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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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    $\begingroup$ Use the primitive element theorem. $\endgroup$ – Qiaochu Yuan Jun 9 '12 at 23:31
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    $\begingroup$ For the second question, apply the first result to $L$ as an extension of $K$ and then to $L$ as an extension of $K(x)$. $\endgroup$ – Qiaochu Yuan Jun 9 '12 at 23:49
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    $\begingroup$ Where exactly 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? $\endgroup$ – DonAntonio Jun 10 '12 at 2:12
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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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  • $\begingroup$ What is the name of this book? $\endgroup$ – Fawzy Hegab May 18 '18 at 20:52
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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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  • $\begingroup$ The ending of this answer is kind of muddled. Edits welcome! $\endgroup$ – Soham Chowdhury Sep 2 '17 at 13:16

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