Download Math 121. Construction of a regular 17-gon 1

Math 121. Construction of a regular 17-gon
1. Introduction
Let K = Q(ζ) with ζ a primitive 17th root of unity, so ζ has minimal polynomial over Q equal
to Φ17 = X 16 + X 15 + · · · + X + 1 ∈ Q[X] and G := Gal(K/Q) = (Z/17Z)× . In particular, the 16
elements
{ζ, . . . , ζ 16 } = {ζ j }1≤j<17
are linearly independent over Q, as any nontrivial Q-linear dependence relation would (upon division
by ζ) give a nontrivial polynomial relation for ζ over Q of degree < 16 = deg Φ17 , a contradiction.
Let σ ∈ G correspond to 3 ∈ (Z/17Z)× (so σ(z) = z 3 for all 17th roots of unity z ∈ K). This
is a generator. Indeed, we just need to check that σ 8 6= 1 (as the order of any proper subgroup
of a cyclic group of order 16 must divide 8), or equivalently that 38 6≡ 1 mod 17. Since 32 = 9, so
34 = 81 ≡ −4 mod 17, we have 38 ≡ 16 ≡ −1 mod 17.
i
Thus, we get a chain of subgroups Gi = hσ 2 i for 0 ≤ i ≤ 4, so G0 = G, G4 = 1, and Gj has
index 2 in Gj−1 for 1 ≤ j ≤ 4, Explicitly,
G0 = G, G1 = h9i, G2 = h−4i, G3 = h−1i, G4 = 1.
Let Ki = K Gi , so K0 = Q, K4 = K, and [Ki : Ki−1 ] = 2 for 1 ≤ i ≤ 4. In particular, as we saw in
class, K3 = K h−1i = Q(ζ + 1/ζ) = Q(cos(2π/17)).
Our aim in this handout is to make explicit the fields K1 , K2 , K3 in the sense that we find
• an explicit xi ∈ Ki ⊂ Q(ζ) for 0 ≤ i ≤ 3 such that Ki = Q(xi ) and x3 = ζ + 1/ζ,
• the quadratic minimal polynomial fi ∈ Ki−1 [T ] = Q(xi−1 )[T ] for xi over Ki−1 (1 ≤ i ≤ 3),
• for the embedding K3 ,→ R carrying ζ + 1/ζ to 2 cos(2π/17) (induced by ι : K = K4 ,→ C
carrying ζ to e±2πi/17 ), the exact sign in the quadratic formula expressing xi in terms of
elements of Q(xi−1 ) with square roots in R.
In this way, we get an explicit formula for x3 via iterated square roots inside R, thereby providing
a construction of a regular 17-gon via straightedge and compass. Exercise 14 in §14.5 of the course
text gives another description of this calculation, and the explicit formula on page 602 of the course
text looks nicer than the one we will obtain. (There is no inconsistency, as there are numerous
ways to give an iterated quadratic tower “formula” for any specific element of K3 .)
2. The primitive elements
Gi
To find an explicit primitive element
P xi ∈ Ki = K over Q for 0 ≤ i ≤ 3, we use the “sum over
an orbit” method: let’s define xi = g∈Gi g(ζ) and hope for the best. By design certainly xi is
Gi -invariant, so xi ∈ Ki . For example,
x0 =
X
g∈G
g(ζ) =
16
X
ζ j = Φ17 (ζ) − 1 = −1 ∈ Q = K0 .
j=0
Also, x3 = g∈G3 g(ζ) = ζ + ζ −1 . The cases of x1 and x2 are more interesting, and we’ll make
them explicit in a moment. But first let’s show by the Fundamental Theorem that the inclusion
Q(xi ) ⊂ Ki is an equality for all 0 ≤ i ≤ 3.
It suffices (why?) to check that if γ 6∈ Gi then γ(xi ) 6= xiP
. (In view of the cyclicity of G,
P it suffices
to check this for just a single γ ∈ Gi−1 − Gi .) But γ(xi ) = g∈Gi γ(g(ζ)) whereas xi = g∈Gi g(ζ),
P
1
2
so if γ(xi ) = xi then
X
(1)
g∈γGi
g(ζ) −
X
g(ζ) = 0.
g∈Gi
But γGi and Gi are disjoint subsets of G = (Z/17Z)× since Gi is a subgroup and γ 6∈ Gi . Since
the G-orbit {ζ, ζ 2 , . . . , ζ 16 } = {ζ j }1≤j≤16 is linearly independent over Q, a relation such as (1) is
impossible (and more generally, no difference among sums of two disjoint subsets of this G-orbit
can vanish). Hence, we have proved that Q(xi ) = Ki for all i.
