# Polynomials and Irreducible polynomials

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

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

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.