that (a) f = qg + r, and (b) deg(r) < deg(g). Let R be a ring and let collection be a collection of subrings of R. Then intersection of subrings S is a subring of R.

Subrings and Quotient Rings

Earlier we introduced groups and then how to manufacture new groups from a given group by forming subgroups and quotient groups. In this section we will describe the analogous process of manufacture for rings. We will study the subrings and quotient rings of a given ring R. We will begin with subrings.

Definition 1: Let R be a ring. A subset S of R is called a subring of R if S is a ring with respect to the operations of addition and multiplication in R.

Example 1: Z is a subring of Q.

Example 2: Q is a subring of R, the field of real numbers.

Example 3: 2Z is a subring of Z.

Example 4: Z[square root of d] is a subring of Q(square root of d).

Example 5: Let M2(Q) denote the ring of 2 cross 2 matrices over Q, and let

T2(Q) = matrix over Q,

U2(Q) = matrix over Q,

We assert that T2(Q) and U2(Q) are subrings of M2(Q). It is not necessary to check the associativity laws or the commutativity of addition, since these already hold in M2(Q). Thus in order to check that T2(Q) and U2(Q) are subrings, it suffices to check that they are subgroups of M2(Q) under addition and that they are closed under multiplication. But this is easy, since


matrix arithmetic

[Note that we need Proposition 2 from the section on subgroups to draw the conclusion that T2(Q) and U2(Q) are subgroups of M2(Q) under addition.]

Example 6: Let R be any ring. Then R and {0} are subrings of R, called the trivial subrings.

The observations of Example 5 lead to

Theorem 2: Let S be a nonempty subset of the ring R. Then S is a subring of R if and only if for a,b element of S, we have a - b element of S, a · b element of S.

Proof: forward implication If S is a subring of R and a,b element of S, we clearly must have a - b element of S, a · b element of S.

backwards implication Suppose that whenever a,b element of S, we have a - b element of S, a · b element of S. Then by Proposition 2 of the section on subgroups, S is a subgroup of R with respect to addition. Moreover, S is closed under multiplication. The associativity of multiplication and distributivity laws hold because they hold in R.

As an easy consequence of the above theorem, we can prove the following proposition.

Proposition 3: Let R be a ring and let collection be a collection of subrings of R. Then

intersection of subrings S

is a subring of R.

Proof: Let

S* = intersection of subrings S,

and let a,b element of S*. It suffices to show that a - belement of S* and a · b element of S*. But, since a,b element of S*, we have a,b element of S for all S element of collection implies that S is a subring of R, we see that a - b, a · b element of S for all S element of collection by Theorem 2. Thus,


a - b, a · b element of intersection of subrings S = S*.

Therefore, by Theorem 2, S* is a subring of R.

Suppose that R is a ring and that C subset of R is any subset. Let us ask the question: What subrings of R contain C? There clearly are such subrings. For example, the trivial subring R is one. Let collection be the collection of all subrings of R which contain C, and let us consider

[C] = intersection of subrings S.

By the above proposition, [C] is a subring of R. Moreover, [C] certainly contains C, since every S element of collection contains C. Further if T is a subring of R which contains C, then T element of collection, and therefore [C] subset of T. Thus we see that [C] is a subring of R which contains C and which is contained in every subring of R which contains [C]. We have therefore proved

Proposition 4: Let C be a subset of the ring R. Then [C] is the smallest subring or R which contains C.

The subring [C] is called the subring generated by the set C. Let us give some examples. If R = Z and C = {2}, then [C] = {..., -4,-2,0,2,4, ...}. That is, [C] is the subring 2Z. Indeed 2Z is a subring of R which contains {2}, and every element of 2Z must automatically be contained in any subring containing {2}. As a second example, let F be a field and let R = F[X], C = F union {X}. Then [C] = F[X], since the smallest subring of R which contains every element of F and X is R itself. Similarly, if C = F union {X2},


[C] = {a0 + a2X2 + a4X4 + ... + a2tX2t | a2telement of F}.

After our brief study of subrings, let us now look at quotient rings. Let R be a ring, S a subring of R. In particular, S is a subgroup of the additive group of R. Since the additive group of R is commutative by the definition of a ring, S is a normal subgroup of the additive group of R. Thus, by our previous group-theoretical results, the set of cosets R/S becomes a group with respect to the law of addition:


(r + S) + (r' + S) = (r + r') + S,

where r + S and r' + S are typical cosets belonging to R/S. It is clear that with respect to this law of addition, R/S is an abelian group. Our problem is: How can we define a law of multiplication on R/S so that R/S becomes a ring? There may be many ways to define such a multiplication, but there is one which has a compelling simplicity:


(1)
(r + S)·(r' + S) = r · r' + S.

