Download Algebra Quals Fall 2012 1. This is an immediate consequence of the

Algebra Quals Fall 2012
1. This is an immediate consequence of the algorithm for putting a matrix
in smith normal form.
2.
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(pn − 1) · · · (pn − pn−1 ). We get that
Let G = SLn (Fp ). WE have |G| = p−1
P
P
n−1
|G| = pk m where k = i=1 i. Let Nk = {A = j≥i+k aij Eij }, matrices that
have 1’s on the diagonal, 00 s below the diagonal, and above the diagonal, there
are only nonzero entries on superdiagonal level k and higher. Notice that using
the relation Eij Ek` P
= δjk Ei` , we get that Nk Nm ⊂ Nk+m . Let Pk = I + Nk .
Using (I − A)−1 =
Ai for nilpotent A, we get that Pk are subgroups. Also,
using this relation we find that Pk is normal in Pk−1 (we reversed the ordering
given in the problem). It is then easy to see that Pi−1 /Pi is abelian.
3.
a. Notice that Symi (M ) is the quotient T i (M )/a where T i (M ) = M ⊗i and
a is the ideal generated by m1 ⊗ . . . ⊗ mi − mσ(1) ⊗ . . . ⊗ mσ(i) , where mj ∈ M
and σ ∈ Si . We claim that given a basis e1 , . . . , en of M , then eJ = ej1 ej2 . . . eji
with jk ≤ jk+1 is a basis of Symi (M ). Indeed, by the quotient, since they
have different indices, we see that they are linearly independent. Also, since the
quotient T i (M ) → Symi (M ) is surjective, given anything in Symi (M ), we may
liftit back to T i (M ), find its expression in the tensor basis, and then find how
it is written in terms of the eJ . Thus Symi (M ) is free.
For the pairing, the linearity of ` ∈ M ∗ implies bilinearity of the pairing.
Also, to show uniqueness, it is easy to check that it is uniquely definined on basis
vectors, e∗I (eJ )∗ is the sum of the indices that are in common, with multiplicity.
So it’s uniquely defined on basis P
vetcors, hence unique.
b. Notice that e∗1 · · · e∗i (eI ) = j (number of copies of ej in I). Also, (e∗j )i (eI ) =
P
i · (number of copies of ej in I). So e∗1 · · · e∗i − j 1j (e∗j )i is 0 on everything. So
it’s a typo. so you need a product. in this case it’s trivial.
c. Choose a basis. Notice that since it’s a sequence of free R-modules, it
splits, so M = M 0 ⊕ M 00 .Then the isomorphism is obvious, writing it in this
basis. For naturality, notice if we have an iso or exact sequences 0 → N 0 →
N → N 00 → 0, then it induces a map that preserves the isomorphism.
4.
n
a. Let ei be the standard
basis vectors of RP
. Then there exists yi such that
P
f (yi ) = ei write yi = j aij ej , and f (ej ) = k bjk ek . Then this gives that f
corresponds to the matrix multiplication by B = (bjk ), and we have an A such
1
that AB = id, so det B 6= 0, so it’s injective.
b. We have an ascending chain ker(f ) ⊂ ker(f 2 ) ⊂ . . . since R is noetherian,
it stabilizes, so ker(f n ) = ker(f n+1 ) for some n, this means that fn : f n R →
f n R has ker fn = 0. but R is surjective, so fn = f , so ker f = 0.
c. R = C[x1 , x2 , . . .], f (x1 ) = 0, f (xi ) = xi−1 for i ≥ 2.
5.
a. We have
Q[21/3 ] ⊗Q Q[21/3 ]
=
Q[21/3 ] ⊗Q Q[X]/(x3 − 2)
= Q[21/3 ] ⊗Q Q[X] ⊗Q[X] ⊗Q[X]/(X 3 − 2)
= Q[21/3 , X] ⊗Q[X] Q[X]/(X 3 − 2)
= Q[21/3 , X]/(X 3 − 2)
= Q[21/3 , X]/(X − 21/3 ) × Q[21/3 , X]/(X 2 + 21/3 X + 41/3 )
Using CRT and various properties of tensor products. We see X 2 + 21/3 X +
4
is irreducible in Q[21/3 ] = Q(21/3 ), so this is a product of fields, so has two
prime ideals, Q[21/3 , X]/(X − 21/3 ) × 0 and 0 × Q[21/3 , X]/(X 2 + 21/3 X + 41/3 ),
and any element generates the entire ideal. Notice that we have
1/3
(21/3 ⊗ 1 − 1 ⊗ 21/3 )(41/3 ⊗ 1 + 21/3 ⊗ 21/3 + 1 ⊗ 41/3 ) = 0
so the generators of the two ideals are (21/3 ⊗ 1 − 1 ⊗ 21/3 ) and (41/3 ⊗ 1 +
2 ⊗ 21/3 + 1 ⊗ 41/3 ).
