Ramification Sequences and Bezoutian Forms

Introduction

Let $F$ be a field of characteristic different from $2$. Over the rational function field $F(t)$ the second Milnor $K$-group is described by local data: at each place of $F(t)$ there is a tame symbol, the tame symbols assemble into the ramification map, and Milnor's exact sequence identifies the image of that map with the group of those sequences in a direct sum of square-class groups whose norms cancel. The image is the group $R_2(F)$ of ramification sequences of $F$, and the exact sequence leaves open which of its elements are the ramification of a single symbol $\{f,g\}$ of $k_2(F(t))$.

The obstruction to representability is a quadratic form over the base field. To a pair $(f,g)$ of polynomials Becher and Raczek attach the Bezoutian form $B(f/g)$, a non-degenerate quadratic form on $F[t]/(g)$ whose class in the Witt group $W(F)$ is computed from the pair by two simple rules, and they relate its triviality to the representability of the sequence of the pair by a symbol. The criterion turns representability into a question about the Witt group, and the worked example of this article answers that question negatively over $\mathbb{Q}$: the sequence has degree $4$, its Bezoutian is a scalar multiple of the Pfister form $\langle\!\langle 2,3\rangle\!\rangle$, that form is anisotropic because $(2,3)_{\mathbb{Q}}$ is a division algebra, and the sequence is therefore not represented by a symbol.

The base is a field $F$ of characteristic not $2$ throughout, and $F(t)$ is its rational function field. The Milnor groups, the symbols and the tame symbol are from Higher Algebraic K-Theory; the polynomial algebra and the resultant are from Polynomial Rings and Rational Functions; the quadratic forms, their discriminant and their Witt group are from Quadratic Forms and Polarisation and The Witt Group and the Grothendieck–Witt Ring, whose Pfister forms $\langle\!\langle a,b\rangle\!\rangle$ are used; the classification of forms over $\mathbb{Q}$ behind the Hasse–Minkowski principle is in Witt's Theorems; the quaternion algebras and the division criterion are from Central Simple Algebras and the Brauer Group and Division Algebras.

The Places of $F(t)$

A place of $F(t)$ is a $\mathbb{Z}$-valued valuation that is trivial on $F$, that is, a surjective map $v : F(t)^\times \to \mathbb{Z}$ with $v(xy) = v(x) + v(y)$ and $v(x + y) \geq \min\{v(x), v(y)\}$, whose restriction to $F^\times$ is zero. There are two kinds of places. Each monic irreducible polynomial $p \in F[t]$ determines the valuation $v_p$ with $v_p(p) = 1$, whose residue field is

$$ \kappa_p = F_p = F[t]/(p), \qquad [F_p : F] = \deg p ; $$

and there is the place at infinity, determined by $v_\infty(h) = -\deg h$, whose residue field is $F_\infty = F$. We write $P$ for the set of monic irreducible polynomials and

$$ P' = P \cup \{\infty\} $$

for the set of all places. The degree of a place is $[\kappa_p : F]$, so a finite place has degree $\deg p$ and the place at infinity has degree $1$.

The Tame Symbol and Milnor's Exact Sequence

The Tame Symbol

Recall from Higher Algebraic K-Theory that a discrete valuation $v$ of a field, with residue field $\kappa_v$ and uniformiser $\pi$, determines the tame symbol

$$ \partial_v : k_2 \longrightarrow k_1(\kappa_v), \qquad \partial_v(\{f,g\}) = (-1)^{v(f)v(g)} u_f^{-v(g)} u_g^{v(f)}, $$

where $f = u_f\pi^{v(f)}$ and $g = u_g\pi^{v(g)}$ are the decompositions into a power of the uniformiser and a unit, and $k_1(\kappa) = \kappa^\times/\kappa^{\times 2}$. That article proves that $\partial_v$ annihilates the defining relations of the symbols, so that it descends to a homomorphism on $k_2$.

At a finite place $p$ of $F(t)$ the uniformiser is $p$ and the decomposition is $f = p^{v_p(f)}u_f$ with $u_f = f/p^{v_p(f)}$, so the entry of a symbol is the class of

$$ (-1)^{v_p(f)v_p(g)}\left(\frac{f}{p^{v_p(f)}}\right)^{-v_p(g)}\left(\frac{g}{p^{v_p(g)}}\right)^{v_p(f)} \quad \text{in } F_p^\times/F_p^{\times 2}. $$

At the place at infinity the uniformiser is $1/t$, and the entry of a symbol lies in $F^\times/F^{\times 2}$.

