Example. Let H be a subgroup of G, and set G/H = {aH|a ∈ G}. ∀x ∈
G, we define
Tx :
G/H
→ G/H
aH
7→ xaH
⇒ G/H is a G−set.
Definition. Let S, S ′ be two G−sets. f : S → S ′ be a map. We say f a
homomorphism of G−sets, if
f (x.s) = x.f (s), ∀x ∈ G, s ∈ S.
Definition. Let S be a G−set. For s ∈ S, we define
G(s) = {x ∈ G|x.s = s}.
Lemma. G(s) is a subgroup of G.
Proof. The identity element e is clearly in G(s) . ∀x, y ∈ G(s) , i.e. x.s =
s, y.s = s ⇒ (xy).s = x.(y.s) = x.s = s ⇒ xy ∈ G(s) . And x−1 .s =
x−1 .(x.s) = (x−1 x).s = e.s = s ⇒ x−1 ∈ G(s) .
2
Definition. G(s) is called the isotropy group of s in G.
Example. Under the conjugation, a group G itself is a G−set. ∀x ∈ G,
the isotropy group of s is
G(s) = {x ∈ G|x.s = s} = {x ∈ G|xsx−1 = s} = {x ∈ G|xs = sx}
called the centralizer of s in G.
Lemma. Let S be a G−set, s, s′ ∈ S, y ∈ G satisfying s′ = y.s. Then
G(s′ ) = yG(s) y −1 .
Proof. G(s′ ) = {x ∈ G|x.s′ = s′ }, G(s) = {x ∈ G|x.s = s}. ∀x ∈
G(s′ ) ⇒ x.s′ = s′ ⇒ x.y.s = y.s ⇒ (y −1 xy).s = s i.e. y −1 xy ∈ G(s)
i.e. x ∈ yG(s) y −1 ⇒ G(s′ ) ⊂ yG(s) y −1 .
1
On the other hand, if x ∈ yG(s) y −1 ⇒ y −1 xy ∈ G(s) ⇒ y −1 xy.s =
s ⇒ x.y.s = y.s i.e. x.s′ = s′ ⇒ x ∈ G(s′ ) ⇒ yG(s) y −1 ⊂ G(s′ ) .
2
Remark. From the lemma, we see that the isotropy groups G(s) , G(s′ ) are
conjugate if s′ = y.s for some y ∈ G.
Proposition. Let S be a G−set, s ∈ S, H = G(s) . If x, y ∈ G, then x, y
are in the same coset of the subgroup H if and only if x.s = y.s.
Proof. Since xH = yH ⇔ x−1 y ∈ H ⇔ (x− 1y)s = s ⇔ y.s = x.s.
2
Definition. Let S be a G−set, s ∈ S. We set
G.s = {x.s|x ∈ G}
and G.s is called the orbit of s under G.
Let G, H and S be the same sa in the previous proposition. Let G/H =
{aH|a ∈ G}. We define a map, for a fixed element s ∈ S ′ ,
f : G/H
7→ a.s
aH
(f is well-defined: since xH = yH
→ S
.
⇒ x−1 y ∈ H = G(s)
⇒ x−1 y.x =
s ⇒ y.s = x.s.)
Lemma. f is a homomorphism of G−sets.
Proof. Note that both G/H and S are G−sets. For a ∈ G, we have
f (a.(xH)) = f ((ax).H) = (ax).s = a.(x.s) = af (xH).
2
Remark. It is clear that Imf = G.s (the orbit of s under G).
Corollary. f is a bijective map from G/H to G.s.
2
Proof. It is clear that f is surjective. To show f is injective, we note
that, from the above proposition x.s = y.s ⇔ x, y are in the same coset of
H = G(s) ⇔ xH = yH.
2
Corollary. Let S be a G−set, s ∈ S. Then the number of elements in
the orbit G.s is equal to [G : G(s) ].
Lemma. Let S be a G−set. ∀s1 , s2 ∈ S, then G.s1 ∩ G.s2 6= ∅ ∼
= G.s1 =
G.s2 .
Proof. s ∈ G.s1 ∩ G.s2 ⇔ s = x.s1 = y.s2 for some x, y ∈ G ⇔ G.s =
G.(x.s1 ) = G.s1 = G.(y.s2 ) = G.s2 .
2
Corollary. Let S be a G−set. Then
1◦ S = union of disjoint orbits, i.e.
S=
]
G.si .
i∈I
where I is an index set, si ∈ S and G.si ∩ G.sj = ∅ if i 6= j.
2◦ If S is a finite set, then
|S| =
X
i∈I
|G.si | =
X
[G : G(si ) ].
(∗)
i∈I
Definition. The identity (∗) is called orbit decomposition formula.
Homework.
1. Let G be a finite group, N G, and |N | and |G/N | be coprime. If a ∈ G
and the order of a divides |N |, then a ∈ N .
2. Let G be a group, c ∈ G, and the order of c is rs, where (r, s) = 1. Prove
that there are elements a and b of G such that c = ab and the order of a and b
are r and s respectively, and ab = ba.
3. Let G be a group of order n, and let F be any field. Prove that G is
isomorphic to a subgroup of GLn (F).
3
Proposition. Let G be a group. Under the conjugation operation, G is
a G−set. Then
|G| =
X
[G : G(x) ]
x∈C
where
G(x) = {y ∈ G|y.x = x} = {y ∈ G|yx = xy} = Zx
the centralizer of x in G, C is the set of representatives for the distinct conjugate
classes.
Proof. By using the orbit decomposition formula, we have
|G| =
X
[G : G(si ) ]
i∈I
where I is an index set of distinct orbits. Note that G.x = G.y ⇔ G.x ∩ G.y 6=
∅ ⇔ ∃ z ∈ G.x ∩ G.y ⇔ z = a.x = b.y for some a, b ∈ G ⇔ axa−1 =
byb−1 ⇔ y = (b−1 a)x(b−1 a)−1 ⇔ y = (b−1 a).x ⇔ G(y) = (b−1 a)G(x) (b−1 a)−1
i.e. G(x) , G(y) are conjugate.
2
Definition. Let S be a subset of a group G, and
N (S) = {a ∈ G|aSa−1 = S},
Z(S) = {a ∈ G|ay = ya, ∀ y ∈ S.}.
Then N (S) is called the normalizer of S in G, and Z(S) is called the centralizer
of S in G.
Proposition. Let N (S), Z(S) be defined as above. Then
1◦ N (S), Z(S) are subgroups of G.
2◦ If S is a subgroup of G, then S N (S).
3◦ The centralizer of G in G is the center of G.
Proof. 1◦ and 3◦ are clear. For 2◦ , ∀ a ∈ N (S) and b ∈ S, then
aba−1 ∈ aS a −1 = S,
i.e. S is a normal subgroup of N (S).
2
4
6.2 Sylow Subgroups
Definition. 1◦ Let G be a group, p a prime number. If |G| = pn for some
integer n ≥ 1, then G is called a p−group.
2◦ If G is a finite group, and H a subgroup of G. We call H a p =subgroup
of G if |H| = pn for some integer n.
3◦ A subgroup H of G is called a Sylow p−subgroup of G if |H| = pn and
pn+1 6 ||G|.
Lemma. If G is a finite abelian group, p is a prime number and p||G|, then
G has a subgroup of order p.
Proof. We prove the lemma by induction on the order of the groups.
Take b ∈ G, and b 6= e(the identity element). Let H =< b > be the cyclic
subgroup generated by the element b. Then
|G| = |H|.[G : H].
Note that p||g|, then p||H| or p|[G : H].
If p||H|, then H has a subgroup of order p, which is also a subgroup of G.
If p 6 ||H|, then p|[G : H], i.e., P ||G/H|. Then by induction, we know that
the quotient group G/H has a subgroup of order p, i.e., there is an element
x ∈ G (x 6∈ H) such that (xH)p = H or xp ∈ H.
Let xp =: y ∈ H and o(y) = q. Then (p, q) = 1. (Since p is prime and
p 6 ||H|.)
We claim that xq 6= e and (xq )p = e i.e. xq is an element of order p, which
generates subgroup of order p.
Indeed, if xq = e, then there exist a, b ∈ Z such that ap + bq = 1 and
x = xap+bq = y a ∈ H,
a contradiction.
2
5
Theorem. (First Sylow Theorem.) Let G be a finite group, p a
prime number and p||G|. Then G has a Sylow p−subgroup.
Proof. We prove the theorem by induction on the order of the groups.
If |G| = p, then G itself is the Sylow p−subgroup of G.
Now we suppose that the statement of the theorem holds for all finite groups
of order less than |G|.
If there is a proper subgroup H of G such that [G : H] is prime to p, then
by the fact that |G| = |H|.[G : H], we know that H has Sylow p−subgroup and
which is also the Sylow p−subgroup of G. In this case, we are done.
Next if each proper subgroup of G has index divisible by p. We consider G
as a G−set under the coonjugate operation. We use the orbit decomposition
formula:
|G| =
X
i∈I
|G.xi | =
X
[G : G(xi ) ],
i∈I
where G.xi ∩ G.xj = ∅ if i 6= j and G(xi ) is the isotropy subgroup of xi in G.
Let Z = Z(G) be the center of G. It is clear that, for a ∈ Z, the orbit
G.a = {a} (i.e. the orbit of the element in the center contains only one elemnt.),
which implies that G.a for a ∈ Z, are distinct orbits. And each such an orbit
contains only one element.
It is clear that if |G.a| > 1, then a 6∈ Z. Thus
|G| =
X
i∈I
|G.a| =
X
xi ∈Z
|G.xi | +
X
xi 6∈Z
|G.xi | = |Z| +
X
xi 6∈Z
|G.xi |,
where |G.xi | = [G : G(xi ) ] > 1 for xi ∈ Z, i.e., G(xi ) is a proper subgroup of G,
by assumption p|[G : G(xi ) ] for xi 6∈ Z.
Therefore p||Z|, which means that the center Z of G is nontrivial. By applying the previous lemma, we know that Z contains a subgroup H of order p.
Then H ⊳ G. Let
f : G → G/H
a
7→
aH
be the canonical hommorphism. If pn ||G| and pn+1 6 ||G|, then pn−1 ||G/H|,
pn 6 ||G/H|. By induction, G/H has a Sylow p−subgroup K ′ . Set K = f −1 (K ′ ),
6
then K is a subgroup of G, which contains H and
f |K : K → K ′
ia a surjective homomorphism. This means
K/H ∼
= K ′.
Thus |K| = |H| · [K : H] = p · |K ′ | = p · pn−1 = pn .
2
Homework.
Let H be a subgroup of G. Then
(a). The number of subgroups of which are conjugate to H is equal to [G :
N (H)].
(b). Z(H) N (H).
(c). N (H)Z(H) is isomorphic to a subgroup of Aut(H).
(d). If G is a finite group and H is an proper subgroup of G then the union of
the subgroup which are conjugate to H is a proper subset of G.
Corollary. If p is a prime number and |G| = pl (l ≥ 1), then the center
Z(G) of G is non-trivial.
Definition. Let S be a G−set. We call s0 ∈ S a fixed point under G if
G.s0 = {s0 }, i.e., a.s0 = s0 , ∀a ∈ G.
Remark. If S = G and the operation is definite by conjugation, then s0
is a fixed point iff s0 ∈ Z(G).
Lemma. Let G be a p−group, and S be a G−set. If |S| = n and (n, p) = 1,
then S contains fixed point under G.
