Exercise 7.12

Proof. We show by induction on the degree $n$ of $f$ that for all polynomials $f\in k[x]$ with $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=n\ge 1$, there exists a field extension $K$ such that $[K:k]$ is finite, and $f(x)$ splits in linear factors on $K$.

If $n=1$, $f(x)=\mathit{ax}+b=a(x-{\alpha }_0)$, where ${\alpha }_0=-b∕a$ : $K=k$ is suitable.

Suppose that the property is true for all polynomials of degree less than $n$ on an arbitrary field $k$.

Let $f(x)\in k[x],\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=n$. From proposition 7.2.1. applied to an irreducible factor of $f$, there exists a field $L,[L:k]<\infty$ and ${\alpha }_1\in L$ such that $f({\alpha }_1)=0$. Then $f(x)=(x-{\alpha }_1)g(x),g(x)\in L[x]$.

Applying the induction hypothesis in the field $L$ on the polynomial $g\in L[x]$ with $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=n-1$, we obtain a field $K,[K:L]<\infty$ such that $g(x)=a(x-{\alpha }_2)\cdots (x-{\alpha }_n)$ with ${\alpha }_i\in K$. So $f(x)=a(x-{\alpha }_1)(x-{\alpha }_2)\cdots (x-{\alpha }_n)$ splits in linear factors in $K$. The induction is done. □