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Let $B

$$\frac{A}{B}\otimes_C D\cong \frac{A\otimes_C D}{B\otimes_C D}$$

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.

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    Presumably, you mean $I\otimes Q$ in the denominator on the right at the end, because $I$ is not a $\mathbb Q-$ submodule of $\mathbb Q[x_1,\dots,x_n]$. – 2017-01-22
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    @ThomasAndrews Ah yes I do. If I'm not mistaken that would be the same as writing $\langle I \rangle_{\mathbb{Q}}$, the ideal generated by $I$ over $\mathbb{Q}$, right? – 2017-01-22
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    Yes, that's also true. – 2017-01-22
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    @ThomasAndrews So is everything I wrote here correct? And in addition, is it also correct that the isomorphism $\frac{A}{B}\otimes_C D\cong \frac{A\otimes_C D}{B\otimes_C D}$ holding for all $A,B$ is equivalent with $D$ being flat? – 2017-01-22
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    The ideal generated by $I$ in $\mathbf Q[x_1,\dots, x_n]$, more exactly. ‘Generated by $I$ in $\mathbf Q$’ is confusional, as $\mathbf Q$ has no non-trivial ideals. And, yes your isomorphism is perfectly correct. – 2017-01-22
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    @Bernard Yes I realised that after writing it. I did indeed mean the ideal generated in $\mathbb{Q}[x_1,\dots,x_n]$ – 2017-01-22
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    Well, in general $B\otimes_CD$ is not (isomorphic to) a submodule of $A\otimes_CD$, so the quotient you're considering doesn't make sense. Saying that the obvious homomorphism $B\otimes_CD\to A\otimes_CD$ is a monomorphism whenever $B$ is a submodule of $A$ (for all $A$) is the same as flatness of $D$. – 2017-01-22
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    @egreg Ah of course, if $D$ is not flat then $B\otimes D\to A\otimes D$ need not be injective and so $B\otimes D$ is not a submodule in a natural way, is that what you mean? – 2017-01-22

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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.

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    Thank you for the answer. I see now that in general we cannot even consider $\frac{A\otimes_CD}{B\otimes_CD}$. Just to be sure though, in case $D$ is flat, we can indeed consider this quotient and in this case we have $\frac{A}{B}\otimes_C D\cong \frac{A\otimes_C D}{B\otimes_C D}$ correct? – 2017-01-22
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    @user2520938 Yes, see the addition I made. In case of flatness, $[B,D]$ can be canonically identified with $B\otimes_CD$, when $B$ is a submodule of $A$. – 2017-01-22
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    And this canonical identification is then just given by $i\otimes 1$, the tensor product map obtained from the inclusion $i:B\to A$ correct? – 2017-01-22
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    @user2520938 Yes, correct. – 2017-01-22