Proof. Applying the orbit decomposition formula, we have
|S| =
X
i∈I
|G.si |,
where G.s1 , · · · , G.sk are distinct orbits. Note that |G.si | = [G : G(si )] and G
is a p−group, we know that |G.si | = pli for some li ≥ 0. If S contains no fixed
7
point, then we have li ≥ 1, for all i. This implies that p||G.si |, ∀i. and therefore
p||S|, which contradicts to the condition (p, n) = 1.
2
Definition. Let G be a finite group, we call G a simple group if G 6= {1}
and G contains no normal subgroup other than {1} and G.
Example. (1) If p is a prime number and |G| = p, then the group G is a
simple group.
(2) The symmetric group Sn (n ≥ 3) is not simple as the alternating group
An is a normal subgroup of Sn .
Example. Let H be a normal subgroup of a group G. Suppose that
|H| = 5 and that |G| is an odd number. Prove that H is contained in the center
of G.
Proof. Since H ⊳ G, then N (H) = G. We apply the result from homework
5.1: G/Z(H) = N (H)/Z(H) is isomorphic to a subgroup of Aut(H). But
|H| = 5 implies H is a cyclic group of order 5 and Aut(H) = 4, while Aut(H)
has only 3 subgroups with order 1,2 and 4. Note that
|G| = |Z(H)| · [G : Z(H)] = |Z(H)| · |G/Z(H)|,
and |G| is an odd number, we have |G/Z(H)| = 1, i.e., Z(H) = G, and therefore
H is in the center of G.
2
Homework.
1. Let H G, | H |= p, where p is the smallest prime that divides | G |.
Prove H ⊂ Z(G).
2. Let G be a group. a, b ∈ G are distinct elements of order 2. If the order
of ab is odd. Prove a and b are conjugate in G.
Remark. The classification of non-abelian groups was completed in 1980’s:
1. The alternative groups An (n ≥ 5);
2. The simple groups of Lie type (Matrix groups);
8
3. 26 sporadic groups.
The monster group M with order 246 · 320 · 59 · 76 · 112 · 133 · 17 · 19 · 23 · 29 ·
31 · 41 · 59 · 71 ≈ 1054 was found by Fischer-Griess (called Friendly Giant).
Mathieu groups (1860-1873): M11 , M12 , M22 , M23 , M24 ;
Janko groups (1965- ): J1 , J2 , J3 , J4 ;
Thompson groups: T h;
Fischer groups: F i22 , F i23 , F i24 ;
The monster group: M .
Recall. 1◦ Isomorphism Theorem for groups: If H, K are subgroups of G,
and K G, then
HK/K ∼
= H/H ∩ G.
2◦ If H, P are subgroups of G and H ⊂ N (P ), then HP is a subgroup of G and
P HP .
Theorem.(2rd Sylow Theorem.) Let G be a finite group.
(i.) If H is a p−subgroup of G, then H is contained in a Sylow p−subgroup
of G.
(ii.) All Sylow p−subgroups are conjugate.
(iii.) The number of Sylow p−subgroups of G is equal to 1 (mod p).
Proof. Let S = {All Sylow p − subgroups of G}, then S is a G−set under
the conjugation operation.
Fix P ∈ S, then P ⊂ G(P ) = {x ∈ G|x.P = P }. Note that
|G| = |G(P )| · [G : G(P )]
where [G : G(P )] = |G.P |. This implies that the order of the orbit G.P =
{x.P |x ∈ G} = {xP x−1 |x ∈ G} is prime to p.
Let H be a p−subgroup of G and |H| > 1. Set S0 = G.P , then S0 is an
H−set under the conjugation operation and (p, |S|) = 1. The previous lemma
implies that S0 contains a fixed point under the group H, i.e.
h.(g.P ) = g.P
9
for some g ∈ G and all h ∈ H, i.e.
h(gP g −1 )h−1 = gP g −1 , ∀h ∈ H
(∗)
where P ′ = gP g −1 is a Sylow p−subgroup which conjugate to P .
(∗) implies that H ⊂ N (P ′ ). Then HP ′ is a subgroup of G and P ′ HP ′ ,
by the Isomorphism Theorem for groups,
HP ′ /P ′ ∼
= H/H ∩ P ′ .
Note that H is a p−subgroup, then |H/H ∩ P ′ | = pr for some r ≥ 0, and
|HP ′ | = |P ′ | · [HP ′ : P ′ ] = |P ′ | · |H/H ∩ P ′ | = pl+r ,
(where we assume that |G| = pl · m and (p, m) = 1).
This gives that r = 0, because that |P ′ | = pl and P ′ is the Sylow p−subgroup
of G. And thus HP ′ = P ′ or H ⊂ P ′ . This proves (i).
For (ii), if we take H to be a Sylow p−subgroup and repeat the argument of
(i), we also have H ⊂ P ′ ⇒ H = P ′ = gP g −1 (since H and P ′ have the same
order). Therefore H and P are conjugate.
Finally, to prove (iii), we first prove a Lemma.
Lemma. Let P be a Sylow p−subgroup of G. Then G has only one Sylow
p−subgroup iff p G.
Proof of the lemma. ” ⇒ ” ∀x ∈ G. Since xP x1 is also a Sylow p−subgroup,
P = xP x−1 , i.e. P G.
” ⇐ ” From (ii), we know all Sylow p−subgroups are conjugate, thus for any
Sylow p−subgroup P ′ , we have P ′ = xP x−1 for some x ∈ G. From P G, we
have P ′ = P.
Now we prove (iii). We take H = P in the argument of (i), then S =
{All Sylow p − subgroups of G} is a H−set under the conjugation operation.
For each element in S ′ , there is a orbit under H. For the element P ∈ S,
the orbit H.P = {P } while for P ′ ∈ S and P ′ 6= P , then the orbit H.P ′ has
more than one element (because, if H.P ′ = {P ′ }, then for any x ∈ H = P ,
xP ′ x−1 = P ′ implies that P ⊂ N (P ′ ), where P, P ′ are all Sylow p−subgroups
10
of N (P ′ ), and P ′ N (P ′ ) ⇒ P = P ′ ). Thus |H.P ′ | = [H : H(P ′ )] implies that
|H.P ′ | is divisible by p. Therefore
|S| =
X
P ′ ∈I
|HP ′ | = |H.P | +
X
P ′ 6=P
|HP ′ | ≡ 1 (mod p).
2
Theorem. Let G be a finite group and p be a prime number and p||G|.
Then the number of Sylow p−subgroups of G is a divisor of |G|.
Proof. Let S = {All Sylow p − subgroups of G}, and S is a G−set under
the conjugation operation. From the 2nd Sylow Theorem, we know that all
Sylow p−subgroups are conjugate, i.e.
S = G.P,
for P ∈ S. Thus |S| = |G.P | = [G : G(P )] which is a divisor of G.
2
Corollary. If |G| = pl · m and (p, m) = 1, then the number of Sylow
p−subgroups is a factor of m.
Proof. Since
pl · m = |G| = |G(P )| · [G : G(P )],
where G(P ) < G and P < G(P ), pl ||G(P )|, thus [G : G(P )]|m.
2
Example. If |G| = 15, prove that G is a cyclic group.
Proof. Since |G| = 15 = 3 · 5, G has Sylow 3-subgroup and Sylow 5-groups.
By Sylow Theorem, we know that there are 3t + 1 Sylow 3-subgroups and
(3t + 1)|5, which implies t = 0 and G has only one Sylow 3-subgroups K. Thus
K G.
Similarly, there are (5t + 1) Sylow 5-subgroups and (5t + 1)|3 . This deduces
that t = 0 and G has only one Sylow 5−subgroups H. And moreover H G.
Set K =< a > and H =< b >. Then o(a) = 3, o(b) = 5 and K ∩ H = {1}.
Note that aba−1 b−1 ∈ H ∩K, ab = ba. Thus o(ab) = 15. So G = HK =< ab > .
2
11
Example. If |G| = 72, prove that G is not simple.
Proof. |G| = 72 = 23 · 32 . Then |G| has (3t + 1) Sylow 3-subgroups and
(3t + 1)|23 ⇒ t = 0 or t = 1.
Case 1. If t = 0, i.e. G has only one Sylow 3-subgroups, which is normal.
Case 2. If t = 1, i.e. G has 4 Sylow 3-subgroups, say P1 , P2 , P3 , P4 . Set
S = {P1 , P2 , P3 , P4 }, then S is a G−set under the conjugation operation. The
Sylow’s Theorem says that all Sylow 3-subgroups are conjugate. Thus ∀g ∈ G
defines a permutation on S
fg : g.{P1 , P2 , P3 , P4 } = {gP1 g −1 , gP2 g−1 , gP3 g −1 , gP4 g −1 }.
Thus there is a group homomorphism:
f : G → S4
g
7→
fg
where fg is not identity map in general. Moreover, |G| = 72 and |S4 | = 24,
which implies that ker f is a non-trivial normal subgroup of G, thus G is not
simple.
2
Homework.
1. Let H be a Sylow p-subgroup of G and K = N (H), prove N (K) = K.
2. If |G| = 224, prove that G is not simple.
Theorem. Suppose |G| = pr m, where r ≥ 1, m > 1 and p 6 |m. If G is
simple, then the number n of Sylow p−subgroups of G satisfying:
|G| divides n!.
Proof. Let P be a Sylow p−subgroup of G, Ω = {All Sylow p-subgroups of G}.
Then Ω is a G−set under the conjugate operation. From Sylow Theorem, we
know that all Sylow p−subgroups are conjugate. Thus
n = |Ω| = |G.P | = [G : G(P )],
12
where the isotropy group
G(P ) = {a ∈ G|a.P = P } = {a ∈ G|aP a−1 = P } = N (P ).
Since G is simple and 1 < P < G, thus P is not normal, and so n > 1.
Set H = N (P ), then [G : H] = n, set T = {a1 H, · · · , an H} be the set of
distinct left cosets of H in G. ∀x ∈ G :
ϕx :
→
T
T
7→ xaH.
aH
Then ϕx defines a permutation of T , i.e. ϕx ∈ Sn . Therefore
ϕ: G
→ Sn
x
7→ ϕx
is a group homomorphism, and the fact that G is simple implies ker ϕ = {1}
(Since not all ϕx are identity map). ⇒ ϕ is injective ⇒ G ∼
= Imϕ < Sn ⇒
|G||n!.
2
Example. If |G| = 224, then G is not simple.
Proof. Suppose G is simple. Note that 224 = 25 · 7, we know that G has
2t + 1 Sylow 2-subgroups and (2t + 1)|7 ⇒ t = 0 or t = 3. That G is simple
implies that t = 3, i.e. G has 7 Sylow 2-subgroups. Then |G||7!, i.e. 25 |6!. But
this is not impossible, a contradiction.
2
Second proof of the 1st Sylow Theorem.
Lemma. If n = pr m, where p is prime, then
n
pr
≡ m (mod p).
Proof. If f (x), g(x) are polynomials with integer coefficients, we say f (x) ≡
g(x) (mod p), if for every j ≥ 0, the coefficient of xj in f (x) and g(x) are
congruent modulo p. For instance, we have
(x + 1)p
2
(x + 1)p
pr
(x + 1)
≡
≡
···
≡
xp + 1 (mod p),
2
xp + 1 (mod p),
r
xp + 1 (mod p).
