Let $f(x)=x^3+3x^2+3\in\mathbb{Z}_5[x]$ and $I=\langle f(x)\rangle$ and $R=\mathbb{Z}_5[x] /I$.

I got stuck in the last exercise:

Find $a\in R$ so $a^3=x^2+1+I$.

What is the right approach of finding such $a$ without guessing?

  • 2
    $\begingroup$ Did you mean $R = \mathbb{Z}_5[x]/I$? $\endgroup$ – Chessanator Jul 12 '19 at 20:25

Note that in $\mathbb{Z}_5[x]/I$, because of the way the polynomial $f(x)$ was chosen, we have that $(x+I)^3= -3x^2 - 3 +I = -3 (x^2 +1 +I)$. So if we were allowed to write $a = \frac{x+I}{\sqrt[3]{-3}}$ we'd have our answer.

Can you find a cube root of $-3$ in $\mathbb{Z}_5$? You can do a number of manipulations working mod $5$ to make this easier.

  • $\begingroup$ how did you know to start using $(x+I)$? $\endgroup$ – vesii Jul 12 '19 at 20:45
  • $\begingroup$ I saw the $3(x^2+1)$ in the polynomial $f$ as something to aim for. $\endgroup$ – Chessanator Jul 12 '19 at 20:48

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