Showing $Tr_{L/K} \circ Tr_{M/L} = Tr_{M/K}$ where $K \subseteq L \subseteq M$ is a tower of field extensions - *Algebraic Number Theory* by Neukerich I'm reading Algebraic Number Theory by Neukirch and I'm having trouble the proof for the statement:

(2.7) Corollary: I a tower of finite field extensions $K \subseteq L \subseteq M$, one has
  $$Tr_{L/K} \circ Tr_{M/L} = Tr_{M/K}, \qquad N_{L/K} \circ N_{M/L} = N_{M/K}.$$

Here we assume $M/K$ is separable.
The part of the proof im having trouble with is the identity
$$Tr_{M/K}(x) = \sum_{i = 1}^{m}\sum_{\sigma \sim \sigma_{i}}\sigma(x) = \sum_{i = 1}^{m}Tr_{\sigma_{i}M/\sigma_{i}L}(\sigma_{i}(x)) = \sum_{i = 1}^{m}\sigma_{i}Tr_{M/L}(x) = Tr_{L/K}(Tr_{M/L}(x))$$
where $m = [L:K]$ and the $\sigma_{i}$ are representatives for equivalence classes under the relation $\sigma \sim \tau \iff \sigma|_{L} = \tau|_{L}$. I believe the first equality, but I don't see how the second and third hold. I understand the $\sigma:M \rightarrow \overline{K}$ in $[\sigma_{i}]$ are $K$-embeddings, and hence restricted to $L$ are also $K$-embeddings, but I dont see how the inner sum corresponds to $Tr_{\sigma_{i}M/\sigma_{i}L}(\sigma_{i}(x))$. We can express $Tr_{\sigma_{i}M/\sigma_{i}L}(\sigma_{i}(x))$ as a sum of $\sigma_{i}L$ embeddings from $\sigma_{i}M$ into $\overline{\sigma_{i}L}$ evaluated at $\sigma_{i}(x)$ but these aren't the $\sigma$ in $\sigma_{i}$ above, are they? They certainly don't look like them as maps. Also what justification do we have to factor the $\sigma_{i}$ out in the third equality? I think these two questions have trivial answers, but I just cant see it.
Any help is appreciated!
 A: Fix an algebraic closure $K^a$ of $K$ such that $M \subseteq K^a$. Let $[L:K] = m$ and let $\{ \sigma_1,\dots,\sigma_m \}$ be a complete set of representatives for the equivalence classes of $K$-embeddings of $M$ into $K^a$, under the equivalence relation
$$
\sigma \sim \tau \iff \sigma |_L = \tau |_L,
$$
where $\sigma, \tau : M \to K^a$ are $K$-embeddings. Let $\tau_1,\dots,\tau_n$ be the $K$-embeddings of $M$ that fix $L$ pointwise.
Extend each $\sigma_i$ and $\tau_j$ to automorphisms of $K^a$; call these maps $\tilde{\sigma}_i$ and $\tilde{\tau}_j$, respectively. Observe that for every $1 \leq i \leq m$, $1 \leq j \leq n$, the map $\tilde{\sigma}_i \circ \tilde{\tau}_j$ when restricted to $M$ gives a $K$-embedding. Conversely, every $K$-embedding $\sigma : M \to K^a$ is a restriction to $M$ of $\tilde{\sigma}_i \circ \tilde{\tau_j}$ for some $1 \leq i \leq m$, $1 \leq j \leq n$.
To prove the claim, let $\sigma : M \to K^a$ be a $K$-embedding. Then there is some $1 \leq i \leq m$ such that $\sigma \sim \sigma_i$. Extend $\sigma$ to an automorphism of $K^a$, and call this extension $\tilde{\sigma}$. Then, the automorphism $\tilde{\sigma}_i^{-1} \circ \tilde{\sigma}$ of $K^a$, when restricted to $M$, gives a $K$-embedding of $M$ that fixes $L$ pointwise. So, there is some $1 \leq j \leq n$ such that $(\tilde{\sigma}_i^{-1} \circ \tilde{\sigma}) |_M = \tilde{\tau}_j |_M$. Hence, there exists $1 \leq i \leq m$ and $1 \leq j \leq n$ such that $\sigma = (\tilde{\sigma}_i \circ \tilde{\tau}_j) |_M$.
In particular, this means that for each $1 \leq i \leq m$,
$$
\{ \sigma \sim \sigma_i : \sigma : M \to K^a \text{ is a $K$-embedding} \} = \{ (\tilde{\sigma}_i \circ \tilde{\tau}_j) |_M : 1 \leq j \leq n \}
$$
So, for every $x \in L$ we have
$$
\sum_{\sigma \sim \sigma_i} \sigma(x) = \sum_{j=1}^n \tilde{\sigma}_i \circ \tilde{\tau}_j (x) = \sum_{j = 1}^n \tilde{\sigma}_i \circ \tilde{\tau}_j \circ \tilde{\sigma}_i^{-1} (\tilde{\sigma}_i(x)).
$$
Now, we observe the following:


*

*$\tilde{\sigma}_i(x) = \sigma_i(x) \in \sigma_i M$ for every $x \in M$;

*The set
$$
\{ (\tilde{\sigma}_i \circ \tilde{\tau}_j \circ \tilde{\sigma}_i^{-1}) |_{\sigma_i M} : 1 \leq j \leq n \}
$$
is the set of all $\sigma_i L$-embeddings of $\sigma_i M$ into $K^a$. That is, every $\sigma_i L$-embedding of $\sigma_i M$ into $K^a$ is a member of the above set, and there are $n$ distinct such embeddings.


