TITLE: Calculating in quotient ring of $\mathbb{R}[X]$
QUESTION [9 upvotes]: Part of an old Oxford exam (1992 A1)
We want to find which elements of the quotient ring $\mathbb{R}[X]/(x^3-x^2+x-1)$ are equal to their own square.
Now, we note first that $x^3-x^2+x-1=(x-1)(x^2+1)$
Let $f(x)=(x-1)(x^2+1)$. Clearly we have $[1+(f(x))]^2=[1+(f(x))]$ but I cannot see how to find others. Any guidance would be appreciated.
Thanks.

REPLY [14 votes]: Your previous question concerns generalized CRT. So you know that
$$\frac{{\bf R}[x]}{(x-1)(x^2+1)}\cong\frac{{\bf R}[x]}{(x-1)}\times \frac{{\bf R}[x]}{(x^2+1)}\cong {\bf R}\times{\bf C}$$
Find the idempotents in $\bf R$ and $\bf C$ to find the idempotents in ${\bf R}\times{\bf C}$, then pull them back through the isomorphisms implicit above.

Indeed the idempotents of ${\bf R}\times{\bf C}$ are $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$. The first and last correspond to $0$ and $1$ in $S={\bf R}[x]/(x-1)(x^2+1)$. To find what the others correspond to, solve e.g.
$$f(x)\equiv\begin{cases}0 & \mod x-1, \\ 1 & \mod x^2+1. \end{cases} $$
without loss of generality with $\deg f(x)\le 2$. Thus we can write $f(x)=(ax+b)(x-1)$ and solve for $a,b$ by distributing out this product, reducing mod $x^2+1$ and setting equal to zero. The other case proceeds in exactly the same way.