b. We use a similar process as above, except we have that X 3 − 2 factors as
(X − α)(X − 2α)(X + 3α) in F7 [α, X], so we have
1/3
F7 [α] ⊗F7 F7 [α] = F7 [α] × F7 [α] × F7 [α]
Then we find
(α ⊗ 1 − 1 ⊗ α)(α ⊗ 1 − 1 ⊗ 2α)(α ⊗ 1 + 1 ⊗ 3α) = 0
and each of the three factors in the product generates a prime ideal.
6
i. Consider the SES 0 → J →α B →β A → 0, where J, B, A are considered
as R modules (not necessarily as algebras). Then by right exactness of tensor
product, we have that J ⊗ R0 →α0 B ⊗ R0 →β 0 A ⊗ R0 → 0 is exact, which
proves the desired stuff. Alternatively, it is easy to see that it is surjective. Let
I = im α0 . It is easy to see that I ⊂ ker β 0 . Define a map A ⊗ R0 → B ⊗ R0 /I,
such that a ⊗ r goes to x ⊗ r(modI), where x → a. It is easy to check that this
2
is well defined, and we have that B ⊗ R0 /I → A ⊗ R0 → B ⊗ R0 is the identity,
hence β 0 /I is injective, so I = ker β 0 .
ii. If A is R-flat, then 0 → J ⊗ R0 → B ⊗ R0 → A ⊗ R0 → 0 is exact (this is a
theorem that can be proved using snake), so the first thing follows immediately.
Let J = (x), B = C[x], A = C[x]/(x2 ) and R0 = C[x]/(x2 ), with J → B
given by x 7→ x2 .
iii.
Say that 0 → J → B → A → 0 is exact, and A is R-flat. Since A is R-flat,
when we tensor with R/mR , we get an exact sequence 0 → J/mR J → B/mR B →
A/mR A → 0. Since f is an isomorphism mod mR , we have that J/mR J = 0.
Notice that we have a surjective map J/mR J → J/mB J, so J/mB J = 0. Thus
by Nakayam’s lemma, J = 0. So it’s an iso.
For a counterexample, let R = C[x](x) , B = C[x, y](x,y) , A = C[x, y](x,y) /(x2 ),
and J = (x)(x) , where J → B is given by x 7→ x2 . Mod mR the map reduces to
C[y](y) → C[y](y) , which is an iso, but the original map is not an iso.
7.
i. Say that f = X q − b is reducible over E, with a a root of g. Then
1, a, . . . , adeg g−1 are a E
of E 0 /E, so [E 0 : E] < q. Conversely, if 1, a, . . . , ad−1
Pbasis
d
i
are a basis, then a − ci a = 0 gives an equation that a satisfies, which must
divide f since a is a root of f .
Consider the equation aq = b, let c = N (a), so cq = bd . Since gcd(q, d) = 1,
there exist r, s such that sq + rd = 1, so b = bsq+rd = bsq crq = (bs cr )q , so b is a
q − th power, and (a/bs cr )q = 1 , it is primitive since a 6∈ E, and bs cr ∈ E, so
a/bs cr 6= 1.
ii. Since K/E is Galois, it is separable, so there exists an α such that
K = E(α). Let f be the minimal polynomial of α over E. Notice that it is
irreducible over E 0 , because if it wasn’t, let β be a root of one of the factors.
Then E(β) is a subfield of K, which is impossible since [K : E] is prime. Then
KE 0 = E 0 (α) and is degree p over E 0 . Since K contains all the roots of f , KE 0
also contains all the roots of f , so KE 0 /E 0 is normal.
To show KE 0 /E 0 is separable, we have K/E is Galois, hence separable, so
f has no multiple roots, so since KE 0 = E 0 (α), KE 0 /E 0 is separable.
