QUADRATIC ELEMENTS IN A CENTRAL SIMPLE ALGEBRA OF DEGREE FOUR

MARKUS ROST

Contents

Introduction 1 1. Preliminaries 2 2. Construction of quadratic elements 2 3. Existence of biquadratic subextensions 4 4. More quadratic elements 5 5. Geometric interpretation 5 Appendix 6 References 7

Introduction

By a theorem of Albert [1, Theorem 11.9] a central simple algebra A of degree 4 contains a biquadratic subalgebra. The major point is to prove the existence of a quadratic subalgebra. See [2] for a short proof.

One can show this also using the fact from projective geometry that, given 4 generic lines in 3-space, there are exactly 2 lines which meet them all. Namely, for any (possibly non-split) A one considers a maximal commutative separable subalgebra H of A and a generic element x ∈ A. Such data determine geometrically 4 lines in projective 3-space: The subalgebra H determines geometrically 4 points and for each of these points there is the line through it and its image under x. The 2 lines which meet all those lines determine a quadratic subalgebra L of A.

It seems to be rather complicated to give explicit generators for L in terms of H and x, in particular in characteristic 2 (cf. Section 4).

However there is a way to construct quadratic elements (described in Sections 1-2) which is algebraically simpler. This construction is symmetric with respect to reversing the product in A (i. e., replacing A with A^{op}). It works smoothly for central simple algebras A of degree 4 with involution of second kind and the base ring can be any local ring. A geometric interpretation is given in Section 5.

Once a quadratic subalgebra L is at hand, it easy to extend it to a biquadratic subalgebra: Let y be a generic element in the orthogonal complement of the centralizer of L. Then L and y^{2} generate a biquadratic subalgebra. See Section 3.

In the Appendix we have added (mainly for fun) some formulas for the coefficients of the characteristic polynomial in a central simple algebra of degree 4. They occurred when trying to establish Lemma 1.

Date: May 30, 2003.

1. Preliminaries

Let F be a field and let A be a central simple algebra of degree 4 over F. For the characteristic polynomial of x ∈ A we use the notation

N(t − x) = t^{4} − T (x)t^{3} + Q(x)t^{2} − S(x)t + N(x)

where N: A → F is the reduced norm of A.

For x ∈ A we consider the linear form

Φ(x): Λ^{3}A → F
Φ(x)(y_{1} ∧ y_{2} ∧ y_{3}) = T (xy_{3}xy_{2}xy_{1} − xy_{1}xy_{2}xy_{3})

Lemma 1. For x, y ∈ A one has

S(xy − yx) = Φ(x)(1 ∧ y ∧ y^{2})

Proof. This amounts to

S(xy − yx) = T (x^{2}y^{2}xy − x^{2}yxy^{2})

To prove this identity it suffices to consider the case A = M_{4}(Q). One has

6S(u) = T (u)^{3} − 3T (u^{2})T (u) + 2T (u^{3})

as can be easily seen from the case of diagonal matrices. Thus, if T (u) = 0 one has

3S(u) = T (u^{3})

and the claim follows from this with u = xy − yx and from T (vw) = T (wv). □

Remark 1. The formula for S in Lemma 4 in the Appendix yields a “denominator free” proof of Lemma 1.

2. Construction of quadratic elements

Let H be a commutative subalgebra of A of dimension 4 and let x ∈ A.

The form Φ(x) restricts to a linear form Λ^{3}H → F. We consider the associated dual map

α(x): Λ^{4}H → H

defined by

(1)  μ ∧ α(x)(ω) = Φ(x)(μ)ω

for μ ∈ Λ^{3}H and ω ∈ Λ^{4}H.

We fix a basis element ω ∈ Λ^{4}H and write α = α(x)(ω).

Corollary 1. Let

ρ = ρ(x) = xα − αx

Then

T (ρ) = S(ρ) = 0

Proof. Clearly T (ρ) = 0 since ρ is a commutator. From Lemma 1 and (1) we get

S(ρ)ω = 1 ∧ α ∧ α^{2} ∧ α = 0
□

Let us write

Δ(x) = Q(ρ)^{2} − 4N(ρ)

Corollary 2. Let

θ = θ(x) = ρ^{2}

One has

θ^{2} + Q(ρ)θ + N(ρ) = 0

The characteristic polynomial of θ is

(t^{2} + Q(ρ)t + N(ρ))^{2}

Let x ∈ A with Δ(x) ≠ 0. Then θ(x) generates a uniform quadratic subalgebra of A. (A quadratic subalgebra L of A is called uniform if it is separable and if A is as L-module of constant rank 8.)

