Aries
Rings, ideals, modules
In this note, we will not recall basic concepts. The definition of ring, ideal,module, homomorphism of rings are not reproduced. In this course, any ring A always be:
1.
c
o
m
m
u
t
a
t
i
v
e
:
x
⋅
y
=
y
⋅
x
for any
x
,
y
∈
A
2.
a
s
s
o
c
i
a
t
i
v
e
:
x
(
y
z
)
=
(
x
y
)
z
for any
x
,
y
,
z
∈
A
3.
u
n
o
t
a
r
y
:
there exists
1
∈
A
such that
1
⋅
x
=
x
for any
x
∈
A
.
1mutative:x\cdot y=y\cdot x \text{ for any }x,y \in A\\ 2.associative:x(yz)=(xy)z \text{ for any } x,y,z\in A\\ 3.unotary:\text{there exists }1\in A\text{ such that }1\cdot x=x\text{ for any }x\in A.
1.commutative:x⋅y=y⋅x for any x,y∈A2.associative:x(yz)=(xy)z for any x,y,z∈A3.unotary:there exists 1∈A such that 1⋅x=x for any x∈A.
So,for any ring homomorphism
f
:
A
→
B
f:A\rightarrow B
f:A→B preserves the units:
f
(
1
)
=
1
f(1)=1
f(1)=1.We denote by
A
×
A^{\times}
A× the multiplicative group of invertible elements of A.
We will say that an element
a
∈
A
a\in A
a∈A is
n
i
l
p
o
t
e
n
t
:
if there exist n
∈
N
s
u
c
h
t
h
a
t
a
n
=
0
d
i
v
i
s
o
r
o
f
z
e
r
o
:
if there is b
∈
A
\
{
0
}
such that ab=0
r
e
g
u
l
a
r
:
if it is not a divider of zero
.
nilpotent:\text{if there exist n}\in \mathbb{N} such that a^{n}=0\\ divisor of zero:\text{if there is b}\in A \backslash\{0\} \text{such that ab=0}\\ regular:\text{if it is not a divider of zero}.
nilpotent:if there exist n∈Nsuchthatan=0divisorofzero:if there is b∈A\{0}such that ab=0regular:if it is not a divider of zero.
We say that A is integral (resp. Is a field) if
A
≠
{
0
}
A \neq\{0\}
A={0} and if any non-zero element of A is regular (resp. Invertible).(The latter conditions are more important)
An ideal
I
⊂
A
I\subset A
I⊂A is principal if there exists
a
∈
A
a\in A
a∈A such that
I
=
A
a
:
{
a
b
:
b
∈
A
}
I=Aa:\{ab:b\in A\}
I=Aa:{ab:b∈A}.We say that
I
I
I is of finite type, if it exits
a
1
,
a
2
,
…
,
a
n
∈
A
a_1,a_2,\dots,a_{n}\in A
a1,a2,…,an∈A(for some
n
∈
N
n\in \mathbb N
n∈N) such that
I
=
A
a
1
+
A
a
2
+
⋯
+
A
a
n
I=Aa_1+Aa_2+\cdots+Aa_{n}
I=Aa1+Aa2+⋯+Aan;in this case, the quotient ring
A
/
I
A /I
A/I is often denoted :
A
/
(
a
1
,
…
,
a
n
)
.
A /\left(a_{1}, \ldots, a_{n}\right).
A/(a1,…,an).
We say that A is principal if all its ideals are principal.
Example 1.1 (1) Ring Z \mathbb Z Z is principal;to show this we must show that any ideal I ⊂ Z I\subset \mathbb Z I⊂Z is principal. if I = 0 I=0 I=0,this is trivial, and if not, let a ∈ I a\in I a∈I be the smallest element which >0.For all b ∈ I b\in I b∈I there exist q , r ∈ Z q,r\in\mathbb Z q,r∈Z such that b = a q + r b=aq+r b=aq+r and 0 ≤ r ≤ a 0\leq r \leq a 0≤r≤a; we can see r ∈ I r\in I r∈I,therefore r = 0 r=0 r=0, by the minimality of a a a, hence I = a Z I=a\mathbb Z I=aZ.
