Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 17F (Galois Theory)

For $\alpha \in L$ with $f(\alpha )=0$, evaluation $K[t]\to L$, $g\mapsto g(\alpha )$, has $(f)$ in its kernel. It therefore induces a $K$-algebra homomorphism ${\sigma }_{\alpha }:K[t]/(f)\to L$. Conversely, such a homomorphism is determined by $\alpha =\sigma (\bar{t})$, since every class is a polynomial in $\bar{t}$. Applying it to $f(\bar{t})=0$ shows $f(\alpha )=0$. These constructions are inverse, giving the root-embedding correspondenceIrreducibility ensures the quotient is a field, so each unital homomorphism is injective.

A splitting field of $f$ over $K$ is an extension generated over $K$ by all its roots in which $f$ splits into linear factors. A nonconstant irreducible polynomial is separable polynomial when its roots in a splitting field are distinct, equivalently when it is coprime to its formal derivative. For a general polynomial, the convention compatible with the printed assertion is that all its irreducible factors are separable; repeated occurrences of the same factor are allowed under this convention.

For a finite field of characteristic $p$, the finite-field Frobenius automorphism $x\mapsto x^p$ is injective and hence surjective. If an irreducible polynomial $g$ had zero derivative, every nonzero exponent would be divisible by $p$. Taking $p$th roots of its coefficients would give $g=h^p$, contradicting irreducibility. Thus $g^{\prime }\ne 0$, and irreducibility gives $\gcd (g,g^{\prime })=1$. Every irreducible factor is therefore separable: finite fields are perfect.

If instead one defines a separable polynomial to mean that the entire polynomial has no repeated roots, the printed claim for every $f$ is false: $(t-1{)}^2$ is a counterexample over any finite field. The proof above establishes the precise intended perfect-field assertion; a polynomial has entirely distinct roots if and only if it is also squarefree.

A finite separable field extension of degree $d$ has exactly $d$ embeddings into an algebraic closure: adjoin finitely many generators successively, and use part (i) to count extensions of each embedding by the distinct roots of the relevant separable minimal polynomial.

For $L=K(\alpha ,\beta )$, label these embeddings ${\sigma }_1,\dots ,{\sigma }_d$. For each pair $i\ne j$, the equalityexcludes at most one $c\in K$. If the images of $\beta$ agree, the images of $\alpha$ cannot agree too, so it excludes none. Since $K$ is infinite, choose $c$ avoiding the finitely many bad values and put $\eta =\alpha +c\beta$. The $d$ images of $\eta$ are distinct, hence its minimal polynomial has degree at least $d$. But $K(\eta )\subset L$ gives degree at most $d$. Therefore $K(\eta )=L$. Combining generators two at a time proves the primitive element theorem for every finite separable extension over an infinite base field.