# Number of embeddings of $\mathbb Q (\alpha)$ into $\mathbb C$ which map $\beta \in \mathbb Q(\alpha)$ to a given conjugate

I'm looking at a proof of the following:

Let $K = \mathbb Q(\alpha)$ and $\beta \in K$ with minimal polynomial $g \in \mathbb Q[X]$, where the roots of $g$ are $\beta_1, \ldots , \beta_m$. Then $d_i = |\{\sigma : K \hookrightarrow \mathbb C \ | \ \sigma(\beta) = \beta_i \}|$ is independent of $i$.

The proof starts by considering $h$, the minimal polynomial of $\alpha$ over $\mathbb Q(\beta)$, and says that $d_i \leq \mathrm{deg}(h) = [\mathbb Q(\alpha) : \mathbb Q (\beta)]$.

Why is this true? I did understand this at one point...

Thanks

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Call $f_i$ the composition $\mathbb Q(\beta) \to \mathbb Q(\beta_i) \to \mathbb C$, where the first morphism is the one sending $\beta$ to $\beta_i$ and the second is the inclusion.
Then $d_i$ is the number of extensions of $f_i$ to $K$ , which is by definition the separable degree $[K:\mathbb Q(\beta)]_{sep}$ (cf. Lang Algebra, Chap.5, §4). So we have $d_i=[K:\mathbb Q(\beta)]_{sep}$.

But that separable degree $d_i=[K:\mathbb Q(\beta)]_{sep}$ is equal to the plain degree $[K : \mathbb Q (\beta)]$ because we are in characteristic zero. So that $d_i=[K : \mathbb Q (\beta)]$, which is indeed independent of $i$ .

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I think I've figured it out - hopefully someone will let me know if this is wrong.

Given a conjugate $\beta_j$ of $\beta$, we want to consider $d_j$, the number of embeddings $\sigma : \mathbb Q(\alpha) \hookrightarrow \mathbb C$ such that $\sigma(\beta) = \beta_j$. If $h$ is the minimal polynomial of $\alpha$ over $\mathbb Q(\beta)$, then denote $h_\sigma$ the image of $h$ under $\sigma$. Since $\sigma$ is a homomorphism, it must map $\alpha$ to a root of $h_\sigma$. There are at most $\mathrm{deg}(h)$ such roots, and each unique root gives rise to a different embedding.

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Looks good to me. Note that you really do have an equality $d_i = \deg h$ here. (In characteristic zero, everything is separable.) – Dylan Moreland Feb 12 '12 at 2:20