Claim: $\Bbb Z[x]\ /\left< x^3-x\right> \simeq \Bbb Z[x]/\left< x\right> \times\ \Bbb Z[x]\ /\left< x^2-1\right> $
My attempt:
Here $\left< x^2-1\right> $ and $\left< x\right> $ are two ideals, where $1$ belongs to $\big(\left< x^2-1\right> +\left< x\right> \big)$, since $1=x^2-(x^2-1)$, so my claim holds true by Chinese Remainder Theorem.
Now I am checking, whether it can be further reducible or not.
Again we can show $\Bbb Z[x]\ /\left< x\right> \simeq \Bbb Z$ by taking the surjective ring homomorphism $\phi\colon \Bbb Z[x]\to \Bbb Z$, such that $\phi(f(x))=f(0)$. Here $\text{ker}(\phi)=\left< x\right>$, so by the first ring isomorphism theorem $\Bbb Z[x]\ /\left< x\right> \simeq \Bbb Z$, and $\Bbb Z[x]\ /\left< x\right> \times\ \Bbb Z[x]\ /\left< x^2-1\right> \simeq\Bbb Z \times\ \Bbb Z[x]\ /\left< x^2-1\right> $.
Now, since $1$ does not belong to $\big(\left< x-1\right> +\left< x+1\right> \big)$, where factorization of $(x^2-1)=(x+1)\cdot(x-1)$, we can't use CRT further on this or we can't break this quotient ring further.
So my ANSWER (after reducing the claim also) to this ring isomorphism is $\Bbb Z[x]\ /\left< x^3-x\right> $ $\simeq$ $\Bbb Z \times\ \Bbb Z[x]\ /\left< x^2-1\right> $. Is my explanation correct? Is it the most reduced version in isomorphism(even after reducing the claim)?
Best Answer
Define ring epimorphisms $\pi_1:\Bbb Z[x]\to\Bbb Z$ and $\pi_2:\Bbb Z[x]\to\Bbb Z\times\frac{\Bbb Z[x]}{\langle x^2-1\rangle}$ by $$\pi_1:f(x)\mapsto f(0)\;\;\;\;\;\pi_2:f(x)\mapsto f(x)+\langle x^2-1\rangle$$ so that $\Psi:\Bbb Z[x]\to\Bbb Z\times\frac{\Bbb Z[x]}{\langle x^2-1\rangle}$ is also a ring homomorphism where $$\Psi:f(x)\mapsto(\pi_1(f(x)),\pi_2(f(x)))$$ $$f(x)\in\ker(\Psi)\iff f(0)=0\;\land\;x^2-1\;|\;f(x)\iff x,x+1,x-1\;|\;f(x)$$ $$\therefore\;f(x)\in\ker(\Psi)\iff x^3-x\;|\;f(x)\;\;\;\;\;\therefore\;\ker(\Psi)=\langle x^3-x\rangle$$ Moreover, $(c,f(x)+\langle x^2-1\rangle)=\Psi(f(x)+(x^2-1)(f(0)-c))$ and thus $\Psi$ is an epimorphism.$$\therefore\;\frac{\Bbb Z[x]}{\langle x^3-x\rangle}\approx\Bbb Z\times\frac{\Bbb Z[x]}{\langle x^2-1\rangle}$$