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I've checked that $R = \{(a,b) \in \mathbb{Z}\times \mathbb{Z}: a\equiv b \pmod 2\}$ is a subring of $\mathbb{Z}\times \mathbb{Z}$. I want to show that $$\mathbb{Z}[X] /(X^2-1) \cong R,$$ where $\mathbb{Z}[X]$ is the polynomial ring with coefficients in $\mathbb{Z}$.

I haven't found a function for which $f\colon \mathbb{Z}[X]\to R$ is surjective, and I think it will be difficult for me to show that $\ker(f) \subset (X^2 - 1)$ since this was the case in other comparable proofs I've tried. Can you help me with this / give me some more direction?

user26857
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TRG
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1 Answers1

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Let $f \in \mathbb{Z}[x]$.

Since $f(x)-f(y)\;$is divisible by $x-y\;$in the ring $\mathbb{Z}[x,y]$, it follows that $(m-n)\mid (f(m) - f(n))$ for all integers $m,n$.

In particular, using $m=1,\;n=-1$, we get $f(1) \equiv f(-1)\;(\text{mod}\;2)$.

Let $I$ denote the ideal $(x^2-1)$ of $\mathbb{Z}[x]$.

Let $R\;$be the subring of $\mathbb{Z}{\,\times\,}\mathbb{Z}$ defined by $$R = \{(a,b) \in \mathbb{Z}{\,\times\,}\mathbb{Z} \mid a \equiv b\;(\text{mod}\;2)\}$$ and let $\phi:\mathbb{Z}[x] \to R\;$be defined by $$\phi(f) = (f(1),f(-1))$$

It's routine to verify that $\phi$ is a ring homomorphism.

Moreover, $\phi$ is surjective, since, if $(a,b) \in R$, then $$\phi(cx + d) = (a,b)$$ where $c,d$ are defined by $$c=\frac{a-b}{2},\;\;d=\frac{a+b}{2}$$

Computing the kernel of $\phi$, \begin{align*} &\phi(f) = (0,0)\\[4pt] \iff\;&(f(1),f(-1))=(0,0)\\[4pt] \iff\;&f(1) = 0\;\,\text{and}\;\,f(-1)=0\\[4pt] \iff\;&(x-1){\mid}f\;\,\text{and}\;\,(x+1){\mid}f\\[4pt] \iff\;&(x^2-1){\mid}f\\[4pt] \iff\;&f \in I\\[4pt] \end{align*} so the kernel of $\phi$ is $I$.

It follows thar $\mathbb{Z}[x]/I \cong R$, as was to be shown.

quasi
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