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For a field $k$ and finitely many generators, this ring consists of arbitrary coefficient families indexed by finite words in the generators. Addition is coefficientwise and multiplication uses concatenation: the coefficient of $w$ in $FG$ is ${\sum }_{uv=w}{F}_u{G}_v$, a finite sum. Finite series form the free associative algebra. If $F,G\ne 0$, their lowest nonzero homogeneous parts multiply to a nonzero polynomial, since a word of fixed length has a unique splitting at a specified position. Thus the ring is a noncommutative domain.

There is one constant path and $r$ loop generators; every other path is a word in those loops. Distinct words are distinct basis paths, so there are no relations among the generators. Thereforethe free associative algebra on $r$ generators, with each loop sent to its corresponding generator. For $r=1$ this is $k[x_1]$; for $r\ge 2$ it is noncommutative. The quiver with one loop is the $r=1$ case of the quiver with r loops.

Use $s=x_1$ and $r=x_2$. A putative left Ore set equality would beThe left side is a single word ending in $x_2$, with coefficient $1$. The right side is a finite linear combination of words ending in $x_1$. Distinct words are linearly independent in the free associative algebra, so equality is impossible. ThereforeThe last letters matter because the denominators in the left Ore set equation occur on the right of the common left multiple.

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.

This quiver has one vertex and $r$ loops. Paths are words in the loops, so its path algebra is a free associative algebra. A representation assigns an arbitrary endomorphism to each loop.