Algebra Test 2

EPFL - Section de Mathématiques
Algebra
A. Bassa
Fall semester 2008-2009
Test 2
Monday, 8th December 2008
Nom : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Prénom : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Section : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
No other documents are to be used during the test. The use of calculators is not allowed. Please
show all your work. GOOD LUCK !
Aucun document n’est autorisé. Les calculatrices sont interdites. Tous les calculs et raisonnements doivent figurer dans le dossier rendu.
Question 1
Question 2
Question 3
Question 4
/20 points
/30 points
/30 points
/20 points
Total / 100
Question 1 (english, 20 points)
1. Find all x ∈ Z, which satisfy the following system of congruences
x ≡ 7 (mod 11)
x ≡ 2 (mod 5)
2. Find all x ∈ Z, which simultaneously satisfy the two congruences above and the conruence
(x − 1) · (x − 2) · (x − 5) ≡ 0 (mod 7).
Question 1 (français, 20 points)
1. Trouver tous les ĺéments x ∈ Z qui vérifient le système de congruences suivant :
x ≡ 7 (mod 11)
x ≡ 2 (mod 5)
2. Trouver tous les éléments x ∈ Z qui satisfont simultanément les deux congruences ci-dessus
et la congruence suivante :
(x − 1) · (x − 2) · (x − 5) ≡ 0 (mod 7).
Solution :
1. We know that gcd(5, 11) = 1 so we can apply the Chinese remainder theorem. By the
extended Euclidean algorithm (or simply by observation) we have
1.11 − 2.5 = 1
Therefore
2(1.11) + 7(−2.5) ≡ 7
≡2
mod 11
mod 5
and we have
x ≡ 2(1.11) + 7(−2.5)
≡ −48 mod 55
≡ 7 mod 55.
mod 55
2. We know that 7 is a prime, so Z/7Z is a field and has no zero-divisors. Therefore
(x − 1) · (x − 2) · (x − 5) ≡ 0 (mod 7)
is equivalent to x ≡ 1 mod 7, or x ≡ 2 mod 7, or x ≡ 5 mod 7. We know that gcd(55, 7) =
1 so we now use the Chinese remainder theorem 3 times.
First, we know from our 7× tables that 7 × 8 = 56, so clearly
−1.55 + 8.7 = 1.
(If you don’t know your 7× tables you could use the extended Eulcidean algorithm instead).
We then have
x ≡ 7.8.7 − 1.55 mod 385 ≡ 337 mod 385,
or
x ≡ 7.8.7 − 2.55
mod 385 ≡ 282
mod 385,
x ≡ 7.8.7 − 5.55
mod 385 ≡ 117
mod 385.
or
Question 2 (english, 30 points)
Let F be a field and let g(x) ∈ F [x].
1. Let a, b ∈ F . Show that
g(x) ≡ b (mod (x − a)) ⇔ g(a) = b.
2. Let a1 , a2 , . . . , an be distinct elements of F , and b1 , b2 , . . . , bn be arbitrary elements of F .
Show that there exists a polynomial p(x) ∈ F [x] such that p(ai ) = bi for each i = 1, 2, . . . n.
Show that moreover this polynomial can be choosen such that deg p(x) < n. (Hint : you
can use the Chinese remainder theorem for polynomials)
3. Find a polynomial p(x) ∈ F5 [x] with deg p(x) < 3 and
p(1) = 1,
p(2) = 3,
p(3) = 1.
Question 2 (français, 30 points)
Soit F un corps et soit g(x) ∈ F [x].
1. Soient a, b ∈ F . Montrer que
g(x) ≡ b (mod (x − a)) ⇔ g(a) = b.
2. Soient a1 , a2 , . . . , an des éléments distincts de F , et b1 , b2 , . . . , bn des éléments quelconques de
F . Montrer qu’il existe un polynôme p(x) ∈ F [x] tel que p(ai ) = bi pour tout i = 1, 2, . . . n.
Démontrer qu’en plus on peut choisir ce polynôme tel que deg p(x) < n. (Indication : vous
pouvez utiliser le théorème des reste chinois pour les polynôme.)
3. Trouver un polynôme p(x) ∈ F5 [x] de degré < 3 et tel que
p(1) = 1,
p(2) = 3,
p(3) = 1.
Solution :
1. ⇒
g(x) ≡ b mod (x − a) ⇒
g(x) = b + (x − a)h(x)
for some h(x) ∈ F [x] ⇒
g(a) = b
⇐
Let g(a) = b.
This implies that g(a) − b = 0, therefore a is a root of g(x) − b.
This means that (x − a)|g(x) − b, and therefore g(x) − b ≡ 0 mod (x − a).
So g(x) ≡ b mod (x − a).
2. From part one we know that this is the same as asking for a p(x) ∈ F [x] such that
p(x) ≡ bi
mod (x − ai )
for all i.
The ai are distinct in F , therefore (x − ai ) are pairwise coprime. We can then use the
Chinese remainder theorem to find a p(x) satisfying the congruences.
