Characteristic 2

Abstract

In this chapter, we extend the results from the previous four chapters to the neglected case where the base field has characteristic 2. Throughout this chapter, let F be a field with algebraic closure $F{}^{al }$.

In this chapter, we extend the results from the previous four chapters to the neglected case where the base field has characteristic 2. Throughout this chapter, let F be a field with algebraic closure $F{}^{al }$.

1 Separability

To get warmed up, we give a different notation (symbol) for quaternion algebras that holds in any characteristic and which is convenient for many purposes.

Definition 6.1.1

Let A be a commutative, finite-dimensional algebra over F. We say A is separable if

$$\begin{aligned} A \otimes _F F{}^{al }\simeq F{}^{al }\times \dots \times F{}^{al }; \end{aligned}$$

otherwise, we say A isinseparable.

Example 6.1.2

If $A \simeq F[x]/(f(x))$ with $f(x) \in F[x]$, then A is separable if and only if f has distinct roots in $F{}^{al }$.

6.1.3

If ${{\,\mathrm{char}\,}}F \ne 2$, and K is a quadratic F-algebra, then after completing the square, we see that the following are equivalent:

• (i) K is separable;

• (ii) $K \simeq F[x]/(x^2-a)$ with $a \ne 0$;

• (iii) K is reduced (K has no nonzero nilpotent elements);

• (iv) K is a field or $K \simeq F \times F$.

6.1.4

If ${{\,\mathrm{char}\,}}F = 2$, then a quadratic F-algebra K is separable if and only if

$$\begin{aligned} K \simeq F[x]/(x^2+x+a) \end{aligned}$$

for some $a \in F$. A quadratic algebra of the form $K=F[x]/(x^2+a)$ with $a \in F$ is inseparable.

Now we introduce the more general notation.

6.1.5

Let K be a separable quadratic F-algebra, and let $b \in F^\times$. We denote by

$$\begin{aligned} \displaystyle {\biggl (\frac{K,b}{F}\biggr )} :=K \oplus Kj \end{aligned}$$

the F-algebra with basis 1, j as a left K-vector space and with the multiplication rules $j^2=b$ and $j\alpha =\overline{\alpha }j$ for $\alpha \in K$, where $\overline{\phantom {x}}$ is the standard involution on K (the nontrivial element of ${{\,\mathrm{Gal}\,}}(K\,|\,F)$ if K is a field). We will also write $({K,b} \mid {F})$ for formatting.

From 6.1.3, if ${{\,\mathrm{char}\,}}F \ne 2$ then writing $K \simeq F[x]/(x^2-a)$ we see that

$$\begin{aligned} \displaystyle {\biggl (\frac{K,b}{F}\biggr )} \simeq \displaystyle {\biggl (\frac{a,b}{F}\biggr )} \end{aligned}$$

is a quaternion algebra over F. The point is that we cannot complete the square in characteristic 2, so the more general notation gives a characteristic-independent way to define quaternion algebras. In using this symbol, we are breaking the symmetry between the standard generators $i,j$, but otherwise have not changed anything about the definition.

2 Quaternion algebras

Throughout the rest of this chapter, we suppose that ${{\,\mathrm{char}\,}}F = 2$. (We will occasionally remind the reader of this supposition, but it is meant to hold throughout.)

Definition 6.2.1

An algebra B over F (with ${{\,\mathrm{char}\,}}F=2$) is aquaternion algebra if there exists an F-basis $1,i,j,k$ for B such that

$$\begin{aligned} i^2+i=a,\ j^2=b,\ \text { and }\ k=ij=j(i+1) \end{aligned}$$

(6.2.2)

with $a \in F$ and $b \in F^\times$.

Just as when ${{\,\mathrm{char}\,}}F \ne 2$, we find that the multiplication table for a quaternion algebra B is determined by the rules (6.2.2), e.g.

$$\begin{aligned} jk=j(ij)=(ij+j)j=bi+b=kj+b. \end{aligned}$$

We denote by $\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ or $[{a,b} \mid {F})$ the F-algebra with basis $1,i,j,ij$ subject to the multiplication rules (6.2.2). The algebra $\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ is not symmetric in a, b (explaining the choice of notation), but it is still functorial in the field F.

If we let $K=F[i] \simeq F[x]/(x^2+x+a)$, then

$$ \displaystyle {\biggl [\frac{a,b}{F}\biggr )} \simeq \displaystyle {\biggl (\frac{K,b}{F}\biggr )} $$

and our notation extends that of Section 6.1.