(2) if K K K is a field, the same argument applies to the ring of polynomials K [ X ] K[X] K[X]:for any non-zero ideal I ⊂ K [ X ] I\subset K[X] I⊂K[X] we choose p ( X ) ∈ I p(X) \in I p(X)∈I which is non zero of minimal degree polynomial. If b ( X ) ∈ I b(X)\in I b(X)∈I,by Euclidean division, we can give q ( X ) , r ( X ) ∈ K [ X ] q(X),r(X)\in K[X] q(X),r(X)∈K[X] such that b ( X ) = p ( X ) ⋅ q ( X ) + r ( X ) b(X)=p(X)\cdot q(X)+r(X) b(X)=p(X)⋅q(X)+r(X) with either r ( X ) = 0 r(X)=0 r(X)=0 or d e g X r ( X ) < d e g X p ( X ) deg_{X} r(X) <deg_{X}p(X) degXr(X)<degXp(X). But r ( X ) ∈ I r(X)\in I r(X)∈I,so finally r ( X ) = 0 r(X)=0 r(X)=0 by the minimality of d e g X p ( X ) deg_{X}p(X) degXp(X), hence I = p ( X ) ⋅ K [ X ] I=p(X)\cdot K[X] I=p(X)⋅K[X].
Algebras
An
A
−
a
l
g
e
b
r
a
A-algebra
A−algebra is dual
(
B
,
f
)
(B,f)
(B,f) consisting of a ring
B
B
B and a homomorphism of rings
f
:
A
→
B
f:A\rightarrow B
f:A→B, called the structural morphism of
B
B
B. A homomorphism of
A
−
a
l
g
e
b
r
a
A-algebra
A−algebra
g
:
(
B
,
f
)
→
(
B
′
,
f
′
)
g:(B, f) \rightarrow\left(B^{\prime}, f^{\prime}\right)
g:(B,f)→(B′,f′)
is a homomorphism of rings
g
:
B
→
B
′
g: B \rightarrow B^{\prime}
g:B→B′ switching the diagram
Obviously, the composition of two homomorphism of
A
−
a
l
g
e
b
r
a
A-algebra
A−algebra
g
:
B
→
B
′
g:B\rightarrow B'
g:B→B′ and
g
′
:
B
′
→
B
′
′
g':B'\rightarrow B''
g′:B′→B′′ is homomorphism of
A
−
a
l
g
e
b
r
a
A-algebra
A−algebra
g
′
∘
g
:
B
→
B
′
′
g^{\prime} \circ g: B \rightarrow B^{\prime \prime}
g′∘g:B→B′′. We denote by
A
−
A
l
g
(
B
,
B
′
)
A-Alg(B,B')
A−Alg(B,B′)
the set of homomorphism of
A
−
a
l
g
e
b
r
a
A-algebra
A−algebra
B
→
B
′
B\rightarrow B'
B→B′.
For example, for all n ∈ N n\in \mathbb N n∈N, the ring of polynomials of n n n variables A [ X 1 , … , X n ] A[X_1,\dots,X_{n}] A[X1,…,Xn] with coefficients in A A A is provided with a canonical structure of A − a l g e b r a A-algebra A−algebra, whose structural morphism is the natural inclusion A → A [ X 1 , … , X n ] A\rightarrow A[X_1,\dots,X_{n}] A→A[X1,…,Xn] which identifies A with the sub-ring of polynomials of total degree 0. Also, for all ideal I ⊂ A I\subset A I⊂A, the canonical projection A → A / I A\rightarrow A /I A→A/I provides the quotient ring A / I A /I A/I with a natural structure of A − a l g e b r a A-algebra A−algebra.
Remark 1.2 (1) Every ring A admits a unique homomorphism
Z
→
A
\mathbb Z\rightarrow A
Z→A, therefore every ring is canonically a
Z
−
a
l
g
e
b
r
a
\mathbb Z-algebra
Z−algebra.
