So some of my thoughts for constructing a finite field of order 27 are making me think of a field with $p^n$ elements, where $p = 3$ and $n = 3$ such that we want a cubic polynomial in $\mathbb{F}_3[X]$ that does not factor.

Could this be thought of as looking for a cubic polynomial in $\mathbb{F}_3[X]$ with no roots in $\mathbb{F}_3$? Could this polynomial work: $x^3 + 2x^2 + 1$ ?

  • $\begingroup$ why is your choice $x^3+2x^2+1$?? Was this a hint or you somehow felt this would work? $\endgroup$ – user87543 Dec 9 '13 at 15:38
  • $\begingroup$ This was a hint $\endgroup$ – user110655 Dec 9 '13 at 15:39
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    $\begingroup$ What can you say about $R/M$ where $R$ is a commutative ring with unity and $M$ is a maximal ideal. In a PID, what can you say about the relationship between irreducible elements, prime ideals and maximal ideals? $\endgroup$ – LASV Dec 9 '13 at 15:42

Yes, it does work: it is irreducible because it has no roots in $\mathbb{Z}_3$ (and $\mathbb{Z}_3$ is a field). Thus, the quotient ring $\mathbb{Z}_3[x]/(x^3 + 2x^2 +1)$ is a field which has $3\cdot 3\cdot 3$ elements.

  • $\begingroup$ what is the point in saying full answer at once... If this is a nice idea, text book would give not just exercises but it would give solutions also instead of giving "Hints"..... Please do not spoil the excitement of OP... Sorry if this bothers you so much.. $\endgroup$ – user87543 Dec 9 '13 at 15:47
  • $\begingroup$ yes yes.. this place is about answering questions.. to be precise this is about OP answering questions with help of other users... $\endgroup$ – user87543 Dec 9 '13 at 15:50

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