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INTRODUCTION TO ALGEBRA II
MIDTERM 1
SOLUTIONS
Do as much as you can. Do not get stuck on a question: if you do not see
a solution after some time, go on to the next question. You are allowed (and
in some cases, you should) use results of the preceding questions to solve a
question (and so, even if you didn’t solve the preceding questions).
1.– (10pts) Solve the congruence 6x ≡ 12 (mod 16)
Answer: x ≡ 2 (mod 16) or x ≡ 10 (mod 16).
2.– (10pts) Compute φ(18). Compute 51000 (mod 18).
Solution: The Euler indicator φ(18) is by definition the number of integers
between 1 and 18 that are prime to 18. Those integers are 1, 5, 7, 11, 13, 17.
So φ(18) = 6.
Other method: 18 = 2 × 32 . Hence φ(18) = φ(2)φ(32 ) (formula seen in
class). Moreover φ(2) = 2 − 1 = 1. And φ(32 ) = 3(3 − 1) = 6. Hence
φ(18) = 6.
By Euler theorem, 5φ(18) = 56 ≡ 1 (mod 18). Since 1000 = 6 × 166 + 4 we
have 51000 ≡ 54 (mod 18) and 54 ≡ 252 ≡ 72 ≡ 49 ≡ 13 (mod 18).
3.– (10pts)
a.– Find a generator of the multiplicative group Z∗19 .
Solution: The powers of 3 modulo 19 are 3, 9, 8, 5, 15, 10, 11, 14, 4, . . . .
No need to go on, since we already see that the order of 3 in the multiplicative
group Z∗19 (which has 18 elements) is at lest 10. Since by Lagrange’s theorem
this order divides 18, it is 18, and 3 is a generator.
b.– How many generators are there in this group ? (it is not necessary to
compute them all, though that’s certainly a method to answer the question).
Solution: We have seen in class, and actually proved again in the last
∗
question, that the group Z19
is cyclic, hence isomorphic to the additive group
Z18 . The generators for this latter group are the integers between 1 and 18
prime to 18. There are 6 of them.
4.– (10pts) Let p be a prime number different from 2 and 5. Show that
p divides an integer with only 9 in its decimal writing (that is 9999 . . . 999.)
You may use Fermat’s little theorem.
Solution:
By Fermat Last theorem, since 10 is prime to p we have
10p−1 − 1 ≡ 0 (mod p). That means that p divides 10p−1 − 1 which is the
number written in decimal notations with p − 1 nines.
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INTRODUCTION TO ALGEBRA II MIDTERM 1 SOLUTIONS
5.– (10pts) Let R be the set of rational numbers that can be written 2an
with a ∈ Z and n ∈ N. Show that R is a subring with unity of Q. Determine
the units of R. What is the fraction field of R ?
0
m
n 0
a
Solution: 2an + 2am = 2 2a+2
so R is closed under the addition of Q.
n+m
0
0
aa
a a
= 2n+m so R is closed under the multiplication Q. Moreover 1 = 210 is
2n 2m
in R so R has a unity. Hence R is a subring with unity of Q.
Let 2an be an element of R. When is it a unit? By definition, it is if one can
aa0
multiply it by an other element and get 1. That is, if 2n+m
= 1 for some a0
and some m. so in particular aa0 = 2n and this implies that a is a power of 2,
and that 2an = 2t where t is some integer (of any sign). Conversely, it is clear
that 2t is a unit, so the unit are all the powers of 2, that is the 2t for t ∈ Z.
Since R ⊂ A and Q ia a field, the fraction field K of R is a subfield of Q.
Since R contains Z, then the fraction field K of R contains the all fractions
of elements of Z (with non-zero denominator), that is Q. So K = Q.
6.– (10pts) Let K be a field. Let P (X) and Q(X) be polynomials in K[X],
with Q(X) 6= 0. We have seen in class that there exist R(X) and S(X) in
K[X] with
(1)
deg R < deg Q
and such that
(2)
P (X) = Q(X)S(X) + R(X)
Complete this result by proving that R(X) and S(X) that satisfies (1) and
(2) are actually unique. (That is, if R2 (X) and S2 (X) are two polynomials
that also satisfies (1) and (2), then R(X) = R2 (X) and S(X) = S2 (X).)
Solution: Substracting
P (X) = Q(X)S(X) + R(X)
and
P (X) = Q(X)S2 (X) + R2 (X)
gives
Q(X)(S(X) − S2 (X)) = R(X) − R2 (X).
The right hand side has degree stricitly less than deg Q. The left hand side
has degree at least deg Q, except if S(X) − S2 (X) = 0. So we see that
S(X) = S2 (X), and from this, that R(X) = R2 (X).
7.– (10pts) Let L be a field, and K be a subfield of L. Let P (X) and Q(X)
be two polynomials in K[X] with Q(X) 6= 0, and assume that there exists
a polynomial S(X) ∈ L[X] such that P (X) = Q(X)S(X). Then show that
S(X) is in K[X].
Solution: We have P (X) = Q(X)S(X) + R(X) with R(X) = 0. On the
other hand, we can perform the euclidean division of P (X) by Q(X) in the
ring K[X], and we get P (X) = Q(X)S2 (X) + R2 (X) with deg R2 < deg Q,
and S2 (X), R2 (x) ∈ K[X]. By the unicity of euclidean division in L(X) (that
is, the question above) we see that S(X) = S2 (X) so S(X) is in K[X].
INTRODUCTION TO ALGEBRA II
MIDTERM 1
SOLUTIONS
3
8.– (20pts) In this question we want to show that a polynomial P ∈ R[X]
with deg P ≥ 3 is never irreductible (in R[X]). We admit the fundamental
theorem of algebra which says that P (X) has a root α in C.
a.– If α ∈ R, show that P (X) is not irreducible.
Solution: If P (X) has a root in R, it is reducible over R since its degree
is > 1.
b.– Assume α 6∈ R now and below. Show that ᾱ is also a root of P (X).
P
Solution: Let P (X) = di=0 ai X i . Then
P (ᾱ) =
d
X
ai ᾱi
i=0
=
d
X
āi ᾱi since ai ∈ R
i=0
= P (α)
= 0
c.– Show that there is a polynomial S(X) ∈ C[X] such that P (X) =
(X − α)(X − ᾱ)S(X).
Since α is a root of P , we have P (X) = (X − α)S1 (X) for some S1 (X) ∈
C[X]. We have 0 = P (ᾱ) = (ᾱ − α)S1 (ᾱ) and since ᾱ − α 6= 0 (α is not real),
ᾱ is a root of S1 , so S1 (X) = (X − ᾱ)S(X) and the results follows.
d.– Show that the polynomial (X − α)(X − ᾱ) is in R[X].
Solution: It is equal to X − (α + ᾱ)X + αᾱ so it is clearly in R.
e.– Conclude that P (X) is not irreducible in R[X]. Solution: It is divisible
by the quadratic (i.e. degree 2) real polynomial (X − α)(X − ᾱ) and it has
degree > 2. It is thus reducible.
9.– (10pts) Let p be a prime number. Show that X p+1 − 9 is not irreducible
in Zp [X].
Solution: (many ways to do that. Here is one.) If p = 3, 3p+1 = 3p 3 ≡ 3 ×
3 ≡ 9 (mod 3) using Fermat Little theorem, so 3 is a root and the polynomial
is not irreducible.