Math 404. Exercises 6. Due Wednesday, June 7, 2013.
The cubic resolvent of x4 + a3 x3 + a2 x2 + a1 x + a0 is the polynomial g(x) = x3 − a2 x2 +
(a1 a3 − 4a0 )x + (4a0 a2 − a21 − a0 a23 ).
1(a). Determine the resolvent cubic and the squared-Jacobian of a polynomial of the form
f (x) = x4 + rx + s.
Solution. Using the formula above, the resolvent cubic g(x) of f (x) = x4 + rx + s, with
a0 = s and a1 = r and a2 = 0 = a3 , is
g(x) = x3 − 4sx − r2 .
The Jacobian-squared Jg2 of g(x), which equals the Jacobian-squared Jf2 of f (x), equals
−4(−4s)3 − 27(−r2 )2 = −27r4 + 256s3 .
(b). Determine the Galois group of x4 + 8x + 12 ∈ Q[x] and of x4 + 8x − 12 ∈ Q[x].
Solution. We will use the following fact: Let f (x) ∈ F [x] have degree 4. Suppose that
you have shown that f (x) has no roots in F , and that the cubic resolvent g(x) of f (x) is
irreducible. Then f (x) is irreducible.
We can prove that as fiollows: If f (x) reduces, it would reduce into a product of ltwo
irreducible factors of degree 2 in F [x] since it has no degree 1 factors by hypothesis. The
splitting field of f (x) would then have dimension either 2 or 4. The splitting field of the
cubic resolvent g(x) is a subfield of the spllitting field of f (x), and since g(x) is irreducible,
a root α of g(x) generates of dimension 3 extension of F . Hence, 3 divides the order of the
splitting field of f (x), which is 2 or 4. That contradiction shows that f (x) is irreducible.
(i). Let f (x) = x4 + 8x + 12 ∈ Q[x]. Check that f (x) has no rational roots (rational roots
would be integers that divide 12). We compute that the cubic resolvent is g(x) = x3 −48x−64,
which is irreducible since it has no rational roots. Hence, by Lagrange’s cubic resolvent
theory, the Galois group is A4 or S4 . To see which it is, we compute the Jacobian-squared:
Jf2 = Jg2 = −4(−48)3 − 27(−64)2 = 84 34 . Hence, Jf = 82 32 ∈ Q, and so, G = A4 .
(ii). Here, f (x) = x4 + 8x − 12 ∈ Q[x], g(x) = x3 + 48x − 64 and again we find that
f (x) and g(x) have no rational roots, so that the Galois group is A4 or S4 . Jf2 = Jg2 =
−4(48)3 − 27(−64)2 < 0. Hence, Jf is not rational, which implies that the Galois group is
S4 .
2. Determine the Galois group of the polynomial x4 + 4x2 + 2 ∈ Q[x]
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Solution. The polynomial is irreducible by Eisenstein criterion. The resolvent cubic for that
polynomial is g(x) = x3 − 4x2 − 8x + 32, which has 4 as a root. Then g(x) = (x − 4)(x2 − 8),
which has only one rational root. Hence, by Lagrange’s theory, G equals C4 or D4 . To see
which it is, we compute the dimension of the splitting field: if it is 4, the Galois group is C4
and if it is 8, the Galois group is D4 .
In considering polynomials of the form x4 + ax2 + b in a previous assignment, we saw that
we can write a polynomial in factored form as
(x − α)(x + α)(x − β)(x + β) = (x2 − α2 )(x2 − β 2 ) = x4 − (α2 + β 2 )x2 + α2 β 2 .
For the polynomial x4 + 4x2 + 2, that formula gives
x4 + 4x2 + 2 = x4 − (α2 + β 2 )x2 + α2 β 2 ,
so that α2 + β 2 = −4 and α2 β 2 = 2.
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Let us√consider α2 and β 2 , the
the quadratic formula,√{α2 , β 2 } =
√ roots of x + 4x + 2. By
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2 2
−2 ± 2; hence, Q[α ] = Q[ 2]. Furthermore, since α β = 2, we have αβ = ± 2, and so,
αβ ∈ Q[α2 ]. Thus, we see that Q[α, β], the splitting field L of x4 + 4x2 + 2, equals Q[α],
since β ∈ Q[α].
Therefore, [L : K] = 4, and the Galois group is C4 .
