Is $\mathbb{F}_{81}$ a field extension of $\mathbb{F}_{27}$?

Question

Is $\mathbb{F}_{81}$ a field extension of $\mathbb{F}_{27}$? If it is, what is $[\mathbb{F}_{81}:\mathbb{F}_{27}]$? In this case, $\mathbb{F}_{81}$ means a field with 81 elements. I know like $\mathbb{F}_{81}$ is the splitting field of $x^{81-1}=1$ over $\mathbb{F}_3$, right? But what next? Or since $81$ is not a power of $27$, $\mathbb{F}_{81}$ is not an field extension of $\mathbb{F}_{27}$?

Answer 1

Suppose it is, then

$$[\Bbb F_{81}:\Bbb F_3]=[\Bbb F_{81}:\Bbb F_{27}]\,[\Bbb F_{27}:\Bbb F_3]$$

Now, $\;[\Bbb F_{p^n}:\Bbb F_p]=n\;$ always, so what can you deduce from the above?

You can also try to prove the general claim: $\;\Bbb F_{p^n}\;$ is a subfield of $\;\Bbb F_{p^m}\;$ iff $\;n\,\mid\,m\;$

Answer 2

Short answer: No.

To be a field extension, you must in particular be a vector space, and the general theory of vector spaces (existence of a basis) tells us that the order of $V$ must be a power of the order of $F$.

Answer 3

Any extension field of $\mathbb{F}_{27}$ is an $\mathbb{F}_{27}$-vector space. In particular, the finite degree extensions must have order a power of 27. Thus $\mathbb{F}_{81}$ is not an extension field.