Subfield and field extensions of a field with $27$ elements

Question

I have constructed the field $\frac{\mathbb{Z}_3[x]}{<2x^3+x+2>}$

$=\{a_o+a_1x+a_2x+....+a_nx^n|a_i \in \mathbb{Z}_3, n \in \mathbb{N}, 2x^3+x+2=0\}$

$=\{a_0+a_1x+a_2x^2:a_i \in \mathbb{Z}_3\}$

Which is a field that has $27$ elements. I would like to find a subfield that has $9$ elements and an extension that has $27^2$ elements. For the extension, could just take $\frac{\mathbb{Z}_3[x]}{<2x^3+x+2>} \times \mathbb{Z}_2$? Looks like a field with $27^2$ elements that has $\frac{\mathbb{Z}_3[x]}{<2x^3+x+2>}$ as a subfield to me.

As far as the finding a subfield with 9 elements, I was thinking of taking the explicit description of our field with $27$ elements:

$=\{a_0+a_1x+a_2x^2:a_i \in \mathbb{Z}_3\}$

And altering it so that it now says:

$=\{a_0+a_1x+a_2x^2:a_i \in \mathbb{Z}_3, x^2=0\}$

$=\{a_0+a_1x:a_i \in \mathbb{Z}_3 \}$

Am I allowed to do this? If not, what is the proper way of going about this. Thank you!

Answer 1

Well, the field ${\Bbb F}_{p^m}$ has ${\Bbb F}_{p^n}$ as subfield if and only if $n$ divides $m$.

In your case, $p=3$, $m=3$ and $n=2$. So it does not work.

Answer 2

There is no subfield with $9$ elements. The set you describe is not a subfield.

In order to find an extension field having $27^2$ elements you certainly cannot take a product of rings, which is never a field.

You rather need to find an irreducible degree $2$ polynomial with coefficients in the given field.

It's easier to find one in $\mathbb{Z}_3[x]$, for instance $x^2+1$. Why is this also irreducible over your field with $27$ elements?