So I'm trying to prove this Proposition:
Every polynomial of degree greater than $0$ is divisible by an irreducible polynomial.
Proof:
Consider a polynomial $P \in \mathbb R [x]$ of degree $\geq 1$
Let $S$ denote the set of non-trivial factors of $P$.
Suppose that $S$ does not contain any irreducible polynomials.
By the well ordering principle let $P_0$ denote the polynomial of least degree in $S$
By our assumption $P_0$ is not irreducible $\therefore P_0=P_1P_2$ for some
$P_1,P_2 \in \mathbb R[x]$
This contradicts the definition of $P_0$ being the polynomial of least degree to be a factor of $P$
Thus we are forced to conclude the polynomial of least degree is irreducible and the claim follows.
$\blacksquare$
My main issue is can I use the well ordering principle in this fashion?
thanks
Best Answer
If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $\mathbb N$.
For that, define $D = \{ n \in \mathbb N : \text{there is a factor of $P$ of degree $n$} \}$. Then $D$ is not empty because $P$ is a factor of $P$. Let $m = \min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.