How Does Polynomial Splitting Occur in Finite Field Extensions?

[mathmari]

Hello

The extension $\mathbb{Z}_p \leq \mathbb{F}_{p^n}$ is normal, as the splitting field of the polynomial $f(x)=x^{p^n}-x$ ($\mathbb{Z}_p$ is a perfect field therefore each polynomial is separable).

So, if $a \in \mathbb{F}_{p^n}$, then $q(x)=Irr(a,\mathbb{Z}_p)$ can be splitted over $\mathbb{F}_{p^n}$ (since all the roots are in $
\mathbb{F}_{p^n}$).

How can I find the splitting of $q(x)$ as an expression of powers of $a$?? (Wondering)

 

[jvicens]

Hello,

Thank you for your question. It is correct that the extension $\mathbb{Z}_p \leq \mathbb{F}_{p^n}$ is normal, as it is the splitting field of the polynomial $f(x)=x^{p^n}-x$. This means that all irreducible polynomials over $\mathbb{Z}_p$ split completely in $\mathbb{F}_{p^n}$.

To find the splitting of $q(x)$ as an expression of powers of $a$, we can use the fact that $q(x)$ can be written as a product of linear factors in $\mathbb{F}_{p^n}[x]$. Let $q(x)=\prod_{i=1}^n (x-a_i)$, where $a_i$ are the roots of $q(x)$ in $\mathbb{F}_{p^n}$. Then we can express $q(x)$ as $\prod_{i=1}^n (x-a_i)=x^n-\sum_{i=1}^n a_i x^{n-i}$.

We can then rewrite this as $q(x)=a^n-\sum_{i=1}^n a_i a^{n-i}$, where $a$ is any root of $q(x)$ in $\mathbb{F}_{p^n}$. This expression shows that the coefficients of $q(x)$ can be expressed as powers of $a$, and we can use this to find the splitting of $q(x)$ in terms of powers of $a$.

I hope this helps answer your question. Let me know if you have any further questions.