Exercise 6.2.7 (Splitting field equivalence)

Proof. Here $f\in K[t],f\ne 0$, and $M:K$ is a field extension.

Consider the propositions

i.

$f$ splits in $M$.

ii.

$M=K({\alpha }_1,\dots ,{\alpha }_n)$, where ${\alpha }_1,\dots ,{\alpha }_n$ are the roots of $f$ in $M$.

ii’.

if $L$ is a subfield of $M$ containing $K$, and $f$ splits in $L$, then $L=M$

We must prove $(i\&\mathrm{ii})⟺(i\&{\mathrm{ii}}^{\prime })$.

$\text{(⇒)}$

Assume (i) and (ii). Let $L$ be a subfield of $M$ containing $K$ such that $f$ splits in $L$. Then $$f(t)=a(t-{\alpha }_1)\cdots (t-{\alpha }_n),a\in L^{\text{∗}},{\alpha }_1,\dots ,{\alpha }_n\in L,$$

where ${\alpha }_1,\dots ,{\alpha }_d$ are the roots of $f$ in $L$ (with possible repetitions).

Since $L\subset M$, ${\alpha }_1,\dots ,{\alpha }_d$ are the roots of $f$ in $M$, and (ii) shows that $M=K({\alpha }_1,\dots ,{\alpha }_n)$.

Then $L\supset K$ and $L\supset \{{\alpha }_1,\dots ,{\alpha }_d\}$, thus $L\supset K({\alpha }_1,\dots ,{\alpha }_d)=M$.

$\text{(⇐)}$

Assume (i) and (ii’).

By (i), $f(t)=a(t-{\alpha }_1)\cdots (t-{\alpha }_n)$, where ${\alpha }_1,\dots ,{\alpha }_n$ are the roots of $f$ in $M$ (with repetitions).

Consider the subfield $L=K({\alpha }_1,\dots ,{\alpha }_n)$. Then ${\alpha }_1,\dots ,{\alpha }_n\in L$, and $a$ is the leading coefficient of $f\in K[t]$, thus $a\in K\subset L$. Therefore $f(t)=a(t-{\alpha }_1)\cdots (t-{\alpha }_n)$ splits in $L$. By (ii’), $L=M$, and this proves that $M=K({\alpha }_1,\dots ,{\alpha }_n)$, so (ii) is true.