Every polynomial of odd degree $\ge 3$ over $\mathbb{R}[x]$ is reducible over $\mathbb{R}$

Question

I need to prove that every polynomial of odd degree $\ge 3$ over $\mathbb{R}[x]$ is reducible over $\mathbb{R}$.

If $p(x)$ is my polynomial, then I just have to prove that $p(x)$ has one real root, right? If I can do it, then I must only apply the division theorem and divide by $x-\alpha$ when $\alpha$ is the root. I've found some arguments like this one but they use too much analysis and I'm on a ring theory course. How can I prove this in a more abstract way?

If you can also prove this for odd degrees $<3$, then you can just say odd degree... — Asaf Karagila, Jan 01 '16 at 17:37
For what it's worth, every real polynomial whose degree is greater than $2$ is $\mathbb R$-reducible. — Git Gud, Jan 01 '16 at 17:59
@Guelando There is no proof of what you want without using analysis (real or complex). — user26857, Jan 01 '16 at 18:57

score 3 · Accepted Answer · answered Jan 01 '16 at 17:39

3

Hint: Try to prove that if $\alpha$ is a complex root of $P(x) \in \Bbb R[x]$ then also $\overline{\alpha}$ is a root of $P(x)$. Then deduce that $P(x)$ must have an even number of non real roots.

answered Jan 01 '16 at 17:39

mrprottolo

3,802
1
15
22

How do you know that $p(x)$ has a complex root? – user26857 Jan 01 '16 at 18:33
Since $\Bbb C$ is the algebraic closure of $\Bbb R$, all the roots of $p(x)$ must be in $\Bbb C$. – mrprottolo Jan 01 '16 at 18:37
Maybe you want to read this and this. – user26857 Jan 01 '16 at 18:46
I assumed that we could use FTA. – mrprottolo Jan 01 '16 at 18:48
1

The OP said that he is not satisfied with another answer since it uses analysis (that is, IVT), and a fair answer should have pointed out that there is no escape. – user26857 Jan 01 '16 at 18:53

Every polynomial of odd degree $\ge 3$ over $\mathbb{R}[x]$ is reducible over $\mathbb{R}$

1 Answers1