Milnor's Exact Sequence

Theorem (Milnor). Let $F$ be a field of characteristic not $2$. The tame symbols at the places of $F(t)$ sum to the ramification map

$$ \partial = \sum_{p \in P'} \partial_p : k_2(F(t)) \longrightarrow \bigoplus_{p \in P'} k_1(\kappa_p), $$

the norm maps of the residue field extensions sum to $N : \bigoplus_{p\in P'}k_1(\kappa_p) \to k_1(F)$, the place at infinity contributing the identity of $k_1(F)$, and the sequence

$$ 0 \longrightarrow k_2(F) \longrightarrow k_2(F(t)) \xrightarrow{\ \partial\ } \bigoplus_{p \in P'} k_1(\kappa_p) \xrightarrow{\ N\ } k_1(F) \longrightarrow 0 $$

is exact.

The first map is extension of scalars from $F$ to $F(t)$. The theorem is Milnor's, and it is quoted here as standard; it appears in the literature with the two kinds of place of the function field in the middle term, the finite places and the place at infinity.

Definition. The group of ramification sequences of $F$ is

$$ R_2(F) = \ker N = \operatorname{im}\partial \subseteq \bigoplus_{p \in P'} k_1(\kappa_p), $$

the equality being exactness of the sequence at the middle term.

Exactness at $k_2(F(t))$ says that a class of $k_2(F(t))$ has trivial ramification sequence exactly when it comes from $k_2(F)$; exactness at the middle term says that the sequences arising are exactly those of norm $1$.

Ramification Sequences

Definition. For a finite set $S \subseteq P'$ its degree is $\deg(S) = \sum_{p \in S}[\kappa_p : F]$. For a sequence $\rho = (\rho_p)_{p \in P'}$ the support is the finite set

$$ \operatorname{Supp}(\rho) = \{p \in P' : \rho_p \neq 1\}, $$

and the degree of $\rho$ is $\deg(\rho) = \deg(\operatorname{Supp}\rho)$. A ramification sequence $\rho \in R_2(F)$ is represented by a symbol if there are $f, g \in F(t)^\times$ with $\rho = \partial(\{f,g\})$.

Since the symbols generate $k_2(F(t))$ and $\partial$ is additive, the sequences of symbols generate $R_2(F)$; the question is whether a given sequence is the sequence of one symbol, and it is the question the Bezoutian form answers.

Remark. The norm condition of the exact sequence is a genuine restriction, and it fixes the entry at the place at infinity once the finitely many entries at the finite places are given. For a finite place $p$ and a constant $c \in F^\times$ the class of $c$ is the class of the constant in $\kappa_p^\times/\kappa_p^{\times 2}$, whose norm to $F$ is $c^{[\kappa_p : F]}$; so the product of the norms of the finite entries must be trivial in $k_1(F)$, and the entry at infinity is then forced.

Definition (the sequence of a pair). Let $g \in F[t]$ be monic and square-free and let $f \in F[t]$ be coprime to $g$. Write $R(f/g)$ for the element of $R_2(F)$ whose entries are

$$ R(f/g)_p = \{f\} \quad \text{for every } p \in P \text{ dividing } g, \qquad R(f/g)_p = 1 \quad \text{for every other finite place } p, $$

the entry at infinity being the one that the norm condition forces. The definition is the source's, and it is the shape a sequence takes when its support is the set of factors of a square-free polynomial.

The Bezoutian Form

The Coefficient Functional $s_g$

Definition. Let $g \in F[t]$ be monic of degree $n$ and square-free, and let $\theta$ be the class of $t$ in the $n$-dimensional $F$-algebra $E_g = F[t]/(g)$, so that $1, \theta, \ldots, \theta^{n-1}$ is an $F$-basis of $E_g$. The coefficient functional is the $F$-linear map

$$ s_g : E_g \longrightarrow F, \qquad s_g(\theta^i) = 0 \ \ (0 \leq i \leq n-2), \qquad s_g(\theta^{n-1}) = 1, $$

so that $s_g(y)$ is the coefficient of $\theta^{n-1}$ in the expansion of $y$ in the basis. For $y \in F[t]$ of degree less than $n$ it is the coefficient of $t^{n-1}$. For $n = 1$ and $g = t + c$ the algebra is $F$ and $s_g(y) = y(-c)$.

