Does the field with 27 elements contain a subfield with 9 elements?

Question

Does the field with $27$ elements contain a subfield with $9$ elements?

I fail to make the extension of $\mathbb{F}_3$ with degree $3$.

Answer 1

If $L/K$ is a field extension, then $L$ is a vector space over $K$. If $L$ and $K$ are finite, $L$ must be finite-dimensional over $K$. Let $K$ have $q$ elements and $L$ have dimension $n$ as a vector space. Then $L$ has $q^n$ elements. Is there an integer $n$ with $27=9^n$?

Answer 2

It is a theorem that if $K\subseteq L\subseteq M$ are algebraic field extensions, then $$ [L:K]\cdot [M:L] = [M:K] $$ In your case, you have a $K$ and an $M$, and you're wondering whether there can be an $L$. What would the numbers in the above equality be? Is that possible?