# Proof Verification: Field extension where every element is a product of elements of a certain form

Let $F$ be a field, and suppose $F$ has no nontrivial field extensions of degree less than $n$. Let $L=F[v]$ be a field extension with $v^n\in F$.

Prove that every element in $L$ is a product of elements of the form $cv+d$ where $c,d\in F$.

Here is what I have done so far, not sure if there are any gaps. Thanks for any critique!

First note that any element in $L$ is of the form $l=\sum_{i=0}^{n-1}c_iv^i$ where $c_i\in F$. This is because $v^n\in F$.

Let $f(x)=\sum_{i=0}^{n-1}c_ix^i$, so that $l=f(v)$. Let $\alpha_1, \dots\alpha_{n-1}$ be the roots of $f$ in the algebraic closure $\bar F$.

$[F(\alpha_i):F]=\deg m_{\alpha_i}(x)\leq\deg f<n$ (where $m_{\alpha_i}$ denotes the minimal polynomial). Since there are no nontrivial field extensions of degree less than $n$, $F(\alpha_i)=F$, which means $\alpha_i\in F$. So $f$ splits over $F$, i.e. $f(x)=\prod_{i=1}^{n-1}(a_ix+b_i)$ where $a_i, b_i\in F$.

Putting $x=v$ yields $l=f(v)=\prod_{i=1}^{n-1}(a_iv+b_i)$, which is what we want to prove.

Is the above proof ok?

• Seems good to me – egreg Sep 5 '16 at 15:35

A slightly cleaner approach. Consider a nonzero polynomial $f(x)\in F[x]$ of degree $d$ with $1<d<n$. Then $f$ is reducible, otherwise a root of it would produce an extension of $F$ of degree $d$. Split $f$ into two factors and apply induction to show that $f$ splits into degree $1$ factors.
Note that this doesn't require the algebraic closure, because you can just consider the extension field $F[T]/(f(T))$, when $f$ is irreducible, and $f$ has a root in this extension.