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What is the order of the factor ring $\mathbb{Z}_{15}\left [ x \right ] / \left \langle 3x^{2}+5x \right \rangle$ ? Since $x$ and $3x+5$ are not relatively prime polynomial in $\mathbb{Z}_{15}$, so I can't use the Chinese Remainder Theorem. Any help please. Thank you.

purecj
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2 Answers2

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Let $f=3x^{2}+5x$ and $I=\langle f \rangle$.

  • $I \ni 3f = 9x^2+15x = 9x^2$.

  • $I \ni 5f = 15x^2+25x = 25x = 10x = -5x$.

  • $I \ni 5xf - 3f = 10x^2 - 9x^2 = x^2$.

Thus, $I=\langle 5x, x^2 \rangle$.

Therefore, for every $g \in \mathbb{Z}_{15}[x]$ we have $g \equiv ax+b \bmod I$, with $a \in \{0,1,2,3,4\}$.

Thus, $\mathbb{Z}_{15}[x] / I$ has $5 \cdot 15 = 75$ elements.

lhf
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  • Nice result. Is there a general notion when a ring $\mathbb{Z}_n[x]/(f)$ is finite? – TomTom314 Sep 25 '19 at 18:49
  • @TomTom314, I don't know. The argument above feels and is ad hoc. – lhf Sep 25 '19 at 18:55
  • Maybe, I have an idea. Consider $\mathbb{Z}{15}[X]/I\to\mathbb{Z}{3}[X]/I $ and $\mathbb{Z}{15}[X]/I\to\mathbb{Z}{5}[X]/I$. I guess this provides an isomorphism $\mathbb{Z}{15}[X]/I\cong \mathbb{F}{25}\times\mathbb{F}_{3}$ – TomTom314 Sep 25 '19 at 20:10
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Hint: Here, $3x^2+5x=x(3x+5)$ both factors are irreducible and relatively prime. Use Chinese reminder theorem for rings to find an isomorphism and to SO get the order of ring.

sabeelmsk
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    There is no Bézout relation for these two polynomials in the ring $\mathbb{Z}_{15}[x]$, so the canonical map involved in Chinese remainder theorem won't be surjective. – GreginGre Sep 25 '19 at 14:20