# Isomorphisms between polynomial quotient rings on infinite base fields.

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.

• 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.) – user26857 May 17 at 8:41
• The key word here is the universal property of quotient rings (homomorphism theorem). – Qi Zhu May 17 at 10:14

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 $$X^2-4$$.

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

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