Proposition (trace description). Let $g \in F[t]$ be monic of degree $n$ and square-free, let $\alpha_1, \ldots, \alpha_n$ be its roots in a splitting field, and let $y \in F[t]$. Then

$$ s_g(y \bmod g) = \sum_{i=1}^n \frac{y(\alpha_i)}{g'(\alpha_i)} = \operatorname{Tr}_{E_g/F}\!\left(\frac{y(\theta)}{g'(\theta)}\right), $$

the last term being the trace of the element $y(\theta)/g'(\theta)$ of $E_g$ over $F$.

Proof. Write $y = qg + r$ with $\deg r < n$. Then $s_g(y \bmod g) = s_g(r)$ is the coefficient of $t^{n-1}$ in $r$, and $y(\alpha_i) = r(\alpha_i)$ for every $i$, because $g(\alpha_i) = 0$. The partial fraction decomposition of $r/g$ is

$$ \frac{r(t)}{g(t)} = \sum_{i=1}^n \frac{r(\alpha_i)/g'(\alpha_i)}{t - \alpha_i}, $$

because $g$ is monic and square-free, so that the residues at its simple poles are $r(\alpha_i)/g'(\alpha_i)$. Expanding $\frac{1}{t-\alpha_i} = t^{-1} + \alpha_it^{-2} + \cdots$ at infinity gives

$$ \frac{r(t)}{g(t)} = t^{-1}\left(\sum_{i=1}^n\frac{r(\alpha_i)}{g'(\alpha_i)}\right) + O(t^{-2}), $$

while $\deg r < n$ and $g$ monic of degree $n$ give $\frac{r(t)}{g(t)} = r_{n-1}t^{-1} + O(t^{-2})$, where $r_{n-1}$ is the coefficient of $t^{n-1}$ in $r$. Comparing the coefficients of $t^{-1}$ gives the first identity, and the second is the definition of the trace of $y(\theta)/g'(\theta)$. $\square$

The Bezoutian Form of a Pair

Definition. Let $g \in F[t]$ be monic of degree $n$ and square-free and let $f \in F[t]$ be coprime to $g$. The Bezoutian form of $f$ modulo $g$ is the quadratic form

$$ q_{f,g} : E_g \longrightarrow F, \qquad q_{f,g}(x) = s_g\big(f(\theta)x^2\big), $$

and its class in the Witt group of The Witt Group and the Grothendieck–Witt Ring is written

$$ B(f/g) = [q_{f,g}] \in W(F). $$

The polar form of $q_{f,g}$ is the bilinear form $b(x,y) = s_g(f(\theta)xy)$, so in the basis $1, \theta, \ldots, \theta^{n-1}$ the Gram matrix is

$$ G = (G_{ij})_{0 \leq i,j \leq n-1}, \qquad G_{ij} = s_g(f\theta^{i+j}), \qquad q_{f,g}\Big(\sum_{i=0}^{n-1}x_i\theta^i\Big) = \sum_{i,j} G_{ij}x_ix_j . $$

The matrix $G$ is symmetric because $i + j = j + i$, so $q_{f,g}$ is a quadratic form in the sense of Quadratic Forms and Polarisation, with $q(x) = B(x,x)$ for its polar form $B$.

Remark (invariance). For $g$ monic and square-free and $f$ coprime to $g$ the Bezoutian depends on the pair only through the class of $f$ modulo $g$ and modulo squares: for $h \in F[t]$ coprime to $g$ and $c \in F^\times$ one has

$$ q_{f h^2, g} \cong q_{f,g} \ \text{ via } x \mapsto hx, \qquad q_{cf,g} = c\,q_{f,g}, \qquad q_{f^{-1},g} \cong q_{f,g} \ \text{ via } x \mapsto fx, $$

on $E_g$. In particular $B(f/g)$ depends on $f$ only through its class in $E_g^\times/E_g^{\times 2}$, and the element $R(f/g)$ of the definition above is therefore attached to the pair in a way compatible with the form.

Theorem (non-degeneracy). Let $g \in F[t]$ be monic and square-free and $f \in F[t]$ coprime to $g$. Then the Bezoutian form $q_{f,g}$ is non-degenerate. It is degenerate for every $f$ sharing a factor with $g$.