13
Thus
r
r
(x + 1)n = ((x + 1)p )m ≡ (xp + 1)m (mod p).
r
We compare the coefficient of xp on both sides to get
n
≡ m (mod p).
pr
2
1st Sylow Theorem. If |G| = pr m, where p is a prime and p 6 |m, then
G has a subgroup of order pr .
Proof. Let Ω = {all subsets of G of order pr }, i.e. Ω = {A ⊂ G||A| = pr }.
r
Then |Ω| = pprm ≡ m (mod p) 6≡ 0.
The set Ω is a G−set with the following operation, ∀g ∈ G
g:
Ω
→
X
7→ gX.
Ω
By applying the orbit decomposition formula
|Ω| =
X
x∈I
|G.X|,
where G.X, X ∈ I are distinct orbits. Since p 6 ||ω|, there exists X0 ∈ I, such
that p ∤ |G.X0 |. Let G(X0 ) be the isotropy subgroup of G under X0 ∈ Ω, then
|G.X0 | = [G : G(X0 )].
Thus pr ||G(X0 )|. Moreover, |G(X0 )| ≥ pr .
On the other hand, for a ∈ X0 , and h ∈ G(X0 ). We have h.a ∈ h.X0 = X0 .
This deduces that G(X0 ).a ⊂ X0 , ⇒ |G(X0 )| = |G(X0 ).a| ≤ |X0 | = pr , ⇒
|G(X0 )| = pr .
2
Homework.
1. If G is a group, and |G| = 1000000, prove G is not simple.
2. Let G be a finite group. H G, and P a Sylow p−subgroup of H. Prove
that G = H · N (P ).
14
Definition. 1◦ Let G be a group. S = {x1 , · · · , xn } is a subset of G. The
subgroup of G generated by the elements x1 , · · · , xn is defined by
{e, y1 · · · , yk |where k ≥ 1, yi ∈ S ∪ S −1 , for all i},
−1
where e is the identity element of G and S −1 = {x−1
1 , · · · , xn }.
2◦ Let G be a group. The subgroup of G generated by all elements of the form
aba−1 b−1 , a, b ∈ G is called the commutator subgroup (or derived subgroup) of
G, and is denoted by G′ .
Lemma. G′ G and G/G′ is an abelian group.
Proof. If ϕ : G → G is a group homomorphism, then ∀a, b ∈ G, f (aba−1 b−1 ) =
f (a)f (b)f (a)−1 f (b)−1 ∈ G′ ⇒ f (G′ ) is a subgroup of G′ .
Now, ∀a ∈ G, we define
→
fa : G
x
G
7→ axa−1 .
Then fa is a group automorphism and fa (G′ ) ⊂ G′ , i.e. aG′ a−1 ⊂ G′ . This
implies that G′ G.
Moreover, a−1 b−1 ab ∈ G′ ⇒ abG′ = baG′ ⇒ (aG′ )(bG′ ) = (bG′ )(aG′ ), ∀a, b ∈
G, i.e. G/G′ is abelian.
2
Definition. Let G be a group.
(1) Set G(1) = G′ , G(i+1) = (G(i) )′ for i ≥ 1, we call G(i) the i−th derived
subgroup of G, then
G G(1) G(2) G(3) · · · .
(2) If there exists n ≥ 1 such that G(n) = {e}, then G is called solvable.
Theorem. If G is a p−group, p is a prime, then G is solvable.
Proof. We assume |G| > 1.
If |G| = p, then G is abelian ⇒ G′ = {e}, i.e. G is solvable.
15
Suppose the result holds for all p−groups with order less than |G|. We
know that the p−group G has a nontrivial center Z ⇒ G/Z is a p−group
with order less than |G|. Thus there is n ≥ 1, such that (G/Z)(n) = {Z} and
(G/Z)(n−1) 6= {Z}. So Gn = Z and Gn+1 = {e}.
2
Recall. If H, K < G, then
|HK| =
|H| · |K|
.
|H ∩ K|
Example. If |G| = mpk , (m, p) = 1, p a prime, and if H is a Sylow
p−subgroup of G and K is a p− subgroup of G such that K * H. Prove that
HK is not a subgroup.
Proof. Since K * H, then K ∩ H is a proper subgroup of K ⇒ |K ∩ H| <
|K|, i.e. p|[K : K ∩ H] (since |K| = |K ∩ H| · [K : K ∩ H]). Therefore
|HK| =
|H| · |K|
= pk · [k : H ∩ K]
|H ∩ K|
⇒ pk+1 ||HK| ⇒ HK is not a subgroup of G.
2
Chapter II. Rings
1.Definition.
Definition. A ring R is a set with two laws of compositions: ” + ” and
” · ”, called addition and multiplication, which satisfy the conditions:
(1) Set R with the addition forms an abelian group, i.e. (R, +) is an aleblian
group. The identity is denoted by 0, and often called the zero element.
(2) Multiplication in R is associative, and R has an identity, denoted by 1,
i.e. 1 ∈ R and 1 · a = a · 1 = a, ∀a ∈ R. The element 1 is often called the identity
element of the ring R.
(3) Distribution laws hold: ∀a, b, c ∈ R,
a · (b + c) = a · b + a · c, (a + b) · c = a · b + b · c.
16
Example. 1◦ (Z, +, ·) is a ring.
2◦ (Z/nZ, +, ·) is a ring, where
R := Z/nZ = {0, 1, · · · , n − 1}
and i = {i + kn|k ∈ Z} = i + nZ, with
a + b = a + b, a + b = ab.
The zero element in the ring is 0, and the identity element is 1.
3◦ R = Mn×n (R) is a ring, with the usual matrix addition and multiplication,
the zero element is
and the identity element is
0=
I =
0
···
0
···
0
···
0
1
···
0
···
0
···
1
.
Remark. If (R, +, ·) is a ring, for a ∈ R, −a is the inverse element in the
group (R, +). For a, b ∈ R, we denote a − b = a + (−b).
Lemma. If (R, +, ·) is a ring with zero element and identity element 1,
then
0 · a = a · 0 = 0, ∀ a ∈ R.
Proof. 0 · a = (0 + 0) · a = 0 · a + 0 · a ⇒ 0 · a = 0.
2
Definition. Let R = (R, +, ·) be a ring and S be a subset of R. We call
S a subring of R if S is closed under the operations of addition, subtraction and
multiplication.
17
Example. (Z, +, ·) is a subring of (R, +, ·).
Remark. 1◦ Sometimes, in the definition of a ring, the existence of the
identity element is not required.
2◦ If (R, +, ·) is a ring and moreover if a · b = ·a, ∀ a, b ∈ R, then R is called
a commutative ring.
3◦ From now on, the word ring, unless we explicitly mention non-commutative,
will mean a ring which is commutative and contains an identity element 1.
Example. (Z, +, ·) is a commutative ring, while Mn×n is not commutative.
Example. C[x] = {all polynomials over the complex field C}
C[a, b] = {all real continuous functions over the inteval [a, b]}.
R = {0} is a ring, called zero ring.
Proposition. If (R, +, ·) is a ring and 1 = 0, then R is zero ring.
Proof. ∀a ∈ R, since 1 = 0, then a = 1 · a = 0 · a = 0.
2
Homework.
1. Describe explicitly the smallest subring of complex numbers which contains the real cube root of 2.
2. In each case, decide whether the given structure forms a ring. If it is not
a ring, determine which of the ring axioms hold and which fail:
(a) U is an arbitrary set, and R is the set of subsets of U . Addition and
multiplication of elements of R are defined by the rules A + B = A ∪ B and
A · B = A ∩ B.
(b) U is an arbitrary set, and R is the set of subsets of U . Addition and
multiplication of elements of R are defined by the rules A+B = (A∪B)−(A∩B)
and A · B = A ∩ B.
(c) R is the set of continuous functions R −→ R. Addition and multiplication
are defined by the rules [f + g](x) = f (x) + g(x) and [f ◦ g](x) = f (g(x)).
18
Definition. Let (R, +, ·) be a ring with zero element 0 and identity element 1. For a ∈ R, if there exists b ∈ R such that ab = ba = 1, i.e. a has an
inverse element with the multiplication, then we say that a is a unit in R.
Example. 1◦ (Z, +, ·) has two units: ±1.
2◦ (R[x], +, ·) is the ring of polynomials. For f (x) ∈ R[x], f (x) is a unit iff
f (x) is a non-zero constant.
2. Polynomials in formal variables.
Let x be a formal variable (i.e. a symbol) and R be a ring. A formal
polynomial with coefficient in R is:
f (x) = an xn + · · · + a1 x + a0 ,
where a0 , · · · , an are coefficients of the polynomial f (x), an 6= 0 is called the
leading coefficient and n is called the degree of the polynomial.
A polynomial of the form: f (x) = an xn is called a monomial polynomial.
We define
R[x] = {all polynomials with variable x and coefficients in R},
then R[x] is a ring with the usual addition and multiplication, called polynomial
ring in one variable.
Similarly, we can define polynomial ring in multi-variables: R[x1 , · · · , xn ].
Example. Z[x] = {all polynomials with integer coefficients} is a ring.
Definition. A complex number α is called an algebraic number if α is
a root of a non-zero polynomial in Z[x], otherwise α is called a transcendental
number.
Example. 1◦
√
2, i are algebraic.
2◦ π, e are transcendental numbers.
3. Homomorphism and ideals.
19
Definition. 1◦ A homomorphism ϕ : R → R′ from a ring R to another
ring R′ is a map satisfying:
ϕ(1) = 1
ϕ(a + b) = ϕ(a) + ϕ(b)
ϕ(ab) = ϕ(a)ϕ(b)
for all a, b ∈ R.
2◦ An isomorphism of rings is a homomorphism and also a bijective map.
3◦ If there is a ring isomorphism from ring R to R′ , we say the two rings are
isomorphic, and denoted by R ∼
= R′ .
Example. Let R[x] be the polynomial ring in variable x over the ring R.
Fix an element a ∈ R and define
ϕ : R[x]
→ R
,
f (x) 7→ f (a)
then ϕ is a ring homomorphism from the polynomial ring to the ring R, which
is indeed a surjective homomorphism.
Proposition. Let ϕ : R → R′ be a ring homomorphism. For a given
element α ∈ R′ , there exists a unique homomorphism:
Φ : R[x] → R′
such that Φ|R = ϕ and Φ(x) = α.
Proof. We define Φ : R[x] → R′ by Φ(an xn + · · · + a0 ) = ϕ(an )αn + · · · +
ϕ(a0 ). It is easy to check that Φ is a ring homomorphism and satisfies the
conditions of the proposition. The uniqueness is clear.
2
Example. Let p be a positive integer. Z/pZ = {0, 1, · · · , p − 1} is a finite
ring. Define
ϕ : Z → Z/pZ
n
7→ n
20
,
then ϕ is a ring homomorphism, and
Φ : Z[x]
an xn + · · · + a0
→ Z/pZ[x]
7→ f (x) = an xn + · · · + a0 = f (x)
,
is also a ring homomorphism. f (x) is called the residue of f (x) modulo p.
Proposition. Let x1 , · · · , xn , y1 , · · · , yn be variables. Then there is a
unique isomorphism from the ring R[x1 , · · · , xn , y1 · · · , yn ] to (R[x1 , · · · , xn ])[y1 , · · · , yn ],
which is identity when restricted on R and set each variable to itself.