To see why the second claim is true, observe that for any $x \in L$,
\begin{align}
&\tilde{\sigma}_i \circ \tilde{\tau}_j \circ \tilde{\sigma}_i^{-1} (\sigma_i(x))\\
={} &\tilde{\sigma}_i ( \tilde{\tau}_j ( \tilde{\sigma}_i^{-1}(\sigma_i(x))))\\
={} &\tilde{\sigma}_i(\tilde{\tau_j}(x))\\
={} &\tilde{\sigma}_i(x).
\end{align}
So, $(\tilde{\sigma}_i \circ \tilde{\tau}_j \circ \tilde{\sigma}_i^{-1}) |_{\sigma_i M}$ is a $\sigma_i L$-embedding. Conversely, let $\tau : \sigma_i M \to K^a$ be a $\sigma_i L$-embedding, and let $\tilde{\tau}$ denote an extension of $\tau$ to an automorphism of $K^a$. Then, $\tilde{\sigma}_i^{-1} \circ \tilde{\tau} \circ \tilde{\sigma}_i$ is an automorphism of $K^a$ that fixes $L$ pointwise, because for any $x \in L$,
\begin{align}
&\tilde{\sigma}_i^{-1} \circ \tilde{\tau} \circ \tilde{\sigma}_i (x)\\
={} &\tilde{\sigma}_i^{-1} ( \tilde{\tau} ( \tilde{\sigma}_i(x)))\\
={} &\tilde{\sigma}_i^{-1}(\tilde{\sigma_i}(x))\\
={} &x.
\end{align}
Hence, there exists $1 \leq j \leq n$ such that $(\tilde{\sigma}_i^{-1} \circ \tilde{\tau} \circ \tilde{\sigma}_i) |_M = \tilde{\tau}_j |_M$. So, $\tau = (\tilde{\sigma}_i \circ \tilde{\tau}_j \circ \tilde{\sigma}_i^{-1}) |_{\sigma_i M}$. Lastly, these $n$ maps are all distinct because the maps $\tau_j$
are distinct.
Hence,
$$
\sum_{j = 1}^n \tilde{\sigma}_i \circ \tilde{\tau}_j \circ \tilde{\sigma}_i^{-1} (\tilde{\sigma}_i(x)) = Tr_{\sigma_i M/ \sigma_i L}(\sigma_i(x)).
$$
This proves the second equality.
The third equality can now be shown easily. For each $x \in M$, we have
\begin{align}
Tr_{M/L}(x) &= \sum_{j=1}^n \tau_j (x)\\
&= \sum_{j=1}^n \tilde{\tau}_j(x) \\
&= \sum_{j=1}^n \tilde{\sigma}_i^{-1} \circ (\tilde{\sigma}_i \circ \tilde{\tau}_j \circ \tilde{\sigma}_i^{-1}) (\tilde{\sigma}_i(x)) \\
&= \tilde{\sigma}_i^{-1} \circ \left( \sum_{j=1}^n \tilde{\sigma}_i \circ \tilde{\tau}_j \circ \tilde{\sigma}_i^{-1} (\tilde{\sigma}_i(x)) \right) \\
&= \tilde{\sigma}_i^{-1} \circ Tr_{\sigma_i M / \sigma_i L}(\sigma_i (x)).
\end{align}
Hence,
$$
\sigma_i(Tr_{M/L}(x)) = \tilde{\sigma}_i(Tr_{M/L}(x)) = Tr_{\sigma_i M / \sigma_i L}(\sigma_i (x)).
$$
This proves the third equality.

Do note that the computations are quite straightforward. The key tool in proving the second equality is the isomorphism extension theorem. The rest is routine checking.
Very often authors use the same notation to denote a $K$-embedding and its extension to an automorphism of $K^a$. Omitting the $\tilde{}$ symbol makes the presentation cleaner. It also helps in simplifying some of the steps provided you're willing to confuse $\sigma$ with $\tilde{\sigma}$ as and when appropriate.
You should look at all the gory details once, and understand the idea behind the proof. This should be enough to reproduce it without any need for this level of elaboration. All the best.

I also concur with @darijgrinberg that there are better proofs of this result. I am quoting from the comment directly:

I suggest reading the proof of Theorem 4.1 in Keith Conrad, Trace and Norm, II (link: www.math.uconn.edu/~kconrad/blurbs/galoistheory/tracenorm2.pdf) instead. What Neukirch is doing is very much a hack. After that, go on to read John R. Silvester, Determinants of Block Matrices, The Mathematical Gazette, Vol. 84, No. 501 (Nov., 2000), pp. 460--467, and figure out on your own why the norm is transitive too.

I also recommend Lang's Algebra: see Theorem 5.1 on page 285 of the third edition.
