Fall 1993 Algebra Prelim Solutions
- It is clear that Re is a left R-submodule of R,
so we need to prove
it is projective. It will be sufficient to show that Re is a direct
summand of a free R-module. Since R(1 - e) is also a left
R-submodule of R, the result will be proven if we can show that
R = Re R(1 - e). If
re R(1 - e), then
re = s(1 - e) for some
s R and so
re = ree = s(1 - e)e = 0. This shows that
Re R(1 - e) = 0. Finally if r R, then
r = r(e + 1 - e) = re + r(1 - e), so
R = Re + R(1 - e).
- Since we are in characteristic zero, everything is separable
so we may use the theorem of the
primitive element. This tells us that there exists
K = F(a). If p is the minimal polynomial of
a over F, then
- Since p is the minimal polynomial, it is monic and irreducible
in F[X]. Also we are in characteristic zero, so p is separable.
This means that when we write
p = p1...pn in K[X] where the
pi are monic irreducible polynomials, the pi are distinct.
c = f[X] + (p) where
f K[X]. Since c2 = 0, we
f2 (p). Thus
f2 = p1...pnq for some
q K[X]. By uniqueness of factorization, we have pi divides f for
all i and we deduce that f (p). Therefore c = 0 as
- Suppose the action of M
2() has a finite orbit with at
least two elements. Using
the formula that the number of elements in an orbit is the index of
the stablizer of any element in that orbit, we see that M
2() has a nontrivial subgroup H of finite index. Then qM
2() H for some positive integer q.
a = q(a/q), which shows that qM
2() = M
conclude that M
2() = H and the result follows.
- To check that we have an action, we must show that
(gh)*l = g*(h*l) for all g, h GL
2(). This is true
||det(gh)l/| det(gh)| = det(g)det(h)l/(| det(g)|| det(h)|)
||det(g)(h*l)/| det(g)| = g*(h*l).
Finally we see that all orbits have the form
so in particular there are finite orbits which are not singletons.
- By Sylow's theorem a group of order 34 has a normal subgroup of
order 17, hence G has a normal subgroup H of order 17. Again by
Sylow's theorem, this subgroup is the unique subgroup of G which has
order 17. Since we
are in characteristic zero, everything is separable and so L is a
Galois extension of LG. Therefore there is a one-one correspondence
between the subfields of L containing LG and the subgroups of
G. This correspondence has the property that if A is a subgroup
of G, then the dimension of L over LA is | A|. The result
follows by setting K = LH.
- Suppose S is not a field. Then it has a nonzero prime ideal
P. Note that S/P is an integral domain.
S[X]/P[X] (S/P)[X], we see that S[X]/P[X] is an
integral domain which is not a field. We deduce that P[X] is a
nonzero nonmaximal prime ideal of S[X]. But nonzero prime ideals in
a PID are maximal and since we are given that S[X] is
a PID, we now have a contradiction and the result
- First H has an identity, namely the zero homomorphism defined
by 0(a) = 0 for all a A. If
f, g H, then
|(f + g)(a + b)
||f (a + b) + g(a + b) = f (a) + f (b) + g(a) + g(b)
||f (a) + g(a) + f (b) + g(b) = (f + g)(a) + (f + g)(b)
which shows that f + g is a homomorphism, and so f + g H.
f, g, h H, then
|((f + g) + h)(a)
||(f + g)(a) + h(a) = f (a) + g(a) + h(a)
||f (a) + (g + h)(a) = (f + (g + h))(a),
(f + g) + h = f + (g + h) which establishes the associative law.
Finally for f H, the inverse of f is - f, where
(- f )(a) = - f (a). Since
(- f )(a + b) = - f (a + b) = - f (a) - f (b) = (- f )(a) + (- f )(b), we see that - f H, and we have established that H is a group.
- We first show that H is torsion free. If f H has order
f (a) 0 for some a A. But then
(nf )(a) = n(fa) 0, a contradiction. Therefore
the subgroup generated by
f1,..., fm is a finitely generated torsion free abelian group, so by
the fundamental structure theorem for finitely generated abelian
groups it is free.
- A 2 by 2 matrix with entries in
be invertible if and only if its columns are linearly independent over
/p. So there are p2 - 1 choices for the
first column (we cannot choose (0, 0) for the first column) and p2 - p choices for the second column (we cannot choose the vector in the
first column). It follows that
| G| = (p2 - 1)(p2 - p).
- Note that H is a Sylow p-subgroup of G, so the number of
conjugates of H is congruent to 1 modulo p. Therefore 1 is
congruent to 8 modulo p, which can only happen if p = 7. In the
case p = 7, we have by Sylow's theorem that the number of conjugates
of H in G is congruent to 1 modulo 7, which is of course congruent
to 8 modulo 7. Thus the answer is p = 7.