Question

the idealizer IR(A) and the eigenring ER(A) for the right ideal A = xR in the free k-ring R =
k<x, y>, and for the right ideal A = xR where x = i + j + k in the ring of quaternions with integer coefficients.

Answer 1

(1) Let $R = k < x, y >$ and $A = xR$. For $r \in R$, we have $r \in \mathrm{IR}(A)$ iff $r \cdot x \in xR$. Writing $r = r_0 + r_1$ where $r_0$ is the constant term of $r$, we see that $r \cdot x = r_0x + r_1x \in xR$ iff $r_1 \in xR$.

This shows that $\operatorname{IR}(A) = k + xR$, from which we get $\operatorname{ER}(A) = (k + xR) / xR \sim k$.

(2) Let $R = \mathbf{Z} \circ \mathbf{Z}i \circ \mathbf{Z}j \circ \mathbf{Z}k$ and $A = xR$, where $x = i + j + k$. Since $x^2 = -3$, we have $3R \subseteq xR$. Writing "bar" for the projection map $R \to \overline{R} = R/3R$, we check easily that the right annihilator of $x$ in $R$ has 9 elements. Since $|\overline{R}| = 3^4$, it follows that

$(\#)[R:xR] = [\overline{R}:x\overline{R}] = 81 / 9 = 9.$

Now $xR$ is not an ideal in $R$, so we have $R \supset \operatorname{IR}(xR) \supseteq xR + Z \supset xR$.

From $(*)$, we see that $\operatorname{IR}(xR) = xR + Z$, and $\operatorname{ER}(xR) = \operatorname{IR}(xR)/xR \sim Z/3Z$.