Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)

I am not able to proceed further.. please help me

share|cite|improve this question
Every real polynomial of degree${}>2$ is reducible. – Marc van Leeuwen Oct 7 '13 at 11:43
@MarcvanLeeuwen : Yes, Yes... :) – Praphulla Koushik Oct 7 '13 at 11:55
up vote 16 down vote accepted

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 $2$.

share|cite|improve this answer
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). – Marc van Leeuwen Oct 7 '13 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 :) – Praphulla Koushik Oct 7 '13 at 11:53


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

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

share|cite|improve this answer
yes, yes.... I have seen this for $x^4+1$...Now, my question is for $x^8+1$.. Is this supposed to be hint??? – Praphulla Koushik Oct 7 '13 at 11:08
Hint: When "Hint" is written, what follows is supposed to be a hint. – Did Oct 7 '13 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 – Praphulla Koushik Oct 7 '13 at 11:10
I have asked this before you have edited.. so i have not seen "hint"... Sorry :) – Praphulla Koushik Oct 7 '13 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 !

share|cite|improve this answer

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 $c_1=c_0=0$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.