Example 6.2.3

The ring ${{\,\mathrm{M}\,}}_2(F)$ of $2\times 2$-matrices with coefficients in F is again a quaternion algebra over F, via the isomorphism

Lemma 6.2.4

An F-algebra B is a quaternion algebra if and only if there exist F-algebra generators $i,j\in B$ satisfying

$$\begin{aligned} i^2+i=a,\ j^2 = b, \ \text { and }\ ij=j(i+1). \end{aligned}$$

(6.2.5)

Proof. Proven the same way as Lemma 2.2.5. $\square$

6.2.6

Let $B=[{a,b} \mid {F})$ be a quaternion algebra over F. Then B has a (unique) standard involution $\overline{\phantom {x}}:B \rightarrow B$ given by

$$\begin{aligned} \alpha =t+xi+yj+zij\mapsto \overline{\alpha }=x+\alpha =(t+x)+xi+yj+zij\end{aligned}$$

since

$$\begin{aligned} \begin{aligned} \alpha \overline{\alpha }&= (t+xi+yj+zij)((t+x)+xi+yj+zij) \\&= t^2+tx+ax^2+by^2+byz+abz^2 \in F. \end{aligned} \end{aligned}$$

(6.2.7)

Consequently, one has a reduced trace and reduced norm on B as in Chapter 3.

We now state a version of Theorem 3.5.1 in characteristic 2; the proof is similar and is left as an exercise.

Theorem 6.2.8

Let B be a division F-algebra with a standard involution that is not the identity. Then either B is a separable quadratic field extension of F or B is a quaternion algebra over F.

Proof. Exercise 6.9. (This theorem is also implied by Theorem 6.4.1.) $\square$

3 $*$ Quadratic forms

We now turn to the theory of quadratic forms over F with ${{\,\mathrm{char}\,}}F=2$. The basic definitions from section 4.2 apply. For further reference, Grove [Grov2002, Chapters 12–14] treats quadratic forms in characteristic 2, and the book by Elman–Karpenko–Merkurjev [EKM2008, Chapters I–II] discusses bilinear forms and quadratic forms in all characteristics.

Let $Q:V \rightarrow F$ be a quadratic form with $\dim _F V=n<\infty$ and associated bilinear form T. Then $T(x,x)=2Q(x)=0$ for all $x \in V$, so one cannot recover the quadratic form from the symmetric (equivalently, alternating) bilinear form.

6.3.1

We begin with the definition of the discriminant. When n is even, we simply define ${{\,\mathrm{disc}\,}}(Q)=\det (T) \in F/F^{\times 2}$—this is equivalent to Definition 4.3.3 when ${{\,\mathrm{char}\,}}F \ne 2$, having absorbed the square power of 2.

When n is odd, the symmetric matrix T always has determinant 0 (Exercise 6.8); we need to “divide this by 2”. So instead we work with a generic quadratic form, as follows. Consider the quadratic form

$$\begin{aligned} Q^{univ }(x_1,\dots ,x_n) :=\sum _{1 \le i \le j \le n} a_{ij} x_i x_j = a_{11} x_1^2 + a_{12} x_1x_2 + \dots + a_{nn} x_n^2 \end{aligned}$$

(6.3.2)

over the field $F^{univ } :=\mathbb Q (a_{ij})_{i,j=1,\dots ,n}$ (now of characteristic zero!) with $a_{ij}$ transcendental elements. We compute its universal determinant

$$\begin{aligned} \det ([T^{univ }])=\det \begin{pmatrix} 2a_{11} &{} a_{12} &{} \cdots &{} a_{1n} \\ a_{12} &{} 2a_{22} &{} \cdots &{} a_{2n} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ a_{1n} &{} a_{2n} &{} \cdots &{} 2a_{nn} \end{pmatrix} \in 2\mathbb Z [a_{ij}]_{i,j} \end{aligned}$$

(6.3.3)

as a polynomial with integer coefficients. We claim all of these coefficients are even: indeed, reducing modulo 2 and computing the determinant over $\mathbb F _2(a_{ij})_{i,j}$, we recall that the determinant of an alternating matrix of odd size is zero (over any field). Therefore, we may let

$$\begin{aligned} \delta (a_{11},\dots ,a_{nn}) :=\det (T^{univ })/2 \in \mathbb Z [a_{ij}]_{i,j} \end{aligned}$$