(2) Let
A
A
A be a ring,
B
B
B is an
A
−
a
l
g
e
b
r
a
A-algebra
A−algebra,
n
∈
N
n\in\mathbb N
n∈N, and
(
b
1
,
…
,
b
n
)
∈
B
n
(b_1,\dots,b_{n})\in B^{n}
(b1,…,bn)∈Bn. Note that there is a unique homomorphism of
A
−
a
l
g
e
b
r
a
A-algebra
A−algebra
f
:
A
[
X
1
,
…
,
X
n
]
→
B
f:A[X_1,\dots,X_{n}]\rightarrow B
f:A[X1,…,Xn]→B such that
f
(
X
i
)
=
b
i
f(X_{i})=b_{i}
f(Xi)=bi for
i
=
1
,
…
,
n
i=1,\dots,n
i=1,…,n:this is the homomorphism defined by
f
(
P
)
:
=
P
(
b
1
,
…
,
b
n
)
∀
P
∈
A
[
X
1
,
…
,
X
n
]
.
f(P):=P(b_1,\dots,b_{n}) \quad \forall P\in A[X_1,\dots,X_{n}].
f(P):=P(b1,…,bn)∀P∈A[X1,…,Xn].
In other words, for all
A
−
a
l
g
e
b
r
a
A-algebra
A−algebra
B
B
B and all
n
∈
N
n\in \mathbb N
n∈N there is a natural bijection
B
n
→
∼
A
−
Alg
(
A
[
X
1
,
…
,
X
n
]
,
B
)
.
B^{n} \stackrel{\sim}{\rightarrow} A-\operatorname{Alg}\left(A\left[X_{1}, \ldots, X_{n}\right], B\right).
Bn→∼A−Alg(A[X1,…,Xn],B).
We will see in section 2.2 how this property characterizes
A
[
X
1
,
…
,
X
n
]
A[X_1,\dots,X_{n}]
A[X1,…,Xn] except for canonical isomorphism.
We say that an A − a l g e b r a A-algebra A−algebra B B B is ring of finite type, if there exists a surjective homomorphism of A − a l g e b r a A-algebra A−algebra π : A [ X 1 , … , X n ] → B \pi:A[X_1,\dots,X_{n}]\rightarrow B π:A[X1,…,Xn]→B, for some n ∈ N n\in \mathbb N n∈N. In view of remark 1.2(2), this amounts to saying that there exists a finite system b ∙ : = ( b 1 , … , b n ) b_{\bullet}:=\left(b_{1}, \dots, b_{n}\right) b∙:=(b1,…,bn) of elements of B B B such that all b ∈ B b\in B b∈B is written under the form b = P ( b 1 , … , b n ) b=P(b_1,\dots,b_{n}) b=P(b1,…,bn) for some polynomial P ∈ A [ X 1 , … , X n ] P\in A[X_1,\dots,X_{n}] P∈A[X1,…,Xn]; We say that b ∙ b_{\bullet} b∙ is a finite system of generators of the A − a l g e b r a A-algebra A−algebra B B B, and we also write B = A [ b 1 , … , b n ] B=A[b_1,\dots,b_{n}] B=A[b1,…,bn]. We say that B B B is an Algebra of finite presentation, if one can find a π \pi π surjection as above, whose kernel π − 1 ( 0 ) \pi^{-1}(0) π−1(0) is an ideal of finite type; In this case B B B is isomorphic to a quotient A [ X 1 , … , X n ] / I A[X_1,\dots,X_{n}] /I A[X1,…,Xn]/I, with I ⊂ A [ X 1 , … , X n ] I\subset A[X_1,\dots,X_{n}] I⊂A[X1,…,Xn] an ideal of finite type. if ( B , f ) (B,f) (B,f) is an A − a l g e b r a A-algebra A−algebra of finite type.(resp. finite presentation), we also say that f f f is a homomorphism of rings of finite type (resp. finite presentation).
Exercise 1.3 If A → f B → g C A \stackrel{f}{\rightarrow} B \stackrel{g}{\rightarrow} C A→fB→gC are two homomorphisms of rings of finite type(resp. finite presentation), show that it is the same for g ∘ f g \circ f g∘f.
