Can $R$ be isomorphic to $R[X]/\langle X^2-a\rangle$ where $a\in R$ and $R$ is an integral domain?

Question

I have proved that if $a=0$, then $R[X]/\langle X^2\rangle\not\cong R$

If they are isomorphic then suppose $f:R[X]/\langle X^2\rangle\to R$ is an isomorphism. Then let $c\in R,c\ne 0$ such that $f(\overline{X})=c$. Again we have $\overline{X}\ \overline{X}=\overline{X^2}=0\implies 0=f(0)=c^2$. As $R$ is integral domain, we have $c=0$, contradiction.

Hence $R[X]/\langle X^2\rangle\not\cong R$.

But I am not sure about if $a\ne 0$. By divison algorithm, we can say that $R[X]/\langle X^2-a\rangle=\{\overline{\alpha X+\beta}|\alpha,\beta\in R\}$.

Can anyone help me in this regard?

Answer 1

If there exists $b$ such that $b^2=a$, $\bar X-b$ has a divisor of zero.

If there does not exists $b$ such that $b^2=a$, $X^2-a$ is irreducible in $R$ but not in $R[X]/(X^2-a)$.

Answer 2

It is possible that $R[X]/\langle X^2-a\rangle\cong R$. Take for example $R=k[t]$ and $a=t$. Then $$R[X]/\langle X^2-t\rangle\cong k[\sqrt{t}]\cong k[t]=R$$ This doesn't work if we would ask if $R$ and $R[X]/\langle X^2-a\rangle$ are isomorphic as $R$-algebras (which is in general more interesting). For example $R$ is a free $R$-module of rank $1$ and $R[X]/\langle X^2-a\rangle$ has rank $2$, hence they cannot be isomorphic as modules and therefore also not as $R$-algebras (assuming that $R$ is commutative). (See also Tsemo's answer for another argument for the $R$-algebra case)