3. Prove that a Galois extension K/F with Galois group D4 is the splitting field of a
polynomial of the form x4 + bx2 + c ∈ F [x].
The following appears to be the ideal solution, using constructibilty.
The group D4 can be constructed as follows:
D4 ⊃ {I, (12)(34), (13)(24), (14)(23)} ⊃ {I, (12)(34)} ⊃ {I},
where each subgroup is normal of index 2 in the preceeding subgroup, but where {I, (12)(34)}
is not normal in D4 . By the construction theorem, the corresponding ladder of fields can be
presented as
K D4 = F ⊂ K {I,(12)(34),(13)(24),(14)(23)} = F (α1 ) ⊂ K {I,(12)(34)} = F (α1 )(α2 ) ⊂ K,
where each step has dimension 2, and where α12 ∈ F and α22 ∈ F (α1 ). Here, α22 is not
an element of F because if it were, K {I,(12)(34)} = F (α1 )(α2 ) would be the splitting field
of (x2 − α12 )(x2 − α22 ), which would violate the non-normality of {I, (12)(34)} ⊂ D4 . Since
[F (α1 : F ] = 2, and α22 ∈ F (α1 ) − F , we conclude that F (α1 ) = F [α22 ]. From there, we
conclude that F (α1 , α2 ) = F [α2 ].
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Since [F [α22 ] : F ] = 2, α22 satisfies an irreducible polynomial x2 + bx + c ∈ F [x], and so, α2
satisfies the polynomial x4 + bx2 + c ∈ F [x], which must be irreducible since [F [α2 ] : F ] = 4.
K is the splitting field of x4 +bx2 +c since the polynomial cannot split in F [α2 ] = K {I,(12)(34)} ,
since {I, (12)(34)} is not normal in D4 .
4. Suppose that K/F is a Galois extension whose Galois group G is a simple group (such as
A5 ), i.e., the only normal subgroups of G are {I} and G. Let α be any element of K that is
not in F , and let f (x) ∈ F [x] be the irreducible polynomial for α. Show that the splitting
field of f (x) is K.
This result indicates that there is not much discrimination among the elements of L when
the Galois group is a simple group.
Solution. Since G is simple, its only normal subgroups are {I} and G. Hence, the only
splitting fields between F and K are F and K. Since α is not an element of F , the splitting
field of its irreducible polynomial in F [x] can only be K.
5(a). Let K/F be a Galois extension with Galois group A4 . Describe the intermediate fields
E/F of dimension 2 in terms of subgroups H of A4 of index 2.
Solution. The 6-element subgroups of S4 are the four subgroups H1 , H2 , H3 , H4 (=S3 ),
where Hj is the staibilizer of the number j. For instance, H4 = {I, (123), (321), (12), (13), (23)}.
There are no other 6-element subgroups. In fact, If H is a 6-element subgroup, it must contain an element of order 3, which is a 3-cycle. If it contains (123), then it contains the
subgroup M = {I, (123), (321)}, which is normal in H since it has index 2. If σ ∈ H is not
in M , then σ ◦ (123)σ = (σ(1), σ(2), σ(3)) is an element of M , by normality. That means in
particular that none of {σ(1), σ(2), σ(3)} equals 4. Hence, H = H4 .
A 6-element subgroup of A4 , as a subgroup of S4 , would be one of the subgroups H1 , H2 ,
H3 , H4 . But none of those subgroups is contained in A4 , since each contains transpositions,
which are odd permutations. Hence, A4 has no 6-element subgroups, and correspondingly,
K/F has no dimension 2 intermediate fields.
5(b). Let K/F be a Galois extension with Galois group D4 . Describe the intermediate fields
E/F of dimension 2 in terms of subgroups H of D4 of index 2. Which of those fields is F [J],
where J is the Jacobian of an irreducible fourth degree polynomial F [x] whose splitting field
is K?
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Solution. The 4-element subgroups of D4 are C4 = {I, (1234), (13)(24), (4321)}, K4 =
{I, (12)(34), (13)(24), (14)(23)}, and C2 × C2 = {I, (13), (24), (13)(24)}. The fields we seek
are K C4 , K K4 , and K C2 ×C2 .
Which is F [J]? Since the Jacobian is fixed by the even permutations only, it must be the
case that F [J] = K K4 .
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