Field of fractions
For any integral ring A we denote by
F
r
a
c
A
Frac\ A
Frac A
the field of fractions of A. Recall that this is the set of fractions
a
/
b
a /b
a/b (also noted
b
−
1
a
b^{-1}a
b−1a), i.e. the equivalence classes of couples
(
a
,
b
)
(a,b)
(a,b) with
a
,
b
∈
A
a,b\in A
a,b∈A and
b
≠
0
b\neq 0
b=0; two such couples
(
a
,
b
)
,
(
a
′
,
b
′
)
(a,b),(a',b')
(a,b),(a′,b′) are equivalent
⇔
a
b
′
=
a
′
b
\Leftrightarrow a b^{\prime}=a^{\prime} b
⇔ab′=a′b. We define the laws of addition and multiplication of
F
r
a
c
A
Frac\ A
Frac A by the obvious formulas:
a
/
b
+
c
/
d
:
=
(
a
d
+
c
b
)
/
b
d
a
/
b
⋅
c
/
d
:
=
a
c
/
b
d
.
a / b+c / d:=(a d+c b) / b d \quad a / b \cdot c / d:=a c / b d.
a/b+c/d:=(ad+cb)/bda/b⋅c/d:=ac/bd.
We can easily verify that these laws do not depend on the choice of the representatives
(
a
,
b
)
,
(
c
,
d
)
(a,b),(c,d)
(a,b),(c,d) for
a
/
b
,
c
/
d
a /b,c /d
a/b,c/d; the identity elements of addition and multiplication are
0
/
1
0 /1
0/1 and
1
/
1
1 /1
1/1, so that
−
(
a
/
b
)
=
(
−
a
)
/
b
-(a /b)=(-a) /b
−(a/b)=(−a)/b and
(
a
/
b
)
−
1
=
b
/
a
(a /b)^{-1}=b /a
(a/b)−1=b/a if
a
≠
0
a\neq 0
a=0. Furthermore, we have the injective homomorphism of
A
→
F
r
a
c
A
:
a
→
a
/
1
A\rightarrow Frac\ A:a\rightarrow a /1
A→Frac A:a→a/1;thus,
F
r
a
c
A
Frac\ A
Frac A is the smallest field containing
A
A
A, which is unique except for isomorphism. We leave the details to the reader, as we will see a more general construction in section 2.3.
Prime and maximum ideals
We recall that an ideal
I
⊂
A
I\subset A
I⊂A is said:
prime
:
if
1
∉
I
and
x
,
y
∉
I
⇒
x
y
∉
I
for all
x
,
y
∈
A
maximal
:
i
f
1
∉
I
and the only ideals of A containing I are I and A.
\text{prime}:\text{if}\ 1\notin I \text{ and } x,y\notin I\Rightarrow x y \notin I \text{ for all }\ x,y\in A\\ \text{maximal}:if\ 1\notin I\ \text{and the only ideals of A containing I are I and A.}
prime:if 1∈/I and x,y∈/I⇒xy∈/I for all x,y∈Amaximal:if 1∈/I and the only ideals of A containing I are I and A.
Proposition 1.4. Let
A
A
A be a ring,
I
⊂
A
I\subset A
I⊂A an ideal. We have:
(i) I is prime if and only if
A
/
I
A /I
A/I is an integral ring.
(ii) I is maximal if and only if
A
/
I
A /I
A/I is a field.
(iii) Any maximum ideal is prime.
Demonstration: (i) Let
x
,
y
∈
A
x,y\in A
x,y∈A, and denote by
x
ˉ
,
y
ˉ
∈
A
/
I
\bar{x}, \bar{y} \in A / I
xˉ,yˉ∈A/I the classes of
x
x
x and
y
y
y.if
x
ˉ
,
y
ˉ
≠
0
\bar{x},\bar{y}\neq 0
xˉ,yˉ=0, we have
x
,
y
∉
I
x,y\notin I
x,y∈/I;if now
I
I
I is prime, we deduce
x
y
∉
I
xy\notin I
xy∈/I,and therefore
x
ˉ
⋅
y
ˉ
≠
0
\bar{x}\cdot\bar{y}\neq 0
xˉ⋅yˉ=0, which shows that
A
/
I
A /I
A/I is integral. Conversely, if
A
/
I
A /I
A/I is integral, we have
x
ˉ
⋅
y
ˉ
≠
0
\bar{x}\cdot\bar{y}\neq 0
xˉ⋅yˉ=0, i.e.
x
y
∉
I
xy\notin I
xy∈/I, and then
I
I
I is prime.