Proof for $n = 1$. Here $g = t + c$ and $\theta = -c$, so $E_g = F$, $s_g$ is the identity of $F$, and $q_{f,g}(x) = f(-c)x^2$. This is non-degenerate exactly when $f(-c) \neq 0$, that is, exactly when $t + c$ does not divide $f$. The general case is the theorem of Becher and Raczek, and is quoted. $\square$

The non-degeneracy is the reason the Bezoutian is taken over the square-free $g$ and the coprime $f$: the form is then a genuine element of the Witt group, and its triviality is a meaningful condition.

Remark (the classical Bezoutian). The name of the form is that of the classical Bezoutian of the pair, the polynomial

$$ \operatorname{Bez}(f,g)(x,y) = \frac{f(x)g(y) - f(y)g(x)}{x-y}, $$

whose coefficient matrix $\mathcal{B}(f,g)$ is symmetric, whose determinant is $\pm\operatorname{Res}(f,g)$, and which is non-degenerate exactly when $\gcd(f,g) = 1$. The form $q_{f,g}$ above is the normalisation of Becher and Raczek, and it is a different quadratic form from the one of the classical matrix in general: for $f = t - 2$ and $g = t^2 - 2t - 2$ the classical form is negative definite, while $q_{f,g}$ is positive definite. The classical Bezoutian and its determinant are treated in the further reading below.

The Computation Rules

The Bezoutian is computable without diagonalising a matrix, by two rules which decompose the modulus and exchange the numerator with the modulus; the rules are those of Becher and Raczek, and they are quoted.

Proposition (first rule). Let $f, g_1, g_2 \in F[t]$ be pairwise coprime, with $g_1$ and $g_2$ monic and square-free. Then in $W(F)$

$$ B\!\left(\frac{f}{g_1g_2}\right) = B\!\left(\frac{fg_2}{g_1}\right) + B\!\left(\frac{fg_1}{g_2}\right). $$

The form on the left lives on $E_{g_1g_2}$, of dimension $\deg g_1 + \deg g_2$, and the two forms on the right live on $E_{g_1}$ and $E_{g_2}$; the two sides have the same dimension, so the equality of Witt classes is an isometry of quadratic forms.

Theorem (second rule). Let $f, g \in F[t]$ be monic, square-free and coprime. Then

$$ B\!\left(\frac{f}{g}\right) + B\!\left(\frac{g}{f}\right) = \begin{cases} 0 & \text{if } \deg f \equiv \deg g \bmod 2, \\[2pt] [1] & \text{if } \deg f \not\equiv \deg g \bmod 2, \end{cases} $$

where $[1]$ is the class of the one-dimensional form $\langle 1\rangle$.

The second rule in a special case. Take $f = t$ and $g = t^2 - 1$. Here $E_g = F[t]/(t^2-1)$ with $\theta^2 = 1$, and $q_{t,g}(x) = s_g(\theta x^2)$ has Gram matrix

$$ G_{ij} = s_g(\theta^{i+j+1}), \qquad G = \begin{pmatrix} s_g(\theta) & s_g(\theta^2) \\ s_g(\theta^2) & s_g(\theta^3)\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}, $$

because $s_g(1) = s_g(\theta^3) = 0$ and $s_g(\theta) = 1$; so $q_{t,g} = \langle 1,1\rangle$. On the other side $E_t = F[t]/(t) = F$, so $q_{g,t}(x) = s_t((\theta^2 - 1)x^2) = s_t(-x^2)$ is the form $\langle -1\rangle$. The sum is $[\langle 1,1,-1\rangle] = [\langle 1\rangle]$ in $W(F)$, because $\langle 1,-1\rangle$ is the hyperbolic plane, in agreement with the rule, the degrees $1$ and $2$ having opposite parity.

A Lemma on Even Degrees

The following computation is the one used in the example; it expresses the Bezoutian of a pair whose modulus is a product of two even-degree factors through a Pfister form.

