Download Section 26 -- Homomorphisms and factor rings

Section 26 – Homomorphisms and factor
rings
Instructor: Yifan Yang
Spring 2007
Overview
• Recall that in the previous semester, we have shown that
• If φ : G → G0 is a group homomorphism, then N = Ker(φ) is
a normal subgroup of G, i.e., gN = Ng for all g ∈ G.
(Corollary 13.20)
• Conversely, if N is a normal subgroup, then there exists a
group homomorphism from G to another group such that
the kernel is N. (Theorem 14.9)
• Let N be a normal subgroup of G. Then the cosets of N
form a group G/N, called a factor group, under the binary
operation ∗ : (aN, bN) 7→ (ab)N. (Theorem 14.5)
• The first isomorphism theorem (Theorem 14.11, or
Theorem 34.2) states that G/Ker(φ) ' φ(G).
• In this section, we will develop an analogous theory for ring
homomorphisms.
Homomorphisms
Definition (26.1)
A map φ from a ring R to another ring R 0 is a (ring)
homomorphism if
1. φ(a + b) = φ(a) + φ(b),
2. φ(ab) = φ(a)φ(b),
for all a, b ∈ R.
Example
1. φ : Z → Zn defined by φ(a) = a mod n is a ring
homomorphism.
2. Let Ri be rings. The projection homomorphisms
πi : R1 × · · · × Rn → Ri are defined by πi (a1 , . . . , an ) = ai .
Properties of homomorphisms
Theorem (26.3)
Let φ : R → R 0 be a ring homomorphism. Then
1. φ(0R ) = 0R 0 ,
2. φ(−a) = −φ(a),
3. If R has a multiplicative identity 1R , then φ(1R ) is the
multiplicative identity of φ(R).
4. If S is a subring of R, then φ(S) is a subring of R 0 .
5. If S 0 is a subring of R 0 , then φ−1 (S 0 ) = {a ∈ R : φ(a) ∈ S 0 }
is a subring of R.
Proof of Theorem 26.3
Proof of (1) and (2). These are proved in Theorem 13.12. Proof of (3).
• We need to prove that for all r 0 ∈ φ(R),
r 0 φ(1R ) = φ(1R )r 0 = r 0 .
• Assume that r 0 ∈ φ(R). Then r 0 = φ(r ) for some r ∈ R.
• Then we have r 0 φ(1R ) = φ(r )φ(1R ).
• Since φ is a ring homomorphism,
φ(r )φ(1R ) = φ(r 1R ) = φ(r ) = r 0 .
• We conclude that φ(1R ) is the multiplicative identity in
φ(R).
Proof of Theorem 26.3, continued
Proof of (4).
• The associativity of multiplication and the distributive laws
for φ(S) are inherited from those for R 0 . Thus, we only
need to show that φ(S) is an abelian group closed under
multiplication.
• The assertion that φ(S) is an abelian group is proved in
Theorem 13.12.
• Now we prove that φ(S) is closed under multiplication.
• Let s10 , s20 ∈ φ(S). Then s10 = φ(s1 ), s20 = φ(s2 ) for some
s1 , s2 ∈ R.
• Then s10 s20 = φ(s1 )φ(s2 ) = φ(s1 s2 ) ∈ φ(S).
Proof of Theorem 26.3, continued
Proof of (5).
• Again, we only need to show that φ−1 (S 0 ) is closed under
multiplication.
• Let s1 , s2 ∈ φ−1 (S 0 ). Then φ(s1 ), φ(s2 ) ∈ S 0 .
• It follows that φ(s1 s2 ) = φ(s1 )φ(s2 ) ∈ S 0 .
• Therefore, s1 s2 ∈ φ−1 (S 0 ).
Remark
• What we showed in (3) above is that φ(1R ) is a
multiplicative identity of the subring φ(R). It may not be a
multiplicative identity for the whole ring R 0 .
• For example, define φ : Z → 2Z by φ(n) = 0 for all n ∈ Z.
This is clearly a ring homomorphism. But φ(1) = 0 is only
the multiplicative identity of φ(Z) = {0}, not of the whole
ring 2Z. In fact, the ring 2Z has no multiplicative identity at
all.
Kernel
Definition
Let φ : R → R 0 be a ring homomorphism. The set
Ker(φ) = {a ∈ R : φ(a) = 0R 0 }
is the kernel of φ.
Example
Let φ2 : R[x] → R be the evaluation homomorphism
φ2 : f (x) 7→ f (2). Then
Ker(φ2 ) = {f (x) ∈ R[x] : (x − 2)|f (x)}
= {(x − 2)g(x) : g(x) ∈ R[x]}.
