Projective dimension of $K[x,y]$ over $K[xy]$ Let $K$ be a field and $K[x,y]$ the polynomial ring in two variables $x$ and $y$ over $K$. Let $R = K[xy]$ be the subring generated as a $K$-algebra by the monomial $xy$. My question is:

What is the projective dimension of $K[x, y]$ over $R$?

I know that $R$ is a ring of global dimension $1$, so the bound $\operatorname{pd}_R K[x,y] \leq 1$ is immediate. Moreover, we have $\operatorname{pd}_R K[x,y] = 0$ if and only if $K[x,y]$ is a projective $R$-module, and since $R$ is projective-free by being a PID, we have that $\operatorname{pd}_R K[x,y] = 0$ if and only if $K[x,y]$ is free over $R$.
I've tried playing around with possible bases for $K[x,y]$ over $R$ to no success. I don't think that $K[x,y]$ is finitely generated over $R$ as the $R$-action can't change the difference between the $x$-degree and the $y$-degree of an element. This makes me think of $\{x^i, y^j\colon i, j \geq 1\} \cup \{1\}$ as a possible basis, but calculations got messy trying to write elements in terms of $R$-combinations of these. I would be very happy if this $R$-module is indeed free, but I'm beggning to have second guesses. Any help or tips are much appreciated!
 A: Let $t = xy$ be the generator of $R$. Any element $p$ of $K[x,y]$ can be uniquely written as $$p = \sum_{i\geq 0} \sum_{j\geq 0} p_{i,j} x^iy^j$$ where the $p_{i,j} \in K$ are almost all zero. Define the numbers $a_{n,m}$, $b_k$ and $c_{n,m}$ for $n \geq 1$ and $m \geq 0$ as follows: $$\begin{cases}a_{n,m} = p_{n+m,m}\,,\\ b_m = p_{m,m}\,,\\ c_{n,m} = p_{n,n+m}\,.\end{cases}$$
We claim that $$\left(\sum_{m \geq 0} b_mt^m\right)+\sum_{n \geq 1}\left( \left(\sum_{m \geq 0} a_{n,m}t^m\right)x^n + \left(\sum_{m \geq 0}c_{n,m}t^m\right)y^n\right)$$ is another expression for $p$. Indeed, any monomial $x^iy^j$ can be uniquely expressed as $t^k$ (if $i = j = k$), $t^mx^n$ (if $i - j = n > 0$ with $m = j$) or $t^my^n$ (if $j - i =n > 0$ with $m = i$). Hence, every summand of the form $p_{i,j}x^iy^j$ appears once and only once on the expression above. This shows that this expression represents $0$ if and only if all $p_{i,j}$ (and hence all $a_{n,m}$, $b_m$ and $c_{n,m}$) are zero. Thus, the set $$\{1,x^i,j^l\colon i\,,j \geq 1\}$$ is a basis of $K[x,y]$ over $R$, showing that $K[x,y]$ is free over $R$ and that $\operatorname{pd}_R K[x,y] = 0$.
