Weil does not imply Cartier on variety $X$. 
Show that the divisor $D$ defined by $a = b = 0$ in the variety $X \subset \mathbb{A}^4$ defined  by $ad - bc = 0$ $($the cone on a smooth quadric surface$)$ is not locally principal.

My attempt at proof is as follows. If $D \cap U = X \cap V(f) \cap U$ for some $f \in k[a, b, c, d]$ and some nonempty open $U \subset \mathbb{A}^4$, then we can assume that $f$ is homogenous. We have a picture in $\mathbb{P}^3$. The resulting $($effective$)$ Cartier divisor should extend to a Cartier divisor on $\mathbb{P}^3$, thus its associated line bundle is a pullback of some $\mathcal{O}(d)$ for $d$ positive. I know that if $S: \mathbb{P}^m \times \mathbb{P}^n \to \mathbb{P}^N$ is the Segre embedding, we have $S^*\mathcal{O}(1) \cong \pi_1^*\mathcal{O}(1) \otimes \pi_2^*\mathcal{O}(1)$. Now, the pullback of the aforementioned is $\mathcal{O}(d) \otimes \mathcal{O}(d)$ on $\mathbb{P}^1 \times \mathbb{P}^1$; this gives us a contradiction because the divisor is $a = b = 0$, which is not really a thing (does not pullback to a divisor of $\mathbb{P}^1 \times \mathbb{P}^1$).
EDIT: An easier way to finish is to just check the tangent space. One of these is $2$, the other one is at least $3$.
I am not completely sure if this approach is completely correct though. Some feedback would be appreciated. Also, if anyone could supply a proof or link to one, that would be nice.
 A: Sorry to resurrect my old question, but for the sake of completeness, I will provide a complete solution.
Consider the tangent space at the origin of our divisor $D$. We have$$k[D] = k[a, b, c, d]/(a, b, ac - bd) \cong k[c, d].$$This Zariski cotangent space has dimension $2$ here (generated by $c$, $d$).
Now, let us consider a tangent space of $X \cap V(f)$. The coordinate ring here is $$k[a, b, c, d]/(f, ac - bd).$$Modding out by $ac - bd$ does not change the dimension of the tangent space at the origin, as it is inside $\mathfrak{m}^2$. However, $(f)$ has height at most $1$, so it can change the tangent dimension by at most $1$. Thus, the dimension of the ring here, and hence the local ring at $0$ (localizing at maximal ideal can not change dimension), must be at least $4 - 1 = 3$, but this is a lower bound for the tangent dimension, so we can do the same thing for that.
Thus, on an open set $U$ containing the origin, it is impossible for $D \cap U$ to agree with $X \cap V(f) \cap U$, as they would have different cotangent dimensions at the origin.
