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.


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.

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    $\begingroup$ By the way, please don't call a commutative ring $k$ if somebody might want to give an example where $k$ is an algebra over a field. :) $\endgroup$ – Jeremy Rickard Feb 28 '16 at 17:36
  • $\begingroup$ Or write $k = \ell[x,y]/\langle x^2,xy,y^2\rangle$ :D $\endgroup$ – Martin Brandenburg Feb 28 '20 at 10:12

Cool, thanks! My apologies for the inconvenient notation. Next time I will call the commutative ring $A$, I promise. ;)

Let me make your counterexample explicit: Let us write module multiplication from the right and for a basis element $b \in \{ 1,x,y \} \subset k$ let $b^* \in k^*$ denote its dual basis vector (sending $b$ to $1$ and the other basis vectors to $0$). The $k$-module structure on $k^*$ is given by composition with right multiplication on $k$, so in particular we have $x^* \cdot x = 1^*$ and $y^* \cdot x = 0$. The map $k \overset i \longrightarrow (k^*)^2$ sends $1$ to $(x^*,y^*)$ and has kernel $k \cdot y \cap k \cdot x = 0$. Moreover we get

$(i \otimes_k i)(1 \otimes x) = (i \otimes_k i)(1 \otimes 1) \cdot x = (x^*,y^*) \otimes (x^*,y^*) \cdot x = (x^*,y^*) \otimes (1^*,0) = (x^*,y^*) \otimes (x^*,0) \cdot x = (1^*,0) \otimes (x^*,0) = (y^*,0) \otimes (x^*,0) \cdot y = (y^*,0) \otimes (0,0) = 0$.

Since $k \otimes_k k \overset \sim \longrightarrow k$ sends $1 \otimes x$ to $x$, this is a non-trivial element in the kernel, proving that $i \otimes_k i$ is not injective.


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