In-class exercise
1. Let φi : R[x] → C be the evaluation homomorphism
φi : f (x) 7→ f (i). Prove that
Ker(φi ) = {f (x) ∈ R[x] : (x 2 + 1)|f (x)}.
2. Let φ√2 : Q[x] → R be the evaluation homomorphism
√
φ√2 : f (x) 7→ f ( 2). Find a polynomial p(x) in Q[x] such
that
Ker(φ√2 ) = {p(x)g(x) : g(x) ∈ Q[x]}.
Cosets
Theorem (10.1)
Let H be a subgroup of G. Let the relation ∼L on G be defined
by
a ∼L b ⇔ a−1 b ∈ H,
and the relation ∼R be defined by
a ∼R b
⇔
ab−1 ∈ H.
Then ∼L and ∼R are both equivalence relations on G.
Cosets
Definition
Let H be a subgroup of a group G. The equivalence class
{b ∈ G : a ∼L b} is called the left coset of H containing a.
Likewise, the equivalence class {b ∈ G : a ∼R b} is called the
right coset of H containing a.
Remark
1. It is straightforward to see that the left coset containing a is
aH = {ah : h ∈ H}, and the right coset containing a is
Ha = {ha : h ∈ H}.
2. When G is an abelian group, the left coset containing a is
the same as the right coset containing a.
3. When G is an additive group (and thus assumed to be
abelian), we use the additive notation a + H.
Kernel and its cosets
Theorem (26.5)
Let φ : R → R 0 be a ring homomorphism. Let H = Ker(φ). Then
for a ∈ R, φ−1 (φ(a)) = a + H = H + a. (In a plain language, this
means that the set of elements of R that take the same value
as a is a + H.)
Proof.
See the proof of Theorem 13.15.
Corollary (26.6)
A ring homomorphism is one-to-one if and only if its kernel is
{0}.
Example
• Let φ2 : R[x] → R be the evaluation homomorphism
φ2 : f (x) 7→ f (2). We have seen earlier that
Ker(φ2 ) = {(x − 2)g(x) : g(x) ∈ R[x]}.
• Now let c ∈ R. Clearly, we have φ2 (c) = c.
• Then Theorem 26.5 predicts that
−1
φ−1
2 (c) = φ2 (φ2 (c)) = c + Ker(φ2 )
= {c + (x − 2)g(x) : g(x) ∈ R[x]}.
• Indeed, suppose that φ2 (f (x)) = c. Applying the division
algorithm (Theorem 23.1), we write f (x) = (x − 2)h(x) + c 0 .
• Then f (2) = c if and only if c 0 = c and f (x) ∈ c + Ker(φ2 ).
Factor rings
Theorem (26.7)
Let φ : R → R 0 be a ring homomorphism with kernel H. Then
the additive cosets of H form a ring R/H with addition and
multiplication given by
+ : (a+H, b +H) 7→ (a+b)+H, × : (a+H, b +H) 7→ (ab)+H.
Proof. We need to prove that
1. The addition is well-defined and R/H is an abelian group
under this operation.
2. The multiplication is well-defined and is associative.
3. The distributive laws hold.
The proof of associativity and distributive law is easy, and will
be skipped.
A lemma
Lemma
The kernel H above satisfies
ah, hb ∈ H
for all a, b ∈ R and h ∈ H.
Proof.
We have φ(ah) = φ(a)φ(h) = φ(a)0R 0 = 0R 0 . Thus,
ah ∈ Ker(φ). The proof of hb ∈ Ker(φ) is similar.
Proof of Theorem 26.7
Proof of (1). This is done in Theorem 14.1.
Proof of well-definedness of multiplication.
• What we need to show is that if a1 + H = a2 + H and
b1 + H = b2 + H, then (a1 b1 ) + H = (a2 b2 ) + H.
• Now a1 + H = a2 + H and b1 + H = b2 + H if and only if
a1 = a2 + h1 and b1 = b2 + h2 for some h1 , h2 ∈ H.
• Then
a1 b1 = (a2 + h1 )(b2 + h2 )
= a2 b2 + h1 b2 + a2 h2 + h1 h2 .
• By the lemma above, h1 b2 , a2 h2 , h1 h2 ∈ Ker(φ).
• Thus, a1 b1 = a2 b2 + h3 for h3 = h1 b2 + a2 h2 + h1 h2 ∈ H.
That is, (a1 b1 ) + H = (a2 b2 ) + H.
Ideals
Theorem (26.9)
Let H be a subring of a ring R. The map
× : (a + H, b + H) 7→ (ab) + H
is a well-defined multiplication of additive cosets of H if and
only if
ah, hb ∈ H
for all a, b ∈ R and all h ∈ H.