Proposition. Let R be a ring with identity element 1R . Then there is a
ring homomorphism
ϕ:Z→R
defined by
ϕ(n) =
1 + · · · + 1R
| R {z
}
n copies
if n ≥ 1
0
if n = 0 .
(−1R ) + · · · + (−1R ) if n ≤ −1
{z
}
|
n copies
Definition. Let R, R′ be rings, ϕ : R → R′ be ring homomorphism, we
define
ker ϕ = {a ∈ R|ϕ(a) = 0}
and called the kernel of the homomorphism ϕ.
Lemma. For a, b ∈ ker ϕ, c ∈ R, we have
a + b ∈ ker ϕ, ac ∈ ker ϕ.
Proof.
ϕ(a + b) = ϕ(a) + ϕ(b) = 0, ⇒ a + b ∈ ker ϕ.
ϕ(a · c) = ϕ(a) · ϕ(c) = 0, ⇒ ac ∈ ker ϕ.
21
2
Remark. If ϕ : R → R′ is a ring homomorphism. ker ϕ in general is not
a subring of R. (Since the identity element 1 6∈ ker ϕ.)
Example. For a ∈ R,
ϕ : R[x]
f (x)
→ R
7→ f (a)
is a ring homomorphism and ker ϕ = {f (x) ∈ R[x]|f (a) = 0} = {(x−a)g(x)|g(x) ∈
R[x]}, where R is the real number field.
Definition. Let I be a non-empty subset of a ring R. If I satisfies the
following conditions, then I is called an ideal of the ring R:
(1) With the condition of R, I is a subgroup of (R, +).
(2) If a ∈ I, b ∈ R, then ab ∈ R.
Corollary. If ϕ : R → R′ is a ring homomorphism, then ker ϕ is an ideal
of R.
Homework.
1. Prove that 7 +
√
√
√
3
2 and 3 + −5 are algebraic numbers.
2. Prove that for all integers n, cos 2π
n is an algebraic number.
3. In each case, decide whether or not S is a subring of R.
(a) S is the set of all rational numbers of the form a/b, where b is not divisible
by 3, and R = Q.
(b) S is the set of functions which are linear combinations of the functions
{1, cos nt, sin nt | n ∈ Z}, and R is the set of all functions R −→ R.
a
b
,
(c) (not commutative) S is the set of real matrices of the form
−b a
and R is the set of all real 2 × 2 matrices.
4. An element of a ring R is called nilpotent if some power of x is zero.
Prove that if x is nilpotent, then 1 + x is a unit in R.
22
5. Let R be a ring. The set of all formal power series p(t) = a0 + a1 t + a2 t2 +
· · · , with ai ∈ R, forms a ring which is usually denoted by R[[t]].
(a) Prove that the formal power series form a ring.
(b) Prove that a power series p(t) is invertible if and only if a0 is a unit of
R.
Lemma. Let R be a ring, a ∈ R. Let
aR{ar|r ∈ R}.
Then aR is an ideal of R.
Proof. It is clear that aR is a subgroup with the operation of addition.
Moreover, ∀b = ar1 ∈ aR, and r2 ∈ R, we have br2 = a(e1 r2 ) ∈ aR, i.e. aR is
an ideal of R.
2
Note. 1. Since R is commutative, for a ∈ R, aR = Ra.
2. If R is a non-commutative ring and I is a non-empty subset of R, such
that (I, +) is a subgroup of (R, +) and (i) ab ∈ I, ∀a ∈ R, b ∈ I, then I is called
a left ideal of R.
(ii.) ab ∈ I, ∀a ∈ I, b ∈ R, then I is called a right ideal of R.
(iii.) ab, ba ∈ I, ∀a ∈ I, b ∈ R, then I is called a (two-sided) ideal of R.
3. For a ∈ R, aR is a right ideal of R, Ra is a left ideal, and RaR =
P
{ ri asi |ri , si ∈ R} is a two-sided ideal of R.
Definition. If R is a ring, the ideal aR is denoted by (a), and called a
principal ideal generated by a.
Example. R[x] = polynomial ring, then
(x − 1)R[x] = {(x − 1)f (x)|f (x) ∈ R[x]}
is a principal ideal generated by x − 1 ∈ R[x].
Define:
ϕ : R[x]
→ R
f (x) 7→ f (a),
23
where a is a fixed real number, then ϕ is a ring homomorphism from R[x] to R,
and ker ϕ = (x − a)R[x], the ideal generated by x − a.
Definition. Let R be a ring, a1 , · · · , an ∈ R, Let
(a1 , · · · , an ) = {
n
X
i=1
ai ri |ri ∈ R}.
Then (a1 , · · · , an ) is an ideal of R, which is called the ideal of R generated by
the elements a1 , · · · , an .
Example. Let R = Z[x], 2, x ∈ Z[x]. Let I = (2, x) be the ideal of Z[x]
generated by the elements 2 and x. Then it is clear that
I = {all polynomials in Z[x] such that the constant term is a multiple of 2}.
Indeed,
ϕ : Z[x]
→ Z/2Z
f (x) 7→ f (0) (mod 2),
then ϕ is a ring homomorphism, and
ker ϕ = {f (x) ∈ Z[x]|f (0) ≡ 0 (mod 2)}
= {f (x) ∈ Z[x]|f (0) and the constant term of f (x) is a multiple of 2}
= I = (2, x).
Remark. If R is a ring ,then R is commutative and contains 1. It is clear
that (0), (1) are principal ideals of R, and (0) is the smallest ideal, and (1) = R
is the largest ideal. We shall often call (0) the zero ideal and (1) = R the unit
ideal of R. (the trivial ideals)
Definition. An ideal I of a ring R is called proper if I 6= (0), R.
Definition. Let R be a ring, and if for each non-zero element a ∈ R,
there exists b ∈ R, such that a · b = 1, then R is called a field.
Fact. The set of units of a field R is {x ∈ R|x 6= 0}.
24
√
Example. R, Q, C are fields, Z is a ring (not a field). Q 2 = {p +
√
√
√
q 2|p, q ∈ Q} is also a field. Since if p + q 2 6= 0, then p − q 2 6= 0 ⇒ (p +
√
√
q 2)(p − q 2) = p2 − 2q 2 6= 0, and
√
√
p
q 2
(p + q 2)( 2
− 2
) = 1.
p − 2q 2
p − 2q 2
Proposition. (a) F is a field, then the only ideals of F are zero ideal and
unit ideal.
(b) Conversely, if a ring R has exactly two ideals, then R is a field.
Proof. (a) is clear. We prove (b).
Assume R has only two distinct ideals, (0) and R, then 0 6= 1 and ∀0 6= a ∈ R,
then (a) is an ideal of R and a ∈ (a) ⇒ (a) = R ⇒ (a) = R ⇒ 1 ∈
(a) ⇒ ∃ b ∈ R, such that 1 = ab, ⇒ R is a field.
2
Corollary. Let F be a field, R′ be a non-zero ring, then each ring homomorphism ϕ : F → R′ is injective.
Proof. We need to show that ker ϕ = (0).
Suppose that ker ϕ 6= (0), then ker ϕ is a non-zero ideal of the field F ,
⇒ ker ϕ = F ⇒ ϕ(1) = 0. But 0 is not the identity element of the ring R′ .
Contradiction. Thus ker ϕ = (0) or ϕ is injective.
2
Proposition. Every ideal in the ring Z of integers is a principal ideal.
Proof. If 0 6= I is an ideal of Z, then (I, +) is a subgroup of (Z, +), then
there exists p ∈ Z, such that I = pZ = (p) is a principal ideal of Z generated by
p.
2
Proposition. Let R be a ring, f, g ∈ R[x]. Assume that the leading
coefficient of f (x) is a unit in R. Then there exist polynomials q, r ∈ R[x], such
that
g(x) = f (x)q(x) + r(x),
25
where r(x = 0) or the degree of r(x) is less than the degree of f (x).
Proof. Omitted.
2
Corollary. Let g(x) be a monic polynomial in R[x], and α ∈ R such that
g(α) = 0. Then x − α divides g(x) in R[x].
Proposition. Let F be a field. Then every ideal in the polynomial ring
F [x] is a principal ideal.
Proof. Let (0) 6= I be an ideal of F [x]. Choose a non-zero polynomial in I
which has the minimal degree in I, say f (x). We claim that I = (f (x)) is an
ideal generated by f (x).
First, since I is an ideal and f (x) ∈ I, (f (x)) ∈ I.
On the other hand, ∀g(x) ∈ I, we divide g(x) by f (x) to get: g(x) =
q(x)f (x) + r(x), where r(x) = 0 or degree of r(x) is less than the degree of f (x).
Note that r(x) = g(x)−q(x)f (x) ∈ I, but f (x) is a non-zero polynomial that has
the minimal degree, thus r(x) = 0 ⇒ g(x) = q(x)f (x) ∈ (f (x)) ⇒ i ⊂ (f (x)).
Thus, I = (f (x)).
2
Corollary. Let F be a field, f (x), g(x) are polynomials in F [x] which
are not both zero, then there exists a unique monic polynomial d(x), called the
greatest common divisor of f (x) and g(x) with the following properties:
(a) d(x) generates the ideals (f (x), g(x)), i.e. (d(x)) = (f (x), g(x)).
(b) d(x)|f (x), d(x)|g(x).
(c) If h(x)|f (x), h(x)|g(x), then h(x)|d(x).
(d) There is polynomials p(x), q(x), such that d(x) = p(x)f (x) + q(x)g(x).
Homework.
1. Let I, J be ideals of a ring R. Show by example that I ∪ J need not be
an ideal, but show that I + J = {r ∈ R | r = x + y, x ∈ I, y ∈ J} is an ideal.
This ideal is called the sum of the ideals I, J.
2. (a) Let I, J be ideals of a ring R. Prove that I ∩ J is an ideal.
26
(b) Show by example that the set of products {xy | x ∈ I, y ∈ J} need not
be an ideal, but that the set of finite sums Σxν yν of products of elements of I
and J is an ideal. This ideal is called the product ideal.
(c) Prove that IJ ⊂ I ∩ J.
(d) Show by example that IJ and I ∩ J need not be equal.
4. Quotient Rings and Relations in a Ring.
Let I be an ideal of a ring R, then with addition +, I is a normal subgroup of
(R, +). Then R/I is the set of cosets, i.e. R/I = {a + I|a ∈ R} a quotient group
with respect for +. We write a+I by a, then a = b iff a+I = b+I ↔ b−a ∈ I.
Moreover,
a · b = (a + I)(b + I) = ab + I = ab
⇒ R/I forms a ring, called the quotient ring.
Theorem. Let I be an ideal of a ring R.
(a)
π : R → R/I
a
7→ a = a + I
is a ring homomorphism, called the canonical homomorphism.
(b) ker π = I.
Proof. (a) Let 1 be the identity element of R. Then 1 = 1 + I is the identity
element of R/I and π(1) = 1, ∀a, b ∈ R
π(a + b) = a + b = a + b + I = (a + I) + (b + I) = a + b
and π(ab) = ab = ab, i.e. π is a ring homomorphism.
(b) is clear.
2
Proposition. Let f : R → R′ be a ring homomorphism with kernel I,
and J is an ideal of R that is contained in I. Then
27
(a) There is a unique homomorphism: f : R/J → R′ such that f ◦ π = f ,
i.e. the following diagram commutes:
f
/ R′ .