Proof. The first claim is clear from Corollary 1. The second claim follows from

N(t^{2} − θ) = N(t − ρ)N(t + ρ) = (t^{4} + Q(ρ)t^{2} + N(ρ))^{2}

The final claim is then clear as well. □

Suppose (A, H) = (M_{4}(F), D_{4}(F)) where D_{4}(F) denotes the 4 × 4 diagonal matrices. Write H = e_{1}F ⊕ e_{2}F ⊕ e_{3}F ⊕ e_{4}F with e_{i} the standard idempotents. Then, with ω = e_{1} ∧ e_{2} ∧ e_{3} ∧ e_{4},

(2)  α(x)(ω) = diag (\begin{matrix}x_{23}x_{34}x_{42} − x_{24}x_{43}x_{32} \\ x_{31}x_{14}x_{43} − x_{34}x_{41}x_{13} \\ x_{41}x_{12}x_{24} − x_{42}x_{21}x_{14} \\ x_{13}x_{32}x_{21} − x_{12}x_{23}x_{31} \end{matrix})

For

(3)  x = (\begin{matrix}0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix})

one finds Δ(x) = 1.

Corollary 3 (Albert). There exists a uniform quadratic subalgebra in A.

Proof. Choose some H, ω as above with H separable. By Corollary 2 it suffices to find x ∈ A with Δ(x) ≠ 0.

As the computation for (3) shows, the polynomial Δ(x) is nonzero. Hence if F is infinite, there exists x with Δ(x) ≠ 0.

If F is finite, then A = M_{4}(F) and the claim is clear anyway. □

Remark 2. Note that we have made no assumption on the characteristic on F. In fact, everything works over an arbitrary local ring.

Remark 3. Everything extends easily to the case of a central simple algebra of degree 4 with involution of second kind.

Remark 4. One has α(x + h) = α(x) for h ∈ H. Thus x ↦ α(x) is a cubic form on A/H with values in Hom(Λ^{4}H, H).

3. Existence of biquadratic subextensions

Let L be a uniform quadratic subalgebra in A and let θ ∈ L be a generator.

Lemma 2. For y ∈ A let

σ = σ(y) = yθ − θy

Then

T (σ) = S(σ) = 0

Let

Δ′(y) = Q(σ)^{2} − 4N(σ)

Let y ∈ A with Δ′(y) ≠ 0. Then θ and σ(y)^{2} generate a biquadratic subalgebra of A.

Suppose A = M_{4}(F),

θ = (\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})

and

y = (\begin{matrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})

Then Δ′(y) = 1.

Proof. This can be shown by inspection in the split case. Suppose A = M_{4}(F) and that L is the subalgebra of matrices of the form

(\begin{matrix}a & 0 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & b & 0 \\ 0 & 0 & 0 & b \end{matrix})

Then a commutator σ = yθ − θy is of the form

σ = (\begin{matrix}0 & U \\ V & 0 \end{matrix})

with U, V ∈ M_{2}(F). The characteristic polynomial of σ is

t^{4} − t^{2} trace(UV) + det(UV)

and one has

σ^{2} = (\begin{matrix}UV & 0 \\ 0 & V U \end{matrix})

The element σ^{2} commutes with the elements of L and generates together with L a biquadratic subalgebra as long as (trace(UV))^{2} ≠ 4 det(UV). The last claim is easy to check. □

Corollary 4 (Albert). There exists a biquadratic subalgebra in A.

Proof. If F is finite, then A = M_{4}(F) and the claim is clear. Otherwise one chooses L and θ as above and picks y with Δ′(y) ≠ 0 (cf. Corollary 3 and its proof). □

4. More quadratic elements

We have seen that for generic x the element θ(x) generates a uniform quadratic subalgebra L. The centralizer Q of L is a quaternion algebra over L.

In the following we give a description of generators for Q. This appears to be somewhat complicated.

We assume char F ≠ 2.

For x ∈ A consider the linear form

φ(x): Λ^{3}H → F

φ(x)(μ) = ∑_{s∈S_{3}} sgn(s)T (x^{2}h_{s(3)}xh_{s(2)}xh_{s(1)}) − 2T (x)Φ(x)(μ)

with μ = h_{1} ∧ h_{2} ∧ h_{3}, h_{i} ∈ H, and let

β(x): Λ^{4}H → H

be the linear map with

μ ∧ β(x)(ω) = φ(x)(μ)ω

for μ ∈ Λ^{3}H and ω ∈ Λ^{4}H.

Let further

ψ(x), ψ′(x): Λ^{4}H → A

ψ(x) = xα(x) + \frac{1}{2}β(x)

ψ′(x) = α(x)x + \frac{1}{2}β(x)

We fix a basis element ω ∈ Λ^{4}H and write ψ = ψ(x)(ω), ψ′ = ψ′(x)(ω).

Let Q be the subalgebra of A generated by ψ and ψ′. Note that ρ = ψ − ψ′ ∈ Q. Let L ⊂ Q be the subalgebra generated by θ = ρ^{2}.

Lemma 3. One has ψ^{2} = ψ′^{2} ∈ F. The algebra L is in the center of Q.

Suppose H is separable. Then for generic x, L is a uniform quadratic subalgebra of A and Q is the centralizer of L in A.

Proof. I know only a tiresome check in the split case (A, H) = (M_{4}(F), D_{4}(F)). □

5. Geometric interpretation

There is an interpretation of our construction in terms of a classical theorem in projective geometry: For 4 generic lines in P^{3} there exist exactly 2 lines meeting them all.

