Let $(e_i)_{i \in I}$ be a basis of M; we endow the set I with a total order. For any subset J of I, set $e_J = e_{i_1} \wedge \cdots \wedge e_{i_r}$, where $(i_1, \ldots, i_r)$ is the sequence of elements of J in increasing order. The L-module $\wedge^{m-p}(M)$ admits as a basis the elements $e_S$, where S runs through the set of subsets of I with $m - p$ elements; $\wedge^m(M)$ has $\{e_I\}$ as a basis. Consequently, there exists a basis of $\operatorname{Hom}_L(\wedge^{m-p}(M), \wedge^m(M))$ consisting of linear mappings $e_J^*$ characterized by the formula

where J runs through the set of subsets of I with p elements. It follows from formula (20) of III, §7, No. 8, p. 519 that for every subset J of I with p elements, we have $h_p(e_J) \in \{e_J^*, -e_J^*\}$; since the elements $e_J$ form a basis of $\wedge^p(M)$, the linear mapping $h_p$ is bijective.

Let us now prove Proposition 1. Let u and $\tilde{u}$ be endomorphisms of M. Relation (3) is equivalent to

the mapping $h_1$ is an isomorphism from M to $\operatorname{Hom}_L(\wedge^{m-1}(M), \wedge^m(M))$ by Lemma 1. Consequently, for every endomorphism u of M, there exists a unique endomorphism $\tilde{u}$ of M satisfying relation (3).

Let $x_1, . . . , x_m$ be elements of M. Let us replace $x$ with $u(x_1)$ and $w$ with $x_2\wedge \cdots \wedge x_m$ in (3); we obtain

$\widetilde{u}(u(x_1))\wedge x_2\wedge \cdots \wedge x_m=u(x_1)\wedge \cdots \wedge u(x_m) =$ det($u$)$x_1\wedge \cdots \wedge x_m$.

Consequently, $h_1(\widetilde{u}\circ u(x_1)) =h_1$(det($u$)$x_1$), which gives the relation $\widetilde{u}\circ u=$ det($u$)$_M$ by Lemma 1.

We denote by U the endomorphism $X-\overline{u}$ of the L[X]-module M[X] (VIII, p. 335). By the above applied to U, there exists an endomorphism $\widetilde{U}$ of the L[X]-module M[X] that satisfies the relations

for $x_1, . . . , x_m$ in M and

(10) $\widetilde{U}\circ U =$ det(X $-\overline{u}$)$_{M[X]}$.

Let us view $\widetilde{U}$ as an element of End(M)[X] (VIII, p. 9); by formula (9) and Lemma 1, it has degree $\leqslant m-1$, so we can write it as

where the $u_r$ are endomorphisms of M. By formula (2), relation (10) gives the equality

in the ring End(M)[X]. By identifying the coefficients of the monomials in X on each side, we obtain the relations

(13) $u_r+u_{r-1}\circ u=c_r(u)_M$ for $1\leqslant r\leqslant m-1$

and

From this, we deduce

Now, when we identify the constant terms, equalities (9) and (11) imply $u_{m-1}=\widetilde{u}$; formula (6) follows.

In particular, $\widetilde{u}$ belongs to the subalgebra of End(M) generated by $u$ and therefore commutes with $u$. We have already established the relation $\widetilde{u}\circ u=$ det($u$)$_M$; formula (4) follows.

Finally, let $x_1, . . . , x_m$ be elements of M. We replace $x$ with $x_1$ and $w$ with $\widetilde{u}(x_2)\wedge \cdots \wedge \widetilde{u}(x_m)$ in formula (3). This gives

= det($u$)$^{m-1}x_1\wedge x_2\wedge \cdots \wedge x_m$

and therefore formula (5).

A) The uniqueness of P follows from Lemma 2 below.

Let $\mathscr{I}$ be the set of irreducible monic polynomials in K[X]. Since P and Q are monic, there exist elements $(a_F)$ and $(b_F)$ of $\mathbf{N}^{(\mathscr{I})}$ such that we have $P =\prod_{F\in\mathscr{I}}F^{a_F}$ and $Q =\prod_{F\in\mathscr{I}}F^{b_F}$. It follows from the equality $P^s= Q^s$ and the uniqueness of the decomposition into irreducible factors that we have $sa_F=sb_F$ for every $F\in \mathscr{I}$. Since $s$ is strictly positive, we consequently have $a_F=b_F$ for every $F\in \mathscr{I}$, and therefore P = Q.

B) Let us now prove the existence of P.