Q
Using the Chinese remainder theorem we will get an answer modulo i (x − ai ). This is a
polynomial of degree n so we can always pick a p(x) satisfying the congruences with degree
less than n.
3. This is an application of part 2.
We first observe that (x − 1) − (x − 2) − 1. Therefore
3(x − 1) − (x − 2) ≡ 1 mod (x − 1)
≡ 3 mod (x − 2).
Therefore any polynomial congruent to 2x − 1 mod (x − 1)(x − 2) will satisfy the first two
conditions.
Now we want a polynomial which is congruent to 2x − 1 ≡ mod (x − 1)(x − 2) and 1
mod (x − 3).
Using the Euclidean algorithm we see that (x − 1)(x − 2) = x2 − 3x + 2 = x.(x − 3) + 2,
therefore we have
1.(x2 − 3x + 2) − x.(x − 3) = 2
and
1 2
1
.(x − 3x + 2) − x.(x − 3) = 1.
2
2
Therefore a solution to all three congruences is
1 2
1
.(x − 3x + 2) − (2x − 1)x.(x − 3)
2
2
mod (x − 1)(x − 2)(x − 3).
Multiplying out we get
1 2
1
.(x − 3x + 2) − (2x − 1)x.(x − 3)
2
2
1
1 2
≡ .(x − 3x + 2) − (2x3 − 6x2 − x2 + 3x)
2
2
1
≡ (−2x3 + 8x2 − 6x + 2)
2
≡ −x3 + 4x2 − 3x + 1
≡ −6x2 + 11x − 6 + 4x2 − 3x + 1
≡ 3x2 + 3x
mod (x − 1)(x − 2)(x − 3)
mod (x3 − 6x2 + 11x − 6)
mod (x3 − 6x2 + 11x − 6)
mod (x3 − 6x2 + 11x − 6)
mod (x3 − 6x2 + 11x − 6)
mod (x3 − 6x2 + 11x − 6) .
Therefore p(x) = 3x2 + 3x. Finally we check that p(1) = 1, p(2) = 3 and p(3) = 1 in F5 .
Question 3 (english, 30 points)
1. Using the Euclidean Algorithm, find the greatest common divisor of the polynomials
f (x) = x5 − 2x4 + 3x3 − 2x2 + 2x
and
g(x) = x4 − 2x3 + 2x2 − 2x + 1
in Q[x]. Find polynomials a(x), b(x) ∈ Q[x] such that
a(x) · f (x) + b(x) · g(x) = gcd(f (x), g(x)).
2. Consider the polynomials x5 − x and x4 + 1 in F3 [x]. Find a polynomial c(x) ∈ F3 [x] such
that
(x5 − x) · c(x) ≡ x2 + 1 (mod x4 + 1).
3. Find all units and zero diviors in the ring
F3 [x]/((x − 1)(x2 + 1)).
Question 3 (français, 30 points)
1. A l’aide de l’algorithme d’Euclide, déterminer le plus grand diviseur commmun des polynômes
f (x) = x5 − 2x4 + 3x3 − 2x2 + 2x
et
g(x) = x4 − 2x3 + 2x2 − 2x + 1
dans Q[x]. Trouver des polynômes a(x), b(x) ∈ Q[x] tels que
a(x) · f (x) + b(x) · g(x) = gcd(f (x), g(x)).
2. On considère les polynômes x5 − x et x4 + 1 dans F3 [x]. Trouver un polynôme c(x) ∈ F3 [x]
tel que
(x5 − x) · c(x) ≡ x2 + 1 (mod x4 + 1).
3. Déterminer toutes les unités et tous les diviseurs de zéro dans l’anneau
F3 [x]/((x − 1)(x2 + 1)).
Solution :
1. We have
f (x) = x · g(x) + x3 + x
g(x) = (x − 2) · (x3 + x) + x2 + 1
x3 + x = x · (x2 + 1).
The last nonzero remainder is x2 + 1, hence
gcd(f (x), g(x)) = x2 + 1.
Substituting backwards we obtain
x2 + 1 = g(x) − (x − 2) · (x3 + x) = g(x) − (x − 2) · (f (x) − x · g(x))
= (x2 − 2x + 1) · g(x) + (2 − x) · f (x).
So a(x) = x2 − 2x + 1 and b(x) = 2 − x is a solution.
2. Applying the Euclidean algorithm to the polynomials x5 − x and x4 + 1 we obtain :
x5 − x = x · (x4 + 1) + x
x4 + 1 = x3 · x + 1
x = x · 1 + 0.
The last nonzero remainder is 1 so the polynomials x5 − x and x4 + 1 are relatively prime.
Substituting backwards, we obtain
1 = x4 + 1 − x3 · x = x4 + 1 − x3 · ((x5 − x) − x · (x4 + 1))
= −x3 · (x5 − x) + (x4 + 1) · (x4 + 1).
So we obtain
(x5 − x) · (−x3 ) ≡ 1 (mod x4 + 1).
Multiplying both sides with the polynomial x2 + 1 we get
(x5 − x) · (−x3 · (x2 + 1)) ≡ x2 + 1 (mod x4 + 1).
