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 $f_1 \otimes_k f_2: X_1 \otimes_k X_2 \longrightarrow Y_1 \otimes_k Y_2$.
1) 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.
2) 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.
Best Answer
Let $F$ be a field, and $k=F[x,y]/(x^2,xy,y^2)$. Since $k$ is a finite-dimensional $F$-algebra, flat=projective, and for $k$-modules $M,N$, there is a natural isomorphism $\operatorname{Hom}_k(M,N^\ast)\cong (M\otimes_kN)^\ast$, where $L^\ast=\operatorname{Hom}_F(L,F)$ denotes $F$-dual.
So we have a counterexample if we can find a monomorphism $i:k\to Y$ and an epimorphism $p:Y'\to k^\ast$ of finite dimensional $k$-modules, and a homomorphism $\alpha:k\to k^\ast$ such that there is no homomorphism $\beta:Y\to Y'$ such that $\alpha=p\alpha i$ (i.e., the map $\operatorname{Hom}_k(Y,Y')\to\operatorname{Hom}_k(k,k^\ast)$ induced by $i$ and $p$ is not surjective).
Let $P=k$ and $I=k^\ast$ be the unique indecomposable projective and injective $k$-modules, and $F=k/(x,y)$ the unique simple $k$-module.
Let $i:k\to I(k)$ be the inclusion of $k$ into its injective hull (where $I(k)\cong k^\ast\oplus k^\ast$) and $p:P(k^\ast)\to k^\ast$ be the surjection from the projective cover of $k^\ast$ (where $P(k^\ast)\cong k\oplus k)$.
Then there are homomorphisms $\alpha:k\to k^\ast$ such that the induced map $\operatorname{Hom}_k(F,k)\to\operatorname{Hom}_k(F,k^\ast)$ is non-zero, but there are no maps $\gamma:k^\ast\to k$ so that the induced map $\operatorname{Hom}_k(F,k^\ast)\to\operatorname{Hom}_k(F,k)$ is non-zero, so $\alpha$ gives a counterexample.
Translating back into the notation of the question, $f_1$ and $f_2$ are both the inclusion $i:k\to I(k)$.
The same works for any finite-dimensional commutative $F$-algebra $k$ that is not self-injective.