(ii) Let
x
∈
A
x\in A
x∈A such that
x
∉
I
x\notin I
x∈/I, therefore
x
ˉ
≠
0
\bar{x}\neq 0
xˉ=0. if
A
/
I
A /I
A/I is a field, there exists
y
∈
A
y\in A
y∈A such that
x
ˉ
⋅
y
ˉ
=
1
\bar{x}\cdot\bar{y}=1
xˉ⋅yˉ=1 in
A
/
I
A /I
A/I, therefore
x
y
−
1
∈
I
xy-1\in I
xy−1∈I, hence
I
+
A
x
=
A
I+Ax=A
I+Ax=A; (Note that the sum of any two ideals is an ideal. )Since
x
x
x is arbitrary, we deduce that the only ideals that contain
I
I
I are
I
I
I and
A
A
A, i.e.
I
I
I is maximal. On the other hand, if
I
I
I is maximal, the hypothesis
x
∉
I
x\notin I
x∈/I Implies that we have
I
+
A
x
=
A
I+Ax=A
I+Ax=A, so there exists
a
∈
I
,
y
∈
A
a\in I,y\in A
a∈I,y∈A such that
x
y
+
a
=
1
xy+a=1
xy+a=1, hence
x
ˉ
⋅
y
ˉ
=
1
\bar{x}\cdot \bar{y}=1
xˉ⋅yˉ=1, which shows that
A
/
I
A /I
A/I is a field. Assertion (iii) follows Immediately from (i) and (ii).
We note:
Max
A
A
A: the set of maximum ideals of
A
A
A(maximum spectrum of A);
Spec
A
A
A: the set of prime ideals of
A
A
A(prime spectrum of
A
A
A).
One of the objectives of this course is to explain why Max
A
A
A and Spec
A
A
A are ‘geometric objects’. According to Proposition 1.4(iii), we have: Max
A
⊂
A\subset
A⊂Spec
A
A
A.
Exercise 1.5. Let A A A be an integral principal ring. Show that Spec A A A={0} ⋃ \bigcup ⋃Max A A A.
This follow prime ideals in integral principal ring is maximal.
Lemma 1.6 If
I
⊂
A
I\subset A
I⊂A is an ideal, we have a canonical bijection:
{ideals
J
J
J of
A
A
A such that
I
⊂
J
I\subset J
I⊂J}
⟷
\longleftrightarrow
⟷{ideals of
A
/
I
A /I
A/I} (
J
⊂
A
J\subset A
J⊂A)
→
\rightarrow
→(
J
/
I
⊂
A
/
I
J /I\subset A /I
J/I⊂A/I).
This bijection induced by restriction of bijections:
{
p
∈
Spec
A
∣
I
⊂
p
}
⟷
Spec
A
/
I
{
m
∈
Max
A
∣
I
⊂
m
}
⟷
Max
A
/
I
.
\{\mathfrak{p} \in \operatorname{Spec} A | I \subset \mathfrak{p}\} \longleftrightarrow \operatorname{Spec} A / I \quad\{\mathfrak{m} \in \operatorname{Max} A | I \subset \mathfrak{m}\} \longleftrightarrow \operatorname{Max} A / I.
{p∈SpecA∣I⊂p}⟷SpecA/I{m∈MaxA∣I⊂m}⟷MaxA/I.