Lemma (Becher). Let $a_1, a_2 \in F^\times$ and let $g_1, g_2 \in F[t]$ be monic of even degree, coprime, and such that $g_1t$ is a square modulo $g_2$. Let $f \in F[t]$ be such that $a_if$ is a square modulo $g_i$ for $i = 1, 2$. Then

$$ B\!\left(\frac{f}{g_1g_2}\right) \sim \big[\langle\!\langle a_1a_2, g_2(0)\rangle\!\rangle\big], $$

where $\sim$ means equality up to a scalar: $\alpha \sim \alpha'$ if and only if $\alpha' = c\,\alpha$ in $W(F)$ for some $c \in F^\times$.

Proof. Write $b = g_2(0)$. By the second rule applied to the monic coprime pair $t, g_2$, whose degrees have opposite parity, and the elementary descriptions $B(g_2/t) = [b]$ and $B(t/g_2) = [1] - [b]$,

$$ B\!\left(\frac{t}{g_2}\right) = [1] - B\!\left(\frac{g_2}{t}\right) = [1] - [b]. $$

Because $g_1t$ is a square modulo $g_2$, the invariance of $q_{f,g}$ under multiplication of $f$ by a square and under inversion gives $B(g_1/g_2) = B(t/g_2)$, so $B(g_1/g_2) = [1]-[b]$; and the second rule applied to the pair $g_1, g_2$, whose degrees have the same parity, gives $B(g_2/g_1) = -B(g_1/g_2) = [b]-[1]$. Since $a_if$ is a square modulo $g_i$, the form $B(fg_j/g_i)$ differs from $a_iB(g_j/g_i)$ by a scalar, so the first rule gives

$$ B\!\left(\frac{f}{g_1g_2}\right) = B\!\left(\frac{fg_2}{g_1}\right) + B\!\left(\frac{fg_1}{g_2}\right) \sim a_1\big([b]-[1]\big) + a_2\big([1]-[b]\big). $$

As forms, the last sum is $\langle a_1b, -a_1, a_2, -a_2b\rangle$, whose multiset of square classes is $\{a_1b,\,-a_1,\,a_2,\,-a_2b\}$. The Pfister form has

$$ a_2\langle\!\langle a_1a_2, b\rangle\!\rangle = \langle a_2, -a_1a_2^2, -a_2b, a_1a_2^2b\rangle \cong \langle a_2, -a_1, -a_2b, a_1b\rangle, $$

the two entries $-a_1a_2^2$ and $a_1a_2^2b$ being the square multiples $-a_1$ and $a_1b$ of the corresponding entries, and the multiset of square classes is the same one. Hence the two forms are isometric up to the scalar, as claimed. $\square$

The Criterion of Becher and Raczek

Theorem (the criterion). Let $g \in F[t]$ be monic and square-free and let $f \in F[t]$ be coprime to $g$, and let $R(f/g) \in R_2(F)$ be the sequence of the pair. Then $R(f/g)$ is represented by a symbol if and only if

$$ B(f/g) = 0 \quad \text{in } W(F). $$

The two implications are the lemma and the theorem of Becher and Raczek, quoted as standard: a symbol whose ramification is the sequence of a pair has trivial Bezoutian, and a pair whose Bezoutian is trivial has its sequence represented by a symbol.

Proposition (the concrete form). Let $a_1, a_2 \in F^\times$ and let $g_1, g_2 \in F[t]$ be monic of even degree and coprime, with $g_1t$ a square modulo $g_2$. Assume that the quadratic form $\langle 1, -a_1a_2\rangle$ over $F(t)$ does not represent $g_2(0)$, and that $a_i$ is a non-square in $\kappa_p$ for every irreducible factor $p$ of $g_i$. Then

$$ \partial\big(\{g_1, a_1\} + \{g_2, a_2\}\big) \neq \partial(\sigma) \qquad \text{for every symbol } \sigma \in k_2(F(t)). $$

Proof. Write $\rho = \partial(\{g_1,a_1\}+\{g_2,a_2\}) \in R_2(F)$. The hypotheses on the $a_i$ give $\operatorname{Supp}(\rho) = \{p \in P : p \text{ divides } g_1g_2\}$ and $\deg(\rho) = \deg g_1 + \deg g_2$. If a symbol $\sigma$ had $\partial(\sigma) = \rho$, then by the structure of the sequences of symbols there would be polynomials $f, h$ and a square-free $g$, pairwise coprime, with $g = g_1g_2$, such that $\sigma = \{f, gh\}$ and $\partial(\sigma) = R(f/g)$; the criterion would give $B(f/g) = 0$. Moreover $R(f/g_1) + R(f/g_2) = R(f/g) = \partial(\sigma) = \rho = R(a_1/g_1) + R(a_2/g_2)$, because the support of $R(f/g)$ splits into the places dividing $g_1$ and those dividing $g_2$; comparing the entries at the places dividing $g_i$ gives $R(f/g_i) = R(a_i/g_i)$, that is, $a_if$ is a square modulo $g_i$. The lemma then gives $B(f/g) \sim [\langle\!\langle a_1a_2, g_2(0)\rangle\!\rangle]$, so $a_2\langle\!\langle a_1a_2, g_2(0)\rangle\!\rangle$ is hyperbolic and the form $\langle 1, -a_1a_2\rangle$ represents $g_2(0)$, against the hypothesis. $\square$

