0

I want to prove that there exists an inclusion $\mathbb{F}_{p^a} \hookrightarrow \mathbb{F}_{p^b}$ iff $a \vert b$.

Suppose that $a \vert b$, then $b =ac$ for some $c \in \mathbb{Z}$. Consider then that $p^b = p^{ac} = (p^{a})^c \implies p^a \subset p^b$. Therefore we can define an inclusion $\mathbb{F}_{p^a} \hookrightarrow \mathbb{F}_{p^b}$.

Conversely, suppose there exists an inclusion $\mathbb{F}_{p^a} \hookrightarrow \mathbb{F}_{p^b}$ and $a \ \not \vert b$. Then $b \neq ac \implies p^b \neq p^{ac} = (p^a)^c \implies p^a \not \subset p^b$ and therefore no inclusion can be defined from $\mathbb{F}_{p^a} \hookrightarrow \mathbb{F}_{p^b}$. This contradiction yields the necessary result.

Note that $p$ represents a prime.

Is this legitimate?

1 Answers1

2

With a little linear algebra, that at this stage is usually well known already, I think it can be pretty easy and short:

For any $\;n\in\Bbb N\;,\;\;\Bbb F_{p^n}\;$ is a linear space over the prime field $\;\Bbb F_p\;$ and of dimension $\;n\;$ , so:

$$\Bbb F_{p^a}\hookrightarrow\Bbb F_{p^b}\iff \Bbb F_{p^a}\;\;\text{is a linear subspace of}\;\;\Bbb F_{p^b}\;\;\text{over}\;\;\Bbb F_p\iff$$

$$b=[\Bbb F_{p^b}:\Bbb F_p]=[\Bbb F_{p^b}:\Bbb F_{p^a}]\cdot[\Bbb F_{p^a}:\Bbb F_p]=[\Bbb F_{p^b}:\Bbb F_{p^a}]\cdot a\implies a\,\mid\,b$$$${}$$

The other direction is completed by the following (which is also a complete proof by its own)

Approach from Field Theory: we know $\;\Bbb F_{p^n}\;$ is the splitting field of $\;x^{p^n}-x\;$ over the prime field $\;\Bbb F_p\;$ , so we have that

$$\Bbb F_{p^a}\hookrightarrow\Bbb F_{p^b}\iff\;\text{every root of}\;x^{p^a}-x\;\;\text{is a root of}\;\;x^{p^b}-x\;\iff $$

$$\left(x^{p^a}-x\right)\,\mid\,\left(x^{p^b}-x\right)\;\iff p^a\,\mid\,p^b\;\iff a\,\mid\,b$$

DonAntonio
  • 211,718
  • 17
  • 136
  • 287