I have a basic question about the cusp in the cuspidal cubic.
Take $X$ to be $y^2=x^3$ in $\mathbb{A}^2$ (over an algebraically closed field for instance), it is well known that the point $(0,0)$ cannot be an effective Cartier divisor (because for instance, blowing up along that point would give an isomorphic curve, but the blow up is smooth and $X$ is not, or because the dimension of the tangent space at $(0,0)$ is 2).
However, I am under the impression that it is a non effective Cartier Divisor, in the following sense: the Weil divisor $[(0,0)]$ is the Weil divisor associated to the Cartier divisor $y/x$ (which is not effective of course), as $\text{ord}_{(0,0)}(y)=3$ and $\text{ord}_{(0,0)}(x)=2$ according to computing the actual order of vanishing.
I do realize it's a bit odd to say that is "is" a Cartier divisor, as it is not the scheme theoretic locus of one.
Is it correct, or am I missing something here?