ON KUMMER CHAINS IN ALGEBRAS OF DEGREE 3

MARKUS ROST

                      preliminary version

Contents

Introduction 1 Kummer elements 1 1. Chains of length 2 2 2. Chains of length 3 2 Appendix 4 References 4

Introduction

This text is a sequel to [1] and we adopt the conventions of that article.

                  These are preliminary notes!

Kummer elements

A is a central simple algebra of degree 3 over k, ζ ∈ k is a primitive cube root of 1.

For the characteristic polynomial of x ∈ A we use the notation

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

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

Lemma 1. For Kummer elements X, Y ∈ A one has

T (XY XY^{−1}) = T (XY)T (XY^{−1})

Proof. By Lemma 6 (see the appendix) one has

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

Taking x = XY, y = Y this yields

N(XY + Y) = N(XY) + Q(XY) · 0 − T (XY)T (XY^{2}) + T (XY XY^{2})
+ T (XY) · 0 − T (XY^{2}) · 0 + 0 + N(Y)

On the other hand

N(XY + Y) = N(X + 1)N(Y) = (N(X) + 1)N(Y) = N(XY) + N(Y)

Date: January 29, 2005.

Hence

T (XY XY^{2}) = T (XY)T (XY^{2})

□

1. Chains of length 2

For Kummer elements X, Y ∈ A^{×} the symbol

X ⟶^{ζ} Y

stands for

Y X = ζXY

Lemma 2. Suppose there exists a chain

X ⟶^{ζ} U ⟶^{ζ} Y

Then XY^{−1} is a Kummer element.

Proof. Let V = XY^{−1}. Then UV U^{−1} = ζ^{2}V and therefore V is a Kummer element. □

Lemma 3. Let X, Y be Kummer elements and suppose XY^{−1} is a Kummer element.

Then XY and Y X commute. Moreover

N(X)X^{−1}Y X^{−1} = N(Y)Y^{−1}XY^{−1} = T (XY) − XY − Y X

Let

U = T (XY) + (ζ − ζ^{2})(ζXY − ζ^{2}Y X)
V = T (XY) + (ζ^{2} − ζ)(ζ^{2}XY − ζY X)

Then

X ⟶^{ζ} U ⟶^{ζ} Y, X ⟶^{ζ^{2}} V ⟶^{ζ^{2}} Y

For generic X, Y, these conditions determine the elements U, V uniquely up to multiplication by scalars.

Proof. ... □

2. Chains of length 3

Let

K = { [X] ∈ P(A) | T (X) = Q(X) = 0, N(X) ≠ 0 }

be the variety of (projective) Kummer elements. Let further

K_{r} = { ([X_{i}])_{i=0,...,r} ∈ P(A)^{r+1} | X_{i−1} ⟶^{ζ} X_{i}, i = 1, …, r }

be the variety of chains of length r and let

h_{r}: K_{r} → K × K
h_{r}(([X_{i}])_{i=0,...,r}) = ([X_{0}], [X_{r}])

be the projections.

Theorem 4. (1) The morphism h_{2} is generically an immersion.

(2) deg(h_{3}) = 2

(3) For r ≥ 4, the morphism h_{r} has a rational section.

(1) follows from Lemma 3, and (3) is shown in [1].

As for the morphism h_{3}, the fibre over the generic point ([X], [Y]) ∈ K × K has the following description:

\begin{matrix}0 = t^{2} − t(3 + ζT (XY X^{−1}Y^{−1}) + ζ^{2}T (Y XY^{−1}X^{−1})) \\ +T (XY)T (X^{−1}Y^{−1})T (XY^{−1})T (X^{−1}Y) \end{matrix}

One finds:

Lemma 5. For Kummer elements X, Y ∈ A one has

\begin{matrix}3 + T (XY X^{−1}Y^{−1}) + T (Y XY^{−1}X^{−1}) = \\ T (XY)T (X^{−1}Y^{−1}) + T (XY^{−1})T (X^{−1}Y) \end{matrix}

Proof. One uses again the formula for N(x + y) in Lemma 6, this time with x = X and y = 1 + Y + Y^{2}. Note here that N(1 + Y + Y^{2}) = 1 − 2N(Y) + N(Y)^{2} for Kummer elements (which can be also deduced formally from Lemma 6). □

The function T (XY X^{−1}Y^{−1}) is not in the function field K generated by T (XY), T (X^{−1}Y^{−1}), T (XY^{−1}), T (X^{−1}Y), because these functions are invariant under reversing the product in the algebra A, while T (XY X^{−1}Y^{−1}) is not. However, if I am not mistaken, T (XY X^{−1}Y^{−1}) satisfies a quadratic equation over K.

  Todo:  Find this relation, and give a nice description.

Perhaps one can deduce it again from Lemma 6 which seems to be really useful. For instance using the last expression (of degree 4) one finds Q(XY) = T (X^{2}Y^{2}) for Kummer elements in a degree 3 algebra.

Maybe it is a good idea to consider the cubic subalgebras

L = k ⊕ Xk ⊕ X^{2}k, H = k ⊕ Y k ⊕ Y^{2}k

of A and to analyze for λ_{i} ∈ L and μ_{i} ∈ H the product

N(λ_{1})N(μ_{1})N(λ_{2}) ⋯ N(μ_{r}) = N(λ_{1}μ_{1}λ_{2} ⋯ μ_{r})

by expanding the right hand side using Lemma 6 with respect to sums of noncommutative monomials in X, Y.

Appendix

This is copied from [2].

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.

Lemma 6. 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. □

References

[1] M. Rost, The chain lemma for Kummer elements of degree 3, C. R. Acad. Sci. Paris Sér. I Math. 328 (1999), no. 3, 185–190.

[2] , Quadratic elements in a central simple algebra of degree four, Preprint, 2003, ⟨www. math.uni-bielefeld.de/˜ rost/lines.html⟩.

Fakultät für Mathematik, Universität Bielefeld, Postfach 100131, 33501 Bielefeld, Germany

   E-mail address: rost@math.uni-bielefeld.de
   URL: http://www.math.uni-bielefeld.de/~rost