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
(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.