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Hello i'm having trouble understanding something. I need to show that

Let $R$ be a ring. $$R[x]/(x^2-1)\cong R[x]/(x^2-4).$$

but i'm pretty lost on how to proceed i've tried a map $f:R[x]/(x^2-1)\to R[x]/(x^2-4)$ $f:(x-1)(x+1)\to(x-2)(x+2)$

but that's as far as i could get.

Thanks in advance!

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    $\begingroup$ The map $f$ has to be defined on the factor rings. In this case, the elements of $\mathbb R[X]/(X^2-1)$ can be written as $a+bx$ (where $x$ is the residue class of $X$) and one defines $f(a+bx)=a+\frac12bx$. This also shows that you can extend the result to commutative rings in which $2$ is invertible, but this extension stops here. (To understand why, see my answer.) $\endgroup$ – user26857 May 17 at 8:41
  • $\begingroup$ The key word here is the universal property of quotient rings (homomorphism theorem). $\endgroup$ – Qi Zhu May 17 at 10:14
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I am pretty sure that $R$ is actually $\mathbb R$.

For instance, $\mathbb Z[X]/(X^2-1)$ is not isomorphic to $\mathbb Z[X]/(X^2-4)$. Suppose the contrary. Then $x$ (the residue class of $X$) from the first ring corresponds to an element $a+bx$ from the second. Since $x^2=1$ we must have $(a+bx)^2=1$, that is, $a^2+2abx+4b^2=1$. It follows that $a^2+4b^2=1$ and $ab=0$. Then $b=0$ ($a=0$ is impossible) and $a=\pm1$. It follows that $x$ corresponds to $\pm1$ and the surjectivity is lost.

For $R=\mathbb R$ one can use the Chinese Remainder Theorem to show that both rings are isomorphic to $\mathbb R\times\mathbb R$.

If one wants to find an explicit isomorphism, then send $X$ to $\frac12X$. In this way, $X^2-1$ corresponds to $\frac14X^2-1$ and this generates the same ideal as $X^2-4$.

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Answer: Let $A:=\mathbb{R}[x]/(x^2-1), B:=\mathbb{R}[x]/(x^2-4), R:=\mathbb{R}[x]$. Since $x^2-1=(x-1)(x+1):=IJ$ where $I=(x-1),J:=(x+1)$ it follows

$$A \cong R/I\oplus R/J \cong \mathbb{R}\oplus \mathbb{R}$$ and similarly

$$B \cong \mathbb{R}\oplus \mathbb{R}\text{ hence } A \cong B.$$

If a similar factorization holds in $R[x]$ you get the same result. You need $2$ to be a unit in $R$: If $I_1:=(x-2), J_1:=(x+2)$

it follows

$$ -4:=(x-2)-(x+2) \in I_1+J_1$$

hence $I_1,J_1$ are coprime and $I_1J_1=x^2-4$. Hence

$$B \cong R[x]/I_1J_1 \cong R[x]/I_1\oplus R[x]/J_1 \cong R\oplus R.$$

Similarly $$ -2=(x-1)-(x+1) \in I +J$$

hence $I+J=(1)$. Hence there is an isomorphism

$$R[x]/IJ \cong R[x]/I \oplus R[x]/J \cong R \oplus R.$$

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