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.
 A: 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 $1\cdot(x-z_0)+(-1)\cdot(x-\overline{z_0})=\overline{z_0}-z_0=-2\Im({z_0})i\ne0$ 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 $\deg f=2$.
A: 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 $(x-z)(x-\bar{z})$?
