# Math 5126 Second Test Review

** Solution to Problem 1 **on Sample Test no. 1 We are
working in the group of units of

(Z/625Z)^{x}, which is a cyclic group of order θ(625) = 500. Since x^{550}
x^{50} mod 625,

we need to prove that 499^{10}
1 mod 625, equivalently 126^{10}1
mod 625. However

126^{5}
(1+5^{3})^{5}
1 mod 5^{4}
and the result follows.

**Solution to Problem 2 on Sample Test no. 1** Write the primes not equal to
the characteristic

p of K as p_{1}, p_{2}, . . . (an infinite number of primes).
For each i, let K_{i} be an extension

of degree p_{i} and let α_{i}∈ K_{i}−K.
Then [K(α_{1}, . . . ,α_{n})
: K] = p_{1} . . . p_{n}, which is not divisible

by p. Therefore K(α_{1}, . . . ,α_{n})
is a finite separable extension of K and hence is generated by

a single element, say α. Since the minimal polynomial
of α over K is irreducible of degree

p_{1} . . . p_{n}, the result is proven.

**Solution to Problem 3 on Sample Test no. 1**

(a) The roots of f are . Since the product of
all the roots of f is 7, if the product

of two of the roots is a rational number, then so is the product of the other
two roots.

Therefore if the product of two of the roots is a rational number , we may assume
that

one of the roots is then we can just go through the three possibilities for the

other root and check that the resulting product is not a rational number .

Now suppose f is not irreducible. Clearly it does not have a linear factor , so
the only

possibility is that it is the product of two quadratic factors ; let g ∈ Q[x] be
one of these

factors . Then the constant coefficient of g is the product of the roots of g,
consequently

the product of two of the roots of f is a rational number, which contradicts the
previous

paragraph.

(b) Suppose. Sincesatisfies
a quadratic over

we may write
where . Squaring , we

obtain

.

From (a), we see that and we deduce that
either a or b = 0. Clearly

we can’t have b = 0 (otherwise , so a = 0 and
we deduce that

with ).
Multiplying both sides by, we conclude

that), which is not the case. This
contradiction finishes the proof of

(b).

(c) Since f has degree 4, its Galois group G will be isomorphic to a subgroup of
S_{4}. Also

we see from (b) that the splitting field of f has degree 8 over Q. We conclude
that G

is isomorphic to a subgroup of order 8 in S _{4}. Since 8 is the order of
a Sylow subgroup

of S_{4}, there is only one such group up to isomorphism (any two Sylow
subgroups are

conjugate), and such a group is nonabelian (in fact isomorphic to D_{8}).

**
Solution to Problem 1 on Sample Test no. 2 **The only way for there to be a
proper

intermediate field is the case f is irreducible, and then [Q(α) : Q] = 4. Suppose K is a

proper subfield, so [K : Q] = 2, and let g denote the minimal polynomial of α over K. Then

g is of the form (x−α) (x− β), where β is one of the other 3 roots of f . If β is complex,

then so are the coefficients of g in degrees 0 and 1, which means that they cannot be in K,

because K ⊆R. We deduce that there is at most one choice for β, namely the other real

root. Since K is generated over Q by the coefficients of g, we conclude that there is at most

one possibility for K and the result follows.

**Solution to Problem 2 on Sample Test no. 2**Let K be the splitting field of x

^{n}−x over

F

_{p}. Then K = F

_{n}, every element of K satisfies x

^{n}−x, and K has degree q over F

_{p}. Let

a∈K−F

_{p}and let f denote the minimal polynomial of α over F

_{p}. Then f has degree q and

divides x

^{n}−x, because α satisfies x

^{n}−x. Thus f is an irreducible polynomial of degree q

which divides x

^{n}−x, as required.

**Solution to Problem 3 on Sample Test no. 2**Let f (x) = x

^{5}+5x

^{3}−20x+5 and let G

denote the Galois group of f over Q. Note that f is irreducible over Q by Eisenstein for the

prime 5, so all the roots of f are distinct. As x→−∞, f (x)→−∞, and as x→∞, f (x)→∞.

Also f (−1) = 19 and f (1) = −9, which shows that f has a least 3 real roots. Moreover

f '(x) = 5x

^{4}+15x

^{2}−20 = 5(x

^{2}+4)(x

^{2}−1), consequently f has only 2 turning points and

we deduce that f has exactly 3 real roots . Thus complex conjugation is an element of G

which fixes 3 of the roots of f and interchanges the other 2. Thus when we consider G

as a subgroup of S

_{5}(i.e. a permutation group on the 5 roots of f ), it has a transposition.

Furthermore 5 | |G| because f is irreducible of degree 5, hence G contains a 5-cycle. Since

S

_{5}is generated by any transposition and any 5-cycle, we deduce that G ≅ S

_{5}.

Solution to Problem 1 on Sample Test no. 3

Solution to Problem 1 on Sample Test no. 3

(a) Since K

^{x}is cyclic, it is generated by a single element, which we will call α. Then

certainly K = F(α), and if n is the order of α, then α

^{n}= 1 and in particular α

^{n}∈ F.

(b) No. One way to see this is that the pth power map π defined by π(k) = k

^{p}is an automorphism

of K, because K is a finite field of characteristic p. Since π(F)⊆ F, we

deduce that π(F) = F and hence π(a) ∉ F for all α ∉ F.

Test on Wednesday, April 2.

Material most of sections 14.1–14.7

One of the problems will be identical to one of the ungraded homework problems and one

of the problems will be identical to one of the sample test problems.

Prev | Next |