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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.

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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 $f_a (X)$ ./.

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  • $\begingroup$ I think the coinciding minimal polynomial is not sketchy at all. In detail: Let $q$ be minimal polynomial of $L_a$ over $F$. Then $q(L_a) \equiv 0$, so $q(L_a)(1) =0$. Hence, $m_a |q$. Conversely, if $m_a(L_a) \equiv 0 \Rightarrow q|m_a$. So the minimal polynomial coincides. What do you think? $\endgroup$ – Bryan Shih Apr 12 '18 at 22:24
  • $\begingroup$ Let's make the definitions and notations 100% clear: $L_a$ is the map "multiplication by $a$" in $E, q (X) \in F[X]$ is the minimal polynomial of this endomorphism $L_a$. Then $q(L_a)$ is an endomorphism of $E$, right ?Let's make the definitions and notations 100% clear: $L_a$ is the map "multiplication by $a$" in $E, q (X) \in F[X]$ is the minimal polynomial of this endomorphism $L_a$. Then $q(L_a)$ is an endomorphism of $E$, right ? You say that $q(L_a)$ is the zero endomorphism , probably because of the Cayley-Hamilton thm. But C-H. applies only to the characteristic polynomial... $\endgroup$ – nguyen quang do Apr 13 '18 at 13:01
  • $\begingroup$ ... Besides, what does $q(L_a)(1)$ mean ? Here you make a confusion between $q(L_a)\in End(E)$ and $q(X)\in F[X]$. Finally, in part (a) of your OP, your assertion " the roots of $\chi_a$ and $p_a$ coincide " as well as its consequence " $\chi_a={p_a}^\frac nm, deg \chi_a=n,deg p_a=m $" must be shown. $\endgroup$ – nguyen quang do Apr 13 '18 at 13:17
  • $\begingroup$ I think you are confusing my argument. $q(X)$ is indeed in $F[X]$, I define it to be the minimal polynomial of $L_a$. So $q(L_a)$ is zero endomorphism, by my definition of $q$. $\chi_a$ is defined to be to characteristic polynomial of $L_a :E \rightarrow E$. By CH, $q|\chi_a$. $q(L_a)(1)$ simply means the endomorphism evaluated at $1 \in E$. $\endgroup$ – Bryan Shih Apr 13 '18 at 13:54
  • $\begingroup$ The fact that $\chi_a=q^r = p_a^r$ follows from the fact that they have same roots and $p_a$ is irreducible over $F$. $\endgroup$ – Bryan Shih Apr 13 '18 at 13:58

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