The Example over $\mathbb{Q}$

Take $F = \mathbb{Q}$, $a = 2$, $b = 3$, and

$$ g_1 = t^2+3t+2 = (t+1)(t+2), \qquad g_2 = t^2+2t+2 = (t+1)^2+1, $$

so that $g_2$ is irreducible with $g_2(0) = a = 2$ and residue field $F_{g_2} = \mathbb{Q}(i)$, and $g_1, g_2$ are monic of degree $2$, coprime and square-free. The hypotheses of the proposition hold: $a_1 = a = 2$ is a non-square modulo each of the factors $t+1$ and $t+2$ of $g_1$, whose residue field is $\mathbb{Q}$, and $a_2 = ab = 6$ is a non-square in $\mathbb{Q}(i)$; and the form $\langle 1,-a_1a_2\rangle = \langle 1, -12\rangle$ does not represent $g_2(0) = 2$, because the quaternion algebra $(12,2)_{\mathbb{Q}} = (3,2)_{\mathbb{Q}}$ is non-split, having Hilbert symbol $-1$ at $2$ and at $3$. The same holds over $\mathbb{Q}(t)$, the map $\operatorname{Br}(\mathbb{Q}) \to \operatorname{Br}(\mathbb{Q}(t))$ being injective.

The class

$$ \xi = \{g_1, a\} + \{g_2, ab\} = \{g_1,2\} + \{g_2,6\} \in k_2(\mathbb{Q}(t)) $$

has the ramification sequence

$$ \rho_{t+1} = \{2\}, \qquad \rho_{t+2} = \{2\}, \qquad \rho_{g_2} = \{6\}, \qquad \operatorname{Supp}\rho = \{t+1, t+2, g_2\}, $$

of degree $1 + 1 + 2 = 4$, and the entry at infinity is $1$: the sum of the norms of the finite entries is $2 \cdot 2 \cdot N_{\mathbb{Q}(i)/\mathbb{Q}}(6) = 2 \cdot 2 \cdot 6^2 = 144$, a square of $\mathbb{Q}$, hence trivial in $k_1(\mathbb{Q})$, so the norm condition forces $\rho_\infty = 1$ and $\rho \in R_2(\mathbb{Q})$ is a genuine ramification sequence of degree $4$.

If $\rho$ were represented by a symbol, the pair would be $(f, g_1g_2)$ for some $f$ with $a_if$ a square modulo $g_i$, by the argument of the proposition. The canonical representative is the solution of the congruences $f \equiv 2 \pmod{g_1}$, $f \equiv 6 \pmod{g_2}$ by an element of degree less than $4$, namely

$$ f = -2t^3-10t^2-16t-6 . $$

Indeed $2f \equiv 4$ modulo $g_1$ and $6f \equiv 36$ modulo $g_2$ are squares, so the pair satisfies the hypotheses of the lemma with $a_1 = 2$, $a_2 = 6$ and $b = g_2(0) = 2$.

The Bezoutian form $q_{f,g_1g_2}$ on $\mathbb{Q}[t]/(g_1g_2)$ has dimension $4$, and in the basis $1, \theta, \theta^2, \theta^3$ its Gram matrix $G_{ij} = s_g(f\theta^{i+j})$ is

$$ G = \begin{pmatrix} -2 & 0 & 4 & -6\\ 0 & 4 & -6 & -2\\ 4 & -6 & -2 & 30\\ -6 & -2 & 30 & -86 \end{pmatrix}, $$

symmetric, of determinant $144$ and signature $(2,2)$. Its Lagrange diagonalisation over $\mathbb{Q}$ is

