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For proving this isomorphism I have tried some things. It seems to be solvable with the Chinese Remainder theorem (for rings). For this I'm trying to show that $$\{q(X^3-X)\ |\ q \in \mathbb{Q}[X] \} =:(X^3+X) = I \cdot J$$ with $I$ and $J$ being relatively prime ideals.

If $I$ were to be $(X)$ and $J$ equal to $(X^2+1)$ this theorem would imply that $$\mathbb{Q}[X]/(X^3+X) \cong \mathbb{Q}[X] \times \mathbb{Q}/(X^2+1).$$ And at last I'd use that $\mathbb{Q}[X]/(X-0) \cong \mathbb{Q} $. But how to write $(X^3+X) = I \cdot J$ formally? Or am I on a totally wrong track anyway?

Xam
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    Let's prove that $(X^3+X) \subset(X)(X^2+1)$. In fact $X^3+X=X\times(X^2+1)\in (X)(X^2+1)$ therefore the ideal generated by $X^3+X$ is contained in $(X)(X^2+1)$. If you can prove the other inclusion: $(X)(X^2+1)\subset(X^3+X)$ then you can deduce that the two ideals are equal. – anonymous67 Oct 01 '17 at 14:16
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    Maybe you haven't seen this result yet, but both this and your previous question follow immediately from the Chinese Remainder Theorem: if $I+J = R$, then $\frac{R}{IJ} \cong \frac{R}{I} \times \frac{R}{J}$. – Viktor Vaughn Oct 01 '17 at 14:28
  • In this particular case let $\varphi(a+bx+cx^2) = (a,a-c+bx)$. Its inverse is $\varphi^{-1}(d,e+fx) = d+fx+(d-e)x^2$. Show $\varphi$ is a ring morphism $\mathbb{Q}[x]/(x^3+x) \to \mathbb{Q}\times \mathbb{Q}[x]/(x^2+1) $. – reuns Oct 01 '17 at 19:40

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