(6.3.4)

be the universal (half-)discriminant. We then define

$$\begin{aligned} {{\,\mathrm{disc}\,}}(Q) :=\delta (Q(e_1),T(e_1,e_2),\dots ,Q(e_n)) \in F/F^{\times 2} \end{aligned}$$

by specialization. Repeating the argument in 4.3.2, if $t_{ij} \in F$ and $e_i' :=\sum _j t_{ij} e_i$ then

$$\begin{aligned} \delta (Q(e_1'),T(e_1',e_2'),\dots ,Q(e_n')) = \delta (Q(e_1),T(e_1,e_2),\dots ,Q(e_n))\det (t_{ij})^2 \end{aligned}$$

(verified universally!) so ${{\,\mathrm{disc}\,}}(Q)$ is well-defined. Moreover, this definition agrees with Definition 4.3.3 when ${{\,\mathrm{char}\,}}F \ne 2$.

Example 6.3.5

For example, ${{\,\mathrm{disc}\,}}(\langle a \rangle )=a$ for $a \in F$, and if

$$\begin{aligned} Q(x,y,z)=ax^2+by^2+cz^2+uyz+vxz+wxy \end{aligned}$$

with $a,b,c,u,v,w \in F$, then

$$\begin{aligned} {{\,\mathrm{disc}\,}}(Q)=4abc+uvw-au^2-bv^2-cw^2 \end{aligned}$$

in all characteristics.

Definition 6.3.6

We say Q is nondegenerate if ${{\,\mathrm{disc}\,}}(Q) \ne 0$.

Next, even though not every quadratic form over F can be diagonalized, so we will also make use of one extra form: for $a,b \in F$, we write [a, b] for the quadratic form $ax^2+axy+by^2$ on $F^2$.

Lemma 6.3.7

There exists a basis of V such that

$$\begin{aligned} Q \simeq [a_1,b_1] \boxplus \dots \boxplus [a_m,b_m] \boxplus \langle c_1, \dots , c_r \rangle \end{aligned}$$

(6.3.8)

with $a_i,b_i,c_j \in F$.

Proof. Exercise 6.11. $\square$

We say that a quadratic form Q isnormalized if Q is presented with a basis as in (6.3.8).

Example 6.3.9

The quadratic forms $Q(x,y,z)=x^2+yz+z^2$ and $Q(x,y,z)=x^2+y^2+z^2$ are normalized over $\mathbb F _2$, but the quadratic form $Q(x,y,z)=xz+yz+z^2$ is not.

Example 6.3.10

For a normalized quadratic form as in (6.3.8),

$$\begin{aligned} {{\,\mathrm{disc}\,}}(Q)&={{\,\mathrm{disc}\,}}([a_1,b_1] \boxplus \cdots \boxplus [a_m,b_m]){{\,\mathrm{disc}\,}}(\langle c_1, \dots , c_r \rangle ) \\&= (a_1 \cdots a_m)^2 {{\,\mathrm{disc}\,}}(\langle c_1,\dots ,c_r \rangle ). \end{aligned}$$

In $F/F^{\times 2}$, we have

Therefore, Q is nondegenerate if and only if $a_1\cdots a_m c_1 \cdots c_r \ne 0$ and $r \le 1$.

Example 6.3.11

Let $B=\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ be a quaternion algebra. Then $1,i,j,ij$ is a normalized basis for B, and by (6.2.7),

$$\begin{aligned} {{\,\mathrm{nrd}\,}}\simeq [1,a] \boxplus [b,ab], \end{aligned}$$

so ${{\,\mathrm{disc}\,}}({{\,\mathrm{nrd}\,}})=b^2$ so ${{\,\mathrm{nrd}\,}}$ is nondegenerate.

4 $*$ Characterizing quaternion algebras

We now consider the characterization of quaternion algebras as those equipped with a nondegenerate standard involution (revisiting Main Theorem 4.4.1, but now with ${{\,\mathrm{char}\,}}F = 2$).

Theorem 6.4.1

Let B be an F-algebra (with ${{\,\mathrm{char}\,}}F=2$). Then B has a nondegenerate standard involution if and only if one of the following holds:

• (i) $B=F$;

• (ii) $B=K$ is a separable quadratic F-algebra; or

• (iii) B is a quaternion algebra over F.

