I am trying to understand the definition of the dimension of a variety using the notion of a transcendental basis. Consider for an algebraically closed field $\mathbb{K}$ the variety $V=\left\{(x,y)\in \mathbb{K}^{2}\mid y^{2}=x^3\right\}$ and coordinate algebra $k\left[V\right]=k\left[\bar{X},\bar{Y}\right]$, where $\bar{X}$ and $\bar{Y}$ are the coordinate function (e.g. $\bar{X}:(\alpha _{1},\alpha _{2})\mapsto \alpha _{1}$). Now, according to my lecture notes$\bar{X}$ is a transcendental element, but I struggled for a while to see what field extension one considers. The way I understand this now is as follows:
One considers the field of fractions $L'=k\left[V\right]\times k\left[V\right]/\sim$ where $(f_{1},g_{1})\sim (f_{2},g_{2}):\Leftrightarrow f_{1}g_{2}=f_{2}g_{1}$. This gives a field extension and we consider the field $L=L'/I$ where $I$ is the ideal generated by $x^3-y^2$. Now the elements of $\mathbb{K}$ are seen as elements of $L$ by considering constant functions. Also, for each element $\alpha \in \mathbb{K}\subset L$ we can consider the polynomial $(p:x\mapsto x-\alpha )$ so that $p(\alpha )=0$ (viewed as the zero function, NOT as $0\in \mathbb{K}$), hence $\alpha \in \mathbb{K}$ is never transcendental. Furthermore, the set $\left\{\bar{X},\bar{Y}\right\}$ is not algebraically independent either, because the polynomial $p:(x,y)\mapsto x^3-y^2$ becomes an element of the Ideal $I$ if evaluated on the restriction functions $x=\bar{X}$ and $y=\bar{Y}$. From this viewpoint, it becomes clear to me that the sets $\{\bar{X}\}$ and $\{\bar{Y}\}$ are algebraically independent.
Is this the correct viewpoint/formalism?