University of Cergy-Pontoise
L1-MI 2nd semester
2010-2011
Test no 1 - Answers and Solutions
Question about course 1. (R, +, ×) is a ring if it satisfies each of the following point
i) R is a nonempty set
ii) + and × are binary operation on R
a+b∈R
2
∀(a, b) ∈ R ,
a×b∈R
iii) (R, +) is an abelian group, that is it satisfies each of the following point
∀(a, b, c) ∈ R3 , (a + b) + c = a + (b + c)
∃0 ∈ R/ ∀a ∈ R, a + 0 = 0 + a = a
∀a ∈ R, ∃a0 ∈ R/ a + a0 = a0 + a = 0
∀(a, b) ∈ R2 , a + b = b + a
iv) the binary operation × is associative
∀(a, b, c) ∈ R3 , (a × b) × c = a × (b × c)
v) the binary operation × is distributive over the binary operation +
a × (b + c) = a × b + a × c
3
∀(a, b, c) ∈ R ,
(b + c) × a = b × a + c × a
Exercise 2. At first, n = 0 gives
62n+1 + 23n+1 = 61 + 21 = 8 = 1 + 7 ≡ 1 [7]
So the claim is true for n = 0. Now assume the claim is true for a given integer n ∈ N. We have
62(n+1)+1 + 23(n+1)+1 = 62n+2+1 + 23n+3+1 = 62 × 62n+1 + 23 × 23n+1 = 36 × 62n+1 + 8 × 23n+1
Using the division algorithm we get
36 = 5 × 7 + 1 ≡ 1 [7]
8 = 1 × 7 + 1 ≡ 1 [7]
Then
62(n+1)+1 + 23(n+1)+1 = 36 × 62n+1 + 8 × 23n+1 ≡ 1 × 62n+1 + 1 × 23n+1
≡ 62n+1 + 23n+1 [7]
[7]
Consequently the claim is also true for n + 1 since it is true for n by inductive assumption. The
conclusion follows by induction.
√
√
2
=
0.
Conversely,
assume
a
+
b
2 = 0. If
Problem 3.
1. Obviously,
if
a
=
b
=
0
then
a
+
b
√
√
b 6= 0 then we get 2 = −a
which
is
a
contradiction
with
2
∈
/
Q.
Consequently
b
= 0 and
b
√
a = −b 2 = 0.
2. A is a nonempty subset of R and (R,
unital ring (and even more a
√
√ +, ×) is a commutative
field). Moreover for every x = a + b 2 ∈ A and y = c + d 2 ∈ A we have
√
√
√
x − y = (a + b√2) − (c + d√2) = (a − c) + (b − d) 2 ∈ √
A
x × y = (a + b 2) × (c + d 2) = (ac + 2bd) + (ad + bc) 2 ∈ A
The first point shows that (A, +) is a subgroup of (R, +) and the second point shows that ×
is a binary operation on A. Consequently (A, +, ×) is a subring
of (R, +, ×) and in particular
√
(A, +, ×) is a commutative ring. Furthermore 1 = 1 + 0 2 ∈ A is the multiplicative identity,
so (A, +, ×) is unital.
Now we need to find
√ a nonzero
√ element in A which has no multiplicative inverse√in A. For
instance, assume 2 = 0 + 1 2 ∈ A has a multiplicative inverse denoted by a + b 2, that is
√
√
√
2 × (a + b 2) = 2b + a 2 = 1
√
In particular we get (2b − 1) + a 2 = 0 and the first point of this problem implies 2b − 1 = 0
and a = 0. But √
2b − 1 = 0 is a contradiction with b ∈ Z (2 has no multiplicative inverse in Z).
It follows that 2 6= 0 has no multiplicative inverse in A and then (A, +, ×) is not a field.