Demonstration: Let
π
:
A
→
A
/
I
\pi:A\rightarrow A /I
π:A→A/I be the canonical projection; the reciprocal bijection associates with any ideal $\bar J
o
f
of
of A /I$, the ideal
π
−
1
(
J
ˉ
)
⊂
A
\pi^{-1}(\bar{J})\subset A
π−1(Jˉ)⊂A. If
p
\frak p
p is an ideal of
A
A
A and
I
⊂
p
I\subset\mathfrak p
I⊂p, we have
A
/
p
=
(
A
/
I
)
/
(
p
/
I
)
A /\mathfrak p=(A /I) /(\mathfrak p /I)
A/p=(A/I)/(p/I), therefore
A
/
p
A /\mathfrak p
A/p is integral (resp. a field) if and only if
(
A
/
I
)
/
(
p
/
I
)
(A /I) /(\mathfrak p /I)
(A/I)/(p/I) is integral (resp. a filed), and with proposition 1.4 we deduce that
p
\mathfrak p
p is prime (resp. maximal) in
A
A
A if and only if
p
/
I
\mathfrak p /I
p/I is prime (resp. maximum) in
A
/
I
A /I
A/I.
Definition 1.7. Let
A
A
A be a ring,
a
∈
A
a\in A
a∈A non-zero element.
(i)We say that
a
a
a is prime if the ideal
A
a
Aa
Aa is prime.
(ii)We say that
a
a
a is reducible, if it is the product of two non-reducible and non-invertible.
(iii)We say that
A
A
A is factorial if it is integral and if any non-zero and non-invertible element of
A
A
A is written as a product of prime elements.
Remark 1.8 (i) Let
A
A
A be a ring. In the following, we will use the notation
⋅
∣
⋅
\cdot|\cdot
⋅∣⋅ for the divisiblity relation in
A
A
A: therefore,
a
∣
b
a|b
a∣b means that
a
,
b
∈
A
a,b\in A
a,b∈A and
b
∈
A
a
b\in Aa
b∈Aa.
(ii) if A is factorial and
a
∈
A
\
0
a\in A\backslash {0}
a∈A\0, the factorization
a
=
u
p
1
⋯
p
t
a=up_1\cdots p_{t}
a=up1⋯pt with
u
∈
A
×
u\in A^{\times}
u∈A× and
p
1
,
⋯
,
p
t
p_1,\cdots,p_{t}
p1,⋯,pt prime elements is essentially unique:if
a
=
v
q
1
⋯
q
s
a=vq_1\cdots q_{s}
a=vq1⋯qs is a other factorization, we have
s
=
t
s=t
s=t and there exists a premutation
σ
:
{
1
,
…
,
t
}
→
∼
{
1
,
…
,
t
}
\sigma:\{1, \ldots, t\} \stackrel{\sim}{\rightarrow}\{1, \ldots, t\}
σ:{1,…,t}→∼{1,…,t} such that
p
i
−
1
q
σ
(
i
)
∈
A
×
p_{i}^{-1}q_{\sigma(i)}\in A^{\times}
pi−1qσ(i)∈A× for
i
=
1
,
…
,
t
i=1,\dots,t
i=1,…,t. For the proof, we reason by induction on
t
t
t: we have
t
=
0
t=0
t=0 if and only if
a
∈
A
×
a\in A^{\times}
a∈A×, and in this case it is clear that
s
=
0
s=0
s=0. Suppose that
t
≥
1
t\geq 1
t≥1 and that the uniqueness of the factors is already known for the products of
t
−
1
t-1
t−1 prime elements; since the ideal
A
p
1
Ap_1
Ap1 is prime, we have
p
1
∣
q
i
p_1|q_{i}
p1∣qi for some
i
≤
s
i\leq s
i≤s, and leaves to permute the factors we can assume that
i
=
1
i=1
i=1. There exists
u
1
∈
A
u_1\in A
u1∈A such that
q
1
=
p
1
u
1
q_1=p_1u_1
q1=p1u1, hence
u
1
∈
A
×
u_1\in A^{\times}
u1∈A× and
a
′
:
=
u
p
2
⋯
p
t
=
u
1
v
⋅
q
2
⋯
q
s
a':=up_2\cdots p_{t}=u_1v\cdot q_2\cdots q_{s}
a′:=up2⋯pt=u1v⋅q2⋯qs; by induction hypothesis, we have the uniqueness of the factorization of
a
′
a'
a′ with premutation of the factors close, hence the same foa
a
a
a.