Proof. If $B=F$, then the standard involution is the identity, and ${{\,\mathrm{nrd}\,}}$ is nondegenerate on F because the reduced (half-)discriminant of the quadratic form ${{\,\mathrm{nrd}\,}}(x)=x^2$ is 1.

If $\dim _F B=2$, then $B=K$ has a unique standard involution (Lemma 3.4.2). By 6.1.4, we see that the involution is nondegenerate if and only if K is separable.

So suppose $\dim _F B > 2$. Since B has a nondegenerate standard involution, there exists an element $i \in B$ such that $T(i,1)={{\,\mathrm{trd}\,}}(i) \ne 0$. We have $i \not \in F$ since ${{\,\mathrm{trd}\,}}(F)=\{0\}$. Rescaling we may suppose ${{\,\mathrm{trd}\,}}(i)=1$, whence $i^2=i+a$ for some $a \in F$, and ${{\,\mathrm{nrd}\,}}|_{F+Fi} = [1,a]$. (We have started the proof of Lemma 6.3.7, and 1, i is part of a normalized basis, in this special case.)

By nondegeneracy, there exists $j \in \{1,i\}^\perp$ such that ${{\,\mathrm{nrd}\,}}(j)=b \ne 0$. Thus ${{\,\mathrm{trd}\,}}(j)=0$ so $\overline{j}=j$ and $j^2=b \in F^\times$. Furthermore,

$$\begin{aligned} 0={{\,\mathrm{trd}\,}}(ij)=ij+j\overline{i}=ij+j(i+1) \end{aligned}$$

so $ij=j(i+1)$. Therefore $i,j$ generate an F-subalgebra $A \simeq [{a,b} \mid {F})$.

The conclusion of the proof follows exactly as in (4.4.3): if $k \in \{1,i,j,ij\}^\perp$ then $k(ij)=k(ji)$, a contradiction. $\square$

Corollary 6.4.2

Let B be a quaternion algebra over F, and suppose that $K \subseteq B$ is a commutative separable F-subalgebra. Then $\dim _F K \le 2$. Moreover, if $K \ne F$, then the centralizer of $K^\times$ in $B^\times$ is again $K^\times$.

Next, we characterize isomorphism classes of quaternion algebras in characteristic 2 in the language of quadratic forms.

6.4.3

Let B be a quaternion algebra over F. We again define

$$\begin{aligned} B^0 :=\{\alpha \in B : {{\,\mathrm{trd}\,}}(\alpha )=0\}=\{1\}^\perp . \end{aligned}$$

(6.4.4)

But now $B^0=F \oplus Fj \oplus Fk$ and in this basis

$$\begin{aligned} {{\,\mathrm{nrd}\,}}(x+yj+zij)=x^2+by^2+byz+abz^2 \end{aligned}$$

(6.4.5)

so ${{\,\mathrm{nrd}\,}}|_{B^0} \simeq \langle 1 \rangle \boxplus [b,ab]$. The discriminant is therefore

$$\begin{aligned} {{\,\mathrm{disc}\,}}({{\,\mathrm{nrd}\,}}|_{B^0})=b^2 = 1 \in F^{\times }/F^{\times 2}. \end{aligned}$$

(6.4.6)

Theorem 6.4.7

Let F be a field with ${{\,\mathrm{char}\,}}F=2$. Then the functor $B \mapsto {{\,\mathrm{nrd}\,}}|_{B^0}$ yields an equivalence of categories between

Quaternion algebras over F,

under F-algebra isomorphisms

and

Ternary quadratic forms over F with discriminant $1 \in F^{\times }/F^{\times 2}$,

under isometries.

Proof. We argue as in Theorem 5.6.8 but with ${{\,\mathrm{char}\,}}F = 2$. The argument here is easier, because all sign issues go away and there is no orientation to chase: by Exercise 6.12, there is a unique $\zeta \in {{\,\mathrm{Clf}\,}}^1 Q \smallsetminus F$ such that $\zeta ^2=1$. The inclusion $\iota :B^0 \hookrightarrow B$ induces a surjective F-algebra homomorphism ${{\,\mathrm{Clf}\,}}^0({{\,\mathrm{nrd}\,}}|_{B^0}) \rightarrow B$, so by dimensions it is an isomorphism; this gives one natural transformation. In the other direction, the map $m_\zeta :V \rightarrow B^0$ by $v \mapsto v\zeta$ is again an isometry by (5.6.10), giving the other.

