Past exam of the non-commutative rings course of the University of Oxford / 2023 / 2 / d / Solution

Take $D=\hat{R}$. We first check that this noncommutative formal power series ring is a noncommutative domain. For nonzero $F,G\in D$, let $p,q$ be the smallest lengths of words with nonzero coefficients in $F,G$, respectively. Their degree-$p$ and degree-$q$ parts $F_p,G_q$ are nonzero homogeneous polynomials in the free associative algebra.

Choose words $u,v$ with nonzero coefficients in $F_p,G_q$. In $F_p{G}_q$, the coefficient of $uv$ is the product of those two coefficients: a word of length $p+q$ has a unique splitting into a prefix of length $p$ and a suffix of length $q$. This product is nonzero in $\mathbb{Q}$. The degree-$p+q$ part of $FG$ is exactly $F_p{G}_q$, since all lower parts of $F$ and $G$ vanish. Thus $FG\ne 0$, proving that $D$ is a noncommutative domain.

By the defining property of the free associative algebra, there is a ring homomorphismThe same substitution defines a ring automorphism of $R$, with inverse $x_1\mapsto x_1-1$, $x_2\mapsto x_2-2$. Composing this ring automorphism with the natural inclusion $R\hookrightarrow D$ proves that $f$ is injective.

Finally,Their constant terms are $3,-1,1,2$, respectively, all nonzero. By the unit criterion for noncommutative formal power series proved in c, all four elements are units. Therefore $D$ and $f$ have every required property. The embedding changes the images of the generators, so the failure of the original monomial left Ore set in b presents no obstruction.