RD
DD
y
y
DD
yy
D
yy f
π DD
y
|y
"
RJ
(b) Imf = {f (a)|a ∈ R} is a subring of R′ , and
R/ ker f ∼
= Imf.
Proof. (a) We define
f:
R′
R/J
→
a+J
7→ f (a)
.
We need to check that f is well-defined. Indeed, if a + J = b + J, ⇒ b − a ∈
J ⊂ I ⇒ f (b − a) = 0 or f (b) = f (a) ⇒ f (a + J) = (f )(b + J), so f is
well-defined. It is easy to see that
f (1 + J) = f (1) = 1′ and
f ((a + J) + (b + J)) = f (a + b + J) = f (a + b) = f (a) + f (b) = f (a + J) + f (b + J)
and f ((a + J)(b + J)) = f (ab + J) = f (ab) = f (a)f (b) = f (a + J)f (b + J)
⇒ f is a ring homomorphism. Moreover, (f ◦ π)(a) = f (a + J) = f (a), for any
a ∈ R ⇒ f ◦ π = f.
To show the uniqueness of such a ring homomorphism, we suppose g is
another such a ring homomorphism such that g ◦ π = f . Then for any a ∈ R
f (a + J) = f (a) = (g ◦ π)(a) = g(a + J)
⇒ f = g.
(b) It is clear to check that Imf is a subring. Let J = ker f , then
f : R/J → Imf ⊂ R′
28
is a surjective ring homomorphism. Moreover, if f (a + J) = 0 ⇒ f (a) =
0 ⇒ a ∈ ker f = J ⇒ a + J = J ⇒ f is injective ⇒ f is an isomorphism.
2
Definition. An isomorphism ϕ from a ring R to itself is called an automorphism. The set Aut(R) of all automorphisms of R forms a group with law
of composition. The group Aut(R) is called the automorphism group of the ring
R.
Example. Determine the group Aut(Q), where Q is the ring of rational
numbers.
Solution. Since ϕ ∈ Aut(Q) ⇒ ϕ(1) = 1, ϕ(0) = 0 ⇒ ϕ(n) = n, ∀ n ∈ Z,
and for n 6= 0, we have
n
1
1
1
1 = ϕ( ) = ϕ(n · ) = ϕ(n) · ϕ( ) = nϕ( ),
n
n
n
n
so ϕ( n1 ) =
1
n.
Thus, ∀ n, m ∈ Z, n 6= 0, we have
ϕ(
m
1
1
1
) = ϕ(m · ) = ϕ(m) · ϕ( ) = m · .
n
n
n
n
⇒ ϕ is an identity map ⇒ Aut(Q) = {1}.
2
Homework.
1. (a) Prove that the kernel of the homomorphism ϕ : C[x, y] → C[t] defined
by x
t2 , y
t3 is the principal ideal generated by the polynomial y 2 − x3 .
(b) Determine the image of ϕ explicitly.
2. Let R be a ring, and let f (y) ∈ R[y] be a polynomial in one variable
with coefficients in R. Prove that the map R[x, y] → R[x, y] defined by x
x + f (y), y
y is an automorphism of R[x, y].
3. Determine the automorphism group Aut(Q[x]).
Proposition. Let J be an ideal of R, R̄ = R/J, and π := R → R̄
be the canonical map.
29
(a) There is a bijective correspondence between the set of ideals
of R which contains J and the set of all ideals in R̄, given by
I → π(I),
¯ ← I¯
π −1 (I)
(b) If I ⊂ R corresponds to I¯ ⊂ R̄, then R/I and R̄/I¯ are
isomorphic rings.
Proof. To prove (a), we need to check the following statements.
(1) If I is an ideal of R which contains J, then π(I) is an ideal
of R̄ = R/J.
¯ is an ideal of R containing
(2) If I¯ is an ideal of R̄, then π −1 (I)
J.
¯ = I.
¯
(3) π −1 (π(I)) = I, and π(π −1 (I))
Indeed, for (1), since π is an additive group homomorphism, π(I)
is an additive subgroup of R̄, so we only need to show: ∀x̄ ∈ π(I), r̄ ∈
R̄, we have x̄ · r̄ ∈ π(I). Since I is an ideal of R, x · r ∈ I and
x̄ · r̄ = x · r ∈ π(I).
Next, since I¯ is an ideal of R̄. Consider the homomorphism
ϕ
π
¯ and the composed homomorphism R →
¯
ϕ : R̄ → R̄/I,
R̄ → R̄/I,
then ϕ ◦ π : R → R̄/I¯ is a surjective homomorphism, and
ker(ϕ ◦ π) = {r ∈ R|ϕ(π(r)) = 0̄}
= {r ∈ R|π(r) ∈ ker ϕ}
¯
= {r ∈ R|π(r) ∈ I}
¯ = π −1 (I)
¯
= {r ∈ R|r ∈ π −1 (I)}
¯ is an ideal of R.
Hence π −1 (I)
Finally, applying the previous proposition, we have
−1 ¯ ∼
¯
R/ ker(ϕ ◦ π) ∼
(I) = R̄/I¯
= R̄/IorR/π
30
¯ is the ideal of R corresponds I.
¯ Now
This prove (b), where π −1 (I)
we prove (3), first, it is clear I ⊂ π −1 (π(I)). On the other hand, if
x ∈ π −1 (π(I)), i.e. π(x) ∈ π(I), then there exists y ∈ I such that
π(x) = π(y) i.e. π(x − y) = 0 and x − y ∈ ker π = J ⊂ I which
imply π −1 (π(I)) ⊆ I. Therefore, we have I = π −1 (π(I)).
¯ = I.
¯
Similarly, one can prove π(π −1 (I))
2
Definition. Elements in Z[i] are called Gauss integers.
Proposition. The ring Z[i]/(1+3i) is isomorphic to the ring Z/10Z
of integers module 10.
Prove. Define
ϕ:
Z → Z[i]/(1 + 3i)
n 7→ n + (1 + 3i)
Since i = 3 in Z[i]/(1 + 3i), for any a + bi ∈ Z, we have a + bi =
a + 3b ∈ Z[i]/(1 + 3i) where a + 3b ∈ Z. Thus, a + bi + (1 + 3i) =
a + 3b + (1 + 3i) = ϕ(a + 3b) and ϕ is surjective.
Next, we show ker ϕ = 10Z. It is clear that ϕ(10n) = 10n + (1 +
3i) = (1 + 3i)(1 − 3i)n + (1 + 3i) ⊂ (1 + 3i), i.e. 10Z ∈ ker ϕ. If
n ∈ ker ϕ, then n = (1 + 3i)(a + bi) for some Gauss integer a + bi.
Hence n = a − 3b + (b + 3a)i, then a − 3b = n, b + 3a = 0. So
n = 10a ∈ 10Z, then ker ϕ = 10Z.
2
Proposition. The ring C[x, y]/(xy) is isomorphic to the subring of
the product ring C[x] × C[y] consisting of the pair (p(x), q(y)) such
that p(0) = q(0).
31
Proof. Define
C[x, y] → C[x]
ϕ:
f (x, y) 7→ f (x, 0)
clearly ϕ is a surjective homomorphism and ker ϕ = (y), then C[x, y]/(x) ∼
=
C[y].
Similarly, we have C[x, y]/(x) ∼
= C[y].
Now consider
ψ:
C[x, y] → C[x] × C[y]
f (x, y) 7→ (f (x, 0), f (0, y))
then ψ is ring homomorphism, and ker ψ = (x) ∩ (y) = (xy). Hence
C[x, y]/(xy) ∼
= imψ = {(p(x), q(y) ∈ C[x] × C[y]|p(0) = q(0)}
.
2
Homework.
1. Describe the ring Z[i]/(2 + i).
5. Adjunction of Elements and
Chinese Remainder Theorem
(5.1)Definition. If R, R′ are rings and R ⊆ R′ , then we call R′
ring extension of R.
(5.2)Example. 1. The ring of complex number is ring extension
of the ring of real numbers.
32
√
2. The ring Q[ 2] is ring extension of the ring of rational numbers.
(5.3)Example. 1. Let R denote the polynomials over the real
number field, I = (x2 + 1) be the principal ideal of R[x] generated
by the element x2 + 1, then R[x]/(x2 + 1) ∼
= C = R[i].
Proof. We define the map
ϕ:
R[x] −→ C
f (x) 7−→ f (i)
then ϕ is a surjective ring homomorphism and we only to show that
ker ϕ = (x2 + 1).
For any f (x) ∈ ker ϕ, there are polynomials p(x), r(x) ∈ R[x]
such that
f (x) = p(x)(x2 + 1) + r(x)
where deg(r(x)) <deg(x2 + 1). Then 0 = f (i) = p(i)(i2 + 1) + r(i)
and r(i) = 0. Now let r(x) = ax + b, a, b ∈ R, then a + bi = 0 which
implies a = b = 0 and r(x) = 0. Therefore f (x) = p(x)(x2 + 1) ∈
(x2 + 1) and ker ϕ = (x2 + 1), as required.
√
2. Z[ 2] ∼
= Z[x]/(x2 − 2).
2
(5.4)Definition. Let R be a ring.
(1) if α ∈ R such that αn = 0 for some positive integer n, then
α is called nilpotent or infinitesimal.
(2) for a ∈ R, if there exists non-zero element b ∈ R such that
ab = 0, then a is called a zero divisor.
(5.5)Example. Let R be a ring, and (x2 ) be the principal ideal of
33
R[x] generated by x2 , then the ring R[x]/(x2 ) has nilpotent element
x̄ := x + (x2 ) ∈ R[x]/(x2 ) such that (x̄)2 = 0.
(5.6)Example. If R is a ring, a ∈ R is nilpotent, then 1 − a
is a unit in R. Since an = 0 for some positive integer n, then
(1 − a)(1 + a + · · · + an−1 = 1 − an = 1.
(5.7)Example. Since 2̄·3̄ = 6̄ = 0 in the ring Z/(6) := {0̄, 1̄, · · · , 5̄},
2̄ is a zero divisor.
(5.7)Example. Let R be a ring and s, t be formal variables. R[s, t]
be the polynomial ring in two variables. Let (st−1) be the principal
ideal of R[s, t] generated by st − 1, then
R[s, t]/(st − 1) ∼
= R[t, t−1 ]
where R[t, t−1 ] = {a−m t−m + · · · + a−1 t−1 + a0 + a1 t + · · · + an tn |
m, n ∈ N, a−m , · · · , an ∈ R} is called the Laurent polynomial ring
or the ring of Laurent polynomials.
(5.8)Chinese Remainder Theorem. Let m1 , · · · , mn be positive
integers such that (mi , mj ) = 1, ∀1 ≤ i 6= j ≤ n. If b1 , · · · , bn ∈ Z,
then there exists m ∈ Z such that
m ≡ b1 (mod m1 )
···············
m ≡ bn (mod mn )
(5.9)Theorem. Let A1 , A2 , · · · , An be ideals of a ring R such that
34
Ai + Aj = R for all i 6= j. If b1 , b2 , · · · , bn ∈ R, then there exists
b ∈ R such that b ≡ bi (mod Ai ) for all i.
Proof. Since A1 + A2 = R, A1 + A3 = R, we have
R = R2 (since ring R has an identity element by definition)
= (A1 + A2 )(A1 + A3 ) ⊆ A1 + A2 ∩ A3 ⊆ R
then R = A1 + A2 ∩ A3 .
