§ 18. Simple Algebras over a Finite Field

§ 18. Simple algebras over a finite field

1. Polynomials over a Finite Field

Theorem 1 (Chevalley–Warning) 00SA

Let K be a finite commutative field of characteristic $p$ . Let $n$ be an integer $\geqslant 1$ and $(f_i)_{i\in I}$ a finite family of nonzero elements of $K[X_1, . . . ,X_n]$ . Denote by Z the set of elements $\mathbf{x}$ of $K^n$ such that we have $f_i(\mathbf{x}) = 0$ for $i\in I$ . If we have $n >\sum_{i\in I}$ deg( $f_i$ ), then the cardinal of Z is divisible by $p$ .

Lemma 1 00LB

Let L be a field, G a finite group, and $\chi$ a nontrivial homomorphism from G to the multiplicative group $L^*$ . We have $\sum_{x\in G}\chi (x) = 0$ .

By assumption, there exists an element $a$ of G such that $\chi (a)\not= 1$ ; since multiplication by $a$ is a permutation of G, we have

\sum_{x\in G}\chi (x) =\sum_{x\in G}\chi (ax) =\chi (a)\sum_{x\in G}\chi (x)

this gives Lemma 1.

Lemma 2 00LC

Let $q$ be the cardinal of K. For any integer $m\geqslant 0$ , set $S_m=$ $\sum_{x\in K}x^m$ . We have $S_m=-1$ if $m$ is a nonzero multiple of $q-1$ and $S_m= 0$ in all other cases.

Recall that $0^0= 1$ (I, §2, No. 1, p. 14). Suppose that $m$ is a multiple of $q-1$ . Since the abelian group $K^*$ has order $q-1$ , we have $x^m= 1$ for every $x\in K^*$ and $S_m= 0^m+ (q-1)\cdot 1$ , which gives the assertion in this case.

Suppose that $m$ is not a multiple of $q-1$ . Set $\chi (x) =x^m$ for $x\in K^*$ . Since the multiplicative group $K^*$ is cyclic of order $q-1$ (V, §12, No. 1, p. 93, Proposition 1), there exists an element $a$ of $K^*$ such that $\chi(a) \neq 1$ . By Lemma 1 applied to $G = K^*$ , we have

S_m = 0^m + \sum_{x \in K^*} \chi(x) = 0;

this gives Lemma 2.

Let us now prove Theorem 1. Let $\mathbf{x} = (x_1, \ldots, x_n)$ be an element of $K^n$ . We have $1 - f_i(\mathbf{x})^{q-1} = 0$ if $f_i(\mathbf{x}) \neq 0$ and $1 - f_i(\mathbf{x})^{q-1} = 1$ if $f_i(\mathbf{x}) = 0$ . Set $P = \prod_{i=1}^r (1 - f_i^{q-1})$ . We have

P(\mathbf{x}) = \begin{cases} 1 & \text{if } \mathbf{x} \in \mathbf{Z}, \\ 0 & \text{if } \mathbf{x} \notin \mathbf{Z}. \end{cases}

Let us expand the polynomial

P

\sum_{\alpha \in \mathbf{N}^n} c_\alpha X^\alpha

; by assumption, it has degree

< (q-1)n

. Let

\alpha

be an element of

\mathbf{N}^n

such that

c_\alpha

is nonzero. Since we have

\alpha_1 + \cdots + \alpha_n < (q-1)n

, there exists an integer

\ell

such that

1 \leq \ell \leq n

and

0 \leq \alpha_\ell < q-1

. By Lemma 2, we then have

\sum_{x \in K} x^{\alpha_\ell} = 0

, and therefore

\sum_{\mathbf{x} \in K^n} \mathbf{x}^\alpha = \prod_{j=1}^n \left( \sum_{x \in K} x^{\alpha_j} \right) = 0.

We consequently have

\sum_{\mathbf{x} \in K^n} P(\mathbf{x}) = \sum_{\alpha \in \mathbf{N}^n} c_\alpha \left( \sum_{\mathbf{x} \in K^n} \mathbf{x}^\alpha \right) = 0.

Now, by formula (1), we have

\sum_{\mathbf{x} \in K^n} P(\mathbf{x}) = \operatorname{Card}(\mathbf{Z}) \cdot 1

, and therefore

\operatorname{Card}(\mathbf{Z}) \cdot 1 = 0

, which means that

\operatorname{Card}(\mathbf{Z})

is divisible by

p

Corollary 00LD

Let $V$ be a vector space of finite dimension $n$ over $K$ and $I$ a finite set, and for each $i \in I$ , let $F_i : V \to K$ be a homogeneous polynomial mapping of degree $d_i > 0$ . If we have $\sum_{i \in I} d_i < n$ , then there exists a nonzero element $x$ of $V$ such that $F_i(x) = 0$ for every $i \in I$ .

