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I'm having trouble following the details of the discussion on pages 9 and 10 of Neukirch's algebraic number theory book.

Suppose $L$ is a separable extension of $K$ with degree n. Consider the set of embeddings of $L$ into $\bar K$, the algebraic closure of $K$, that fix $K$ (K-embeddings). Why are there $n$ embeddings in this set?

EDIT: Also, consider some element $x\in L$. Let $d$ be the degree of $L$ over $K(x)$ and $m$ be the degree of $K(x)$ over $K$. Why are the $K$-embeddings of $L$ partitioned by the equivalence relation

$ \sigma\sim\tau\ \Leftrightarrow\ \sigma x = \tau x $

into $m$ equivalence classes of $d$ elements each?

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    Where *e$x$actly* are you in that page 11 in Neukirch's book? I've the 1999 edition of the book and in page 11 he talks about discriminant, proposition 2.8 ...so where are you?2012-06-10

3 Answers 3

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The idea behind the proof is that for a field $K$ and an element $\alpha \in \bar{K}$, the roots of the minimal polynomial of $\alpha \in \bar{K}$ are exactly the conjugates of $\alpha$ over $K$. Then taking $L = K(\alpha)$ each conjugate of $\alpha$ defines a unique embedding from $L$ to $\bar{K}$. Since $[L: K] = n$, there are $n$ distinct embeddings.

For the full details of this proof look at Lemma 5.17 and Theorem 5.18 of this.

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    What is the name of this book?2018-05-18
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I'll provide a slightly different solution to your second question. Assuming that $\#\text{Hom}(L,\bar K)=n$, let $x=x_1,\dots,x_m\in \bar K$ be the conjugates of $x$ (i.e., the roots in $\bar K$ of the minimal polynomial $m_{x,K}(t)$ for $x$ over $K$). Recall that each of the $m$ isomorphisms $K(x)\rightarrow K(x_j)$ extend to $d$ isomorphisms $L=K(x,\theta)\rightarrow K(x_j,\theta_k)$ where $\theta\in \bar K$ is a primitive element for $L$ over $K(x)$ (i.e. $L=K(x,\theta)$) and $\theta=\theta_1,\dots,\theta_d$ are its conjugates. (This is Theorem 8 in section 13.1 of Dummit and Foote.) This accounts for all $n=md$ embeddings $L\rightarrow \bar K$ via $L=K(x,\theta)\rightarrow K(x_j,\theta_k)\rightarrow\bar K$, where the last map is inclusion. Thus, what we find is that for each $1\leq j \leq m$ there are $d$ embeddings $\sigma:L\rightarrow \bar K$ with $\sigma: x\mapsto x_j$. Hence there are $m$ equivalence classes, each with $d$ elements.

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I ran up against this while reading Neukirch, so let me expand Qiaochu Yuan's comment into a short answer.

From the statement that there are $n$ $K$-embeddings of a separable extension $L$ of $K$ into $\bar K$, we find that there are

  • $md$ embeddings $L \hookrightarrow \bar K$ that fix $K$

  • $d$ embeddings $L \hookrightarrow \bar K$ that fix $K(x)$; equivalently, these are $K$-embeddings that also fix the element $x\in L$

Each of the $md$ $K$-embeddings $\sigma \in {\mathrm{Hom}}_K(L,\bar K)$ is a member of an equivalence class corresponding to a choice $x_i = \sigma x$ of a conjugate of $x$ to generate $K(x)$ with (there are $m$ of those). For every such conjugate, the equivalence class comprises $d$ $K(x_i)$-embeddings of $L$ into $\bar K$.

Note that the second point uses the fact that the algebraic closure ${K(x)}^{\mathrm{alg}} = K^{\mathrm{alg}} = \bar K$.

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    The ending of this answer is kind of muddled. Edits welcome!2017-09-02