√
√
3. (a) Let x = a + b 2 ∈ A and y = c + d 2. Then
√
√
√
√
N (x) = x × ϕ(x) = (a + b 2) × (a − b 2) = a2 − ab 2 + ab 2 + 2b2 = a2 − 2b2 ∈ Z
√ √
and N (xy) = N (a + b 2)(c + d 2)
√ = N (ac + 2bd) + (ad + bc) 2
=
=
=
=
=
(ac + 2bd)2 − 2(ad + bc)2
a2 c2 + 4abcd + 4b2 d2 − 2a2 d2 − 4abcd − 2b2 c2
a2 c2 − 2b2 c2 − 2a2 d2 + 4b2 d2
(a2 − 2b2 )(c2 − 2d2 )
N (x)N (y)
(b) Assume x ∈ A has a multiplicative inverse in A denoted by x−1 . Using the previous
results we get
N (x)N (x−1 ) = N (xx−1 ) = N (1) = 1
In particular, N (x) ∈ Z has a multiplicative inverse N (x−1 ) ∈ Z. That implies either
N (x) = 1 or N (x) = −1 (only these two integers have multiplicative inverse in Z).
Conversely, if N (x) = x × ϕ(x) is equal to 1 or −1 then x has a multiplicative inverse
in A which is respectively ϕ(x) or −ϕ(x).
Problem 4. Consider the following map for a given number λ ∈ R∗
2
R2 −→ R
y
(x, y) 7−→ λx,
λ
2
0 0
1. (R , +) is an abelian group. Let (x, y) and (x , y ) be two points in R2 . Then
fλ :
fλ ((x, y) + (x0 , y 0 )) = fλ ((x + x0 , y + y 0 ))
0
0 y+y
=
λ(x + x ),
λ
0
y
0 y
= λx,
+ λx ,
λ
λ
= fλ ((x, y)) + fλ ((x0 , y 0 ))
So fλ is a group endomorphism from (R2 , +) to itself.
2. (a) Let (x, y) be a point in R2 . Then
y
y
fλ ◦ fµ (x, y) = fλ (fµ (x, y)) = fλ µx,
= λµx,
= fλµ (x, y)
µ
λµ
Consequently, fλ ◦ fµ = fλµ .
(b) Since the product of two numbers in R∗ stays in R∗ , we have
◦ : F × F −→ F
(fλ , fµ ) 7−→ fλ ◦ fµ = fλµ ∈ F
In other words, ◦ is a binary operation on F.
(c) The associativity of (F, ◦) comes from that one of (R∗ , ×)
∀(λ, µ, ν) ∈ (R∗ )3 , (fλ ◦ fµ ) ◦ fν = fλµ ◦ fν = f(λµ)ν = fλ(µν) = fλ ◦ fµν = fλ ◦ (fµ ◦ fν )
f1 : (x, y) 7→ (x, y) is obviously the identity element of (F, ◦). Moreover for every λ ∈ R∗
we have
fλ ◦ f1/λ = fλ/λ = f1 and f1/λ ◦ fλ = fλ/λ = f1
In particular, every fλ ∈ F has an inverse element (which is f1/λ ∈ F). Finally (F, ◦) is
a group.
3. (a) Let λ and µ be two numbers in R∗ . We have
h(λµ) = fλµ = fλ ◦ fµ = h(λ) ◦ h(µ)
Then h is a group homomorphism from (R∗ , ×) to (F, ◦). Moreover h is a surjective map
by definition of F and an injective map since
∀(λ, µ) ∈ (R∗ )2 , h(λ) = h(µ) ⇒ fλ = fµ ⇒ fλ (1, 0) = fµ (1, 0) ⇒ (λ, 0) = (µ, 0) ⇒ λ = µ
Finally, h is a bijective map and hence a group isomorphism from (R∗ , ×) to (F, ◦).
(b) Since (R∗ , ×) is an abelian group, we have
∀(λ, µ) ∈ (R∗ ), fλ ◦ fµ = h(λ) ◦ h(µ) = h(λµ) = h(µλ) = h(µ) ◦ h(λ) = fµ ◦ fλ
Then the binary operation ◦ is commutative on F or equivalently, (F, ◦) is an abelian
group.