Past exam of the number theory course of the University of Oxford / 2010 / A / 1 / iii

First, the number of embeddings extending a prescribed base embedding is at most the extension degree. Prove this for any finite extension by choosing finitely many algebraic generators and applying part (i) successively. At each step the number of extensions is at most the degree of that simple step; multiplication and the tower law for field extensions give the total bound.

If $\tilde{K}/K$ is not separable, choose $\alpha$ with inseparable minimal polynomial of degree $e=[K(\alpha ):K]$. Its image under $i$ has strictly fewer than $e$ distinct roots in any target field: inseparability is preserved under the coefficient embedding. Thus there are $r<e$ embeddings of $K(\alpha )$ extending $i$. Each extends to $\tilde{K}$ in at most $[\tilde{K}:K(\alpha )]$ ways, giving at mostIf instead $i(m_{\alpha })$ does not split in $L$, it again has fewer than $e$ distinct roots in $L$, and exactly the same estimate applies. Hence either obstruction makes the embedding count strictly less than $d$.

Set $K_r=K({\alpha }_1,\dots ,{\alpha }_r)$, beginning with $K_0=K$. Suppose $j:K_{r-1}\to L$ extends $i$. The minimal polynomial of ${\alpha }_r$ over $K_{r-1}$ divides its minimal polynomial over $K$. Its image under $j$ therefore divides $i(m_{{\alpha }_r})$, which splits into distinct linear factors in $L$. It consequently has exactly $[K_r:K_{r-1}]$ distinct roots in $L$. Part (i) supplies exactly this many extensions of $j$ to $K_r$.

Starting with the unique prescribed map on $K$, multiply these branching counts:For the final assertion, let $M$ be a splitting field of a separable polynomial and embed it in an algebraic closure $\Omega$. It is generated by roots whose minimal polynomials are separable and split in $\Omega$. Applying the just-proved count with $L=\Omega$ gives $[M:K]$ embeddings. Were $M/K$ inseparable, part (a) would give strictly fewer. Thus the splitting extension is separable. This works for the factorwise convention as well, since the generators' minimal polynomials divide the separable irreducible factors.