By Theorem 1 of VIII, p. 252, there exist a Galois extension L of K of finite degree and an isomorphism of L-algebras $\theta : A_{(L)}\rightarrow \mathbf{M}_n(L)$. By formula (12) of III, §9, No. 1, p. 542, the polynomial Pc($a; X$) is also the characteristic polynomial of the element $1\otimes a$ of the L-algebra $A_{(L)}$, hence of the element $\theta (1\otimes a)$ of the L-algebra $\mathbf{M}_n(L)$.

Set P(X) = det(X$I_n-\theta (1\otimes a)$); it is a monic polynomial in L[X]. By Example 3 of III, §9, No. 3, p. 545, we have

(19) Pc($a; X$) $= P(X)^n$.

Let G be the Galois group of L over K. For $\sigma \in G$, denote by $\overline{\sigma}$ the automorphism of the ring L[X] that coincides with $\sigma$ on L and fixes X. Then K[X] is the set of polynomials Q of L[X] such that we have $\overline{\sigma}(Q) = Q$ for every $\sigma \in G$ (V, §10, No. 1, p. 56, Theorem 1). Since the polynomial Pc($a; X$) $= P(X)^n$ belongs to K[X], we have $\overline{\sigma}(P)^n= P^n$ for every $\sigma \in G$. By Lemma 2, we therefore have $\overline{\sigma}(P) = P$ for every $\sigma \in G$, so P belongs to K[X].

An element $a$ of A is invertible if and only if its norm is nonzero (III, §9, No. 4, p. 545, Proposition 3). Proposition 3 therefore follows from formula (24).

(28) Pcrd$_{A_{(L)}/L}(1\otimes a; X) =$ Pcrd$_{A/K}(a; X)$,

(29) Trd$_{A_{(L)}/L}(1\otimes a) =$ Trd$_{A/K}(a)$,

(30) Nrd$_{A_{(L)}/L}(1\otimes a) =$ Nrd$_{A/K}(a)$

(“invariance under extension of scalars”).

The two sides of equality (28) have the same $n$-th power by the definition (formula (20) of VIII, p. 340) and the relation

Pc$_{A_{(L)}/L}(1\otimes a; X) =$ Pc$_{A/K}(a; X)$ (III, §9, No. 3, p. 544, formula (21)). Equality (28) therefore follows from Lemma 2 of VIII, p. 339. Formulas (29) and (30) follow from (28), (21), and (22).

(31) Pcrd$_{A/K}(a; X) =\chi_{\theta(a)}(X)$,

(32) Trd$_{A/K}(a) =$ Tr($\theta (a)$),

(33) Nrd$_{A/K}(a) =$ det($\theta (a)$).

Let $\theta '$ be the L-algebra homomorphism from $A_{(L)}$ to End$_L(V)$ such that $\theta '(\lambda \otimes a) =\lambda \theta (a)$ for $\lambda \in L$ and $a\in A$. The algebra $A_{(L)}$ is simple by Corollary 2 of VIII, p. 222; the homomorphism $\theta '$ is therefore injective. But the algebras $A_{(L)}$ and End$_L(V)$ over the field L have the same degree, equal to $n^2$, so $\theta '$ is an isomorphism. Corollary 1 then follows from Proposition 4 and Example 2 above.

(34) Pcrd$_{A/K}(a;a) = 0$,

(35) Pcrd$_{A/K}(\lambda a;\lambda X) =\lambda^n$ Pcrd$_{A/K}(a; X)$,

(36) Trd$_{A/K}(a+a') =$ Trd$_{A/K}(a) +$ Trd$_{A/K}(a')$,

Trd$_{A/K}(\lambda a) =\lambda$ Trd$_{A/K}(a)$,

(37) Trd$_{A/K}(aa') =$ Trd$_{A/K}(a'a)$,

(38) Nrd$_{A/K}(aa') =$ Nrd$_{A/K}(a)\cdot$ Nrd$_{A/K}(a')$,

Nrd$_{A/K}(\lambda a) =\lambda^n$ Nrd$_{A/K}(a)$,

(39) Trd$_{A/K}(1) =n$, Nrd$_{A/K}(1) = 1$.

Since A is central simple and has reduced degree $n$ over K, there exist an extension L of K and a vector space V of dimension $n$ over L such that $A_{(L)}$ is isomorphic to the algebra End$_L(V)$ (VIII, p. 252, Theorem 1). Corollary 2 then follows from Corollary 1 and the properties of the trace and determinant of an endomorphism. In particular, formula (34) follows from the Cayley– Hamilton theorem (III, §8, No. 11, p. 541, Proposition 20, cf. also VIII, p. 338, Remark 2).

(40) Pcrd$_{A^o/K}(a; X) =$ Pcrd$_{A/K}(a; X)$,

(41) Trd$_{A^o/K}(a) =$ Trd$_{A/K}(a)$,

(42) Nrd$_{A^o/K}(a) =$ Nrd$_{A/K}(a)$.

