WEIL q-NUMBERS OF DEGREE 4 OVER Q

Let π ∈ C be a Weil q-number of degree 4 over Q. We argue that:

(1) For no embedding σ : Q(π) → C is the image σ(π) real. This is true because we saw in the talk that in this case π has degree 1 or 2 over Q.

(2) Let P (X) = X^{4} + aX^{3} + bX^{2} + cX + d be the minimal polynomial over Q. Note that we can write

P (X) = X^{4} + aX^{3} + bX^{2} + cX + d = (X − α_{1})(X − α_{2})(X − α_{3})(X − α_{4})

where α_{1}, α_{2}, α_{3} and α_{4} are all 4 conjugates of π in C.

(3) In particular, for each i we have \overline{α_{i}} ∈ \{α_{1}, α_{2}, α_{3}, α_{4}\} (where z ↦ \overline{z} is complex conjugation). Since none of the α_{i} are real we may choose the numbering such that \overline{α_{1}} = α_{2} and \overline{α_{3}} = α_{4}.

(4) Note that a, b, c, d ∈ Z because π is an algebraic integer.

(5) Because |α_{i}| = \sqrt{q} we see that \overline{α_{i}} = q/α_{i}. In particular we see that the roots of the polynomial X^{4}P (q/X) are the same as the roots of P (X). By looking at the leading coefficient we deduce that

X^{4}P (q/X) = q^{2}P (X).

Writing this out we obtain

q^{4} + aq^{3}X + bq^{2}X^{2} + cqX^{3} + dX^{4} = q^{2}X^{4} + aq^{2}X^{3} + bq^{2}X^{2} + cq^{2}X + dq^{2}.

We conclude that d = q^{2} and that c = aq. Thus we conclude that

P (X) = X^{4} + aX^{3} + bX^{2} + aqX + q^{2}.

(6) Note that a = −α_{1} − α_{2} − α_{3} − α_{4}. In particular we have

|a| ≤ 4\sqrt{q}

(7) Note that b = α_{1}α_{2} + α_{1}α_{3} + α_{1}α_{4} + α_{2}α_{3} + α_{2}α_{4} + α_{3}α_{4}. In particular we

have

|b| ≤ 6q

(8) Let β = π + \overline{π} = π + q/π. What is the minimal polynomial of β over Q? To see this note that π is a root of P (X)/X^{2}. We write this as

P (X)/X^{2} = X^{2} + aX + b + aq/X + q^{2}/X^{2}
= (X + q/X)^{2} − 2q + a(X + q/X) + b
= (X + q/X)^{2} + a(X + q/X) + b − 2q
= Y^{2} + aY + (b − 2q).

where Y = X + q/X. In other words, β satisfies the equation Y^{2} + aY + (b − 2q) = 0.

(9) In order for π to be a Weil q-number we know by general theory that β has to be totally real. In other words the discriminant Δ of the quadratic polynomial Y^{2} + aY + (b − 2q) has to be positive. We compute Δ = a^{2} − 4(b − 2q) = a^{2} + 8q − 4b. The resulting inequality is a^{2} + 8q > 4b, or

b < a^{2}/4 + 2q.

Note that, in case b > 0, this is a stronger inequality than our previous inequality for the magnitude of b.

(10) Note that π is a solution to the equation X^{2} − βX + q = 0. Thus, in order for Q(π) to be a CM field of degree 4 over Q, we need β^{2} − 4q under any embedding of Q(β) into R to be negative. In other words we need

(\frac{−a ± \sqrt{Δ}}{2})^{2} − 4q < 0

(11) We work out what this means:

((−a ± \sqrt{Δ})/2)^{2} − 4q < 0 ⇔ ((−a ± \sqrt{Δ})/2)^{2} < 4q
⇔ (−a ± \sqrt{Δ})/2 < 2\sqrt{q} and (−a ± \sqrt{Δ})/2 > −2\sqrt{q}
⇔ (−a ± \sqrt{Δ}) < 4\sqrt{q} and (−a ± \sqrt{Δ}) > −4\sqrt{q}
⇔ ±\sqrt{Δ} < a + 4\sqrt{q} and ± \sqrt{Δ} > a − 4\sqrt{q}
⇔ \sqrt{Δ} < a + 4\sqrt{q} and − \sqrt{Δ} > a − 4\sqrt{q}
⇔ Δ < a^{2} + 8a\sqrt{q} + 16q and Δ < a^{2} − 8a\sqrt{q} + 16q
⇔ a^{2} − 4b + 8q < a^{2} − 8|a|\sqrt{q} + 16q
⇔ −4b < −8|a|\sqrt{q} + 8q
⇔ −b < −2|a|\sqrt{q} + 2q
⇔ b > 2|a|\sqrt{q} − 2q

Note that one of the conclusions of this sequence of inequalities is also that |a| ≤ 4\sqrt{q} which we saw before.

(12) So a comlete set of inequalities is the following

|a| ≤ 4\sqrt{q}
b < a^{2}/4 + 2q
b > 2|a|\sqrt{q} − 2q