Let $(e_1, \ldots, e_n)$ be a basis of $V$ over $K$ . By the definition of homogeneous polynomial mappings (IV, §5, No. 9, p. 55, Definition 3), for every $i \in I$ , there exists a homogeneous polynomial $f_i$ in $K[X_1, \ldots, X_n]$ of degree $d_i$ such that we have $F_i(\xi_1 e_1, \ldots, \xi_n e_n) = f_i(\xi_1, \ldots, \xi_n)$ . Let $Z$ be the set of elements $x$ of $V$ such that $F_i(x) = 0$ for every $i \in I$ . By Theorem 1, the cardinal of $Z$ is divisible by $p$ , and since $0$ belongs to $S$ , we have $\operatorname{Card}(Z) \geq p > 1$ .

2. Simple Algebras over Finite Fields

Theorem 2 (Wedderburn) 00SB

Every finite field is commutative.

Let D be a finite field, and let K be its center. The K-algebra D is central simple of degree $m^2$ , where $m$ is a strictly positive integer. The reduced norm is a homogeneous polynomial mapping Nrd $: D\rightarrow K$ of degree $m$ (VIII, p. 345, Proposition 6), and we have Nrd( $a$ ) $\not= 0$ for every $a\not= 0$ in D (VIII, p. 340, Proposition 3). The corollary above implies that $m\geqslant m^2$ , and therefore $m= 1$ . So we have D = K.

Corollary 1 00LE

Every finite simple ring is isomorphic to a matrix ring $\mathbf{M}_n(L)$ , where $n$ is a strictly positive integer and L is a finite commutative field.

This follows from Theorem 2 and the structure theorem for simple rings (VIII, p. 120, Theorem 1).

Corollary 2 00LF

Let K be a finite commutative field. Every central simple algebra over K is isomorphic to a matrix algebra $\mathbf{M}_n(K)$ , where $n$ is a strictly positive integer.

This follows from Theorem 2 and the structure theorem for central simple algebras (VIII, p. 252, Theorem 1).

Remark 1 00LG

Here is another proof of Theorem 2. Let D be a finite field, let K be its center, and let L be a maximal commutative subfield of D. Let $x$ be an element of $D^*$ . It belongs to a maximal commutative subfield $L_1$ of D. We have the equality

[D : K] = [L : K]^2= [L_1: K]^2

by Corollary 2 of VIII, p. 265, and therefore $[L : K] = [L_1: K]$ . By Proposition 3 of V, §12, No. 2, p. 94, the extensions L and $L_1$ of K are isomorphic. By VIII, p. 263, Corollary, there exists an element $a$ of $D^*$ such that $aLa^{-1}= L_1$ , so $a^{-1}xa$ belongs to L. We then have $(ay)^{-1}x(ay) =a^{-1}xa$ for every $y\in L^*$ . Consequently, if S is a set of representatives of the left cosets of $D^*$ modulo $L^*$ , then every element of $D^*-\{1\}$ can be written as $sxs^{-1}$ with $s\in S$ and $x\in L^*-\{1\}$ . We denote the order of $D^*$ by $d$ and that of $L^*$ by $\ell$ . Since the cardinal of S is equal to $d/\ell$ , we have $d-1\leqslant (d/\ell )(\ell -1) =d-d/\ell$ . It follows that $\ell =d$ , and therefore L = D, which proves that the field D is commutative.

Remark 2 00LH

Let L be a commutative field with the following property: $(C_1)$ Let V be a vector space of finite dimension $n$ over the field L, and let $d$ be such that $0< d < n$ . For every homogeneous polynomial mapping $F : V\rightarrow L$ of degree $d$ , there exists a nonzero element $x$ of V such that $F(x) = 0$ .

The proof of Theorem 2 shows that every field of finite degree over L with center L is equal to L.

By the corollary of VIII, p. 356, every finite field has property $(C_1)$ . We can prove (VIII, p. 360, Exercise 7) that the following fields have property $(C_1):$

– every algebraic extension of a finite field

– every field of rational fractions in one variable with coefficients in an algebraically closed field (Tsen’s theorem).

– $*$ every field endowed with a discrete valuation for which it is complete and

whose residue field is algebraically closed (VIII, p. 332, Exercise 17) $.*$

Remark 3 00LI

Suppose that the field K satisfies the following condition:

– If L is an extension of K of finite degree, then it is cyclic and the norm mapping $N : L^*\rightarrow K^*$ is surjective.

This condition is satisfied, in particular, when the field K is finite (V, §12, No. 2, p. 95, Proposition 4). We can then prove that every field of finite degree over K with center K is equal to K (Exercise 10 of VIII, p. 329).

Exercises

See the exercises for § 18.