Let V be a 4-dimensional vector space, let A = End(V), let V = V_{1} ⊕ V_{2} ⊕ V_{3} ⊕ V_{4} be a decomposition into 1-dimensional subspaces and H be the subalgebra of elements h ∈ A with h(V_{i}) ⊂ V_{i}, i = 1, 2, 3, 4. Let further x ∈ A be a generic element.

In the projective space P(V) let L_{i} be the line through the points [V_{i}], [xV_{i}] and let K_{1}, K_{2} be the lines meeting every L_{i}.

The lines K_{1}, K_{2} are in fact given by the 2 2-dimensional eigenspaces of the element ψ(x) above.

Let V^{∗} = V_{1}^{∗} ⊕ V_{2}^{∗} ⊕ V_{3}^{∗} ⊕ V_{4}^{∗} be the dual decomposition of the dual space (i.e., V_{i}^{∗}(V_{j}) = δ_{ij}F).

In the projective space P(V^{∗}) let L′_{i} be the line through the points [V_{i}^{∗}], [V_{i}^{∗}x] and let K′_{1}, K′_{2} be the lines meeting every L′_{i}.

The lines K′_{1}, K′_{2} are in fact given by the 2 2-dimensional eigenspaces of the element ψ′(x) above.

Every line L in P(V^{∗}) defines a line L^{∗} in P(V). Let S_{1}, S_{2} be the two lines which meet all of K_{1}, K_{2}, (K′_{1})^{∗}, (K′_{2})^{∗}.

The lines S_{1}, S_{2} are in fact given by the 2 2-dimensional eigenspaces of the element θ(x) above.

There is also an interpretation of the element ρ = ρ(x). Since ρ^{2} is the identity on S_{i}, the element ρ is an involution on each S_{i}. An involution on a line is given by a pair of points. How to get a pair of points on S_{i}? We have two pairs of points already: the intersections of S_{i} with K_{1}, K_{2} and with (K′_{1})^{∗}, (K′_{2})^{∗}, respectively. Now a pair of points on a line gives a point in the 2-fold symmetric product of the line, a P^{2}. The diagonal embedding of the line to that P^{2} identifies it with a conic. Thus the two pairs of points determine a line in P^{2}, which intersects the conic in a pair of points. This yields certainly the pair given by ρ (I haven’t really checked this).

Appendix

Lemma 1 follows also from the additivity rule for S in the following Lemma.

Lemma 4. For x, y ∈ A one has

T (x + y) = T (x) + T (y)
Q(x + y) = Q(x) + T (x)T (y) − T (xy) + Q(y)
S(x + y) = S(x) + Q(x)T (y) − T (x)T (xy) + T (x^{2}y)
+ T (x)Q(y) − T (xy)T (y) + T (xy^{2}) + S(y)
N(x + y) = N(x) + S(x)T (y) − Q(x)T (xy) + T (x)T (x^{2}y) − T (x^{3}y)
+ Q(x)Q(y) − T (x)T (xy)T (y) + T (x)T (xy^{2}) + T (x^{2}y)T (y)
+ Q(xy) − T (x^{2}y^{2})
+ T (x)S(y) − T (xy)Q(y) + T (xy^{2})T (y) − T (xy^{3}) + N(y)

Proof. In the power series ring A[[t]] one has

1 + t(x + y) = (1 + tx)[1 − t^{2} \frac{x}{1 + tx} \frac{y}{1 + ty} ](1 + ty)

The middle term expands as follows:

1 − t^{2} \frac{x}{1 + tx} \frac{y}{1 + ty} = 1 − t^{2}xy + t^{3}x(x + y)y − t^{4}x(x^{2} + xy + y^{2})y + ⋯

Taking norms gives in F [[t]]/(t^{5})

\begin{matrix}N(1 + t(x + y)) = N(1 + tx)N(1 + ty)[1 − t^{2}T (xy) + t^{3}T (x^{2}y + xy^{2}) \\ + t^{4}(Q(xy) − T (x^{3}y + x^{2}y^{2} + xy^{3}))] \end{matrix}

Multiplying out yields the claims. □

Remark 5. This argument works for central simple algebras of any degree. By further expansion one gets expressions for the coefficients of the characteristic polynomial of x + y as integral polynomials in the coefficients of the characteristic polynomials of noncommutative monomials in x, y.

References

[1] A. A. Albert, Structure of algebras, American Mathematical Society Colloquium Publications, vol. 24, American Mathematical Society, Providence, RI, 1961, Revised printing of the 1939 edition.

[2] D. Haile, A useful proposition for division algebras of small degree, Proc. Amer. Math. Soc. 106 (1989), no. 2, 317–319.

Department of Mathematics, The Ohio State University, 231 W 18th Avenue, Columbus, OH 43210, USA

   E-mail address: rost@math.ohio-state.edu
   URL: http://www.math.ohio-state.edu/~rost