I am reading the Hungerford, Algebra, Chapter V, proof of Theorem 7.3 and can't prove the transitivity of norm until now.
Proposition 6.12. Let $F$ be a finite dimensional extension field of $K$ and $N$ a normal extension field of $K$ containing $F$. The number of distinct $K$-monomorphisms $F\to N$ is precisely $[F:K]_s$, the separable degree of $F$ over $K$.
Definition 7.1. Let $F$ be a finite dimensional extension field of $K$ and $\bar{K}$ an algebraic closure of $K$ containing $F$. Let $\sigma_1,\dots, \sigma_r$ be all the distinct $K$-monomorphisms $F\to \bar{K}$ ( c.f. Proposition 6.12 above ). If $u\in F$, the norm of $u$, denoted, $N_K^F(u)$ is the element
$$ N_K^F(u):=(\sigma_1(u)\sigma_2(u) \cdots \sigma_r(u))^{[F:K]_i}$$
( $[F:K]_i$ is the inseparable degree. ) The trace of $u$, denoted $T_K^F(u)$, is the element
$$ T_K^F(u):= [ F: K]_i(\sigma_1(u) + \sigma_2(u) + \cdots +\sigma_r(u)).$$
Theorem 7.3. Let $F$ be a finite dimensional extension field of $K$. Then for all $u,v \in F$ :
(i) $N_K^F(u)N_K^F(v)=N_K^F(uv)$ and $T_K^F(u) + T_K^F(v) = T_K^F(u+v)$ ;
(ii) if $u\in K$, then $N_K^F(u)=u^{[F:K]}$ and $T_K^F(u)=[F:K] u$ ;
(iii) $N_K^F(u)$ and $T_K^F(u)$ are elements of $K$. More precisely, $N_K^F(u)= ((-1)^na_0)^{[F:K(u)]} \in K$ and $T_K^F(u) = – [F : K(u)] a_{n-1} \in K$, where $f=x^n+a_{n-1}x^{n-1} + \cdots + a_0 \in K[x]$ is the irreducible polynomial of $u$ ;
(iv) if $E$ is the intermediate field, then $$ N_K^E(N_E^F(u))=N_K^F(u) \ and \ T_K^E(T_E^F(u))=T_K^F(u).$$
SKETCH OF PROOF OF THEOREM 7.3. (i) and (ii) follow directly from Definition 7.1 above and the facts that $r= [ F:K]_s$ ( $\because$ Proposition 6.12 above ) and $[F:K]_s [F:K]_i=[F:K]$.
(iii) Let $E:= K(u)$. And algebraic closure $\bar{K}$ of $K$ which contains $F$ is also an algebraic closure of $E$. The proof of Lemma 6.11 shows that the distinct $K$-monomorphisms $F \to \bar{K}$ are precisely the maps $\sigma_k \tau_i ( 1 \le k \le t ; 1 \le j \le r )$, where the $\sigma$'s are all the $K$-automorphisms of $\bar{K}$ whose restrictions to $E$ are distinct and the $\tau$'s are all the distinct $E$-monomorphisms $F\to \bar{K}$. We accept this fact. Thus by proposition 6.12 above, $t= [ E:K]_s$ ( $\because$ By proposition 6.12 above, $tr = [F:K]_s = [F : E]_s [E:K]_s$. And also by using the proposition 6.12, $r= [F:E]_s$ ), whence $n= [E:K] = [E:K]_s [E:K]_i = t[E:K]_i$ ( C.f. see Ramarks after Definition 6.10 ).
Use (ii) and Corollary 6.13 ( The transitivity of the separable and inseparable degrees ) to show that $N_K^F(u) = ( \prod_{k=1}^{t} \sigma_k(u))^{[F:E][E:K]_i}$ and $T_{K}^F(u) = [F:E] [E:K]_i ( \Sigma_{k=1}^{t}\sigma_k(u))$. We accept this calculation. If needed I will upload proof.
Since $\sigma_i : K(u) \cong K(\sigma_i(u))$, Corollary 1.9 implies that $\sigma_1(u) , \dots , \sigma_t(u)$ are all the distinct roots of $f$ ( $\because$ by the choice, restrictions of $\sigma$'s to $E:=K(u)$ are distinct so each $\sigma_1(u) , \dots , \sigma_t(u)$ are all distinct. And since $t=[E:K]_s$, they are all roots of $f$ by the Corollary 6.14 below ).
Note that
Corollary 6.14. Let $f \in K[x]$ be an irreducible monic polynomial over a field $K$, $F$ a splitting field of $f$ over $K$ and $u_i$ a root of $f$ in $F$. Then (i) every root of $f$ has multiplicity $[K(u_1) : K]$ , so that in $F[x]$, $$ f= [ (x-u_1) \cdots (x-u_n)]^{[K(u_1) : K]_i},$$ where $u_1 , \dots, u_n$ are all the distinct roots of $f$ and $n= [K(u_1) : K]_s$ ;
(ii) $u_1^{[K(u_1) : K]_i}$ is separable over $K$.
By Corollary 6.14,
$$ f= [ (x-\sigma_1(u))(x-\sigma_2(u)) \cdots ( x- \sigma_t(u))]^{[E:K]_i}
= [ x^t – ( \Sigma_{k=1}^{t} \sigma_k(u))x^{t-1} + \cdots + ( (-1)^t \prod_{k=1}^{t}\sigma_k(u))]^{[E:K]_i}.$$
If $[E:K]_i =1$, then $n=t$ and the conclusion is immediate. If $[E:K]_i >1$, then $[E:K]_i$ is a positive power of $p = \operatorname{char}K$. ( $\because$ Corollary 6.5 ). It is easy to calculate $a_0$ and to see that $a_{n-1}=0 = T_K^F(u)$ ; use Exercise III.1.11 ( The Freshman's Dream ).