So a possible solution is
c(x) = −x3 · (x2 + 1) = −x5 − x3 .
3. By the division theorem, in every class of F3 [x]/((x − 1)(x2 + 1)) there is a polynomial of
degree less than 3. Moreover [p(x)] is a unit if and only if gcd(p(x), (x − 1)(x2 + 1)) = 1.
All nonzero elements in F3 [x]/((x − 1)(x2 + 1)), which are not units, are zero divisors. So
the units are precisely the classes of polynomials of degree less than 3 which are relatively
prime to (x − 1)(x2 + 1). The polynomials x − 1 and x2 + 1 are both irreducible in F3 [x]
(they have degree 1 and 2 and do not have any root in F3 ). So a polynomial is relatively
prime to (x − 1)(x2 + 1) if and only if it is not divisible by x − 1 and x2 + 1. The multiples
of x − 1 of degree less than 3 are the following :
(x − 1), 2 · (x − 1), x · (x − 1), 2x · (x − 1), (x + 1) · (x − 1),
(x + 2) · (x − 1), (2x + 1) · (x − 1), (2x + 2) · (x − 1).
Similarly the multiples of x2 + 1 of degree less than 3 are the following :
x2 + 1, 2 · (x2 + 1).
So the zero divisors of the ring F3 [x]/((x − 1)(x2 + 1)) are given by
[(x − 1)], [2 · (x − 1)], [x · (x − 1)], [2x · (x − 1)], [(x + 1) · (x − 1)],
[(x + 2) · (x − 1)], [(2x + 1) · (x − 1)], [(2x + 2) · (x − 1)], [x2 + 1], [2 · (x2 + 1)].
All remaining nonzero polynomials of degree less than 3 are relatively prime to (x−1)(x2 +1)
and hence units. So the units of F3 [x]/((x − 1)(x2 + 1)) are :
[1], [2], [x], [2x], [x + 1], [2x + 2], [x2 ], [2x2 ], [x2 + x], [2x2 + 2x], [x2 + x + 2], [2x2 + 2x + 1],
[x2 + 2x + 1], [2x2 + x + 2], [x2 + 2x + 2], [2x2 + x + 1].
Question 4 (english, 20 points)
1. Show that x2 + x + 1 is the only irreducible polynomial in F2 [x] of degree 2.
2. Construct a field K with 16 elements.
3. Show that the multiplicative group K ∗ is cyclic.
Question 4 (français, 20 points)
1. Montrer que x2 + x + 1 est le seul polynôme irréductible de degré 2 dans F2 [x].
2. Construire un corps K à 16 éléments.
3. Montrer que le groupe multiplicatif K ∗ est cyclique.
Solution :
1. All polynomials of degree 2 in F2 [x] are given as follows :
f1 (x)
f2 (x)
f3 (x)
f4 (x)
=
=
=
=
x2 ,
x2 + 1,
x2 + x,
x2 + x + 1.
A polynomial of degree 2 over F2 is irreducible if and only if it has no root in F2 . We have
f1 (0) = f3 (0) = f2 (1) = 0. So the polynomials f1 , f2 and f3 are reducible in F2 [x]. On the
other hand, f4 (0) = f4 (1) = 1 6= 0, so the polynomial f4 has no root in F2 and hence is
irreducible over F2 . We also see that it is the only irreuducible polynomial of degree 2 in
F2 [x] since all other polynomials of degree 2 are reducible.
2. To construct a field with 16 = 24 elements, we have to find an irreducible polynomial in
F2 [x] of degree 4. Consider the polynomial p(x) = x4 + x + 1. We want to show that p(x) is
irreducible. If p(x) would be reducible, then it necessarily would have an irreducible factor
of degree 1 or 2. So it suffices to show that no irreducible polynomial of degree 1 or 2 divides
p(x). The irreducible polynomials of degree 1 are x and x + 1. From part 1. we know that
the only irreducible polynomial of degree 2 is x2 + x + 1. We can check easily that p(x) is
not divisible by x, x + 1 and x2 + x + 1 (it has remainder 1 when divided by anyone of
them), so hence it has to be irreducible over F2 . A field with 16 elements is therefore given
by
K = F2 [x]/ < x4 + x + 1 > .
3. |K ∗ | = 16 − 1 = 15. To show that the multiplicative group K ∗ is cyclic, we have to find
an element of order 15. By Lagrange’s theorem, the order of any element of K ∗ has to
divide |K ∗ | = 15, so the possibilities for the orders of an element are 1, 3, 5 and 15. Hence it
suffices to find an element which has order different from 1, 3 and 5. Consider the element
a = [x]. We have a4 + a + 1 = 0, hence a4 = a + 1. We compute the consecutive powers of
a:
a1
a2
a3
a4
a5
=
=
=
=
=
a
a2
a3
a+1
a4 · a = (a + 1) · a = a2 + a.
Since a, a3 and a5 are all different from 1, the element a does not have order 1, 3 or 5. So
it necessarily has order 15 and hence the group K ∗ is cyclic.