Lemma 6

Let $K'$ be a commutative algebra of finite degree $d$ over K and

a monic polynomial with coefficients in $K'$. Then the polynomial Q = $N_{K'[X]/K[X]}(P)$ in K[X] is monic of degree $sd$, the coefficient of $X^{sd-1}$ in Q(X) is equal to Tr$_{K'/K}(a_1)$, and the constant term of Q is $N_{K'/K}(a_s)$.

We denote the $K'$-algebra $K'[T]/(P(T))$ by $K''$ and the canonical class of T in $K''$ by $t$. The sequence $(1, t, . . . , t^{s-1})$ is a basis of $K''$ over $K'$, and the matrix of multiplication by $t$ in this basis is of the form

The determinant of $XI_n - \tau$ is calculated by induction on $s$, by expanding along the first row. We obtain $\det(XI_n - \tau) = P(X)$. In other words, we have $P(X) = Pc_{K''/K'}(t; X)$. In particular, $\operatorname{Tr}_{K''/K'}(t) = -a_1$ and $N_{K''/K'}(t) = (-1)^s a_s$. By the transitivity formula (III, §9, No. 4, p. 548, Corollary), we have

On the other hand, $[K'' : K] = [K'' : K'][K' : K] = sd$, so $Q(X)$ is a monic polynomial of degree $sd$. Lemma 6 follows.

Let us prove Proposition 8. Since the ring $B$ is simple, its center $L$ is a field (VIII, p. 121, Corollary 1). By Theorem 5 of VIII, p. 259, the commutant $B'$ of $B$ in $A$ is a simple ring with center $L$, and we have the equality $[A : K] = [B : K][B' : K]$. We denote the reduced degree of $B'$ over $L$ by $r$, that of $B$ over $L$ by $s$, and the degree of $L$ over $K$ by $d$. We have

and therefore $n^2 = r^2 s^2 d^2$, that is, $n = rsd$.

Let $b$ be an element of $B$, and let $P(X)$ be its reduced characteristic polynomial over the $L$-algebra $B$; it is monic of degree $s$. By Lemma 6, the polynomial $Q = N_{L[X]/K[X]}(P)$ is monic of degree $sd$. The polynomial $R = Q^r$ is therefore monic of degree $rsd = n$. Again by Lemma 6, the coefficient of $X^{n-1}$ in $R(X)$ is equal to $-r \operatorname{Tr}_{L/K}(\operatorname{Trd}_{B/L}(b))$, and the constant term of $R(X)$ is $(N_{L/K}((-1)^s \operatorname{Nrd}_{B/L}(b)))^r = (-1)^n N_{L/K}(\operatorname{Nrd}_{B/L}(b))^r$.

Since $[A : K] = r^2 d[B : K]$, the left $B$-module $A$ is free of rank $r^2 d$ (VIII, p. 124, Proposition 5). We therefore have

By the corollary of III, §9, No. 4, p. 548, we have

and since P(X) is the reduced characteristic polynomial of $b$ over the L-algebra B, we have

(62) Pc$_{B/L}(b; X) = P(X)^s$.

Finally, by formulas (60)—(62) and the definition of R(X), we have

(63) Pc$_{A/K}(b; X) = N_{L[X]/K[X]}(P(X))^{dr^2s}= Q(X)^{dr^2s}= R(X)^{rsd}= R(X)^n$,

so R(X) is the reduced characteristic polynomial of $b$ over the K-algebra A.

We have proved formula (56). Formulas (57) and (58) follow immediately from formula (56) and Lemma 6 because the coefficient of $X^{n-1}$ in Pcrd$_{A/K}(b; X)$ is equal to $-$ Trd$_{A/K}(b)$ and the constant term is $(-1)^n$ Nrd$_{A/K}(b)$.