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For the ring $\mathbb{Z}_{15}[x]$ and its ideal $<3x^2 + 5x>$

Find the order of the $\mathbb{Z}_{15}[x]/<3x^2 + 5x>$


In my answer sheet it said

$\mathbb{Z}_{15}[x]/<3x^2 + 5x> \simeq (\mathbb{Z}_{3}[x]/<5x>) \times (\mathbb{Z}_{5}[x]/<3x^2>)$

Hence the order is 75.

I don't understand Why does $\mathbb{Z}_{15}[x]/<3x^2 + 5x> \simeq (\mathbb{Z}_{3}[x]/<5x>) \times (\mathbb{Z}_{5}[x]/<3x^2>)$.

Why those are isomorphic each other?

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As Chris hinted, it is easy to lift up $\,\Bbb Z_{15} \cong \Bbb Z_3\times \Bbb Z_5\,$ by CRT. Let's examine the idea more closely.

Notice in $\,R = \Bbb Z/15\!:\ (3)+(5)=(1)\,\Rightarrow\, (3)\cap (5) = (3)(5) = (0)$ $\smash{\overset{\small\rm CRT}\Rightarrow}\, R^{\phantom{|^|}}\!\!\! \cong R/3\times R/5$

The above ideal equalities extend to $\,E = R[x]/(3x^3+5x),\,$ thus also $\smash{\overset{\small\rm CRT}\Rightarrow}\, E^{\phantom{|^|}}\!\!\! \cong E/3\times E/5$

Ring isomorphism theorems $\Rightarrow E/3 \cong \Bbb Z_3[x]/5x,\,$ $\,E/5 \cong \Bbb Z_5[x]/3x^2$


For variety here's another way: we apply CRT in $R=\Bbb Z_{\color{#c00}{15}}[x]\,$ with $\,I+J=(3,5x) + (5,3x^2) \supseteq (5,3)= (1).\,$ By ring isomorphism theorems $\,R/I = R/(3,5x)^{\phantom{|^|}}\!\!\! \cong \Bbb Z_3[x]/(5x),\,$ $\,R/J = R/(5,3x^2) \cong \Bbb Z_5[x]/(3x^2)$

$I\!+\!J=(1)\,\Rightarrow\,I\cap J = IJ = (3,5x)(5,3x^2) =(9x^2,-5x)= (3x^2,5x)\,$ by $\,2(9x^2)=3x^2$

hence $\,IJ =(3x^2,5x)=(3x^2\!+\!5x)\ $ by $\ (6,-5)(3x^2\!+\!5x)=(3x^2,5x)\,$ by $\,\color{#c00}{15=0}\,$ in $\,R$.

Conclude $\ R/(3x^2\!+\!5x) = R/(I\cap J)\overset{\rm\small CRT_{\phantom |}\!} = R/I\times R/J = \Bbb Z_3[x]/(5x)\times \Bbb Z_5[x]/(3x^2)$

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  • $\begingroup$ Does your $I$ and $J$ are $I \lhd \mathbb{Z}_3[x]$ and $J \lhd \mathbb{Z}_5[x]$? $\endgroup$ – se-hyuck yang Jan 11 '20 at 5:29
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    $\begingroup$ @se-hyuckyang $\ I=(3,5x)\,$ and $\,J = (5,3x^2)\,$ denote ideals in $\,R = \Bbb Z_{15}[x]\ \ \ $ $\endgroup$ – Bill Dubuque Jan 11 '20 at 5:31
  • $\begingroup$ Hmm.. I see. I'll try again. :) $\endgroup$ – se-hyuck yang Jan 11 '20 at 5:33
  • $\begingroup$ Mr. @Bill Dubuque, How did you find the $(3,5x) (5,3x^2) = (x^2, 5x) = (3x^2 + 5x)$?[I mean what motivation or hint make you find those?] Plus could you more explain for me the reason why $(x^2,5x) = (3x+5x)$? $\endgroup$ – se-hyuck yang Jan 11 '20 at 5:53
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    $\begingroup$ @se-h Let $\,f=3x^2\!+\!5x,\,$ so $\,(3,f)=(3,5x).\,$ By the Second Isomorphism Theorem we have, $$(R[x]/f)/(3,f)/f \,\cong\, R[x]/(3,f) \,\cong\, \Bbb Z_{15}[x]/(3,5x) \,\cong\, \Bbb Z_3[x]/5x\,$$ where the final isomorphism is by $\,\Bbb Z_{15}[x] \to \Bbb Z_3[x] \to \Bbb Z_3[x]/5x\,$ is onto with kernel $\,(3,5x)$ $\endgroup$ – Bill Dubuque Jan 16 '20 at 3:13

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