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Let $p$ be a prime number. I know that $\mathbb{F}_{p^n}$ is a finite field with $p^n$ Elements, which is the splitting field for $x^{p^n}-x$ over $\mathbb{F}_{p}$. I also know that the extension $\mathbb{F}_{p^n}/\mathbb{F}_{p}$ is normal and separable, and hence Galois.

What I dont see is why its degree is n.

Miriam
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    This is more or less unrelated to any of the rest of what you write. It is a vector space over $\mathbb{F}_p$ with $p^n$ elements, hence it has dimension $n$. – Tobias Kildetoft Mar 13 '16 at 19:01
  • I believe the answers to this old question address your question also. Not closing this as a duplicate without a second opinion though. In other words, this is at a simpler level than e.g. normality that you already figured out. (+1) – Jyrki Lahtonen Mar 13 '16 at 19:05

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Let $d$ be the dimension of $\mathbb{F}_{p^n}$ over $\mathbb{F}_{p}$. Then, you have an isomorphism as vector spaces: $$\mathbb{F}_{p^n} \cong \mathbb{F}_{p}^d$$ The cardinality of the first one is $p^n$, while the cardinality of the second one is $p^d$. Hence $d=n$.

Watson
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