The other day on this site I came across a question about $\mathbb Z_3 [x]/ \langle x^2 + 1 \rangle$.

At the time I misread the question and thought it was

Find all the irreducible polynomials in the field $\mathbb Z_3 [x]/ \langle x^2 + 1 \rangle$

I was about to post an answer saying that since there are only nine elements, 3 of which are constants, one can determine which elements are irreducible by computing all possible products of the non-constant elements.

As I was typing I realised that this is a stupid answer as, of course, anyone would figure to use brute force to solve the problem and it's not insightful.

Then I recalled that for polynomials over a field of degree 2 and 3 the polynomial is reducible if and only if it has a root in the field.

Unfortunately, I don't know if this can be modified to also apply to equivalence classes of polynomials.

So my question is:

What methods are there to test elements for irreducibility?


Can this theorem about polynomials be somehow adapted to be also applicable to equivalence classes of polynomials?

  • $\begingroup$ None of the elements of the ring $\mathbb{Z}_3[x]/\langle x^2+1\rangle$ are irreducible (and its elements are not appropriately called "polynomials"). An irreducible element of a ring is necessarily not 0 and not a unit, whereas as you have mentioned yourself, the ring $\mathbb{Z}_3[x]/\langle x^2 + 1\rangle$ is a field, so every non-zero element is a unit. $\endgroup$ – Zev Chonoles Sep 1 '15 at 2:57
  • $\begingroup$ @ZevChonoles I see. What if we consider a polynomial ring quotiented by some ideal generated by an reducible polynomial? Then there are some non-units (equivalence classes of polynomials). Then if we had $p$ is irreducible and $q \sim p$ then we could still ask the question, right? $\endgroup$ – learner Sep 1 '15 at 3:05
  • $\begingroup$ When I was teaching Algebra, I would partition the elements of a commutative ring $R$ thus: $\{0\}$; nonzero zero-divisors: units; reducible elements, namely things writable as the product of two nonunit nonzero-divisors; and everything else, which are the irreducibles. If you take a field $k$ and form $k[x]/(f)$ where $f$ is reducible, then there are no irreducibles nor reducibles: everything is either a zero-divisor or a unit. This is not obvious, would need a proof. $\endgroup$ – Lubin Sep 3 '15 at 1:25

Polynomials over the field $\mathbb F_3[x]/\langle x^2 + 1\rangle$ are elements of the polynomial ring $(\mathbb F_3[x]/\langle x^2 + 1\rangle)[z]$. Note that irreducible elements in a ring are by definition nonunits, so the fact that $\mathbb F_3[x]/\langle x^2 + 1\rangle$ is a field makes the question, as you have interpreted it, trivial.

  • $\begingroup$ My question is not about polynomials over the field. $\endgroup$ – learner Sep 1 '15 at 3:18

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