Definition
An additive subgroup I of a ring R satisfying aI ⊂ I, Ib ⊂ I for all
a, b ∈ R is an ideal of R.
Proof of Theorem 26.9
• Assume first that ah, hb ∈ H for all a, b ∈ R and all h ∈ H.
Then the argument in the proof of Theorem 26.7 (the
well-definedness part) shows that × is well-defined.
• Conversely, assume that × is well-defined. We now show
that ah ∈ H for all a ∈ R and h ∈ H.
• Consider (a + H)(0 + H). By the definition of multiplication,
this is equal to (a0) + H = 0 + H.
• On the other hand, we also have 0 + H = h + H for all
h ∈ H. Thus,
0 + H = (a + H)(0 + H) = (a + H)(h + H) = ah + H. This
implies that ah ∈ H.
• The proof of hb ∈ H is similar.
Factor rings (quotient rings)
Corollary (26.14)
Let I be an ideal of a ring R. The the additive cosets of I form a
ring R/I with addition and multiplication given by
+ : (a + I, b + I) 7→ (a + b) + I,
× : (a + I, b + I) 7→ (ab) + I.
Definition
The ring R/I is called the factor ring (or quotient ring) of R by I.
Examples
Examples of ideals.
1. Let R = Z. Then nZ are ideals of Z for all n ∈ Z. The factor
ring Z/nZ are the familiar Zn .
2. Let R = R[x]. Then {(x − 2)g(x) : g(x) ∈ R[x]} and
{(x 2 + 1)g(x) : g(x) ∈ R[x]} are ideals of R[x].
3. Let R be a commutative ring, and a be an element. Then
the set {ra : r ∈ R} is an ideal, called the principal ideal
generated by a, and is denoted by hai.
Examples of subrings that are not ideals.
1. The set Z is a subring of Q, but not an ideal.
2. Let R = R[x], and S be the set of all constant polynomials.
Then S is a subring of R, but not an ideal.
Ideals as kernels of homomorphisms
Theorem (26.16)
Let I be an ideal of a ring R. The map γ : R → R/I defined by
γ : a 7→ a + I is a ring homomorphism with kernel I.
Proof.
Considering I as an additive subgroup of R, Theorem 14.9
already shows that γ is a one-to-one and onto additive group
homomorphism. It remains to prove that γ(ab) = γ(a)γ(b),
which however follows easily from the definition of γ.
Properties of homomorphisms
Theorem (26.3)
Let φ : R → R 0 be a ring homomorphism. Then
1. φ(0R ) = 0R 0 ,
2. φ(−a) = −φ(a),
3. If R has a multiplicative identity 1R , then φ(1R ) is the
multiplicative identity of φ(R).
4. If S is a subring of R, then φ(S) is a subring of R 0 .
5. If S 0 is a subring of R 0 , then φ−1 (S 0 ) = {a ∈ R : φ(a) ∈ S 0 }
is a subring of R.
6. If I is an ideal of R, then φ(I) is an ideal of φ(R).
7. If I 0 is an ideal of R 0 or φ(R), then φ−1 (I 0 ) is an ideal of R.
Isomorphism theorem
Theorem (analogue of the first isomorphism theorem for
groups)
Let φ : R → R 0 be a ring homomorphism. Then we have
R/Ker(φ) ' φ(R).
Proof.
• Define φ∗ : R/Ker(φ) → φ(R) by φ∗ (a + I) = φ(a).
• In Theorem 14.11, we have already proved that φ∗ is a
well-defined additive group isomorphism.
• It remains to show that
φ∗ ((a + I)(b + I)) = φ∗ (a + I)φ∗ (b + I). This again follows
easily from the definition of φ∗ .
Example
• Let φi : R[x] → C be the evaluation homomorphism
φi : f (x) 7→ f (i).
• We have seen earlier that
Ker(φi ) = {(x 2 + 1)g(x) : g(x) ∈ R[x]} = I.
• The image of φi is clearly C.
• Thus, R[x]/I ' C. Note that coset representatives of R[x]/I
can be taken to be ax + b, where a, b ∈ R. Then the
isomorphism is φ∗i : ax + b + I 7→ ai + b.
• Observe that the evaluation homomorphism
φ−i : R[x] → C given by φ−i : f (x) 7→ f (−i) has the same
kernel and image.
• Thus, we have another isomorphism φ∗−i : R[x]/I → C
defined by φ∗−i : ax + b + I → −ai + b.
• This type of arugments is the starting point for the Galois
theory.
Homework
Problems 4, 9, 12–15, 18, 20, 22, 26 of Section 27.