Are the rings $\mathbb{Z}[x]/(x^2-1)$ and $\mathbb{Z}[x]/(x^2-4)$ isomorphic?

Question

Are the rings $\mathbb{Z}[x]/(x^2-1)$ and $\mathbb{Z}[x]/(x^2-4)$ isomorphic?

I think they are not...? Well if there was $\mathbb{F}_3$ instead of $\mathbb{Z}$ then they would be...

Answer 1

Hint: In the former, there are exactly two things that satisfy $X^2=4$. In the latter, there are at least four elements that satisfy $X^2=4$.

You're right about $F_3$, of course, because then those two polynomials are identical.

Answer 2

Suppose that we have an isomorphism $$ \phi: \mathbb Z[x]/(x^2 - 1) \to \mathbb Z/(x^2 - 4). $$ Then $\phi$ must map $\overline x \in \mathbb Z[x]/(x^2 - 1)$ to the equivalence class of some linear polynomial $p(x) = ax + b$, for otherwise it would not be surjective. Now $\overline{x-1}$ is a zero divisor in $\mathbb Z[x]/(x^2 - 1)$ by the binomial formula, so $\phi(\overline{x-1}) = \overline{ax + b - 1}$ is a zero divisor in $\mathbb Z/(x^2 - 4)$. For the same reason, $\phi(\overline{x+1}) = \overline{ax + b + 1}$ is a zero divisor in $\mathbb Z/(x^2 - 4)$, and even $$ (ax + b + 1)(ax + b - 1) = q(x)(x^2 - 4) $$ for a polynomial $q \in \mathbb Z[x]$. But by the degree formula, $q \in \mathbb Z$, and we get $q = a^2$, $0 = a(b-1) + a(b+1) = 2ab$, so $b=0$, and a contradiction.

Answer 3

Ok I think I solved this... In the ring $\mathbb{Z}[x]/(x^2-1)$ we have two elements $a=x+1,1-x$ that satisfy $a^2=2a$. But in the other ring only $2$ is such an element... So the rings are not isomorphic

Answer 4

If the rings were isomorphic, their groups of units would be isomorphic too. The former ring has nontrivial units as $\pm x$. One can prove that the latter ring only has the trivial units $\pm1$.

A clean way to prove the last statement uses the norm map: put $A=\mathbf Z[x]/(x^2-4)$ and let $\sigma\colon A\rightarrow A$ be the ring morphism determined by $\sigma(x)=-x$. The norm map is the map $$ N\colon A\rightarrow \mathbf Z $$ defined by $N(\alpha)=\alpha\cdot\sigma(\alpha)$, analoguous to the module-squared of a complex number. Explicitly, one has $$ N(ax+b)=(ax+b)(-ax+b)=b^2-4a^2, $$ for all integers $a,b$. Since $N$ is multiplicative and $N(1)=1$, the map $N$ induces a morphism of groups from the multiplicative group $A^\times$ of $A$ to the multiplicative group $\{\pm1\}$ of $\mathbf Z$. In particular, if $ax+b$ is invertible in $A$, one has $b^2-4a^2=\pm1$ in $\mathbf Z$. It follows that $$ (b-2a)(b+2a)=\pm1 $$ in $\mathbf Z$. Hence $$ b-2a,\ b+2a=\pm1. $$ It follows that $a=0$ and $b=\pm1$, as required.