The principal difficulty with this definition is that the product of r + S and r' + S depends on the choice of the coset representative r and r', not just on the cosets r + S, r' + S. In order for our law of multiplication to be well defined, it is both necessary and sufficient that


(2)
(r + s + S) · (r' + s' + S) = (r + S) · (r' + S),

for all s,s' element of S. But the left hand side equals (r + s) · (r' + s') + S, while the right hand side equals r · r' + S. Therefore, (2) holds if and only if


r · r' + s · r' + r · s' + s · s' + S = r · r' + S

Since s · s' element of S, this is equivalent to saying that

(3)
s · r' + r · s' element ofS,

where (3) must hold for all s,s' element of S and r,r' element of R. Putting s' = 0, we see that we must have s · r' element of S for s element of S, r' element of R. Similarly, putting s = 0, we see that we must have r · s' element of S for r element of R, s' element of S. Conversely, if s · r' element of S and r · s' element of S r,r' element of R, s,s' element of S, then (3) holds. Thus, to sum up, we have shown that the consistency condition (2) is equivalent to

(4)
r · s' element of S,     s · r' element of S

for r,r' element of R, s,s' element of S. The condition (4) may or may not be satisfied for a given subring S. But whenever it is satisfied, we have shown that it is possible to use (1) to define a law of multiplication on R/S.

Let ~ denote the equivalence relation on R defined by: a ~ bequivalent toa - b element of S. Then R/S is the set of equivalence classes with respect to ~ . The condition (2) merely states that the binary operation · on R is compatible with ~. The law of multiplication (1) on R/S is then the binary operation introduced in the section on equivalence relations.

Definition 5: Let R be a ring. An ideal of R is a subring I of R such that if a element of I, r element of R, then a · r element of I and r · a element of I.

Note that if R is commutative, then the two conditions a · r element of I, r · a element of I can be replaced by the single condition r · a element of I. Note that also that an ideal of R is a subring, but there may be subrings which are not ideals. For example, Z is a subring of Q, but Z is not an ideal of Q since 1/2 element of Q, but 1/2 · 1 not an element of Z. Lets take a look at some examples of ideals.

Example 7: Let R be any ring. then {0} and R are subrings of R and it is clear that these subrings are ideals. These ideals are called the trivial ideals.

Example 8: Let R = Z. We have seen that nZ = {n · r | r element ofZ} is a subring of Z. If s element of Z and n · r element of nZ, then,

(n · r) · s = n · (r · s) element of nZ.

Therefore, nZ is an ideal of Z. In general, we see that every subgroup of Z is a subring.

Example 9: Let F be a field. X an indeterminate over F, f element of F[X], In analogy with Example 8, set

fF[X] = {f · g | g element of F[X]}.

Then fF[X] is an ideal of F[X].

Example 10: Let F be a field, F[X,Y] the ring of polynomials in two indeterminates X and Y over F. Set

(X,Y) = {X · f + Y · g | f,g element of F[X,Y]}.

Then (X,Y) consists of all polynomials with zero constant term and is an ideal of F[X,Y].

Example 11: Let R be any commutative ring, a element of R. Then set

aR = {a · r | r element of R}.

Then aR is an ideal of R. If R = Z, then the example reduces to Example 8. If R = F[X], then this example reduces to example 9.

Before delving any further into the theory of ideals, lets complete our construction of quotient rings. Let R be a ring and let us recall how we were led to the notion of an ideal. We started with a subring S of R and asked whether the multiplication (1) actually makes sense. We found that this is the case if and only if S is an ideal of R.

Proposition 6: Let R be a ring, I an ideal of R, R/I is the set of cosets of the form a + I(a element of R). Define addition of cosets by

(5)
(a + I) + (b + I) = (a + b) + I    (a,b element of R),

and we define multiplication of cosets by

(6)
(a + I) · (b + I) = ab + I.

Then with respect to these operations, R/I becomes a ring, called the quotient ring of R with respect to I.

Proof: From our discussion above, R/I is an abelian group with respect to the law of addition (5). Moreover, we showed that the law of multiplication (6) makes sense, since I is an ideal. Multiplication is associative: For if a,b,c element of R,

(a + I) · [(b + I) · (c + I)] = (a + I) · (b · c + I)
= a · (b · c) + I
= (a · b) · c + I
= (a · b + I) · (c + I)
= [(a + I) · (b + I)] · (c + I).