$$ q_{f,g_1g_2} \cong \langle -2, 4, -3, 6\rangle \cong 6\,\langle 1,-2,-3,6\rangle = 6\,\langle\!\langle 2,3\rangle\!\rangle, $$

so the Bezoutian is a scalar multiple of the Pfister form $\langle\!\langle a_1a_2, g_2(0)\rangle\!\rangle = \langle\!\langle 12, 2\rangle\!\rangle = \langle\!\langle 2,3\rangle\!\rangle$ of the lemma, as it must be. That Pfister form is the norm form of the quaternion algebra $(2,3)_{\mathbb{Q}}$, and that algebra is a division algebra, so the form is anisotropic; hence the Bezoutian form is anisotropic, in particular not hyperbolic, and $B(f/g_1g_2) \neq 0$ in $W(\mathbb{Q})$. By the criterion, the sequence $\rho$ is not represented by a symbol.

The Biquaternion Division Algebras

The same computation, over a general field, gives the biquaternion division algebras over $F(t)$ of Division Algebras. With $g_1 = t^2+(a+1)t+a$ and $g_2 = t^2+at+a$ the class $\{g_1,a\}+\{g_2,ab\}$ corresponds, under the dictionary between $k_2$ and the two-torsion of the Brauer group of Central Simple Algebras and the Brauer Group, to the biquaternion algebra

$$ B = (g_1, a)_{F(t)} \otimes_{F(t)} (g_2, ab)_{F(t)}, $$

and the ramification of $B$ differs from the ramification of every quaternion algebra over $F(t)$ exactly when the sequence is not represented by a symbol. Becher's theorem states the equivalence.

Theorem (Becher). Let $a, b \in F^\times$ with $a \notin F^{\times 2}$ and $b \notin aF^{\times 2} \cup (a-4)F^{\times 2}$. Then the following are equivalent:

$\{a,b\} = 0$ in $k_2(F)$, that is, the quaternion algebra $(a,b)_F$ is split;
$\partial(\{g_1,a\}+\{g_2,ab\}) = \partial(\sigma)$ for some symbol $\sigma \in k_2(F(t))$, where $g_1 = t^2+(a+1)t+a$ and $g_2 = t^2+at+a$.

In particular, if $(a,b)_F$ is not split, then the ramification of $B$ is not the ramification of any quaternion algebra over $F(t)$, a situation the source records by saying that $B$ has Faddeev index $4$; then $B$ is a division algebra over $F(t)$ that contains no quaternion algebra defined over $F$.

Theorem (Becher, existence over a general field). Assume that $k_2(F) \neq 0$ and that $F$ is not real euclidean, the set of squares of $F$ not being an ordering. Then there is a field element $a$ and a $b$ as above such that the ramification sequence $\rho = \partial(\{g_1,a\}+\{g_2,ab\})$ of degree $4$ is not represented by a symbol, and the algebra $B$ above is a division algebra that does not contain any quaternion algebra defined over $F$.

The example of this article is the case $F = \mathbb{Q}$, $a = 2$, $b = 3$ of the two theorems.

Summary

Let $F$ be a field of characteristic not $2$. The places of the rational function field $F(t)$ are the monic irreducible polynomials $p \in F[t]$, with residue field $\kappa_p = F[t]/(p)$, and the place at infinity, with residue field $F$. The tame symbol at a place is the homomorphism $\partial_p(\{f,g\}) = (-1)^{v(f)v(g)}u_f^{-v(g)}u_g^{v(f)}$ into $k_1(\kappa_p) = \kappa_p^\times/\kappa_p^{\times 2}$, and by Milnor's exact sequence

$$ 0 \to k_2(F) \to k_2(F(t)) \xrightarrow{\ \partial\ } \bigoplus_{p \in P'} k_1(\kappa_p) \xrightarrow{\ N\ } k_1(F) \to 0 $$

the ramification map is injective on $k_2(F)$ and its image is the group of ramification sequences $R_2(F)$, the kernel of the norm. The support and the degree of a sequence measure the places carrying it, and a sequence is represented by a symbol when it is $\partial(\{f,g\})$ for a single symbol.

