Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having trouble following the details of 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 $x\in K$. 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 relations of $d$ elements?

share|cite|improve this question
Use the primitive element theorem. – Qiaochu Yuan Jun 9 '12 at 23:31
For the second question, apply the first result to $L$ as an extension of $K$ and then to $L$ as an extension of $K(x)$. – Qiaochu Yuan Jun 9 '12 at 23:49
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 where are you? – DonAntonio Jun 10 '12 at 2:12
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.