(iii) Let
a
,
b
,
d
∈
A
\
{
0
}
a,b,d\in A\backslash \{0\}
a,b,d∈A\{0}; we say that
d
d
d is a greatest common divisor (abbreviated gcd) of
a
a
a and
b
b
b, if
a
,
b
∈
d
A
a,b \in dA
a,b∈dA and if for all
c
∈
A
c\in A
c∈A with
a
,
b
∈
c
A
a,b\in cA
a,b∈cA, we have
d
∈
c
A
d\in cA
d∈cA. Symmetrically,
e
∈
A
e\in A
e∈A is a smaller common multiple (abbreviated lcm) of
a
a
a and
b
b
b if
e
∈
A
a
⋂
A
b
e\in Aa\bigcap Ab
e∈Aa⋂Ab and if for all
x
∈
A
a
⋂
A
b
x\in Aa\bigcap Ab
x∈Aa⋂Ab we have
x
∈
A
e
x\in Ae
x∈Ae;i.e.
A
e
=
A
a
⋂
A
b
Ae=Aa\bigcap Ab
Ae=Aa⋂Ab. Note that if
A
A
A is integral, the gcd and the lcm of
a
a
a and
b
b
b, when they exist, are determined to the nearest multiplication of invertible elements: because if
d
d
d and
d
′
d'
d′ are gcd of
a
a
a and
b
b
b, we have
d
′
∣
d
d'|d
d′∣d and
d
∣
d
′
d|d'
d∣d′, i.e.
d
=
d
′
u
,
d
′
=
d
v
d=d'u,d'=dv
d=d′u,d′=dv for some
u
,
v
∈
A
u,v \in A
u,v∈A hence
d
=
d
u
v
d=duv
d=duv, and therefore
u
v
=
1
uv=1
uv=1, because
A
A
A is integral; we reason the same for the lcm.
(iv) if $ A$ is integral and if lcm
(
a
,
b
)
(a,b)
(a,b) exists, then lcm
(
a
,
b
)
(a,b)
(a,b) exists, and we have:
g
c
d
(
a
,
b
)
⋅
l
c
m
(
a
,
b
)
=
a
b
gcd(a,b)\cdot lcm(a,b)=ab
gcd(a,b)⋅lcm(a,b)=ab
with multiplication of invertible elements near, i.e. if
e
e
e is a lcm of
a
a
a and
b
b
b, then
e
−
1
a
b
∈
A
e^{-1}ab\in A
e−1ab∈A is gcd of
a
a
a and
b
b
b: indeed, there exists
d
∈
A
d\in A
d∈A such that
a
b
=
e
d
,
ab=ed,
ab=ed, and since on the one hand
a
∣
e
a|e
a∣e, it follows that
d
∣
b
d|b
d∣b, and on the other hand,
b
∣
e
b|e
b∣e, therefore
d
∣
a
d|a
d∣a too. Now, if
c
c
c divides
a
a
a and
b
b
b, say
a
=
c
x
,
b
=
c
y
,
a=cx,b=cy,
a=cx,b=cy, we see that
c
x
y
cxy
cxy is a common multiple of
a
a
a and
b
b
b. therefore
e
∣
c
x
y
e|cxy
e∣cxy, and of
e
d
=
c
⋅
c
x
y
ed=c\cdot cxy
ed=c⋅cxy it comes that
c
∣
d
c|d
c∣d, as wished. (On the other hand, the existence of the gcd does not entail that of the lcm: see Remark 1.111 ).