To show that the restriction of Galois groups is an isomorphism. Notice that
if σ|K was the identity, then it fixes α and E 0 , so fixes KE 0 /E 0 , hence is the
identity, so the map is injective, and since they have the same order it’s an iso.
iii. Say that we can. Let K 0 be the preimage of the first field above E in the
radical tower, wlog, replace K by K 0 . Since it’s a radical extension, it’s galois,
so apply part ii with E 0 = E(ζp ) (since [Q(ζp ) : Q] = p − 1, we can’t embed K
into E(ζp ). This implies that [KE(ζp ) : K] = [E(ζp ) : E], but by part i we have
that ζp ∈ K so the left side is 1, so ζp ∈ E, which is impossible since E ⊂ R.
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8.
i. Let a be an ideal in A[[x]]. Let f1 ∈ a of minimal degree, and let fi be in
a of minimal degree that is not in hf1 , . . . , fi−1 i, and let ci denote the leading
coefficient of fi . We assume that a is not finitely generated for a contradiction.
Since A is noetherian, hc1 , c2 , . . .i is finitely generated, say by c1 , . . . cn (reindex
so it all works out nicely).P Consider f ∈Pa not in hf1 , . . . , fn i, with leading
coefficient d1 . Then d1 =
r1i ci , so f − r1i fi xe1i has a higher degree than
f and is in a. Then we consider the leading term of this new polynomial, d2 ,
and similarly get coefficients r2,i and exponenets e2,i . continue
this procedure.
P P
eji
= 0. Thus
Notice that ej,i > ej−1,i , so that we have that f − i fi
j rji x
a = hf1 , . . . , fn i.
ii. By induction, R = A[[x1 , . . . , xn ]] is noetherian. We have a map R → Â
sending xi 7→ ai . This is the map we want. It’s easy to check that it’s surjective.
Since  is a quotient of a noetherian ring, it’s noetherian.
9.
i. Let A ∈ EndG V . Pick an eigenvector v1 of A with eigenvalue λ. Consider
v2 ∈ ker(A − λ)2 , etc so we have a basis of a subsace of V in which A has λ
on the diagonals and all nonzero entries are above the diagonal. If this is all
of V , good, otherwise repeat. Say that when we repeat, we get an eigenvector
W with eigenvalue µ 6= λ. Then V = Vλ ⊕ Vµ (action by g commutes with
(A − λ)n , so stays in the same eigenspace), a contradiction, so A = S + N . If
V is decomposable, V = V1 ⊕ V2 , then the projection of V onto V1 is not of the
form S + N , and it’s in EndG V .
ii. Notice that (I − N )−1 = I + N + N 2 + . . .. Thus if S is not zero, we have
a formula for the inverse of S + N . If S is zero, it’s nilpotent, so everything
is 0 or nilpotent. If A is nilpotent and B is in EndG V , we have that BA has
nonzero kernel (the kernel contains the kernel of A, which is nonzero), hence is
nilpotent since it isn’t invertible. If B is nilpotent, then AB is nilpotent, if B is
invertible, and v ∈ ker A, then B −1 v ∈ ker AB, so it’s nilpotent. Say C = A + B
is invertible. Then Id = C −1 A + C −1 B, so Id − C −1 B = C −1 A, so C −1 A is
invertible, a contradiction.
iii. Let A = EndG V . It is a finite dimensional algebra. Let R = Rad(A) be
the set of elements of act by 0 in all irreducible reps. We have that I ⊂ R, and
Rn = 0 for some n, so I n = 0.
10.
a. We have |G| = (p2 + p + 1)(p + 1)(p − 1)2 .
Notice that we have an inclusion
3
F×
p3 o Gal(Fp /Fp ) ,→ GL3 (Fp )
4
notice that in the inclusion, (x, σ) 7→ Mx Mσ , which represents the matrices
of x, σ. Choosing a basis 1, α, α2 of Fp3 /Fp , we see that σ is a permutation
matrix with determinant 1. Notice that det Mx = N m(x)1, so letting N ⊂ F×
p3
denote the elements of norm 1, we have an inclusion
N o Gal(Fp3 /Fp ) ,→ GL3 (Fp )
3
−1
Notice that |N | = pp−1
= p2 + p + 1, and that N is cyclic.
It is easy to check that gcd(p2 +p+1, p+1) = 1 and gcd(p2 +p+1, p−1) = 3.
So this implies that the ` sylow subgroups are contained in N o Gal(Fp3 /Fp ),
hence are cyclic, because N is cyclic.
b. We saw that gcd(p2 + p + 1, p − 1) = 3, so 3|p − 1. Consider elements of
the form


a


b
−1
(ab)
where a, b have order 3 in F×
p . Notice that this gives that the 3-sylow subgroups contains a copy of Cm × Cm , where 3m = p − 1, so is not cyclic.
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