Choose an extension L of K, a vector space V of dimension $n$ over L, and a homomorphism $\theta$ from A to End$_L(V)$ (VIII, p. 252, Theorem 1). Let $V^*$ be the dual vector space of V. The mapping that sends an element $a$ of A to the endomorphism $^t\theta (a)$ of $V^*$ is a K-algebra homomorphism from $A^o$ to End$_L(V^*)$. Corollary 3 then follows from Corollary 1 and Corollary 3 of III, §8, No. 4, p. 528.

The trace, norm, and characteristic polynomial of $a$ are therefore the same whether we view $a$ as an element of A or as an element of $A^o$. This property does not always hold when A is not assumed central simple (III, §9, p. 644, Exercise 1).

a) The mapping $t:x\mapsto t_x$ is an isomorphism of $(A,A)$-bimodules from A to its dual Hom$_K(A,K)$.

b) Let $h$ be a linear form on A. The following properties are equivalent:

(i) There exists an element $\lambda$ of K such that $h(x) =\lambda$ Trd$_{A/K}(x)$

for every $x\in A$.

(ii) We have $h(xy) =h(yx)$ for all $x, y$ in A.

Recall (II, §1, No. 14, p. 225–226) that the $(A$, A)-bimodule structure on $A^*=$ Hom$_K(A,K)$ is defined by the relation

for $a, b, c\in A$ and $t\in A^*$. In particular, for every $x$ in A, we have

$\langle at_xb, c\rangle =\langle t_x, bca\rangle =$ Trd$_{A/K}(xbca)$,

$\langle t_{axb}, c\rangle =$ Trd$_{A/K}(axbc)$,

and these two elements are equal by formula (37) of VIII, p. 342. We therefore have $at_xb=t_{axb}$, which means that $t$ is a homomorphism of $(A$, A)-bimodules from A to $A^*$.

We choose an extension L of the field K and an isomorphism $\theta$ from the L-algebra $A_{(L)}$ to the matrix algebra $\mathbf{M}_n(L)$ (VIII, p. 252, Theorem 1). We identify the vector space $(A^*)_{(L)}$ with the dual of the vector space $A_{(L)}$ over the field L. By Proposition 4 of VIII, p. 341, with these conventions, we have

(44) Trd$_{A_{(L)}/L}= 1_L\otimes$ Trd$_{A/K}$.

Let $t_{(L)}$ be the L-linear mapping from $A_{(L)}$ to $A^*_{(L)}$ deduced from $t$ by extension of scalars; by formula (44) and Corollary 1, we have

(45) $\langle t_{(L)}(x), y\rangle =$ Trd$_{A_{(L)}/L}(xy) =$ Tr($\theta (x)\theta (y)$)

for $x, y$ in $A_{(L)}$. By Proposition 7 of II, §10, No. 11, p. 358, the mapping $t_{(L)}$ is bijective; it follows that $t$ is bijective. We have proved a).

Let $h$ be in $A^*$; by the above, there exists an element $a$ of A such that $h$ is equal to $t_a$. By a), we have

Consequently, the relation “$h(xy) =h(yx)$ for $x, y$ in A” is equivalent to “$t_{ax-xa}= 0$ for every $x\in$ A”, and by part a) of the proof, this means that $a$ belongs to the center K of A. This proves b) in view of formula (36).

Let P and Q be elements of $K[\mathbf{T}]$ with $Q\not= 0$. The coefficients of the polynomials R in $L[\mathbf{T}]$ such that P = QR are the solutions of a system of linear equations with coefficients in K. Consequently, if there exists a polynomial $R\in L[\mathbf{T}]$ such that P = QR, then there also exists one in $K[\mathbf{T}]$ (II, §8, No. 5, p. 321, Proposition 6). This proves the inclusion $K(\mathbf{T})\cap L[\mathbf{T}]\subset K[\mathbf{T}]$; the reverse inclusion is obvious.

Recall that the reduced characteristic polynomial of an element $a$ of A can be written as

(46) Pcrd$_{A/K}(a; X) =\sum_{r=0}^n(-1)^rb_r(a) X^{n-r}$

and that we have

$b_0(a) = 1,b_1(a) =$ Trd$_{A/K}(a),b_n(a) =$ Nrd$_{A/K}(a)$.

Let $(e_i)_{i\in I}$ be a basis of A over K and $\mathbf{T}= (T_i)_{i\in I}$ a family of variables.

