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I am trying to find the number of ideals in $R:=\Bbb Z[x]/(x^3+1, 7)$ and $S:=\Bbb Z[x]/(x^3+1, 3)$.

I started with $R$ and tried to write it in terms of familiar rings, by using fundamental homomorphism theorem and CRT: \begin{eqnarray} R &\cong&(\Bbb Z[x]/(7))/((x^3+1,7)/(7))\\ &\cong&\Bbb F_7[x]/(x^3+1)\\ &\cong&\Bbb F_7[x]/(x+1)(x^2-x+1)\\ &\cong&\Bbb F_7 \times \Bbb F_7[x]/(x^2-x+1). \end{eqnarray}

I have a couple of questions regarding this.

  1. When I used the CRT, I checked that the two ideals are coprime in $\Bbb F_7[x]$. There are some different notions of coprimality: two ideals being coprime in $\Bbb C[x]$, which I am familiar with; two ideals being coprime in $\Bbb F_7[x]$; and two generators of the ideals being coprime in each of two rings. Do these concepts coincide?

  2. I am stuck with the second component of the direct product. If the coefficient ring were $\Bbb Z$ or $\Bbb R$, I would consider a homomorphism of substitution of some complex number, to make it appear more familiar. But since $\Bbb F_7$ is not a subfield of $\Bbb C$, I don't know if this works.

  3. Is this problem substantially different between $S$ and $R$?

I would appreciate your help. I would be grateful if you put it in elementary terms (I am not familiar with algebraic number theory).

Pteromys
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    In $\mathbb F_7[x]$, $x^3 + 1 = (x + 1)(x + 2)(x + 4)$. – Tunococ Sep 10 '13 at 07:46
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    @Tunococ: You're right about mod $7$, but mod $3$ we have $x^3+1=(x+1)^3$ by the so called Freshman's dream. In other words, the difference between $S$ and $R$ is that in mod $3$ case we have a repeated factor. This affects the tally of ideals in the senses that the decisions to include or not to include a factor need to take multiplicity into account. – Jyrki Lahtonen Sep 10 '13 at 07:49
  • @JyrkiLahtonen Does that mean I can't use CRT? – Pteromys Sep 10 '13 at 07:52
  • @Tunococ That helps! What is a good way of factoring a polynomial in a finite field? – Pteromys Sep 10 '13 at 07:53
  • Using CRT when repeated factors are present is trickier, indeed. Here CRT will go out of the window (when looking at $S$), because the repeated factor is the only irreducible factor. Re: factoring polynomials over a finite field this question has some discussion and links. Still you should just check for zeros as the first step. Here $x^3+1\mid x^6-1$, and as all the non-zero elements of $\Bbb{F}_7$ are sixth roots of unity, you can immediately tell that the polynomial splits into linear factors. – Jyrki Lahtonen Sep 10 '13 at 07:59

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There is no conceptual difference between finite fields and fields in general. Every theorem which you know for fields in particular also applies to finite fields. For example, $F[x]$ is Euclidean if $F$ is a field, in particular a principle ideal domain. It follows that the ideals of $F[x]/(f)$ correspond to the monic divisors of $f$. So in your examples you just have to compute the number of monic divisors of $x^3+1 \in \mathbb{F}_7[x]$ and of $x^3+1 \in \mathbb{F}_3[x]$. Of course the factorizations mentioned by Tunococ are helpful. I wouldn't use CRT here.

Martin Brandenburg
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