Here is a second direct proof. By 6.4.3, the quadratic form ${{\,\mathrm{nrd}\,}}|_{B^0}$ has discriminant 1. To show the functor is essentially surjective, let $Q:V \rightarrow F$ be a ternary quadratic form with discriminant $1 \in F^{\times }/F^{\times 2}$. Then $Q \simeq \langle u \rangle \boxplus [b,c]$ for some $u,b,c \in F$. We have ${{\,\mathrm{disc}\,}}(Q)=ub^2=1 \in F^{\times 2}$ so $b \in F^\times$ and $u \in F^{\times 2}$. Rescaling the first variable, we obtain $Q \sim \langle 1 \rangle \boxplus [b, c]$. Thus by 6.4.3, Q arises up to isometry from the quaternion algebra $\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ with $a=cb^{-1}$.

For morphisms, we argue as in the proof of Proposition 5.2.4 but with ${{\,\mathrm{char}\,}}F=2$. In one direction, an F-algebra isomorphism induces an isometry by uniqueness of the standard involution. Conversely, let $f:B^0 \rightarrow (B')^0$ be an isometry. Let $B \simeq \displaystyle {\biggl [\frac{a,b}{F}\biggr )}$. Extend f to an F-linear map $B \rightarrow B'$ by mapping $i \mapsto b^{-1} f(ij)f(j)$. The map f preserves 1: it maps F to F by Exercise 6.15, since $F=(B^0)^\perp =((B')^0)^\perp$, and $1={{\,\mathrm{nrd}\,}}(1)={{\,\mathrm{nrd}\,}}(f(1))=f(1)^2$ so $f(1)=1$. We have $f(j)^2={{\,\mathrm{nrd}\,}}(f(j))={{\,\mathrm{nrd}\,}}(j)=b$ and similarly $f(ij)^2=ab$ since $j,ij\in B^0$. Thus

$$\begin{aligned} 1&={{\,\mathrm{trd}\,}}(i)=b^{-1}{{\,\mathrm{trd}\,}}((ij)j)=b^{-1}T(ij,j)= \\&=b^{-1}T(f(ij),f(j))={{\,\mathrm{trd}\,}}(b^{-1}f(ij)f(j))={{\,\mathrm{trd}\,}}(f(i)) \end{aligned}$$

and similarly ${{\,\mathrm{nrd}\,}}(f(i))={{\,\mathrm{nrd}\,}}(i)=a$, thus $f(i)^2+f(i)+a=0$. Finally,

$$\begin{aligned} f(i)f(j)=b^{-1}f(ij)f(j)^2=f(ij) \end{aligned}$$

and

$$\begin{aligned} f(j)f(i)&= b^{-1}f(j)f(ij)f(j) = b^{-1} f(j)(f(j)f(ij)+T(f(j),f(ij))) \\&=f(ij)+f(j)=(f(i)+1)f(j)=\overline{f(i)}f(j) \end{aligned}$$

so f is an isomorphism of F-algebras. Therefore the functor is full and faithful, yielding an equivalence of categories. $\square$

Corollary 6.4.8

The maps $B \mapsto Q={{\,\mathrm{nrd}\,}}|_{B^0} \mapsto C=V(Q)$ yield bijections

that are functorial with respect to F.

Proof. The remaining parts of the bijection follow as in the proof of Corollary 5.2.6. $\square$

We now turn to identifying the matrix ring in characteristic 2.

Definition 6.4.9

A quadratic form $H :V \rightarrow F$ is a hyperbolic plane if $H \simeq [1,0]$.

Recall that $[1,0]:F^2 \rightarrow F$ is given by the quadratic form $x^2+xy=x(x+y)$, visibly isometric to the quadratic form xy. Definition 6.4.9 agrees with Definition 5.4.1 after a change of basis.

Lemma 6.4.10

If Q is nondegenerate and isotropic then $Q \simeq H \boxplus Q'$ with H a hyperbolic plane.

Proof. We repeat the proof of Lemma 5.4.2. $\square$

We may again characterize division quaternion algebras by examination of the reduced norm as a quadratic form as in Main Theorem 5.4.4 and Theorem 5.5.3.