Inductively, we assume
R = A1 + A2 ∩ A3 ∩ · · · ∩ Ak−1
then
R = R2 = (A1 + A2 ∩ A3 ∩ · · · ∩ Ak−1 ) · (A1 + Ak )
⊆ A1 + A2 ∩ A3 ∩ · · · ∩ Ak ⊆ R
therefore, we have
R = A1 + A2 ∩ A3 ∩ · · · ∩ Ak = A1 +
Similarly,
R = Ak +
n
\
Ai ,
i=1
i6=k
n
\
Ai
i=2
∀1 ≤ k ≤ n
Therefore, for each bk ∈ R, there exists ak ∈ Ak and rk ∈
such that bk = ak + rk .
n
T
Ai
i=1
i6=k
We take b = r1 + r2 + · · · = rn , then for any i,
b ≡ ri (modAi ) ≡ (bi − ai )(modAi ) ≡ bi (modAi )
2
35
Homework. 1. Let I, J be ideals in a ring R. Prove that the
residue of any element of I ∩ J in R/IJ is nilpotent.
2. Let I, J be ideals of a ring R such that I + J = R. Prove that
IJ = I ∩ J.
6. Integral Domains and Fraction Fields
(6.1)Definition. An integral domain is a ring R having on nonzero zero divisors. In other words, if a, b ∈ R such that ab = 0, then
a = 0 or b = 0.
(6.2)Lemma. If R is an integral domain, then the cancelation law
holds, i.e. if ab = ac and a 6= 0, then b = c.
Proof. Since ab = ac, we have a(b − c) = 0, but a 6= 0 ,then
b − c = 0 i.e. b = c.
2
(6.3)Proposition. If R is an integral domain, then the polynomial
ring R[x] is also an integral domain.
Proof. Let f (x) = an xn +· · ·+a1 x+a0 , g(x) = bm xm +· · ·+b1 x+
b0 ∈ R[x] such that an 6= 0, bn 6= 0. Then f (x) · g(x) = an bm xm+n +
lower terms. Since an bm 6= 0, we have f (x) · g(x) 6= 0 and R[x] has
no non-zero zero divisors.
2
(6.4)Proposition. If R is an integral domain and | R |<∝, then
R is a field.
36
Proof. We only need to show that each non-zero element in R
has an inverse.
Indeed, let 0 6= a ∈ R, then aR = {ar | r ∈ R} ⊆ R and since
ar1 = ar2 iff r1 = r2 . Therefore, aR = R and 1 ∈ R. i.e. there is
b ∈ R such that 1 = ab which implies a is invertible. Hence R is a
field.
2
(6.5)Theorem. Let R be an integral domain. Then there exists
an embedding of R into a field F , that is, there is an injective ring
homomorphism from the integral domain R to a field F .
Proof. For a, b ∈ R and b 6= 0, we define a symbol a/b, called a
fraction. Two fractions a1 /b1 , a2 /b2 where b1 , b2 6= 0 are said related,
denoted by a1 /b1 ≈ a2 /b2 , if a1 b2 = a2 b1 .
It is easy to see that the relation ≈ is an equivalence relation on
the set R0 of all fractions of R. Let F = R0 / ≈ = {a/b | a, b ∈
R, b 6= 0 and a1 /b1 = a2 /b2 iff a1 b2 = a2 b1 }. We claim that F is a
field, where the addition and multiplication are given by
a1 /b1 · a2 /b2 = a1 a2 /b1 b2
a1 /b1 + a2 /b2 = (a1 b2 + a2 b1 )/b1 b2
Now we can define a map
ϕ:
R −→ F
a 7−→ a/1
Clearly, ϕ is an injective ring homomorphism.
2
(6.6)Remark. The field F is often called the field of fraction of
37
the integral domain R.
(6.7)Example. Z is an integral domain, the field of fraction of Z
is the field Q of rational numbers.
(6.8)Example. Let R[x] be the ring of real polynomials, then the
field of fractions of R[x] is rational function field is:
F = {f (x)/g(x) | f (x), g(x) ∈ R and g(x) are not zero polynomial}
where the operations are:
f1 (x)/g1 (x) · f2 (x)/g2 (x) = f1 (x)f2 (x)/g1 (x)g2 (x)
f1 (x)/g1 (x) + f2 (x)/g2 (x) = f1 (x)g2 (x) + f2 (x)g1 (x)/g1 (x)g2 (x).
(6.9)Proposition.(Universal property of the field of fractions) Let R be an integral domain with field of fractions F . If
ϕ : R −→ K is an injective ring homomorphism from R to a field
K. Then there is a unique extension of ϕ to a ring homomorphism
φ : F −→ K such that φ |R = ϕ.
Proof. We define:
φ:
F −→ K
a/b 7−→ ϕ(a)ϕ(b)−1
for a, b ∈ R and b 6= 0.
First we check that φ is well-defined. Indeed, if b 6= 0, then
ϕ(b) 6= 0 since ϕ is injective. Hence ϕ(b) is invertible. Moreover, if
a1 /b1 = a2 /b2 , then a1 b2 = a2 b1 ; hence ϕ(a1 )ϕ(b2 ) = ϕ(a2 )ϕ(b1 ), and
ϕ(a1 /b1 ) = ϕ(a1 )ϕ(b1 )−1 = ϕ(a2 )ϕ(b2 )−1 = ϕ(a2 /b2 ), as required.
38
Next, we show φ is a ring homomorphism:
φ(a1 /b1 · a2 /b2 ) = φ(a1 a2 /b1 b2 ) = φ(a1 a2 )φ(b1 b2 )−1
= φ(a1 )φ(a2 )φ(b1 )−1 φ(b2 )−1 = φ(a1 /b1 )φ(b1 /b2 ).
Similarly, we have
φ(a1 /b1 + a2 /b + 2) = φ(a1 /b1 ) + φ(a2 /b2 )
and φ(a) = φ(a/1) = ϕ(a)ϕ(1) = ϕ(a), hence φ |R = ϕ.
Finally, we show the uniqueness of φ. If ψ is another such a ring
homomorphism, then for any a ∈ R, we have ψ(a) = ϕ(a) = φ(a).
Moreover, if a 6= 0, we have 1 = ψ(1) = ψ(a/a) = ϕ(a)ψ(1/a), so
ψ(1/a) = ϕ(a)−1 and φ(1/a) = ϕ(a)−1 . Now for any a/b ∈ F , we
have ψ(a/b) = ψ(a)ψ(1/b) = ϕ(a)ϕ(b)−1 = φ(a/b).
2
Homework.
1. Let R̄ = Z[x]/(2x). Prove that every element of R̄ has a
unique expression in the form a0 + a1 x + · · · + an xn , where ai are
integers and a1 , · · · , an are either 0 or 1.
2. Is there an integer domain containing exactly 10 elements?
3. A subset S of an integer domain R which is closed under multiplication and which does not contain 0 is called a multiplicative set.
Given a multiplicative set S, we define S-fractions to be elements
of the form a/b, where b ∈ S. Show that the equivalence classes of
S-fractions from a ring.
39
7. Maximal Ideals
(7.1)Definition. Let R be a ring. If M 6= R is an ideal of R such
that whenever N is an ideal of R containing M, then N = M or
N = R. Then M is called an maximal ideal of R.
(7.2)Lemma. Let M be an ideal of R. Then
(a) M is a maximal ideal iff R/M is a field.
(b) the zero ideal is maximal iff R is a field.
Proof. (a) If M is maximal. We need to show each non-zero
element in R/M is invertible. Indeed, if 0 6= a ∈ R/M, i.e. a ∈ R
and a ∈
/ M, then aR + M is an ideal of R and containing M as a
proper subset. Thus aR + M = R and there is some r ∈ R and
m ∈ M such that 1 = ar + m. Hence ar = (a + M)(b + M) =
ar + M = 1 + M = 1, i.e. a is invertible, as required.
On the other hand, if N is an ideal of R and containing M as a
proper subset. Then there is some a ∈ N \ M, i.e. a = a + M(6=
0) ∈ R/M. Then R/M is a field implying there is r ∈ R/M such
that a · r = 1. Therefore, 1 = ar + m for some m ∈ M, then 1 ∈ N
and N = R, i.e. M is maximal.
(b) follows from (a).
2
(7.3)Proposition. The maximal ideals of the ring Z are principal
ideals generated by prime integers.
Proof. Let M be a maximal ideal of Z. Then M = (n) for some
positive integer n. It is clear that M is maximal implying that n is
40
prime, otherwise n = p · q, p > 1, q > 1 implies (n) $ (p) $ Z, a
contradiction.
2
(7.4)Proposition. The maximal ideals of the polynomial ring C[x]
are principal ideals generated by polynomials the form x−a for some
a ∈ C.
Proof. Let M be a maximal ideal of C[x], then there is a monic
polynomial f (x) ∈ C[x] such that M = (f (x)) and deg(f (x)) ≥ 1,
then f (x) has a root in C, say a. Hence, f (x) = (x − a)g(x) for
some polynomial g(x) and M = (f (x)) ⊂ (x − a) $ C[x], thus
M = (x − a).
2
(7.5)Corollary. If ϕ : C[x] −→ C is defined by ϕ(f (x)) = f (a) for
some fixed element a ∈ C, then ker ϕ = (x − a).
(7.6)Hilert’s Theorem. The maximal ideals of the polynomial
ring C[x1 , · · · , xn ] are in bijective correspondence with points of
complex n-dimension space. And for α = (a1 , · · · , an ) ∈ Cn
ϕα :
C[x1 , · · · , xn ] −→ C
f (x1 , ·, xn ) 7−→ f (a1 , · · · , an )
is a surjective ring homomorphism with ker ϕα a maximal ideal generated by polynomials:
(x1 − a1 ), (x2 − a2 ), · · · , (xn − an )
and all maximal ideals of C[x1 , · · · , xn ] are of the above form.
(7.7)Example. Let C[a, b] be the set of real continuous functions
41
over the closed interval [a, b]. Then C[a, b] is a ring with the usual
addition and multiplication of functions. For c ∈ [a, b], set
Ic = {f (x) ∈ C[a, b] | f (c) = 0}.
(1) Prove that Ic is a maximal ideal of C[a, b].
(2) If I is a maximal ideal of C[a, b], prove that there exists
c ∈ [a, b] such that I = Ic .
Proof. (1) It is clear that Ic is an ideal of C[a, b]. Suppose I is
an ideal of C[a, b] such that Ic $ I ⊆ C[a, b]. We take f (x) ∈ I \ Ic ,
then f (c) 6= 0. Let g(x) = f 2 (x) + (x − c)2 , then g(x) ∈ I and
g(x) 6= 0 for any x ∈ [a, b]. Hence
1
g(x)
∈ I and 1 =
1
g(x)
∈ I, then
I = C[a, b]. i.e. Ic is maximal.
