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Let $R$ be the ring $\Bbb Z[x]/((x^2+x+1)(x^3 +x+1))$ and I be the ideal generated by $2$ in $R$. What is the cardinality of the ring $R/I$?

  1. 27

  2. 32

  3. 64

  4. infinite

Now I was thinking $R$ could be written as $(\mathbb Z[x]/(x^2 +x+1)/(x^3 + x+1) = \Bbb Z[i]/(x^3+x+1)$. Now if I quotient it with the $2$ of this ring, $2$ is not irreducible anymore. $2=(1+i)(1-i)$. So I don't think I can write it as $\Bbb Z_2[i]/(x^3+x+1)$. So what is going wrong?

P.S. Thanks for all your help . It is done now.I have another doubt here. The ideal we quotiented the ring $Z_2[x] by is not generated by an irreducible polynomial . So,can the generator being reducible or irreducible make any difference or not?

user118494
  • 5,837

2 Answers2

7

$$R/2R\simeq\mathbf Z/2\mathbf Z[x]/\bigl((x^2+x+1)(x^3+x+1)\bigr)$$ Now in $\mathbf Z/2\mathbf Z[x]$, $x^2+x+1$ and $x^3+x+1$ are coprime, hence by the Chinese Remainder Theorem, $R/2R$ is the product of two fields: $$R/2R\simeq \mathbf Z/2\mathbf Z[x]/\bigl(x^2+x+1)\times\mathbf Z/2\mathbf Z[x]/(x^3+x+1)\simeq \mathbf F_4\times \mathbf F_8. $$ The factors have $4$ and $8$ elements, respectively.

Bernard
  • 175,478
  • No need to use CRT in order to find that $R/2R$ has $32$ elements: it is an $\mathbb F_2$-vector space of dimension $5$. – user26857 Jul 17 '15 at 19:06
  • You're right. My intention was the show it's the product of $\mathbf F_4$ by $\mathbf F_8$. – Bernard Jul 17 '15 at 19:12
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Hint: The order in which you quotient does not matter. So first consider $\mathbb{Z}[x]/(I)$, where $I$ is defined above. What does this give you?

$$ \mathbb{Z}[x]/(I)=\mathbb{F}_2[x]$$ where $\mathbb{F}_2$ is the finite field of two elements.

Now quotient off by the ideal $((x^2+x+1)(x^3+x+1))$. Let $S$ be the ring obtained. Note that $(x^2+x+1)(x^3+x+1)$ has degree $5$, so every polynomial in $((x^2+x+1)(x^3+x+1))$ has degree $\geq 5$. Thus in $S$, we have some relation that allows us to write $$ \overline{x}^5=p(\overline{x}), $$ where $p(\overline{x})$ is a polynomial of degree $\leq 4$ and $\overline{x}$ is the residue of $x$. Therefore every polynomial in $S$ has degree at most $4$, at $S$ has scalars from $\mathbb{F}_2$. Can you finish?

TomGrubb
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