MHB What Is the Splitting Field for \(x^{p^{80}}-1\) Over \(\mathbb{F}_p\)?

[mathmari]

Hey!

Could you give me some hints how to find a splitting field for the polynomial $$x^{p^{80}}-1$$ where $p$ is prime?? (Wondering)

 

[Euge]

Hey!

Could you give me some hints how to find a splitting field for the polynomial $$x^{p^{80}}-1$$ where $p$ is prime?? (Wondering)

Over what field are viewing this polynomial? Is it over the rationals?

 

[mathmari]

Over what field are viewing this polynomial? Is it over the rationals?

The polynomial is in $\mathbb{Z}_p$.

 

[mathbalarka]

So you are looking at splitting field of $x^{p^n} - 1$ over $\Bbb F_p$. Note that $(a + b)^p = a^p + b^p$ in $\Bbb F_p$ so $x^{p^n} - 1 = \left ( x^{p^{n-1}} \right )^p - 1 = (x^{p^{n-1}} - 1)^p$. Continue in this fashion to derive $x^{p^n} - 1 = (x - 1)^{p^n}$. What should the splitting field look like then?

 

[mathmari]

So you are looking at splitting field of $x^{p^n} - 1$ over $\Bbb F_p$. Note that $(a + b)^p = a^p + b^p$ in $\Bbb F_p$ so $x^{p^n} - 1 = \left ( x^{p^{n-1}} \right )^p - 1 = (x^{p^{n-1}} - 1)^p$. Continue in this fashion to derive $x^{p^n} - 1 = (x - 1)^{p^n}$. What should the splitting field look like then?

Is it as followed?? (Wondering)

$$x^{p^{80}}-1=\left ( x^{p^{79}} \right )^p-1= \left ( x^{p^{79}}-1\right )^p=\left ( \left ( x^{p^{78}}\right )^p -1 \right )^p=\left ( \left ( x^{p^{78}}-1\right )^p \right )^p=\left ( x^{p^{78}}-1\right )^{p^2}= \dots = \left ( x^p-1 \right )^{p^{79}}=\left ( \left ( x-1 \right )^p \right )^{p^{79}}=\left ( x-1 \right )^{p^{80}}$$The only root is $x=1$, right??

Is then the splitting field $\mathbb{Z}_p$ ?? (Wondering)

 

[mathbalarka]

Yes, the splitting field of $x^{p^{80}} - 1$ over $\Bbb F_p$ is the same as that of $x - 1$, which is just $\Bbb F_p$.

 

[mathmari]

Yes, the splitting field of $x^{p^{80}} - 1$ over $\Bbb F_p$ is the same as that of $x - 1$, which is just $\Bbb F_p$.

I see... Thank you very much! (Happy)