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I'm studing theory about polynomial ring, I have this exercise:

For what prime number does the polynomial $f=x^3+\overline{2}x +\overline{2}\in\mathbb{Z}_p$ admit $\overline{3}$ as a root (I hope the term is correct)?

I work in this way $f(\overline{3}) = \overline{27}+\overline{6}+\overline{2}=\overline{35}$ But this is correct if $\overline{35}=\overline{0}$; this happens in $\mathbb{Z}_p$ if and only if: $35\equiv0\pmod p$ So I can say $p=5$. The exercise continues, but I don't know how to proceed:

Write $f$ as product of irreducible factors in $\mathbb{Z}_p$

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    You could also have $p=7$.2012-03-14

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Some hints:

Since $\deg f = 3$, we know that either

  • $f$ is irreducible
  • $f = g_1 g_2$ for two factors with (w.l.o.g.) $\deg g_1 = 1$ and $\deg g_2 = 2$, where $g_2$ is irreducible.
  • $f = g_1 g_2 g_3$ for three factors with $\deg g_i = 1$ for $i = 1,2,3$.

Now, you may assume that the $g_i$ are monic (i.e., their leading coefficient is 1), so they correspond to roots of $f$. In other words, if you know the roots of $f$, you can easily find its irreducible factors.

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    Yes, i know the theory. I would see some examples, cause this theorem is the only one that I can use.2012-03-15