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How many elements are in the quotient ring $\displaystyle \frac{\mathbb Z_3[x]}{\langle 2x^3+ x+1\rangle}$ ?

I guess I should be using the division algorithm but I'm stuck on how to figure it out.

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    $\begingroup$ Is $2x^3 + x +1$ irreducible in $\mathbb{Z}_3 [x]$? What does this say about the kind of ideal generated by the polynomial? $\endgroup$
    – tamefoxes
    Apr 29, 2015 at 3:20
  • $\begingroup$ The quotient ring has nothing to do with division so you shouldn't write it using the $\frac{...}{...}$ notation! $\endgroup$
    – user21820
    Apr 29, 2015 at 3:21
  • $\begingroup$ my professor writes it like this all the time. In a sense it is a type of "division" $\endgroup$ Apr 29, 2015 at 3:30

3 Answers 3

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More generally, $$\left|\frac{\mathbb F_p[x]}{\text{ <an irreducible polynomial of degree $n$ over } \mathbb F_p>}\right|=p^n$$

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Any element of this quotient ring is of the form $a_0+a_1x+a_2x^2+\langle 2x^3+x+1\rangle $ where $a_i\in \mathbb Z_3$

Thus we have $27$ elements

Note:whenever there is a polynomial $f(x)$ of degree $\geq 3 \in \mathbb Z_3[x]$ then by division algorithm we will get two polynomials $q(x),r(x)$ such that $f(x)=(2x^3+x+1)q(x)+r(x) $ where $r(x)=0 $ or $\deg r(x)<\deg (2x^3+x+1)$ which is $2$

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  • $\begingroup$ So, we have $a_0+a_1x+a_2x^2+\langle 2x^3+x+1\rangle $ and $a_i\in \mathbb Z_3$ How exactly do we get 27 out of that? Is it because $a_0, a_1$ and $1_2$ can each be 0,1,2 so we have 3*3*3? $\endgroup$
    – Harry
    Apr 29, 2015 at 23:39
  • $\begingroup$ yes exactly you are right @Harry $\endgroup$
    – Learnmore
    Apr 30, 2015 at 2:18
  • $\begingroup$ gotchya! Thanks for the help! $\endgroup$
    – Harry
    Apr 30, 2015 at 2:34
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Using the division algorithm, you can show that each element in the quotient ring is represented by a unique polynomial in $\mathbb{Z}_3[x]$ of degree $\leq 2$. How many such polynomials are there?

It's exactly like counting the number of elements of $\mathbb{Z}/n\mathbb{Z}$.

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