$\frac{A}{B}\otimes_C D\cong \frac{A\otimes_C D}{B\otimes_C D}$? Let $B<A$ and $D$ be $C$-modules. Am I correct in saying that $D$ being flat over $C$ means that
$$\frac{A}{B}\otimes_C D\cong \frac{A\otimes_C D}{B\otimes_C D}$$
To me this seems an immediate consequence of the exact sequence
$$0\to B\to A\to A/B\to 0$$
and then by exactness
$$0\to B\otimes D\to A\otimes D\to (A/B)\otimes D\to 0$$
In particular it would follow, I believe, that for $I$ an ideal in $\mathbb{Z}[x_1,\dots,x_n]$ we have that
$$\frac{\mathbb{Z}[x_1,\dots,x_n]}{I}\otimes_{\mathbb{Z}}\mathbb{Q}\cong \frac{\mathbb{Q}[x_1,\dots,x_n]}{I\otimes \mathbb{Q}}$$
I'm not very confident about my understanding of tensor products, so some confirmation/correction would be appreciated.

The quotient
$$
\frac{A\otimes_CD}{B\otimes_CD}
$$
doesn't make sense in general. The tensor of the embedding $B\to A$ is in general not injective, so there's no way to identify $B\otimes_CD$ with a submodule of $A\otimes_CD$.
Flatness of $D$ is precisely the statement that

for every module $A$ and every submodule $B$ of $A$, the tensor $B\otimes_CD\to A\otimes_CD$ of the inclusion map $B\to A$ is injective.

In general, for every exact sequence $X\xrightarrow{f}Y\xrightarrow{g}Z\to0$ of $C$-modules, if you denote by $[X,D]$ the image of the map $X\otimes_CD\to Y\otimes_CD$, then
$$
Z\cong \frac{Y\otimes_CD}{[X,D]}
$$
which is an easy consequence of right exactness of the tensor product.
