1
$\begingroup$

Do you have a example of field $K$ s.t. $|K|=p^n$ with $n\geq 2$ ? I know that $\mathbb Z/n\mathbb Z$ is a field iff $n$ is prime, but since $p^n$ is not prime, what kind of field can be such that $|K|=p^n$. Would $$\mathbb Z/p\mathbb Z\times \mathbb Z/p\mathbb Z$$ a field ? If yes, does a field $K$ s.t. $|K|=p^n$ is necessarily isomorphic to $$\mathbb Z/p\mathbb Z\times ...\times \mathbb Z/p\mathbb Z\ \ ?$$

$\endgroup$
5
$\begingroup$

The way you generate a field of order $p^n$ is to find an polynomial of degree $n$ that is irreducable over $\mathbb Z / p\mathbb Z [x]$, then mod out by the ideal generated by it. This is a maximal ideal, so modding out gives you a field.

$\endgroup$
2
$\begingroup$

The answer is no. In $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$, we have $$(0,1)\times (1,0)=(0,0)$$ and a field cannot have zero divisors.

$\endgroup$
  • $\begingroup$ Good example. Thank you :-) $\endgroup$ – Rick Nov 16 '15 at 21:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.