(iv) Use the notation in the first paragraph of the proof of (iii), with $E$ any intermediate field. Apply the appropriate definitions and use Corollary 6.13 ( transitivity of the separable and inseparable degrees ).
I'm trying to follow the bold statement and don't know how to breakthrough.
1. First attempt : Let $E$ be an intermediate field of $F/K$. Note that as in the proof of (iii), we have
$$ N_K^F(u) = (\prod_{k=1}^{t} \sigma_k(u))^{[F:E][E:K]_i} , t= [E:K]_s \tag{1}$$
$$ v:=N_E^F(u) = (\prod_{k=1}^{t'} \sigma'_k(u))^{[F:E'][E' : E]_i} , t'=[E':E]_s \tag{2}$$
$$ N_K^E(v) = ( \prod_{k=1}^{t''} \sigma''_k(v))^{[E:K'][K':K]_i} , t''=[K':K]_s \tag{3} $$
, where 1) $\sigma$'s are all the $K$-automorphisms of $\bar{K}$ whose restrictions to $E$ are distinct, 2) $\sigma'$'s are all the $E$-automorphims of $\bar{E}= \bar{K}$ whose restriction to $E':=E(u)$ are distinct, 3) $\sigma''$'s are all the $K$-automorphisms of $\bar{K}$ whose restriction to $K':=K(v)$ are distinct. Then after inserting (2) into (3), try to check that it output (1). Sorry for messy notation. After trying to some calculation ( involving transitivity of separable and inseparable degree, multiplicativity of norm etc..) , I reached to conclusion that if
$$ ( (\sigma_1''(\sigma_1'(u)) \cdots \sigma_{t''}''(\sigma_1'(u)) \cdots ( \sigma_1''(\sigma_{t'}'(u)) \cdots \sigma_{t''}''(\sigma_{t'}'(u)) ) = ( \sigma_1(u) \cdots \sigma_t(u) )^{[E' : E]_s / [ E: K']_s} $$
, then we can check our goal.. And it is so brutal force.. Could we adopt a similar approach?
2. Second attempt : Let $v:=N_E^F(u) \in E$. Let $g:= x^{m} + b_{m-1}x^{m-1} _ \cdots + b_0 \in K[x]$ be the irreducible polynomial of $v\in E$. Then by the Theorem 7.3-(iii) above,
$$ N_K^E(N_E^F(u)) = N_K^E (v) = ( (-1)^m b_0 )^{[E: K(v)]}$$
This is same as $N_K^F(u) = ( (-1)^na_0)^{[F:K(u)]}$, where $f:=x^n+ a_{n-1}x^{n-1} + \cdots + a_0 \in K[x]$ be the irreduicible polynomial of $u$. ? In this approach, I don't know what kind of relationship to catch
P.s. imgae of the proof of the Theorem 7.3 is as follows :
How can we follow the underlined statement?
Or is there other simpler route to show the transitivity of norms, following hungerford's intention? Any help would be appreciated.
Best Answer
Let $E$ be an intermediate field. Let $\overline{K}$ is algebraic closure of $K$ containing $F$. Then $\overline{E}= \overline{K}$, since $\overline{K}$ is algebraic over $E$ and $\overline{K}$ is algebraically closed. The proof of lemma $6.11$ shows that the distinct $K$-monomorphisms $F \to \overline{K}$ are precisely the maps $\sigma_k \tau_j ( 1 \le k \le t ; 1 \le j \le r )$, where the $\sigma$'s are all the $K$-automorphisms of $\overline{K}$ whose restrictions to $E$ are distinct and the $\tau$'s are all the distinct $E$-monomorphisms $F\to \overline{K}=\overline{E}$.
By definition of norm, $$ N_E^F(u)=(\tau_1(u)\cdots \tau_r(u))^{[F:E]_i}= \left(\prod_{j=1}^r \tau_j (u)\right)^{[F:E]_i}.$$
Proof of lemma $6.11$ shows that $\sigma |_E$'s are all the distinct $K$-monomorphisms $E\to \overline{K}$. Note $N_K^E(N_E^F (u))$ make sense because $N_E^F (u)\in E$ by (iii). By definition of norm and corollary $6.13$, $$\begin{align} N_K^E (N_E^F (u)) &= \left[ \sigma_1\left(N_E^F(u)\right)\cdots \sigma_t \left(N_E^F(u)\right) \right]^{[E:K]_i} \\ &= \left[ \sigma_1\left(\prod_j \tau_j(u)\right)^{[F:E]_i} \cdots \sigma_t \left( \prod_j \tau_j(u) \right)^{[F:E]_i} \right]^{[E:K]_i} \\ &= \left[ \left( \prod_j \sigma_1 \tau_j(u)\right)^{[F:E]_i} \cdots \left( \prod_j \sigma_t \tau_j(u) \right)^{[F:E]_i} \right]^{[E:K]_i} \\ &= \left( \prod_{i,j} \sigma_i\tau_j (u) \right)^{[F:E]_i[E:K]_i} \\ &= \left( \prod_{i,j} \sigma_i\tau_j (u) \right)^{[F:K]_i} \\ &= N_K^F(u). \end{align}$$
where $1\leq i\leq t$ and $1\leq j\leq r$. Proof is similar for trace.