Explicitly, since G1 = {1, σ 2 , σ 4 , σ 6 , . . . , σ 14 } = {1, 9 = −8, −4, −2, −1, 8, 4, 2}, we have
x1 = ζ + σ 2 (ζ) + σ 4 (ζ) + σ 6 (ζ) + · · · + σ 14 (ζ)
= ζ + ζ −8 + ζ −4 + ζ −2 + ζ −1 + ζ 8 + ζ 4 + ζ 2
= (ζ + ζ −1 ) + (ζ 2 + ζ −2 ) + (ζ 4 + ζ −4 ) + (ζ 8 + ζ −8 ).
Hence, for θ = 2π/17, the embedding ι : K ,→ C via ζ 7→ e±2πi/17 = e±iθ carries x1 to
(2)
2(cos θ + cos(2θ) + cos(4θ) + cos(8θ))
√
which is positive (proof: cos(θ), cos(2θ) > 1/ 2 since 2/17 < 1/8, and cos(4θ) > 0 since 4/17 <
1/4). We will use this positivity to identity ι(x1 ) as an explicit quadratic irrationality over Q inside
R below. (Recall that [K1 : Q] = [K1 : K0 ] = 2, so the field K1 = Q(x1 ) is real quadratic.)
Likewise, since G2 = {1, σ 4 , σ 8 , σ 12 } = {1, −4, −1, 4}, we have
x2 = ζ + σ 4 (ζ) + σ 8 (ζ) + σ 12 (ζ) = ζ + ζ −4 + ζ −1 + ζ 4 = (ζ + ζ −1 ) + (ζ 4 + ζ −4 ).
Hence, the embedding ι as above carries x2 to 2(cos θ+cos(4θ)). The nontrivial Galois conjugate for
x2 over K1 is σ 2 (x2 ) = (ζ 2 + ζ −2 ) + (ζ 8 + ζ −8 ) since σ 2 generates Gal(K2 /K1 ) = hσ 2 i/hσ 4 i (or more
explicitly, {2, −2, 8, −8} = G1 − G2 ), so ι carries the K1 -conjugate σ 2 (x2 ) to 2(cos(2θ) + cos(8θ)).
But for θ = 2π/17 we have seen cos θ, cos(4θ) > 0, so
cos θ + cos(4θ) > 0 > cos(2θ) + cos(8θ)
(since (1/2) − 8/17 < 2/17, or stare at a 17-gon). Hence, once we compute the minimal polynomial
f2 for x2 over Q(x1 ) = K1 , ι(x2 ) must be the larger root of ι(f2 ) ∈ R[T ]; i.e., ι(x2 ) is given by the
positive square root in the quadratic formula for the roots of ι(f2 ) ∈ ι(K1 )[T ] ⊂ R[T ] in R.
3. Quadratic polynomials
Now we compute the minimal quadratic polynomial fi ∈ Ki−1 [T ] for xi with 1 ≤ i ≤ 3. Since
i−1
Gi−1 is generated by σ 2 , so
i−1
Gal(Ki /Ki−1 ) = Gi−1 /Gi = hσ 2
i
i/hσ 2 i,
i−1
this quotient group of order 2 is represented by {1, σ 2 }. Hence, the Gal(Ki /Ki−1 )-orbit of the
i−1
primitive generator xi of Ki over Ki−1 is {xi , σ 2 (xi )}. It follows that
i−1
fi (T ) = (T − xi )(T − σ 2
i−1
(xi )) = T 2 − (xi + σ 2
i−1
(xi ))T + xi σ 2
2i−1
(xi ) ∈ Ki−1 [T ].
By definition, xi is the sum over the Gi -orbit of ζ, so σ
(xi ) is the sum over the orbit of ζ under
i−1
the subset σ 2 Gi ⊂ Gi−1 that is exactly the nontrivial Gi -coset in Gi−1 , Hence,
X
i−1
xi + σ 2 (xi ) =
g(ζ) = xi−1 ,
g∈Gi−1
so
i−1
fi = T 2 − xi−1 T + xi σ 2
(xi ) ∈ Ki−1 [T ].