Theorem 6.4.11

Let $B=\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ (with ${{\,\mathrm{char}\,}}F=2$). Then the following are equivalent:

• (i) $B \simeq \displaystyle {\biggl [\frac{1,1}{F}\biggr )} \simeq {{\,\mathrm{M}\,}}_2(F)$;

• (ii) B is not a division ring;

• (iii) The quadratic form ${{\,\mathrm{nrd}\,}}$ is isotropic;

• (iv) The quadratic form ${{\,\mathrm{nrd}\,}}|_{B^0}$ is isotropic;

• (v) The binary form [1, a] represents b;

• (vi) $b \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$ where $K=F[i]$; and

• (vii) The conic $C :=V({{\,\mathrm{nrd}\,}}|_{B^0}) \subset \mathbb P ^2$ has an F-rational point.

Proof. Only condition (v) requires significant modification in the case ${{\,\mathrm{char}\,}}F = 2$; see Exercise 6.13. $\square$

Lemma 6.4.12

Let $K \supset F$ be a quadratic extension of fields. Then K is a splitting field for B if and only if there is an injective F-algebra homomorphism $K \hookrightarrow B$.

Proof. If $\iota :K \hookrightarrow B$ and $K=F(\alpha )$, then $1 \otimes \alpha - \iota (\alpha ) \otimes 1$ is a zerodivisor in $B \otimes _F K$, since

$$\begin{aligned} \left( 1 \otimes \alpha - \iota (\alpha ) \otimes 1\right) \left( 1 \otimes \alpha - \iota (\overline{\alpha }) \otimes 1\right) =0, \end{aligned}$$

(6.4.13)

and so $B \otimes _F K \simeq {{\,\mathrm{M}\,}}_2(K)$ and K is a splitting field.

Conversely, let $K=F(\alpha )$ and suppose $B \otimes _F K \simeq {{\,\mathrm{M}\,}}_2(K)$. If $B \simeq {{\,\mathrm{M}\,}}_2(F)$, we can take the embedding mapping $\alpha$ to a matrix with the same rational canonical form. So we suppose that $B=\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ is a division ring. By Theorem 6.4.11(iv) (over K) and 6.4.3, there exist $x,y,z,u,v,w \in F$ not all zero such that

$$\begin{aligned} (x+u\alpha )^2 + b(y+v\alpha )^2 + b(y+v\alpha )(z+w\alpha ) + ab(z+w\alpha )^2 = 0; \end{aligned}$$

(6.4.14)

expanding and rewriting into the powers of $\alpha$ gives

$$\begin{aligned} (u^2+bv^2+bvw+abw^2)\alpha ^2 + (vz+wy)b\alpha + (x^2+by^2+byz+abz^2) = 0. \end{aligned}$$

(6.4.15)

Let $\beta :=x+ yj+zij$ and $\gamma :=u+ vj+wij$. Then $\gamma \in B^\times$, since $\gamma =0$ implies ${{\,\mathrm{nrd}\,}}(\beta )=0$ and yet B is a division ring. Then the equation (6.4.15) can be written

$$\begin{aligned} {{\,\mathrm{nrd}\,}}(\gamma )\alpha ^2 + {{\,\mathrm{trd}\,}}(\beta \gamma )\alpha + {{\,\mathrm{nrd}\,}}(\beta ) = 0; \end{aligned}$$

then a direct calculation shows that the element

$$\begin{aligned} \mu =\beta \gamma ^{-1} = {{\,\mathrm{nrd}\,}}(\gamma )^{-1}\beta \gamma \end{aligned}$$

satisfies the same equation as (6.4.15) in the variable $\alpha$, so there is an embedding $K \hookrightarrow B$ defined by $\alpha \mapsto \mu$. $\square$

Exercises

Throughout these exercises, we let F be a field (of any characteristic, unless specified).

• 1. Recall the primitive element theorem from Galois theory: if $K \supseteq F$ is a separable field extension of finite degree, then there exists $\alpha \in K$ such that $K=F(\alpha )$—and hence $K \simeq F[x]/(f(x))$ where $f(x) \in F[x]$ is the minimal polynomial of $\alpha$.

  Extend this theorem to algebras as follows. Let B be a separable, commutative, finite-dimensional F-algebra. Show that $B \simeq F[x]/(f(x))$ for some $f(x) \in F[x]$.

$\triangleright$ 2.:

Let B be a quaternion algebra over F and let $K \subset B$ be a separable quadratic F-algebra. Show that there exists $b \in F^\times$ such that $B \simeq \displaystyle {\biggl (\frac{K,b}{F}\biggr )}$ (as in 6.1.5).

