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Fix an algebraic closure $\overline{\mathbf{Q}}$ of the rational numbers.

Let $\mathbf{Q}\subset K$ be a number field.

I know that the degree $[K:\mathbf{Q} ]$ equals the number of embeddings of $K$ into the complex numbers.

Does $[K:\mathbf{Q}]$ also equal to the number of embeddings of $K$ into $\overline{\mathbf{Q}}$?

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1 Answer 1

up vote 7 down vote accepted

I suppose that by $\overline{\Bbb Q}$ you mean the subfield of $\Bbb C$ consisting of the algebraic numbers.

If $F\subset L$ and $F\subset L^\prime$ are fields, any map of fields $L\rightarrow L^\prime$ which is the identity on $F$ will send elements that are algebraic over $F$ to elements that are algebraic over $F$. This is a straightforward consequence of the definition of homomorphism.

In an extension ${\Bbb Q}\subset K$ of finite degree, all elements are algebraic. Thus the image of an embedding $K\rightarrow\Bbb C$ must be entirely included in $\overline{\Bbb Q}$.

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This is true even if the degree is not finite, as long as the extension is algebraic. But number field usually means finite extension. –  lhf Oct 8 '11 at 14:48

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