3
We need to make the constant term explicit in terms of the description Q(xi−1 ) of Ki−1 (for
1 ≤ i ≤ 3) in order to give a “quadratic formula” expression for xi over Q(xi−1 ) (and then bring
in positivity work in R from the previous section to pin down the correct sign in the quadratic
formula when working with the embedding ι : K → C sending ζ to e±2πi/17 = e±iθ ).
We have already seen that x0 = −1, so
f1 = T 2 + T + (x1 σ(x1 )) ∈ K0 [T ] = Q[T ].
Hence, we have to identity x1 σ(x1 ) as an explicit rational number. It was seen above that
x1 = ζ + ζ −8 + ζ −4 + ζ −2 + ζ −1 + ζ 8 + ζ 4 + ζ 2 ,
so since σ(ζ) = ζ 3 we see that σ(x1 ) is given by multiplying these exponents by 3. That is,
σ(x1 ) = ζ 3 + ζ −7 + ζ 5 + ζ −6 + ζ −3 + ζ 7 + ζ −5 .
Now expand the product x1 σ(x1 ) of two 8-fold sums as a sum of 8 × 8 = 64 terms by massive
application of the distributive law. This yields by a miracle (check!) that each ζ j (1 ≤ j ≤ 16)
appears exactly 4 times, so
x1 σ(x1 ) = 4(ζ + ζ 2 + ζ 3 + · · · + ζ 16 ) = −4.
√
To summarize, f1 = T 2 + T − 4, so its root ι(x1 ) in R has the form (−1 ± √
17)/2 for a sign to be
determined. (In particular, the unique quadratic subfield K1 of Q(ζ) is Q( 17).) Our positivity
work with cosines showed that ι(x1 ) > 0, so
√
ι(x1 ) = (−1 + 17)/2
(which numerically agrees with the expression in (2), as it must).
Moving on to f2 = T 2 − x1 T + (x2 σ 2 (x2 )) ∈ K1 [T ], since σ 2 (ζ) = ζ 9 = ζ −8 we deduce from the
formula
x2 = ζ + ζ −4 + ζ −1 + ζ 4
that σ 2 (x2 ) = ζ −8 + ζ −2 + ζ 8 + ζ 2 . Thus, expanding the product x2 σ 2 (x2 ) of two 4-fold sums as a
sum of 4 × 4 = 16 terms yields
x2 σ 2 (x2 ) = ζ + ζ 2 + ζ 3 + · · · + ζ 16 = −1.
In other words, f2 = T 2 − x1 T − 1, so by the quadratic formula
p
x1 ± x21 + 4
x2 =
.
2
But recall from the previous section that under ι : K → C (carrying K3 onto Q(cos(2π/17)) ⊂ R),
the root ι(x2 ) of ι(f2 ) ∈ R[T ] is the larger of the two roots (the other root being ι(σ 2 (x2 ))). Hence,
ι(x2 ) is given by the quadratic formula using the positive square root:
p
ι(x1 ) + ι(x1 )2 + 4
ι(x2 ) =
2
√
in R with ι(x1 ) = (−1 + 17)/2.
Finally, we come to the crux of the matter: identifying the constant term of
f3 = T 2 − x2 T + (x3 σ 4 (x3 )) ∈ Q(x2 )[T ]
which has as one of its roots x3 = ζ + 1/ζ, with ι(x3 ) = 2 cos(θ) for θ = 2π/17. Since x3 = ζ + ζ −1
and σ 4 (x3 ) = ζ −4 + ζ 4 (as 34 ≡ −4 mod 17), so ι(x3 ) = 2 cos(θ) and ι(σ 4 (x3 )) = 2 cos(4θ), the easy
geometric (or numerical) fact that cos(θ) > cos(4θ) shows that ι(x3 ) is the larger of the two roots
of ι(f3 ). Hence, once we explicitly compute x3 σ 4 (x3 ) as an element of K2 = Q(x2 ) we can conclude
4
that the real number ι(x3 ) = 2 cos(2π/17) is given by the quadratic formula for the roots of ι(f3 )
using the positive square root. It all now comes down to computing x3 σ 4 (x3 ) in
hσ 4 i
K3
= K3G2 = K2 = Q(x2 ) = Q ⊕ Qx2 ⊕ Qx22 ⊕ Qx32 .