• 3. Let $F{}^\text{sep }$ be a separable closure of F and let B be a quaternion algebra over F. Show that $B \otimes _F F{}^\text{sep }\simeq {{\,\mathrm{M}\,}}_2(F{}^\text{sep })$. [More generally, see Exercise 7.24.]

$\triangleright$ 4.:

Let K be a separable quadratic F-algebra and let $u,b \in F^\times$. Show that $\displaystyle {\biggl (\frac{K,b}{F}\biggr )} \simeq \displaystyle {\biggl (\frac{K,ub}{F}\biggr )}$ if and only if $u \in {{\,\mathrm{nrd}\,}}(K^\times )={{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$.

• 5. Let B be a quaternion algebra over F, and let $K_0 \supseteq F$ be a quadratic field. Prove that there exists a separable extension $K \supseteq F$ linearly disjoint from $K_0$ over F (i.e., $K \otimes _F K_0$ is a domain) such that K splits F.

• 6. Suppose ${{\,\mathrm{char}\,}}F = 2$ and let $a \in F$ and $b \in F^\times$.

  • (a) Show that $\displaystyle {\biggl [\frac{a,b}{F}\biggr )} \simeq \displaystyle {\biggl [\frac{a,ab}{F}\biggr )}$ if $a \ne 0$.

  • (b) Show that if $t \in F$ and $u \in F^\times$, then $\displaystyle {\biggl [\frac{a,b}{F}\biggr )} \simeq \displaystyle {\biggl [\frac{a+(t+t^2),bu^2}{F}\biggr )}$.

• 7. Let ${{\,\mathrm{char}\,}}F = 2$ and let $B=\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ be a quaternion algebra over F. Compute the left regular representation $\lambda :B \rightarrow {{\,\mathrm{End}\,}}_K(B) \simeq {{\,\mathrm{M}\,}}_2(K)$ where $K=F[i]$ as in 2.3.8.

$\triangleright$ 8.:

Suppose ${{\,\mathrm{char}\,}}F = 2$. Let $M \in {{\,\mathrm{M}\,}}_n(F)$ be a symmetric matrix with n odd, and suppose that all diagonal entries of M are zero. Show that $\det M = 0$.

$\triangleright$ 9.:

Let ${{\,\mathrm{char}\,}}F=2$ and let B be a division F-algebra with a standard involution. Prove that either the standard involution is the identity (and so B is classified by Exercise 3.9), or that the conclusion of Theorem 3.5.1 holds for B: namely, that either $B=K$ is a separable quadratic field extension of F or that B is a quaternion algebra over F. [Hint: Replace conjugation by i by the map $\phi (x)=ix+xi$, and show that $\phi ^2=\phi$. Then diagonalize and proceed as in the case ${{\,\mathrm{char}\,}}F \ne 2$.]

$\triangleright$ 10.:

Let ${{\,\mathrm{char}\,}}F=2$. Show that the even Clifford algebra ${{\,\mathrm{Clf}\,}}^0 Q$ of a nondegenerate ternary quadratic form $Q:V \rightarrow F$ is a quaternion algebra over F.

$\triangleright$ 11.:

Prove Lemma 6.3.7, that every quadratic form over F with ${{\,\mathrm{char}\,}}F=2$ has a normalized basis.

$\triangleright$ 12.:

Let ${{\,\mathrm{char}\,}}F=2$ and let $Q:V \rightarrow F$ be a quadratic form over F with discriminant $d \in F^\times /F^{\times 2}$ and $\dim _F V=n$ odd. Show that $Z({{\,\mathrm{Clf}\,}}Q) \simeq F[x]/(x^2-d)$ and that there is a unique $\zeta \in Z({{\,\mathrm{Clf}\,}}Q) \cap {{\,\mathrm{Clf}\,}}^1 Q$ such that $\zeta ^2=1$.

$\triangleright$ 13.:

Prove Theorem 6.4.11.

$\triangleright$ 14.:

Let $Q :=Q' \boxplus Q''$ be an orthogonal sum of two anisotropic quadratic forms over F (with F of arbitrary characteristic). Show that Q is isotropic if and only if there exists $c \in F^\times$ that is represented by both $Q'$ and $-Q''$.

$\triangleright$ 15.:

Let B be a quaternion algebra over F (with F of arbitrary characteristic). Show that $F=(B^0)^\perp$.

$\triangleright$ 16.:

Prove Wedderburn’s little theorem in the following special case: a quaternion algebra over a finite field with even cardinality is not a division ring. [Hint: See Exercise 3.16.]