# Is $x^8+1$ irreducible in $\mathbb{R}[x]$

Question is to check if :

$x^8+1$ is irreducible over $\mathbb{R}[x]$.

even before this I tried to see $x^4+1$ and $x^2+1$.

for $x^2+1$, it does not have a root in $\mathbb{R}$ So, it is irreducible.

for $x^4+1$, checking for roots does not imply anything.

So I tried to solve for $a,b,c,d \in \mathbb{R}$ in $x^4+1=(x^2+ax+b)(x^2+cx+d)$ and concluded that $x^4+1$ is reducible.

It is becoming more difficult when the power of $x$ is getting bigger.

The only thing I can say about $x^8+1$ is it does not have a factor of degree $1,3,5,7$ (if not, it would have a real number as a root which is not possible)

• Every real polynomial of degree${}>2$ is reducible. Oct 7, 2013 at 11:43
• @MarcvanLeeuwen : Yes, Yes... :)
– user87543
Oct 7, 2013 at 11:55

Do you know that $\mathbb C$ is the algebraic closure of $\mathbb R$ and it has dimension $2$ as $\mathbb R$-vector space?

This implies that every irreducible polynomial over $\mathbb R$ has degree less or equal to $2$.

More details: every polynomial $p \in \mathbb R[x]$ has root in $\mathbb C$ and the $\mathbb R[\alpha]$, the subring of $\mathbb C$ generated by $\alpha$ a root of $p$, is field and so a subfield of $\mathbb C$. Its dimension as $\mathbb R$-vector space is equal to the degree of the irriducible polynomial having root $\alpha$, but such dimension must also be less or equal then the dimension of $\mathbb C$, which is $2$. So the irriducibile polynomial of $\alpha$ can either have degree $1$ (the element belong to $\mathbb R$) or $2$.

• Somewhat easier: $1,\alpha,\alpha^2$ are $\Bbb R$-linearly dependent, which gives a monic polynomial of degree $1$ or $2$ with $\alpha$ as root, and (if taken of minimal degree) dividing $p$ (consider the remainder in Euclidean division). Oct 7, 2013 at 11:47
• yes, yes.. I know that $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$ and it has dimension $2$ as $\mathbb{R}$-vector space... your explanation is very useful... Thank you :)
– user87543
Oct 7, 2013 at 11:53

Hints:

(1) $z^4+1=(z^2+1)^2-2z^2=(z^2+\sqrt2z+1)(z^2-\sqrt2z+1)$.

(2) $x^4\pm\sqrt2x^2+1=(x^2+1)^2-(2\mp\sqrt2)x^2=(x^2+\sqrt{2\mp\sqrt2}x+1)(x^2-\sqrt{2\mp\sqrt2}x+1)$.

• yes, yes.... I have seen this for $x^4+1$...Now, my question is for $x^8+1$.. Is this supposed to be hint???
– user87543
Oct 7, 2013 at 11:08
• Hint: When "Hint" is written, what follows is supposed to be a hint.
– Did
Oct 7, 2013 at 11:09
• $x^8+1=(x^4+1)^2-2x^4=(x^4+1)^2-(\sqrt{2}x)^2=(x^4+1-\sqrt{2}x)((x^4+1+\sqrt{2}x)$????? IS this you want me to try .... :) THank you than you
– user87543
Oct 7, 2013 at 11:10
• I have asked this before you have edited.. so i have not seen "hint"... Sorry :)
– user87543
Oct 7, 2013 at 11:10

If you have found that $x^4+1$ is reducible, then $x^4+1 = f(x)g(x)$ where $f$ and $g$ are nonconstant polynomials. But then, $x^8+1 = f(x^2)g(x^2)$ is reducible too !

To complement Giorgios answer here another argument, that also helps in finding factors:

If p is a real polynomial and has a root $a\in\mathbb C\setminus\mathbb R$, then $p(x)$ will be divisible by the real polynomial $f(x)=(x-a)(x-\bar a)=x^2 -2\Re a\,x+|a|^2$. Indeed, there must be a real polynomial $q$ and $c_1,c_0\in\mathbb R$ such that $$p(x)=q(x)f(x)+c_1 x+c_0,$$ and setting $x=a$ we obtain $$0 = c_1 a+c_0,$$ which, since $a\notin\mathbb R$, is only possible if $c_1=c_0=0$.

Here's a pretty straightforward solution. I'll denote principal ideal generated by A as (A). In R[x], (x^8+1) is maximal iff x^8+1 is irreducible in R[x] (as R is a field). We know that principal ideal (x^8+1) is strictly contained in (x^8+1)+(x^7) which is strictly contained in R[x]. So, (x^8+1) is not maximal. Therefore x^8+1 is not irreducible.