Consider $\mathbb{Z}[x]$, we know that this is not a PID. Now consider the ideal $J = \langle x^2-2, 5\rangle$. I need to prove that $\mathbb{Z}[x]/J \cong \mathbb{Z}_5[x]/\langle x^2-2\rangle$.
Looking for a hint, will complete the proof.
Ring isomorphism $\mathbb{Z}[x]/J \cong \mathbb{Z}_5[x]/\langle x^2-2\rangle$
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abstract-algebra
ring-theory
ideals
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2Hint: there is nothing special about this case: in general, $R/(I + J) \cong (R/I)/(J/I)$. – 2017-02-25
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0@Magdiragdag could you specify R,I and J in the question... is R, Z or Z[x]? – 2017-02-25
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0In context: $R = {\mathbb Z}[x]$, $I = \langle 5\rangle$, $J = \langle x^2 - 2\rangle$. – 2017-02-25
1 Answers
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Hint : Consider the composition of surjections $$\Bbb Z[x]\to \Bbb Z_5[x]\to \dfrac{\Bbb Z_5[x]}{\langle x^2-2\rangle}$$and find its kernel. Then apply the first isomorphism theorem.
Additional hint : To find the kernel of a composition $f\circ g$, notice that $$x\in \ker (f\circ g) \Leftrightarrow f(g(x))=0\Leftrightarrow g(x)\in \ker f.$$ By the way this proof strategy works for the general case mentioned by Magdiragdag in the comments.
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0how do you find the kernel of a composition could you pls expand on your answer – 2017-02-25