Given a commutative ring $k$ and for $i = 1,2$ a homomorphism of $k$-modules $X_i \overset {f_i} \longrightarrow Y_i$ with $X_i$ flat over $k$.
Is the following conclusion true for general $k$? If $f_1$ and $f_2$ are injective, so is their tensor product $f_1 \otimes_k f_2: X_1 \otimes_k X_2 \longrightarrow Y_1 \otimes_k Y_2$.
It is certainly true, if also $Y_1$ (or $Y_2$) is flat using the factorization $X_1 \otimes_k X_2 \longrightarrow Y_1 \otimes_k X_2 \longrightarrow Y_1 \otimes_k Y_2$: By flatness of $X_2$ and $Y_1$ both maps are injections and so is their composite.
It is also true, if $k$ is integral. The map $X_1 \otimes_k X_2 \longrightarrow Y_1 \otimes_k Y_2 \longrightarrow Y_1 \otimes_k Y_2 \otimes_k Q(k)$ factors as $X_1 \otimes_k X_2 \longrightarrow X_1 \otimes_k X_2 \otimes_k Q(k) \longrightarrow Y_1 \otimes_k Y_2 \otimes_k Q(k)$ and the first map is injective by flatness of $X_1 \otimes_k X_2$, because $k$ being integral injects into $Q(k)$. The second map can be considered as the tensor product of the maps $X_i \otimes_k Q(k) \longrightarrow Y_i \otimes_k Q(k)$. These are injective by the flatness of $Q(k)$ and their domain and codomain are $Q(k)$-vector spaces hence flat over $k$. So the second map is injective by 1) again.
By a local-global argument one may reduce the problem to the case where $k$ is local and $X_1,X_2$ are free. I could neither find a proof for this case, nor could I construct a counter-example. I would be very grateful if anyone has an idea to solve this problem.