(2) Let I be a maximal ideal of C[a, b] and not the form given
in (1). i.e. ∀c ∈ [a, b], there is a function fc (x) ∈ I such that
fc (c) 6= 0. Note that fc (x) is continuous on [a, b], there exists an
open neighborhood Oc of c on [a, b] such that fc (x) 6= 0, ∀x ∈
Oc ∩ [a, b]. Then {Oc | c ∈ [a, b]} is an open covering of the closed
set [a, b]. Thus by the Theorem of finite covering, we know that there
exist a finitely many open sets in {Oc | c ∈ [a, b]}. Say Oc1 , · · · , Ock
k
k
S
P
such that [a, b] ⊂
Oci . We set f (x) =
fc2i (x), then f (x) 6=
i=1
0, ∀x ∈ [a, b] and f (x) ∈ I. Thus,
1
f (x)
i=1
∈ C[a, b] and 1 =
1
f (x)
· f (x) ∈
I. i.e. I = C[a, b] a contradiction. Then there is an element c ∈ [a, b]
such that f (c) = 0 for all f (x) ∈ I, hence Ic ⊂ I and Ic = I.
42
(7.8)Remark. To prove Ic is maximal, we may define
ϕ:
C[a, b] −→ R
f (x) 7−→ f (c)
then ϕ is a surjective ring homomorphism with kernel
ker ϕ = {f (x) | f (x) ∈ C[a, b], f (c) = 0} = Ic .
Thus C[a, b]/Ic ∼
= R and Ic is maximal.
Homework.
1. Prove that the ring F5 [x]/(x2 + x + 1) is a field.
2. Let R be a ring, with M an ideal of R. Suppose that every
element of R which is not in M is a unit of R and that moreover it
is the only maximal ideal of R.
3. Let P be an ideal of a ring R. Prove that R̄ = R/P is an
integer domain if and only if P 6= R, and that if a, b ∈ P , then
a ∈ P or b ∈ P . (An ideal P satisfying these conditions is called a
prime ideal.)
4. Let ϕ : R → R′ be a ring homomorphism, and let P ′ be a
prime ideal of R′ .
(a) Prove that ϕ−1 (P ′ ) is a prime ideal of R.
(b) Given an example in which P ′ is a maximal ideal, but ϕ−1 (P ′ )
is not maximal.
8. Algebraic Geometry
43
(8.1)Definition. Let f1 (x1 , · · · , xn ), · · · , fk (x1 , · · · , xn ) ∈ C[x1 , · · · , xn ],
and let V = {(a1 , · · · , an ) | fj (a1 , · · · , an ) = 0, ∀1 ≤ j ≤ k} be the
set of common zeros of the polynomials. Then V is called an algebraic variety.
(8.2)Example.1. A complex line in C2 is a variety as it is the zeros
of the linear equation ax + by + c = 0.
2. A singleton point (a, b) is a variety as it is the common zeros
of the equations x − a = 0, y − b = 0.
3. The special linear group SL(2, C) is a variety in C4 as it is
the common zeros of the equation x11 x21 − x12 x22 − 1 = 0.
Recall Hilbert’s Theorem: The maximal ideals in the polynomial ring C[x1 , · · · , xn ] are in one-to-one correspondence with points
in Cn .
(8.3)Theorem. Let f1 , · · · , fr be polynomials in C[x1 , · · · , xn ], and
V be the variety defined by the system of the equations f1 (x1 , · · · , xn ) =
0, · · · , fr (x1 , · · · , xn ) = 0. Let I be the ideal generated by the polynomials f1 , · · · , fr . Then the maximal ideals of the quotient ring
R = C[x1 , · · · , xn ]/I are in bijective correspondence with points in
V.
−
Proof. If M→
a is a maximal ideal containing I corresponding to
→
a point −
a ∈ Cn . We define
ϕ :C[x1 , · · · , xn ] −→ C
→
→
f (−
x ) 7−→ f (−
a)
44
→
→
→
then ker ϕ = M( −
a ) and fi (−
a ) = 0, ∀ 1 ≤ i ≤ r, i.e. −
a ∈V.
→
On the other hand, for any −
a ∈ V , define
ϕ :C[x1 , · · · , xn ] −→ C
→
→
f (−
x ) 7−→ f (−
a)
then f1 , · · · , fr ∈ ker ϕ which is a maximal ideal of C[x1 , · · · , xn ]
containing I.
2
(8.4)Definition. Let A be a nonempty set, ≤ be a relation on A
such that
1.(reflexive) a ≤ a, ∀ a ∈ A;
2.(antisymmetric) a ≤ b and b ≤ a, then a = b;
3.(transitive) if a ≤ b, b ≤ c, then a ≤ c.
then the relation ≤ is called a partial order and A is a partially
ordered set.
(8.5)Definition. 1. Let (A, ≤) be a partially ordered set, T be a
subset. If for any a, b ∈ T , we always have a ≤ b or b ≤ a, then T is
called a totally ordered subset.
2. Let S be a subset of a partially ordered set A. If there is
a ∈ A such that s ≤ a, ∀s§, then a is called a upper bound of S.
3. Let A be a partially ordered set, an element a ∈ A is called a
maximal element if whenever a ≤ x for some x ∈ A, one has x = a.
(8.6)Zorn’s Lemma. Let (A, ≤) be a partially ordered set. If
each totally ordered subset of A has a upper bound, then A has a
maximum element.
45
(8.7)Theorem. Let R be a ring, I is an ideal of R and I 6= R.
Then I is contained in a maximal ideal of R.
Proof. Let A be the set of proper ideals of R that contains I as
a subset. Then the inclusion relation ⊆ is a partial ordering on A.
If {Bi } is a totally ordered subset of A, then B = ∪Bi is also an
ideal of R and 1 ∈
/ B, i.e. {Bi } has a upper bound. Thus by Zorn’s
Lemma, a has a maximal element which is a maximal ideal of R. 2
(8.8)Corollary. The only ring R having no maximal ideal is the
zero ring.
(8.9)Corollary. Let f1 , · · · , fr ∈ C[x1 , · · · , xn ]. If the system of
equations f1 = 0, · · · , fr = 0 has no solution in Cn , then there exist
g1 , · · · , gr ∈ C[x1 , · · · , xn ] such that
1=
r
X
i=1
fi · gi
.
Proof. Let I = (f1 , · · · , fr ). By Theorem8.3, there is a bijective
correspondence between the maximal ideal of C[x1 , · · · , xn ]/I and
the common zeros of f1 = 0, · · · , fr = 0. Since f1 = 0, · · · , fr = 0
has no common zero, C[x1 , · · · , xn ]/I has no maximal ideal. Hence
C[x1 , · · · , xn ]/I is a zero ring and I = C[x1 , · · · , xn ].
2
(8.10)Example. Let f1 = x2 + y 2 −1, f2 = x2 −y + 1, f3 = xy −1 ∈
C[x, y]. Determine the ideal of C[x, y] generated by f1 , f2 , f3 .
46
Solution. Solve the system of equations
x2 + y 2 − 1 = 0
x2 − y + 1 = 0
xy − 1 = 0
We see that it has no solution, then (f1 , f2 , f3 ) = C[x, y].
2
(8.11)Theorem.(Hilbert) Let f1 , · · · , fr and g be polynomials in
C[x1 , · · · , xn ]. Let V be the variety of zeros of f1 , · · · , fr , and let I be
the ideal generated by f1 , · · · , fr . If g = 0 for all (x1 , · · · , xn ) ∈ V ,
then there exists a positive integer n such that g n ∈ I.
Proof. Suppose g(x1 , · · · , xn ) = 0, ∀(x1 , · · · , xn ) ∈ V . We consider polynomials f1 (x1 , · · · , xn ), · · · , fr (x1 , · · · , xn ), g(x1, · · · , xn )y−
1 in C[x1 , · · · , xn , y], where x1 , · · · , xn , y are variables. It is clear
that these polynomials have no common zero in Cn+1 . Therefore,
the above Corollary implies
C[x1 , · · · , xn ] = (f1 , · · · , fr , gy − 1)
. i.e. there exist polynomials pi (x1 , · · · , xn , y), 1 ≤ i ≤ r and
q(x1 , · · · , xn , y) ∈ C[x1 , · · · , xn , y] such that
r
X
→
→
→
p i (−
x , y) + q(−
x , y)(g(−
x )y − 1) = 1
i=1
→
Taking y = g(−
x )−1 in the above identity:
r
X
→
→
→
p i (−
x , g(−
x )−1 )fi (−
x)=1
i=1
→
Now we clear the denominators in Pi (−
x , y) by multiplying both
sides of the equation by a sufficiently large power of g. This yields
r
X
→
→
n
g (x) =
hi (−
x )fi (−
x)∈I
i=1
47
→
→
→
where hi (x) = g n (−
x )pi (−
x , g(−
x )−1 ).
2
Homework.
1. Let U, V be varieties in Cn . Prove that U ∪ V and U ∩ V are
varieties.
2. Let f1 , · · · , fr ; g1 , · · · , gs ∈ C[x1 , · · · , xn ], and let U, V be the
zeros of {f1 , · · · , fr }, {g1, · · · , gs } respectively. Prove that if U and
V do not meet, then (f1 , · · · , fr ; g1 , · · · , gs ) is the unit ideal.
3. Modules
(3.1)Definition. Let R be a ring with identity. An abelian group
V = (V, +) is called an R-module if there is a map:
R × V −→ V
(r, v) 7−→ r.v
such that
1.
1.v = v,
2.
(rs).v = r.(s.v),
3.
(r + s).v = r.v + s.v,
4.
r.(v1 + v2 ) = r.v1 + r.v2 .
for all v, v1 , v2 ∈ V and r, s ∈ R, where 1 is the identity of the ring
R.
48
(3.2)Example. R = Mn×n (R) the ring of n × n matrices. Let
V = Rn the Euclidean space. We define the action of R in V by:
→
→
M.−
x = M−
x ∈V
x
1
.
→
for any M ∈ R, −
x = .. ∈ Rn = V . Then Rn is a Mn×n (R)
xn
module.
a
1
..
(3.3)Example. Let R be a ring with identity and V = { . |
an
ai ∈ R, 1 ≤ i ≤ n}, then V is an abelian group with the usual
addition, and with the following action:
ra
a
1 1
. .
r. .. = ..
ran
an
a
1
..
for r ∈ R, . ∈ V . Then V is an R-module.
an
(3.4)Definition. The R-module defined in (3.3)Example. is
called a free module of R.
(3.5)Example. If V is an abelian group written additively, then V
49
is a Z-module. Indeed,
v + v + · · · + v(n copies)
n≥1
n.v = 0
n=0
(−v) + · · · + (−v)(| n | copies) n ≤ −1
for any n ∈ Z, v ∈ V .
(3.6)Remark. 1. An R-module not necessarily a free R-module.
For example, let R = Z, then the free R-module V has the form:
a
1
..
V = { . | ai ∈ Z, 1 ≤ i ≤ n}
an
Thus V contains infinitely many elements, but Z/pZ = {0̄, 1̄, · · · , p − 1}
is a Z-module which is a finite set, thus is not a free Z-module.
2. If R is a field, then an R-module is a vector space over R,
thus is a free R-module.
(3.7)Definition. If V is a R-module, a non-empty subset M of V
is called a submodule of the R-module V if
1. M is an additive subgroup of V ,
2. ∀r ∈ R, v ∈ M implies r.v ∈ M.
(3.8)Example. Let R be a ring. Taking V = R, then R can be
regarded sa an R-module. In which the action is the multiplication
in the ring R.
(3.9)Proposition. The submodules of the R-module R are nothing but the ideals of R.
50
(3.10)Definition. 1. If V is an R-module, then {0}, V are submodules of V , called trivial submodules of V .