(v) if
A
A
A is factorial, any pair
(
a
,
b
)
(a,b)
(a,b) of non-zero elements of
A
A
A admits a gcd and lcm. Indeed, thanks to (iv) it suffices to show the existence of the lcm; so let
a
=
u
p
1
ν
1
⋯
p
k
ν
k
a=u p_{1}^{\nu_{1}} \cdots p_{k}^{\nu_{k}}
a=up1ν1⋯pkνk and
b
=
v
p
1
μ
1
⋯
p
k
μ
k
b=v p_{1}^{\mu_{1}} \cdots p_{k}^{\mu_{k}}
b=vp1μ1⋯pkμk factorization with
u
,
v
∈
A
×
u,v\in A^{\times}
u,v∈A×,
v
i
,
u
i
∈
N
v_{i},u_{i}\in \mathbb N
vi,ui∈N and
p
i
p_{i}
pi for all
i
=
1
,
…
,
k
i=1,\dots,k
i=1,…,k, with
A
p
i
≠
A
p
j
Ap_{i}\neq Ap_{j}
Api=Apj for
i
≠
j
i\neq j
i=j. Given (iii), we can easily see that
∏
i
=
1
k
p
i
max
(
ν
i
,
μ
i
)
\prod_{i=1}^{k} p_{i}^{\max \left(\nu_{i}, \mu_{i}\right)}
∏i=1kpimax(νi,μi) is a lcm of
a
a
a and
b
b
b; it is also deduced that
∏
i
=
1
k
p
i
min
(
ν
i
,
μ
i
)
\prod_{i=1}^{k} p_{i}^{\min \left(\nu_{i}, \mu_{i}\right)}
∏i=1kpimin(νi,μi) is a gcd of
a
a
a and
b
b
b: the details are left to the reader.
Exercise 1.9. (i) Show that every integral and main ring is factorial.
(ii) Show that the ring
Z
[
−
5
]
\mathbb Z[\sqrt{-5}]
Z[−5
(iii) Let
A
A
A be an integral ring; show that any pair of elements of
A
\
{
0
}
A\backslash \{0\}
A\{0} admits a gcd
⇔
\Leftrightarrow
⇔ any pair of elements of
A
\
{
0
}
A\backslash \{0\}
A\{0} admits a lcm.
With exercise 1.9 (i) and example 1.1 we see that Z \mathbb Z Z is factorial, and similarly for K [ X ] K[X] K[X], if K K K is an arbitrary field. The prime elements of Z \mathbb Z Z are obviously the usual integers; the first K [ X ] K [X] K[X] are the irreducible polynomials, i.e. the P ∈ K [ X ] P \in K[X] P∈K[X] with d : = d e g X P > 0 d:= deg_X P> 0 d:=degXP>0, which are not products of polynomials of d e g r e e s < d degrees<d degrees<d. In both cases, the key point of the proof is the existence of a Euclidean division for any pair of non-zero elements; the following problem axiomatizes the properties required to reproduce this argument:
Problem 1.10. Ring
A
A
A is said to be Euclidean if it is integral and there is an application
∣
⋅
∣
:
A
\
{
0
}
→
N
|\cdot|: A \backslash\{0\} \rightarrow \mathbb{N}
∣⋅∣:A\{0}→N
satisfying the following condition:
–for all $a, b \in A \backslash {0} $ there exists $q,r\in A $ such that $a = bq + r $ and either
r
=
0
r = 0
r=0, or
∣
r
∣
<
∣
b
∣
| r | <| b |
∣r∣<∣b∣ (Euclidean division of
a
a
a by
b
b
b).
(i) Show that any Euclidean ring is principal (and therefore, factorial).
(ii) Show that
Z
[
i
n
]
\mathbb Z [i \sqrt{n} ]
Z[in
(iii) The rest of the problem is devoted to a classical arithmetic application of the ring
Z
[
i
]
\mathbb{Z}[i]
Z[i] of Gauss integers. First, let
p
∈
N
p \in \mathbb{N}
p∈N be a prime number with
p
≡
1
(
m
o
d
4
)
p \equiv 1(\bmod 4)
p≡1(mod4). Show that there exists
x
∈
Z
x \in \mathbb{Z}
x∈Z such that
x
2
≡
−
1
(
m
o
d
p
)
x^{2} \equiv-1(\bmod p)
x2≡−1(modp).
(iv) Deduce from (ii) and (iii) the following Fermat Theorem: Let
p
∈
N
p \in \mathbb{N}
p∈N be an odd prime number; then p is the sum of two squares
p
=
a
2
+
b
2
p=a^{2}+b^{2}
p=a2+b2 of integers
a
,
b
∈
N
a, b \in \mathbb{N}
a,b∈N if and only if
p
≡
1
(
m
o
d
4
)
p \equiv 1(\bmod 4)
p≡1(mod4).
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