Two definitions of characteristic polynomial for fintie separable extension Let $E/F$ be a finite separable extension of degree $n$. Let $\sigma_i:E \rightarrow F^a$ be the embeddings of $E$ to an algebriac closure $F^a$ which contains $E$. I have seen two definitions for characteristic polynomial of $a \in E$:

Definition 1: Let $L_a:{_F}E \rightarrow {_F}E$, denote the $F$-linear map induced by $x \mapsto xa$, then
$$\chi_a(x):= \det (xI-L_a).$$
Definition 2: $\chi_a(x)= \prod_i (x - \sigma_i(a))$.

How does one show these two are the same?

My thoughts (I am unsure if this is correct)
Let $g$ denote the polynomial in 2.
(a) In 1, $\chi_a(x) \in F[x]$. Also there is an injective $E \hookrightarrow End_F(E)$, $a \mapsto L_a$. Thus, the minimal poylnomial of $L_a$ over $F$ coincides with $a$, denoted by $p_a=\min(F,a)$. By Cayley Hamilton, $p_a|\chi_a$ over $F$. As the roots of $\chi_a$ and $p_a$ coincide, $\chi_a=p_a^{n/m}$, $\deg \chi_a =n, \deg p_a= m$.
(b) In 2 If I show $g$ has same roots as $p$, and $g \in F[x]$, then $g(a)=0$ implies $p|g$ and same argument shows $g=p_a^{n/m}=\chi_a$.
(c) Let $K$ be the splitting field of $p$ over $F$. $K$ is  Galois. If $\tau \in Gal(K/F)$, then we may extend this to a map between algebraic closures, $\tau:F^a \rightarrow F^a$. Observe now for each $i$, $\tau \sigma_i = \sigma_j$. Also, if $\tau \sigma_i = \tau \sigma_j$, then $\sigma_i = \sigma_j$. So coefficients of $g$ are fixed under $Gal(K/F)$, hence lies in $F$.
(d) For each $i$, $\sigma_i(a)$ is also a root of $p$. Conclusion (b) holds.
 A: Your argument contains many obscurities. In (a) for example, what is the meaning of "the minimal polynomial of $m_a$ coincides with $a$, denoted by $p_a= min (a, F)$" ? I propose the following approach which gives more information than necessary
 (at the price of some computational complications).
1° Suppose first that $a$ is a primitive element, i.e. $E=F(a)$. If $f_a=X^n + a_{n-1} X^{n-1} +...+ a_0\in F[X]$ is the minimal polynomial of $a$ over $F$,   then $E\cong F[X]/(f_a)$ admits a basis $(1, a,..., a^{n-1})$ over $F$, and the matrix of the endomorphism $m_a$ (= multiplication by $a$) wrt. to this basis is the so called "companion matrix" with lines $(0, 0,..., 0, -a_0), (1, 0, ..., 0, -a_1), (0, 1,.., 0,..., -a_2),..., (0, 0,..., 1, -a_{n-1})$. It is then easily checked that det ($X.Id_E - m_a$) = $f_a$.
2° In the general case, putting $[E:F(a)]$, of degree $r$, let us show that the characteristic polynomial $P_a (X)$ of $m_a$ is equal to the $r$-th power of the minimal polynomial $f_a(X)$ of $a$ (this is stronger than a simple consequence of Cayley-Hamilton). Let $(y_i)$, $1\le i\le q$, be a basis of $F(a)$ over $F$, and $(z_j)$, $1\le j \le r$, be a basis of $e$ over $F(a)$. Then $(y_i z_j)$is a basis of $E$ over $F$, and $[E:F]:=n=qr$. Let $M=(b_{ih})$ the matrix of $m_a$ in $F(a)$ with respect to the basis $(y_i)$, so that $ay_i = \sum b_{ih}y_h$. Then $a(y_i z_j)=(\sum b_{ih} y_h)z_j = \sum b_{ih}(y_h z_j)$. Ordering lexicographically the basis $(y_i z_j)$ of $E$ over $F$, we see that the matrix $M'$ of $m_a$ in $E$ with respect to this basis is a diagonal table of matrices  $M_1$ of the form $M$ on the diagonal and $0$ elsewhere. The matrix $X.Id - M_1$ is thus a diagonal table of matrices $X. Id_q -M$, hence det ($X. Id_n - M'$) = (det ($X.Id_q - M))^r$. But the LHS is the characteristic polynomial $P_a (X)$, and the RHS is $f_a(X)^r$ according to the special case 1° .
3° To meet your line of thought, it only remains to interpret 2° in Galois terms. Enlarging $E$ if necessary, we can replace it by the normal closure of $F(a)$ over $F$. Then $E/F$ is Galois, and every $F$-embedding $s$ of $F(a)$ into $E$ admits exactly $r$ extensions to elements $s_i$ of $Gal(E/F)$. So we get back on the tracks of your def.2, and your $\chi_a (X)$ is the $r$-th power of $f_a (X)$  ./.
