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

Any extending field homomorphism $j$ is determined by $j(\alpha )$, since every element of $K(\alpha )$ is represented by a polynomial in $\alpha$ with coefficients in $K$. Applying $j$ to the minimal-polynomial equation shows $i(m_{\alpha })(j(\alpha ))=0$. Thus there are at most $m$ choices.

For each listed root ${\beta }_r$, define $K[x]\to L$ by $\sum a_k{x}^k\mapsto \sum i(a_k){\beta }_r^k$. Its kernel contains $(m_{\alpha })$, so it induces a unital map from the field $K[x]/(m_{\alpha })\cong K(\alpha )$ into $L$. A unital map from a field is injective. This gives an extending homomorphism $j_r$ with $j_r(\alpha )={\beta }_r$. Distinct roots give distinct maps, proving exactly $m$ extensions.

An irreducible polynomial is separable if its roots in a splitting field are distinct, equivalently if it is coprime to its formal derivative. For a general nonzero polynomial, the factorwise separability convention means that every irreducible factor is separable. This is the convention under which the printed splitting field implication holds without an extra hypothesis.

If instead separability of an arbitrary polynomial is defined to require all its roots to be distinct, that implication needs the polynomial to be squarefree. For example, $(x-1{)}^2$ has separable splitting extension $K/K$, but a repeated root. These two conventions agree for irreducible polynomials; repeated factors are the entire distinction here.

An algebraic field extension $E/K$ is separable if every $\alpha \in E$ has a separable minimal polynomial over $K$.

For an inseparable example, take $K={\mathbb{F}}_p(t)$ and $f(X)=X^p-t$. Eisenstein's criterion at the prime $t$ in ${\mathbb{F}}_p[t]$, followed by Gauss lemma for polynomials, proves irreducibility over its fraction field. Its derivative is zero, and in a splitting field it is $(X-t^{1/p}{)}^p$, so it is inseparable.

Now let $M/K$ be a separable splitting extension of a nonzero polynomial $f$. Every irreducible factor $g$ of $f$ has a root $\alpha \in M$, and $g$ is a scalar multiple of the minimal polynomial of $\alpha$. Separability of $M/K$ makes this polynomial separable. Therefore every irreducible factor of $f$ is separable, which proves the requested conclusion in the factorwise convention. The squarefree qualification under the alternative convention is explained in part (a).

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.