Algebra Prelim Solutions, Fall 2007
 Let G be a simple group of order 168.
The number of Sylow 7subgroups of G is congruent to 1 mod 7 and
divides
168/7 = 24. This number cannot be 1 because that would
mean that G has exactly one Sylow 7subgroup, consequently G
would have a normal Sylow 7subgroup and we would deduce that G is
not simple, contrary to the hypothesis. It follows that G has
exactly 8 Sylow 7subgroups. Also by Lagrange's theorem, two
distinct Sylow 7subgroups must intersect in the identity. Since any
element of order 7 is contained in a Sylow 7subgroup and there are 6
elements of order 7 in each Sylow 7subgroup, we deduce that there
are 8*6 = 48 elements of order 7 in G.
 Note that
ρ = i. The following are easy to check:
Q(√2)≠Q√3. Thus
Q(√2,√3) is a Galois extension of degree 4
over
Q.
Let K be the splitting field of x^{4}  2 over
Q and let
L be the splitting field of x^{2}  3 over
Q.
The roots of x^{4}  2 are
± ^{4}√2,
±i ^{4}√2, hence
K is a Galois extension of degree 8 over
Q, and has
maximal real subfield of degree 4 over
Q, namely
Q( ^{4}√2). Since this subfield is not normal over
Q, we deduce that
Q(√2,√3) is not
contained in K. Therefore
K∩L = Q, and we deduce
that
K∩L(i) = Q(i). The Galois
group of
L/Q has order two and is therefore isomorphic
Z/2Z. Also the Galois group of
K/Q is
a group of order 8 and not every subgroup is normal, because
Q(^{4}√2) is not normal over
Q. and we
deduce that this group is isomorphic to the dihedral group D_{8} of
order 8. Finally
Gal(K/Q(i))≌Z/4Z, being generated by the automorphism
determined by
^{4}√2  > i ^{4}√2.
 (a)
 The Galois group of
(x^{4} 2)(x^{2}  3) over
Q is
Gal(K/Q)×Gal(L/Q)≌D_{8}×Z/2Z. The Galois group of
(x^{4} 2)(x^{2}  3)
over
Q(i) is
Gal(K/Q(i))×Gal(L(i)/Q(i))≌Z/4Z×Z/2Z.
 (b)

Q(i) is Galois over
Q because it is the
splitting field of x^{2} + 1 over
Q.
 (c)
 Yes, because
Gal(LK/Q(i)) has nontrivial normal
subgroups.
 Since
0 > A  ^{f}> B  ^{g}> C > 0 is split
exact, there exists
h : B > A such that hf = 1_{A}, the
identity map on A. Then
(1_{D} h)(1_{D} f )= 1_{D} hf = 1_{D} 1_{A} = 1. Thus if
x∈D A and
(1_{D} f )(x) = 0, then
(1_{D} f )(1_{D} h)(x) = 0, consequently 1(x) = 0 and we conclude that x = 0, as required.
 Certainly S^{1}R is an integral domain, since it is a subring of
the field of fractions of R, so we need to prove that every ideal
of S^{1}R is principal.
Let
I < S^{1}R and let
J = I∩R. Then
J < R,
so J = xR for some
x∈R. Obviously
xS^{1}R⊆I, so
it remains to prove that
xS^{1}R⊇I. However if
y∈I, then
sy∈I∩R = J where
s∈S and hence we may
write sy = xr for some
r∈R. Therefore
y = s^{1}(sy) = xs^{1}r∈xS^{1}R and the result is proven.
 Let G be a group of order
2^{4}·11^{2}. The number of Sylow
11subgroups is congruent to 1 mod 11 and divides 16, consequently
there is exactly one Sylow 11subgroup; call this Sylow 11subgroup
H. Then
H < G. Now G/H and H are pgroups for p = 2
and 11 respectively, and pgroups are solvable (even nilpotent).
However the property of being solvable is closed under extensions,
that is H and G/H solvable implies G is solvable, which is the
required result.
 (a)
 Apply Eisenstein's criterion for the prime 3.
 (b)
 We know that f is irreducible (from (a)) and that g is
irreducible (use Eisenstein for the prime 2). Since
Q[x]
is a PID, we see that (f ) and (g) are maximal ideals of
Q[x]. Furthermore
(f )≠(g), because f and g are
not scalar multiples of each other. It now follows from the Chinese
remainder theorem that
Q[x]/(fg)≌Q[x]/(f )×Q[x]/(g), a product of two fields. The dimension
over
Q of these two fields are the degrees of the
polynomials f and g, that is 4 and 2 respectively.
 (a)
 Since
1·0 = 0, we see that
0∈t(X). Next suppose that
x, y∈t(X). Then there exist
r, s∈R\ 0 such that
rx = 0 = sy and we have
(rs)(x + y) = 0. Since
rs≠ 0 because
R is an integral domain, we conclude that
x + y∈t(X). Finally
suppose that
x∈t(X) and
r∈R. Then there exists
s∈R\ 0 such that sx = 0 and consequently s(rx) = 0. This
shows that
rx∈t(X) and we have established that t(X) is an
Rsubmodule of X.
 (b)
 Write T = t(X) and let
x∈t(T); we want to prove that
x∈T. Since
x∈t(T), there exists
s∈R\ 0 such that
sx∈T, and then there exists
t∈R\ 0 such that
t(sx) = 0. It follows that (st)x = 0 and since
st≠ 0
because R is an integral domain, we conclude that
x∈T as
required.
 (c)
 Because t(X/t(X)) is cyclic,
t(X/t(X))≌R/I for some
I < R. But
t(X/t(X)) = 0 by (b), hence I = 0 and we
deduce that
X/t(X)≌R. Since R
is a projective Rmodule,
0 > t(X) > X > X/t(X) > 0 splits, in particular
X≌t(X)⊕R, as required.
Peter Linnell
20070811