Normal closure as the compositum of conjugates

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An excerpt from Serge Lang's Algebra Chapter V $\S4$ p. 242.

Let $E$ be a finite extension of $k$. The intersection of all normal extensions $K$ of $k$ (in an algebraic closure $E^\text{a}$) containing $E$ is a normal extension of $k$ which contains $E$, and is obviously the smallest normal extension of $k$ containing $E$. If $\sigma_1,\dots,\sigma_n$ are the distinct embeddings of $E$ in $E^\text{a}$, then the extension $$ K = (\sigma_1E)\dots (\sigma_nE) $$ which is the compositum of all these embeddings, is a normal extension of $k$, because for any embedding of it, say $\tau$, we can apply $\tau$ to each extension $\sigma_iE$. Then $(\tau\sigma_1,\dots,\tau\sigma_n)$ is a permutation of $(\sigma_1,\dots,\sigma_n)$ and thus $\tau$ maps $K$ into itself.

I just want to confirm that my understanding of the domains and codomains of these maps $\sigma_i$ and $\tau$ are correct.

First, we are fixing some unstated embedding $\gamma: k\to E^\text{a} \cong k^\text{a}$. Then, we can extend $\gamma$ into $n=[E:k]_s$ embeddings of $E$ into $E^\text{a}$ given by $\sigma_i:E\to E^\text{a}$ such that $\sigma_i|_k=\gamma$. (Here, $[E:k]_s$ is the separable degree of the extension $E/k$.) Then $K=\sigma_1(E)\dots\sigma_n(E)$ is a subfield of $E^\text{a}$ and $\tau$ is an embedding $ \tau: K\to E^\text{a} $ such that $\tau|_{\gamma(k)} = \text{Id}_{\gamma(k)} $.

Is this correct? I'm interpreting the $\sigma_i$ as embeddings of the extension $E/k$ into $E^\text{a}$ where $E/k$ is not a priori considered to be subfields of the fixed algebraic closure $E^\text{a}$. This is why I think we must choose a particular embedding $\gamma: k\to E^\text{a}$ since such a choice is non-canonical (but they didn't mention this choice in the text). Then, I'm interpreting $\tau$ as an embedding of $K$, which is now considered to be a subfield of $E^\text{a}$, back into $E^\text{a}$, while keeping $\gamma(k)$ fixed. "Keeping $\gamma(k)$ fixed" is now a well-defined and sensible notion.

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Choose an algebraic closure $E^a$ of $E$ (this includes the choice of an embedding $E \to E^a$!).

Then $k$ embeds into $E^a$ canonically via $E$, i.e. $k \to E \to E^a$, where $k \to E$ is the map corresponding to the extension $E/k$.

Note that the quoted text does not use $k^a$ at all.