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Suppose that $K$ is a finite extension of $\mathbb Q$, say of degree $n$. By the primitive element theorem, $K=\mathbb Q(\alpha)$. Then $\alpha$ has $n$ conjugates and we correspondingly get $n$ embeddings of $K$ into $\mathbb C$. But I believe that all these embeddings need to be have the same images in $\mathbb C$ (as an example, $\mathbb Q(\sqrt 2)$ and $\mathbb Q(-\sqrt 2)$ are really the same field). Is there any relation between the number of embeddings with distinct images and the order of the group $Aut(K/\mathbb Q)$? On working a few examples, it seems that if $|Aut(K/\mathbb Q)|=a$ and there are $b$ embeddings with distinct images in $\mathbb C$, then $ab=n$, the degree of the extension. Is this true? If so, how can I prove it? I don't seem to be having much success. I'd appreciate some help.

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  • $\begingroup$ The $n$ embeddings are all distinct maps. Their range may be the same. Have you been introduced to the concepts of normal extension, and normal closure of a finite field extension? Without them, it is hard to explain how things work. $\endgroup$ – Crostul Feb 2 '17 at 9:41
  • $\begingroup$ Yes, I know about normal extension. I think I may not remember much about what a normal closure is though. $\endgroup$ – adrija Feb 2 '17 at 9:58
  • $\begingroup$ Seing the algebraic numbers as complex numbers, if $K$ is Galois then every automorphism/embedding $K \to \mathbb{C}$ sends $K$ to $K$ and of course are the identity on $\mathbb{Q}$, so they are given by the Galois group $Gal(K/\mathbb{Q})$, having $n = [K:\mathbb{Q}]$ elements $\endgroup$ – reuns Feb 2 '17 at 11:14
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For an algebraic extension $K/\mathbb{Q}$ of degree $n$ there are $n$ embeddings of $K$ into $\mathbb{C}$, see (The number of) embeddings of an algebraic extension of $\mathbb{Q}$ into $\mathbb{C}$. For example, let $K=\mathbb{Q}(\sqrt[4]{2})$. Then $a=|Aut_{\mathbb{Q}}(K)|=2$, $b=4$ and $n=4$. Hence $ab\neq n$.

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  • $\begingroup$ Sorry, I think I couldn't get myself understood properly. What I meant was that there are $b$ embeddings with distinct images. I realize that the embeddings as maps are distinct. Sorry for the confusion. I've edited the question to make my point clear. $\endgroup$ – adrija Feb 2 '17 at 9:54
  • $\begingroup$ @adrija $i\sqrt[4]{2}$ is a conjugate of $\sqrt[4]{2}$ but $\mathbb{Q}(i\sqrt[4]{2})$ and $\mathbb{Q}(\sqrt[4]{2})$ are different subsets of $\mathbb{C}$. Is this the kind of example you wanted? $\endgroup$ – Arun Kumar Feb 2 '17 at 10:07
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Yes, of course there is, and this is one of the main results in fields extensions-Galois Theory.

Theorem: If $\;K/F\;$ is a fields extension of degree $\;n\;$ and $\;S,\,F\subset S\subset K\;$ is the separable closure of this extension, then there are $\;[S:F]\;$different embeddings $\;K\mapsto\overline{\Bbb F}\;$ , when this last is any algebraically closed field containing $\;F\;$ ( we can take the algebraic closure of $\;F\;$) .

If we assume the extension $\;K/F\;$ is finite and separable, then the number of embeddings equals $\;n=[K:F]\;$ iff $\;K/F\;$ is normal, which would mean the extension $\;K/F\;$ is Galois. In this last case, any such embedding is an automorphism of $\;K/F\;$ , meaning $\;\sigma K=K\;$ for any embedding $\;\sigma:K\to\overline F\;$, and $\;\sigma f=f\;,\;\;\forall\,f\in F\;$ .

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Maybe an example is appropriate. Let $p$ be a polynomial of degree $3$ with rational coefficients and one real root $r$ and two complex (conjugate) roots ($c$ and $\overline{c}$). The splitting field $F$ of $p$ is an extensioin of degree $6$ and the Galois group of $p$ is the symmetric group $S_3$, the group that permutes the three roots. The Galois group consists of field automorphisms of $F$. On the other hand the field $M$ generated by the root $r$ is a subfield of $F$ is of degree $3$ and is not normal. The only elements of the Galois group that leave this field invariant (and so define automorphisms on $M$) are thos that permute the other two root. This makes that $M$ has two other different embeddings in $F$ that are images of automorphisms of $F$.

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