The Bezoutian form of $f$ modulo a monic square-free $g$ is the form $q_{f,g}(x) = s_g(f(\theta)x^2)$ on $E_g = F[t]/(g)$, where $s_g$ is the coefficient of $\theta^{n-1}$ in the basis $1,\theta,\ldots,\theta^{n-1}$; equivalently $s_g(y\bmod g) = \operatorname{Tr}_{E_g/F}(y(\theta)/g'(\theta))$. It is non-degenerate exactly when $f$ is coprime to $g$, and its class $B(f/g)$ in the Witt group is computed by the two rules

$$ B\!\left(\frac{f}{g_1g_2}\right) = B\!\left(\frac{fg_2}{g_1}\right) + B\!\left(\frac{fg_1}{g_2}\right), \qquad B\!\left(\frac{f}{g}\right) + B\!\left(\frac{g}{f}\right) = 0 \ \text{or}\ [1], $$

the second value occurring when the degrees of $f$ and $g$ have opposite parity. For monic $g_1, g_2$ of even degree with $g_1t$ a square modulo $g_2$ and $a_if$ a square modulo $g_i$, $B(f/g_1g_2)$ is a scalar multiple of the Pfister form $\langle\!\langle a_1a_2, g_2(0)\rangle\!\rangle$. The criterion of Becher and Raczek reads the representability of the sequence $R(f/g)$ by a symbol from the vanishing of $B(f/g)$, and the concrete form of the criterion is the proposition above: under its hypotheses no symbol has the same ramification as $\{g_1,a_1\}+\{g_2,a_2\}$.

Over $\mathbb{Q}$ the class $\{g_1,2\}+\{g_2,6\}$ with $g_1 = (t+1)(t+2)$ and $g_2 = t^2+2t+2$ has a ramification sequence of degree $4$ with entries $2$, $2$, $6$ and norm $144$, and the Bezoutian form of the pair with $f = -2t^3-10t^2-16t-6$ has determinant $144$, signature $(2,2)$ and diagonalisation $\langle -2,4,-3,6\rangle \cong 6\langle\!\langle 2,3\rangle\!\rangle$. The Pfister form is the norm form of the division algebra $(2,3)_{\mathbb{Q}}$, so it is anisotropic and the sequence is not represented by a symbol. The corresponding biquaternion algebra over $\mathbb{Q}(t)$ is a division algebra that contains no quaternion algebra defined over $\mathbb{Q}$, and Becher's theorems give the same conclusion over every field that is not real euclidean and has a non-split quaternion algebra.

Summary of Notation

Symbol	Meaning
$F$, $F(t)$	Field of characteristic not $2$; its rational function field
$P$, $P'$	Monic irreducible polynomials of $F[t]$; the places $P \cup \{\infty\}$
$v_p$, $v_\infty$	Valuation of a finite place; valuation $-\deg$ of the place at infinity
$\kappa_p$, $F_p$	Residue field of the place $p$; for $p \in P$ it is $F[t]/(p)$
$[\kappa_p : F]$	Degree of a place
$k_2$, $k_1$	Milnor groups modulo $2$ of Higher Algebraic K-Theory
$\partial_p$, $\partial$	Tame symbol at $p$; the ramification map, the sum of the tame symbols
$N$	Sum of the norm maps of the residue extensions
$R_2(F)$	Group of ramification sequences, $\ker N = \operatorname{im}\partial$
$\operatorname{Supp}\rho$, $\deg\rho$	Support and degree of a ramification sequence
$R(f/g)$	The sequence with entry $\{f\}$ at each divisor of $g$ and $1$ elsewhere
$E_g = F[t]/(g)$	The $F$-algebra of a monic square-free $g$, of degree $n$
$\theta$	The class of $t$ in $E_g$
$s_g$	Coefficient functional of $\theta^{n-1}$; $\operatorname{Tr}_{E_g/F}(\cdot/g'(\theta))$
$q_{f,g}$, $B(f/g)$	Bezoutian form of $f$ modulo $g$; its class in $W(F)$
$G_{ij} = s_g(f\theta^{i+j})$	Gram matrix of the Bezoutian form
$\langle\!\langle a,b\rangle\!\rangle$	Pfister form $\langle 1,-a,-b,ab\rangle$
$\sim$	Equality of Witt classes up to a scalar
$\operatorname{Bez}(f,g)$, $\mathcal{B}(f,g)$	Classical Bezoutian and its coefficient matrix
$g_1$, $g_2$	The polynomials $t^2+(a+1)t+a$ and $t^2+at+a$ of the application