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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!

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    $\begingroup$ Your "extension" has $27\cdot2$ elements, and isn't a field because $(1,0)(0,1)=(0,0)$. $\endgroup$ – Arthur May 11 at 15:38
  • $\begingroup$ Right, good catch. $\endgroup$ – Mathematical Mushroom May 11 at 15:43
  • $\begingroup$ How do you usually extend a field? (How did you extend $\Bbb Z_3$ to your field with $27$ elements?) The same thing ought to work here. $\endgroup$ – Arthur May 11 at 15:44
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    $\begingroup$ Mod out by an irreducible polynomial of degree 2 with coefficients in my field of 27 elements? Interesting. So since it's degree 2, it will be irreducible if it has no roots... Still ,there are obvious complications for me in that the coefficients are are pretty complicated in themselves, being cosets. But I'm sure I do'nt need to dig into messy details, I just need to adjoin an element that has minimal polynomial of order 2 over my field of 27 elements... But i'm not exactly sure how to do this... $\endgroup$ – Mathematical Mushroom May 11 at 15:49
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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?

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  • $\begingroup$ "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?" Well, say $f(x)=a(x)b(x)$ where $a(x),b(x) \in \frac{\mathbb{Z}_3}{(p)}$. Then $a(x) = g(x) + \mathbb{Z}_3$ and $b(x) = h(x) + \mathbb{Z}_3$ as cosets, but we could see $a(x) = g(x) + r(x)$ and $b(x) = h(x) + s(x)$... thus $(g(x)+r(x))(h(x)+s(x))$ would be a possible reduction of $f(x)$ in $\mathbb{Z}_3$, but $f(x)$ is irreducible there so nope. Is there a better way to see this? $\endgroup$ – Mathematical Mushroom May 11 at 18:12
  • $\begingroup$ @MathematicalMushroom Yes, of course: if it is reducible over your field $F$ with $27$ elements, then it has a root $\alpha\in F$; then $\mathbb{Z}_3[\alpha]$ would be a $9$ element subfield of $F$, which doesn't exist. $\endgroup$ – egreg May 11 at 19:38
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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.

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