# Write down all the possible values for the degree of an irreducible polynomial in $\mathbb{R}[x]$.

Write down all the possible values for the degree of an irreducible polynomial in $\mathbb{R}[x]$.

One of my friends told me that the answer is $1$ but he could not explain it. I believe that the answer would be any even number. e.g. $x^{2}+1$ is irreducible. Please help me where is my fault.

-

First, your friend is wrong. But so are you.

Hint: it's enough to consider non-constant polynomials of even degree, for polynomials in $\mathbb{R}[x]$ of odd degree are reducible (why?)

Now if $z \in \mathbb{C}$ is a root of some $f \in \mathbb{R}[x]$ then $\bar{z}$ is also a root of $f$. What can you conclude about $(x-z)(x-\bar{z})$?

-
i am very sorry that i couldn't understand your hint properly. – poton Sep 16 '12 at 12:22
@poton These are very good hints, so stick with it! The first hint points out that real polynomials of odd degree always have a root (and hence you can factor that root out). The second hint is that real polynomials always have complex roots in conjugate pairs. How about you multiply out $(x-z)(x-\overline{z})$ and see what it looks like? – rschwieb Sep 16 '12 at 12:26
i know that odd degrees are reducible. and (x−z)(x−z¯) are irreducible as i already mentioned that e.g. x^2+1=(x-i)(x+i) is irreducible . but i am confused then what should be the exact answer. – poton Sep 16 '12 at 12:34
@poton: Some off-topic first: you should enclose your math formulae with dollar-signs which formats it automatically and makes it more legible. Compare x^2+1 with $x^{2}+1$. As for your question: if $z$ is a root of $f$ then $(x-z)$ divides $f$. Hence $(x-z)$ and $(x-\bar{z})$ divide $f$, and so does their product $(x-z)(x-\bar{z})=...$ (DO this last calculation and maybe you'll see the light). What can you conclude now about irreducibility of $f$ over $\mathbb{R}[x]$? – Nils Matthes Sep 16 '12 at 12:50
@poton So you noticed that it is an irreducible polynomial over the reals? Can you see that means that every complex number is a root of a polynomial of degree 2 or less over the reals?! It's right there in front of you... – rschwieb Sep 16 '12 at 12:50

Assume $f\in \mathbb R[x]$ is irreducible. By the fundamental theorem of algebra, every polynomial $f(x)\in\mathbb C[x]$ has a root $z_0$. If $z_0\in \mathbb R$, we conclude that $f(x)=(x-z_0)g(x)$ with $g\in \mathbb R[x]$. As $f$ is irreducible, $g$ must be a unit (i.e. a nonzero constant) and $\deg f=1$. If on the other hand $z_0\notin\mathbb R$, then $f(\overline{ z_0})=\overline{f(z_0)}=\bar0=0$ because the coefficients of $f$ are real. Thus (in $\mathbb C[x]$) $f$ is divisible both by $(x-z_0)$ and by $(x-\overline{z_0})$. Because $1\cdot(x-z_0)+(-1)\cdot(x-\overline{z_0})=z_0-\overline{z_0}=2\Im({z_0})i\ne0$ is a unit in $\mathbb C[x]$, we conclude that $f$ is in fact also divisible by their product $(x-z_0)(x-\overline{z_0})=x^2-2\Re(z_0)+|z_0|^2\in \mathbb R[x]$, i.e. $f(x) =(x^2-2\Re(z_0)+|z_0|^2)g(x)$ for some $g\in\mathbb R[x]$. As above $g$ must be a nonzero constant and $\deg f=2$.

-