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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

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  • $\begingroup$ $S$ must be the set of monic (non-trivial factors of $P$). With this, you can say that $P_1,P_2$ are monic (and have degree $\geq1$) $\endgroup$ – Martín Vacas Vignolo Dec 26 '18 at 11:26
  • $\begingroup$ Why must the factors be monic? $\endgroup$ – PolynomialC Dec 26 '18 at 11:28
  • $\begingroup$ No. $S$ is a subset of set of all factors. The key is that in your proof you can not say that $\deg P_i \geq 1$ $\endgroup$ – Martín Vacas Vignolo Dec 26 '18 at 11:30
  • $\begingroup$ Is this to make sure the set is of finite size and thus bounded below? Cheers for the input. $\endgroup$ – PolynomialC Dec 26 '18 at 11:33
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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.

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