= Solution
Use $q\in k^\times$ and $YX=qXY$ for the <quantum plane>. Its <monomials> $X^iY^j$, $i,j\geq0$, form a <basis>, with multiplication
$$
(X^iY^j)(X^uY^v)=q^{ju}X^{i+u}Y^{j+v}.
$$
One way to verify the <basis> assertion without assuming it is to define this multiplication on the <vector space> with the displayed formal <basis>. This defines an <associative algebra>: for a third <monomial> $X^wY^z$, the two products of three <monomials> have the same exponent $ju+jw+vw$ of $q$, and its generators satisfy the required relation. Conversely, the relation puts every word into this form, establishing the presentation. Order exponent pairs by the <lexicographic order>. The largest <monomials> of two nonzero finite sums give the uniquely largest <monomial> of their product, with coefficient $c d q^{ju}\ne0$. Therefore \b[the quantum plane is a domain]. If one allows $q=0$, the assertion fails because $YX=0$ with both factors nonzero.
A <uniform module> is a nonzero <module> in which any two nonzero <submodules> have nonzero intersection. Suppose the right regular <module> of a <right Noetherian domain> $A$ were not a <uniform module>. Choose nonzero $a,b$ from two <right ideals> with zero intersection. Then $aA\cap bA=0$. The <right ideals>
$$
aA,\ baA,\ b^2aA,\ldots
$$
form a <direct sum>. For if $\sum_{i=0}^m b^ia c_i=0$, then $ac_0\in aA\cap bA$ is zero, so $c_0=0$ by the <noncommutative domain> property. Cancel the nonzero factor $b$ on the left and repeat to obtain every $c_i=0$. Each summand is nonzero, so their finite partial sums form a strictly ascending chain of <right ideals>. This contradicts the <ascending chain condition> of a <right Noetherian ring>. Hence \b[$A$ is a uniform right module].
It follows that $aA\cap sA\ne0$ whenever $a,s\ne0$: there are nonzero $u,v$ with $au=sv$. This is the <right Ore condition> for the multiplicative set $S=A\setminus\{0\}$; zero numerators cause no difficulty. The <Ore localization> theorem therefore constructs the <ring> of right fractions
$$
\boxed{Q=AS^{-1}=\{as^{-1}:a\in A,\ s\ne0\},\qquad A\hookrightarrow Q.}
$$
The map is injective because an element mapping to zero is annihilated on the right by some nonzero denominator, impossible in a <noncommutative domain>. To make the denominator convention concrete, if $su=tv\ne0$ then
$$
as^{-1}+bt^{-1}=(au+bv)(su)^{-1}.
$$
For multiplication, choose $bu=sc$ with $u\ne0$; then
$$
(as^{-1})(bt^{-1})=ac(tu)^{-1}.
$$
Common right multiples make these operations independent of the chosen representatives. Every nonzero $as^{-1}$ has inverse $sa^{-1}$, so $Q$ is a <division ring>. The original <field> $k$ is central in $A$ and therefore in the inverses as well, making $Q$ a <division algebra> over $k$. No commutative <fraction field> construction is being assumed.
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