(Note that [K2 : Q] = [G : G2 ] = 4.) This is ultimately just a bit of linear algebra over Q, given
our knowledge of the Q-linear dependence relations among powers {ζ j }0≤j<17 .
We seek to compute the unique a0 , a1 , a2 , a3 ∈ Q such that
?
a0 + a1 x2 + a2 x22 + a3 x32 = x3 σ 4 (x3 ) = (ζ + ζ −1 )(ζ −4 + ζ 4 ) = ζ −3 + ζ 5 + ζ −5 + ζ 3 .
(3)
By definition, x2 = ζ + ζ −4 + ζ −1 + ζ 4 . Hence, with some help from the distributive law,
x22 = 4 + (ζ 2 + ζ −2 + ζ 8 + ζ −8 ) + 2(ζ 3 + ζ −3 + ζ 5 + ζ −5 ).
Multiplying this against another factor of x2 = ζ + ζ −4 + ζ −1 + ζ 4 and using
P16
j=1 ζ
j
= −1 gives
x32 = −1 + 8(ζ + ζ −1 + ζ 4 + ζ −4 ) + 2(ζ 2 + ζ −2 + ζ 6 + ζ −6 + ζ 7 + ζ −7 + ζ 8 + ζ −8 ).
Plugging these descriptions of x2 , x22 , x32 ∈ Q(ζ) into the required relation
a0 + a1 x2 + a2 x22 + a3 x32 = ζ 3 + ζ −3 + ζ 5 + ζ −5
from (3) yields the reformulation (upon collecting common appearances of each ζ j )
0 = (a0 − a3 + 4a2 ) + (8a3 + a1 )(ζ + ζ −1 + ζ 4 + ζ −4 ) + (2a3 + a2 )(ζ 2 + ζ −2 + ζ 8 + ζ −8 )
+(2a2 − 1)(ζ 3 + ζ −3 + ζ 5 + ζ −5 ) + 2a3 (ζ 6 + ζ −6 + ζ 7 + ζ −7 ).
Observe that every ζ j for 1 ≤ j ≤ 16 appears exactly once on the right side. By replacing the
“constant term” a0 − a3 + 4a2 with the equivalent expression
−(a0 − a3 + 4a2 )(ζ + ζ 2 + ζ 3 + · · · + ζ 16 ),
the Q-linear independence of the set {ζ, ζ 2 , ζ 3 , . . . , ζ 16 } implies that all of the coefficients
8a3 + a1 , 2a3 + a2 , 2a2 − 1, 2a3
are equal to a0 − a3 + 4a2 . In other words, we arrive at the system of linear equations
a0 − a3 + 4a2 = 8a3 + a1 = 2a3 + a2 = 2a2 − 1 = 2a3 .
Elementary manipulations show that this system has the unique solution
(a0 , a1 , a2 , a3 ) = (−3/2, 3, 0, −1/2).
Hence,
f3 = T 2 − x2 T + ((−1/2)x32 + 3x2 − 3/2) ∈ Q(x2 )[T ].
Since we have seen that ι(x3 ) is the larger of the two roots of ι(f3 ) ∈ Q(ι(x2 ))[T ] ⊂ R[T ], the
quadratic formula and some elementary simplifications of the discriminant give the formula
p
ι(x2 ) + 2ι(x2 )3 + ι(x2 )2 − 12ι(x2 ) + 6
2 cos(2π/17) = ι(x3 ) =
2
p
√
using the positive square root, where ι(x2 ) = (ι(x1 ) + ι(x1 )2 + 4)/2 with ι(x1 ) = (−1 + 17)/2.
This is a formula inside R for 2 cos(2π/17) via iterated square roots; check it numerically!
The interested reader is invited to adapt this method to the case of the regular 257-gon (PhD
thesis of F. Richelot in 1832) and the regular 65537-gon (first done by J. Hermes in 1894, after 10
years of effort). To get you started, conveniently the cyclic groups (Z/257Z)× and (Z/65537Z)×
7
15
each admit 3 as a generator (since 32 ≡ −1 mod (28 + 1) and 32 ≡ −1 mod (216 + 1), as one
checks by successive squaring modulo 28 + 1 = 257 and 216 + 1 = 65537 respectively).