2. If V 6= 0 is an R-module, and V has no non-trivial submodule, then V is called a simple module or irreducible module.
(3.11)Definition. Let R be a ring and V, W be R-modules.
1.
a map ϕ : V −→ W is called a R-homomorphism if
ϕ(v + v ′ ) = ϕ(v) + ϕ(v ′ )
ϕ(r.v) = r.ϕ(v)
where v, v ′ ∈ V, r ∈ R.
2.
if ϕ is an R-module homomorphism, and ϕ is also a bijective
map, then ϕ is called an R-module isomorphism.
3.
if ϕ is an R-module homomorphism from V to W , set
ker ϕ = {v ∈ V | ϕ(v) = 0}
im ϕ = {ϕ(v) | v ∈ V }
and ker ϕ,im ϕ are called the kernel and image of the module homomorphism ϕ.
(3.12)Lemma. If ϕ : V −→ W is an R-module homomorphism,
then ker ϕ,im ϕ are respectively R-submodules of V and W .
(3.13)Proposition. Let R be a commutative ring with identity,
Rm , Rn be the free R-modules. If ϕ : Rm −→ Rn is an R-module
homomorphism, then there exists a matrix
a
. . . a1m
11
Mϕ = . . . . . . . . . . . . . ∈ Mn×m (R).
an1 . . . anm
51
v
1
..
such that, for any v = . ∈ Rm ,
vm
P
v1
a1i vi
a11 . . . a1m
. .
ϕ(v) = Mϕ v = (. . . . . . . . . . . . . .. = .. ).
P
an1 . . . anm
vm
ani vi
.
0
..
.
Proof. Set ǫi = 1 ←− i-th entry, 1 ≤ i ≤ m. Since ϕ(ǫi ) ∈
..
.
0
n
R , ∀1 ≤ i ≤ m, we set
a
1i
..
ϕ(ǫi ) = . , 1 ≤ i ≤ m
ani
v
1
..
and for any v = . ∈ Rm , we have v = v1 ǫ1 + · · · + vm ǫm . Thus,
vm
P
a1i vi
a1i
P
P
P
..
..
ϕ(v) =
ϕ(vi ǫi ) =
vi ϕ(ǫi ) =
vi . = . =
P
ani vi
ani
Mϕ v.
2
(3.14)Corollary. The R-module homomorphisms from the free
module Rm to free module Rn are in one-to-one correspondence
52
with the n × m matrices with entries from R.
(3.15)Definition. Let R be a commutative ring with identity, W
be a submodule of R-module V , then the quotient group:
V /W = {a + W =: ā | a ∈ V }
is an R-module by defining r.ā = r.a for r ∈ R, a ∈ V . Called a
quotient module of V factored by the submodule W .
(3.16)Proposition. (a) The canonical map
π:
V −→ V /W
v 7−→ v̄ := v + W
is a surjective R-module homomorphism with kernel W .
(b) Let f : V −→ V ′ be an R-module homomorphism with
kernel W . Then there is a unique homomorphism f¯ : V /W −→ V ′
such that f = f¯ ◦ π.
(c) (The First Isomorphism Theorem) If ϕ : V −→ V ′ is
an R-module homomorphism, then V / ker ϕ ∼
= im ϕ.
(d) If S, W are submodule of V such that W ⊂ V , then
V /W/S/W ∼
= V /S as R-modules.
Proof. (a) is clearly. For (b), we define
f¯ :
V /W −→ V ′
ā := a + W 7−→ f (a)
We first show f¯ is well-defined, if ā = b̄, then b − a ∈ W = ker f .
Thus f (b) − f (a) = f (b − a) = 0 and f (b) = f (a), as required.
53
Second, for any r ∈ R, ā, b̄ ∈ V /W , we have f¯(r.ā) = f¯(r.a) =
¯ + b̄) = f¯(a + b) = f (a + b) =
f (r.a) = r.f (a) = r.f¯(ā) and f(ā
f (a) + f (b) = f¯(ā) + f¯(b̄). i.e. f¯ is an R-module homomorphism.
Third, for any a ∈ V , (f¯ ◦ π)(a) = f¯(ā) = f (a), thus, f = f¯ ◦ π.
Finally, the uniqueness is clear.
2
Homework.
1. Prove (c) and (d) of the last Proposition.
4. Fields
(4.1)Definition. A field F is a set together with two laws of compositions +, ×:
+ : F × F −→ F
× : F × F −→ F
(a, b) 7−→ a + b
(a, b) 7−→ ab
satisfying the following conditions:
(1) F is an abelian group with +, the identity element is denoted
by 0.
(2) F \ {0} is a commutative group with ×, and its identity
element is denoted by 1.
(3) Distributive law holds:
(a + b)c = ac + bc,
54
∀a, b, c ∈ F.
√
(4.2)Example. 1. Q ⊂ Q( 2) ⊂ R ⊂ C.
2. Fp = {0̄, 1̄, · · · , p − 1} where p is a prime number, and
ī = i + pZ, then Fp is a field.
3. C(x) = {p(x)/q(x) | p(x), q(x) ∈ C[x], and q(x) 6= 0} is a
field, called function field.
(4.3)Definition. A field having only finitely many elements is
called a finite field.
(4.4)Definition. Let F ⊂ K be fields. We say F is a subfield of
K, and K is called an extension field of F .
(4.5)Definition. Let K be an extension of a field F .
(1) α ∈ K is called an algebraic element over F if α is a root
of a non-zero polynomial with coefficients from F .
(2) α ∈ K is called transcendental over F if α is not a root of
any such a polynomial.
(4.6)Example. i =
√
−1 is an algebraic over the field Q, while π
is transcendental over Q.
(4.7)Lemma. Let F ⊂ K be fields and α ∈ K. We define homomorphism
ϕ :F [x] −→ K
f (x) 7−→ f (α)
Then, (1) α is transcendental over F if ϕ is injective.
(2) α is algebraic over F if ϕ is not injective.
(4.8)Definition. Let R be a ring, if
55
(a) R is integral domain
(b) every ideal of R is principal
then R is called a principal ideal domain (PID).
(4.9)Example. Z is a PID.
(4.10)Definition. A polynomial p(x) with coefficient in a field F
is called irreducible if it is not a constant and its only divisors of
lower degree in F [x] are constants.
(4.11)Example. x2 − 2 is irreducible in Q, while is not irreducible
in R.
(4.12)Remark. Let F ⊂ K be fields, and α be algebraic over F .
If
ϕ :F [x] −→ K
g(x) 7−→ g(α)
is a surjective ring homomorphism, then
F [x]/ ker ϕ ∼
=K
where ker ϕ is a principal ideal since F [x] is a PID. Thus ker ϕ =
(f (x)) for some f (x) ∈ F [x] a monic polynomial.
(4.13)Lemma. f (x) in (4.12)Remark. is irreducible over F .
Proof. Since K is a field, thus ker ϕ is a maximal ideal of
F [x]. Therefore f (x) is irreducible, otherwise f (x) = g(x)h(x)
where g(x0, h(x) has lower degrees than that of f (x), then (f (x)) &
(g(x)) $ F [x] a contradiction.
2
56
(4.14)Definition. We call f (x) the irreducible polynomial for the
algebraic element α over F .
(4.15)Definition. 1. Let F ⊂ K be fields and α ∈ K. The smallest subfield which contains F and α is denoted by F (α), and called
a field extension of F generated by α.
2. If α1 , · · · , αn ∈ K, then F (α1 , · · · , αn ) denotes the smallest
subfield of K containing α1 , · · · , αn and F .
Recall. If F is a ring (or a field), then the ring generated by α over
R is the set
{an αn + · · · + a1 α + a0 | n ≥ 0, ai ∈ F, 1 ≤ i ≤ n}
which is denoted by F [α].
(4.16)Lemma. If F is a field, then the field F (α) is isomorphism
to the field of fractions over F [α].
(4.17)Proposition. If α is transcendental over F , then the map
ϕ :F [x] −→ F [α]
f (x) 7−→ f (α)
is an isomorphism, and the field F (α) is isomorphism to the rational
field F (x).
(4.18)Proposition If α is algebraic over F , and f (x) is the irreducible polynomial for α, then
ϕ̄ :
F [x]/(f (x)) −→ F [α]
g(x) := g(x) + (f (x)) 7−→ g(α)
57
is an isomorphism, and F [α] is a field, thus F [α] = F (α).
Proof. Let
ϕ:
F [x] −→ F (α)(⊇ F [α])
g(x) 7−→ g(α)
then ker ϕ = (f (x)). Since f (x) is irreducible, (f (x)) is a maximal
ideal of F [x], thus F [x]/(f (x)) is a field which is isomorphism to
im ϕ = F [α], thus F [α] is a field and F [α] = F (α).
2
(4.19)Corollary. If α1 , · · · , αn be algebraic elements over F , then
F [α1 , · · · , αn ] = F (α1 , · · · , αn )
Proof. Since F [α1 , · · · , αn ] = (F [α1 , · · · , αn−1 ])[αn ], thus by
induction we know the corollary holds.
2
(4.20)Proposition. Let α be an algebraic element over F , and
let f (x) be its irreducible polynomial. Suppose f (x) has degree n.
Then {1, α, · · · , αn−1} is a basis for F [α] as a vector space over F .
Proof. Since F [x]/(f (x)) ∼
= F [α]. We only to show {1, α, · · · , αn−1}
is linearly independent over F . Note that (f (x)) = ker ϕ, where
ϕ:
F [x] −→ F [α]
g(x) 7−→ g(α)
We need to show that no nonzero polynomial g(α) over F has degree less than n such that g(α) = 0. Indeed, if g(α) = 0, then
g(α) ∈ ker ϕ = (f (x)). Thus g(x) = f (x)h(x) for some h(x) ∈ F [x],
58
while deg(g(x)) =deg(f (x))+deg(h(x)) ≥ n, a contradiction.
2
(4.21)Proposition. Let α ∈ K, β ∈ L, where K, L are extension
fields of F , and α, β are algebraic over F . Then there is an isomorphism of fields σ : F (α) −→ F (β) which is the identity map
when restricted on the subfield F and send α to β if and only if the
irreducible polynomials for α and β are equal.
Proof. “⇐=” Suppose f (x) is the irreducible polynomial for
both α and β over F . Then
F [x]/(f (x)) ∼
= F [α] = f (α),
F [x]/(f (x)) ∼
= F [β] = F (β)
Suppose the two isomorphism are respectively ϕ and ψ, then ψϕ−1
is the required isomorphism, from F (α) to F (β).
“=⇒” Let σ be such an isomorphism. Let f (x) = an xn + · · · +
a1 x + a0 be the irreducible polynomial of α over F . i.e.
an αn + · · · + a1 α + a0 = 0
Note that σ is an isomorphism and σ |F =id., σ(α) = β. Then
0 = σ(0) = σ(an αn + · · · + a0 ) = an β n + · · · + a0 which implies f (x)
is also the irreducible polynomial for β over F .
2
Homework.
1. Let R be an integral domain containing a field F as a subring
and which is finite dimensional when viewed as a vector space over
F . Prove that R is a field.
2. Let α be a complex root of the irreducible polynomial x3 −
3x + 4. Find the inverse of α2 + α + 1 in F (α) explicitly in the form
a + bα + cα2 , a, b, c ∈ Q.
59
3. Let K = F (α), where α is a root of the irreducible polynomial
f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 . Determine the element α−1
explicitly in terms of α and of the coefficients ai .
60
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