# how many embeddings into $\overline{\mathbf{Q}}$ does a given number field have

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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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 $\overline{\Bbb Q}$.