We choose an extension L of K and an L-algebra isomorphism $\theta$ from $A_{(L)}$ to $\mathbf{M}_n(L)$. We denote by $\overline{\theta}: A_{(L(\mathbf{T}))}\rightarrow \mathbf{M}_n(L(\mathbf{T}))$ the isomorphism of $L(\mathbf{T})$-algebras deduced from $\theta$ by extension of scalars. By Corollary 1 of VIII, p. 342, we have

(47) $P(X) =\chi_{\overline{\theta}(u)}(X) =$ det(X$I_n-\overline{\theta}(u)$) $=$ det$(XI_n-\sum_{i\in I}T_i\theta (1\otimes e_i))$.

Since the matrices $\theta (1\otimes e_i)$ belong to $\mathbf{M}_n$(L), this formula shows that P is a homogeneous polynomial of degree $n$ in $L[\mathbf{T},X]$. It also belongs to $K(\mathbf{T})[X]$ and can be written as $P(X) =\sum_{j\geqslant 0}c_jX^j$, where each $c_j$ belongs to the intersection $K(\mathbf{T})\cap L[\mathbf{T}]$. By Lemma 3, each of the elements $c_j$ belongs to $K[\mathbf{T}]$; Lemma 4 follows.

(48) Pcrd$_{A_{(K')}/K'}(\sum_{i\in I}t_i\otimes e_i)= P((t_i)_{i\in I},X)$.

Let $\varphi : K[\mathbf{T}]\rightarrow K'$ be the unique K-algebra homomorphism that sends $T_i$ to $t_i$ for every $i\in I$; it defines on $K'$ the structure of a $K[\mathbf{T}$]-algebra. The $K'$-algebra $A_{(K[\mathbf{T}])(K')}$ can be identified with $A_{(K')}$ (transitivity of extension of scalars), where the element $1\otimes (\sum T_i\otimes e_i)$ is identified with the element $\sum t_i\otimes e_i$ of $A_{(K')}$. We denote by $\overline{\varphi}: K[\mathbf{T}][X]\rightarrow K'[X]$ the K-algebra homomorphism deduced from $\varphi$. By formula (21) of III, §9, No. 3, p. 544, the characteristic polynomial of $\sum t_i\otimes e_i$ with respect to the $K'$-algebra $A_{(K')}$ is the image by $\overline{\varphi}$ of the characteristic polynomial of $\sum T_i\otimes e_i$ with respect to the $K[\mathbf{T}$]-algebra $A_{(K[\mathbf{T}])}$, that is, of $P^n$. In other words, we have

(49) Pc$_{A_{(K')}/K'}(\sum_{i\in I}t_i\otimes e_i; X)= P((t_i)_{i\in I},X)^n$;

Lemma 5 then follows from Lemma 2 of VIII, p. 339.

Consider the specific case $K'= K$ of Lemma 5. We have

(50) Pcrd$_{A/K}(\sum_{i\in I}t_ie_i; X)= P((t_i)_{i\in I},X)$

for every element $(t_i)_{i\in I}$ of $K^I$. Since the polynomial P in $K[\mathbf{T},X]$ is homogeneous of degree $n$, it can be expanded uniquely as

where $B_r$ is a homogeneous polynomial of degree $r$ in $K[\mathbf{T}]$. By formulas (46), (50), and (51), we have

for every element $(t_i)_{i\in I}$ of $K^I$. Proposition 6 follows.

(52) Pcrd$_{A/K}(a; X) =$ Pc$_{L/K}(a; X)$,

(53) Trd$_{A/K}(a) =$ Tr$_{L/K}(a)$,

(54) Nrd$_{A/K}(a) = N_{L/K}(a)$.

By Proposition 3 of VIII, p. 262, the L-modules A and $L^n$ are isomorphic; we therefore have the relation

Pc$_{A/K}(a; X) =$ Pc$_{L/K}(a; X)^n$.

Since the polynomial Pc$_{L/K}(a; X)$ is monic, it is therefore equal to the reduced characteristic polynomial Pcrd$_{A/K}(a; X)$ (VIII, p. 339, Lemma 2); this gives formula (52). By comparing the coefficients of $X^{n-1}$ (resp. the constant terms) on each side of (52), we obtain formula (53) (resp. (54)).

Pcrd$_{D/K}(a; X) =$ Pc$_{L/K}(a; X)$,

(55) Tr$_{D/K}(a) =$ Tr$_{L/K}(a)$,

Nrd$_{D/K}(a) = N_{L/K}(a)$.

Indeed, a maximal commutative subfield L of D is a maximal commutative semisimple subalgebra of D by Corollary 2 of VIII, p. 265.

(56) Pcrd$_{A/K}(b; X) = N_{L[X]/K[X]}$(Pcrd$_{B/L}(b; X)$)$^r$,

(57) Trd$_{A/K}(b) =r$ Tr$_{L/K}$(Trd$_{B/K}(b)$),

(58) Nrd$_{A/K}(b) = N_{L/K}$(Nrd$_{B/L}(b)$)$^r$.