How to show that it is irreducible?

Asked Jan 26 '19 at 12:40

Active Jan 26 '19 at 12:58

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Let $\gcd(p,n)=1$. Consider $\;x^n-1$ over $\Bbb F_p[x]\;$. If its splitting field is $K$ find $\;[K:\Bbb F_p]$.

Now $K=\Bbb F_p(e^{2\pi i/n})$ Also the polynomial of which $e^{2\pi i/n}$ is a root is $1+x+x^2+\ldots +x^{n-1}$

But How to show that it is irreducible?

Do we need to use the fact that $\gcd(p,n)=1$.

Can I get some help please?

edited Jan 26 '19 at 12:58

reuns

77,999

asked Jan 26 '19 at 12:40

@DietrichBurde;$\Bbb Z_p$ means the ring $\Bbb Z_n$ where $n=p$ and it is a field – Jan 26 '19 at 12:51
This finite field is usually denoted by $\Bbb F_p$. – Dietrich Burde Jan 26 '19 at 12:53
@DietrichBurde;i meant that only – Jan 26 '19 at 12:53
Possible duplicate of Irreducible cyclotomic polynomial over a field $\mathbb{F}_q$? – Dietrich Burde Jan 26 '19 at 12:54
You can use $\mathbb{Z}_p$ for the cyclic group but in Galois field theory $\mathbb{Z}_p$ always refers to the $p$-adic integers (it is not a field but your question still made sense there) – reuns Jan 26 '19 at 12:58
@reuns,i don't understand why the assumption $\gcd(p,n)=1$ is needed? – Jan 26 '19 at 13:01
@reuns,how is that useful in order to show it for the case here – Jan 26 '19 at 13:01
Say $n=p$ odd. You know that $(a+b)^p = ?$ thus $x^p-1 = ?$. Trying to define what is the equivalent of $e^{2i \pi /n}$ in $\mathbb{F}_p$ is what you are doing here. $\mathbb{Z}[e^{2i \pi /n}]/(p)$ contains $\mathbb{Z}/(p)=\mathbb{F}_p$ but it doesn't have to be a field. – reuns Jan 26 '19 at 13:03
@reuns,$n=kp\implies x^{kp}-1=(x^k)^p-1=(x^k-1)^p$ – Jan 26 '19 at 13:04
the polynomial $1+x+...+x^n $ is irreducible over $F_p$ iff p is of order $n-1$ $\pmod{n}.$ This is not always true. take $p=2$ and $n=15$ when that polynomial is divisble by both $x^2+x+1$ and $x^4+x^3+x^2+x+1.$ If $n$ is itself a prime the iff condition is that $p$ is primitive (ie of order $n-1$) mod $n.$ – Patrick Sole Jan 26 '19 at 13:05
@reuns;How do you know that $x^k-1$ splits in $\Bbb Z_p$ – Jan 26 '19 at 13:09

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