# Calculating in quotient ring of $\mathbb{R}[X]$

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.

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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.

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This is a really clever approach; had not realised the two parts of the question tied in. So we have $\{(0,0),(1,0),(0,1),(1,1) \}$ as the idempotents of $\mathbb{R} \times \mathbb{C}$. Can I ask how we know that the ideals are comaximal, as is needed for the CRT? –  Mathmo May 17 '13 at 22:28
@Mathmo Distinct prime ideals are coprime, always fulfilling the hypothesis of CRT. More computationally, notice $(x-1)(x+1)=x^2-1$ which helps us see $$-\frac{x+1}{2}\color{Blue}{(x-1)}+\frac{1}{2}\color{Purple}{(x^2+1)}=1.$$ (General commutative algebra facts: An ideal contains $1$ iff it is improper, so $(a)+(b)=R$ iff $ax+by=1$ for some $x,y\in R$.) Just as importantly, can you see how to pull $(1,0)$ and $(0,1)$ back to ${\bf R}[x]/(x-1)(x^2+1)$? –  anon May 17 '13 at 22:37
I'm sorry but I cannot see how to do this. I really struggle to conceptualise what elements of this type of quotient ring actually look like. –  Mathmo May 17 '13 at 22:51
@Mathmo I have provided some details on how to pull the elements $(1,0)$ and $(0,1)$ back to the original quotient ring. –  anon May 18 '13 at 1:28
@anon I realize you were thinking this, but for other readers I think the statement "Distinct prime ideals are coprime" needs to be qualified to "distinct nonzero ideals of this ring are coprime." Of course in general rings it's not true. In $\Bbb Z [x]$, $(x)$ and $(2)$ are prime, but $(x)+(2)\subseteq (x,2)\neq \Bbb Z[x]$. –  rschwieb May 18 '13 at 14:15