Similarly, the distributive laws in R imply the corresponding laws in R/I. Thus R/I is a ring.

Proposition 7: Let R be a ring, I an ideal of R. Then

(1) If R is commutative, then R/I is commutative.

(2) If R is a ring with identity 1, then R/I is a ring with identity 1 + I.

Let us return to the examples of ideals which we gave above and lets describe the corresponding quotient rings.

Example 12: R = Z, I = nZ,, n > 0. In this case, R/I consists of the residue classes modulo n. Moreover, it is immediate that the operations defined by (5) and (6) are just addition and multiplication of residue classes, as defined previously. Therefore, in this case, R/I = Zn.

Example 13: Let R be any ring, I a trivial ideal of R - that is, I = {0} or I = R. I the case I = {0}, then R/I consists of the cosets

a + {0}     (a element of R).

Two such cosets a + {0}, a' + {0} are equal if and only if a - a' element of {0}; that is a = a'. Thus, the elements in R/{0} are in the one-to-one correspondence with the elements of R, via the mapping

amaps a + {0}.

Addition and multiplication in R/{0} corresponds to addition and multiplication in R under this correspondence. Thus, R/{0} is "essentially" R. (Strictly speaking, R/{0} is isomorphic to R.) Assume now that I = R. Then the elements of R/I are the cosets a + I(a element of R) and two cosets a + I, a' + I are equal if and only if a - a' element of I. Therefore, since I = R, any two cosets of R/I are the same and R/I is the trivial ring.

Example 14: Let F be a field, R = F[X], f element of F[X] a polynomial such that n = deg(f) > 1, I = fF[X]. The elements of R/I are the cosets g + I(g element of F[X]). Note, however, that many different g can represent the same coset. In fact

g + I = g + f · h + I,    helement of F[X].

From all these different representations for the coset g + I, let us pick a "natural" one. By the division algorithm in F[X], there exist polynomials q,r element of F[X] such that g = f · q + r and either r = 0 or 0 < deg(r) < deg(f) = n. However, since f · q element of I,


g + I = f · q + r + I = r + I.

Therefore, each coset g + I of R/I can be written in the form r + I, where

(7)
r = a0 + a1X + ... + an-1Xn-1   (ai element of F).

Moreover, if r + I = r' + I, then r - r' is a multiple of f. But since r and r' have degree less than deg(f), this implies r = r'. We have therefore proved that every coset g + I can be written uniquely in the form r + I, where r is given by (7).

Let us consider a special case. Let R = Q[X], f = X2 - 2. Then the elements of R/I are the cosets

a0 + a1X + I,   a0,a1 element of Q.

The addition of cosets is given by


(a0+a1X+I) + (b0+b1X+I) = (a0+b0) + (a1+b1)X + I

The multiplication of cosets is defined by

(a0 + a1X + I) · (b0 + b1X + I)
= a0b0 + (a1b0)X + a1b1X2 + I
= (a0b0 + 2a1b1) + (a1b0 + a0b1)X + a1b1(X2 - 2) + I
= (a0b0 + 2a1b1) + (a1b0 + a0b1)X + I

Note that the above addition and multiplication are very similar to the corresponding operations on the ring Q(square root of 2). In fact, if we let correspond to the coset a0 + a1X + I the element a0 + a1square root of 2 of Q(square root of 2), then addition and multiplication in the two rings Q[X]/(X2 - 2)Q[X] and Q(square root of 2) correspond to one another. Thus essentially, the ring Q[X]/(X2 - 2)Q[X] is the "same" as the ring Q(square root of 2). In the next section we will see that Q[X]/(X2 - 2)Q[X] and Q(square root of 2) are isomorphic to each other.

Let us close this section by determining all the subrings of a quotient ring R/I.

Proposition 8: Let R be a ring, I an ideal of R. If S is a subring of R containing I, then S/I is a subring of R/I. Conversely, every subring of R/I is of the form S/I, where S is a subring of R which contains I.

Proof: S/I = {s + I | s element of S}, and therefore S/I subset of R/I. Let s + I, t + I element of S/I. Then s - t, s · t element of S since S is a subring of R. Therefore,

(s + I) - (t + I) = (s - t) + I element of S/I,
(s + I) · (t + I) = s · t + I element of S/I.

Thus, by Proposition 3, S/I is a subring of R/I. Conversely, let U subset of R/I be a subring. Let S = {s element of R | s + I element of U}. Then S is a subring of R containing I. And it is cleat that S/I = U.