What is the kernel of the tensor product of two maps?

Question

Assume that $f_1\colon V_1\to W_1, f_2\colon V_2\to W_2$ are $k$-linear maps between $k$-vector spaces (over the same field $k$, but the dimension may be infinity). Then the tensor product $f_1\otimes f_2\colon V_1\otimes V_2\to W_1\otimes W_2$ is defined, and it's obvious that $\ker f_1\otimes V_2+ V_1\otimes \ker f_2 \subseteq \ker (f_1\otimes f_2)$. My question is whether the relation $\subseteq$ is in fact $=$.

If this does not hold, how about assuming all these vector spaces are commutative associated $k$-algebras with identity and that all the maps are $k$-algebra homomorphisms? Or can you give a "right" form of the kernel $\ker (f_1\otimes f_2)$?

Answer 1

Yes, that's true. Let $f_i : V_i \to W_i$ be two linear maps. Since $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $W_1 \otimes W_2$, we may assume that $f_1,f_2$ are surjective. But then they are split, so that we can assume that $V_i = W_i \oplus U_i$ and that $f_i$ equals the projection $V_i \to W_i$, with kernel $U_i$. Then $V_1 \otimes V_2 = W_1 \otimes W_2 \oplus W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2$ and $f_1 \otimes f_2$ equals the projection of $V_1 \otimes V_2$ onto $W_1 \otimes W_2$. Hence the kernel is $W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2 = U_1 \otimes V_2 + V_1 \otimes U_2$.

This shows even more: The kernel is the pushout $(\ker(f_1) \otimes V_2) \cup_{\ker(f_1) \otimes \ker(f_2)} (V_1 \otimes \ker(f_2))$.

By the way, this argument is purely formal and works in every semisimple abelian $\otimes$-category. What happens when we drop semisimplicity, for example when we consider modules over some commutative ring $R$? Then we only need some flatness assumptions:

Let $f_1 : V_1 \to W_1$ and $f_2 : V_2 \to W_2$ be two morphisms in an abelian $\otimes$-category (for example $R$-linear maps between $R$-modules). If $f_1,f_2$ are epimorphisms, then we have exact sequences $\ker(f_1) \to V_1 \to W_1 \to 0$ and $\ker(f_2) \to V_2 \to W_2 \to 0$. Applying the right exactness of the tensor product twice(!), we get that then also

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2 \to 0$

is exact. If $f_1,f_2$ are not epi, we can still apply the above to their images and get the exactness of

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to \mathrm{im}(f_1) \otimes \mathrm{im}(f_2) \to 0.$

Now assume that $\mathrm{im}(f_1)$ and $W_2$ are flat. Then $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $\mathrm{im}(f_1) \otimes W_2$ which embeds into $W_1 \otimes W_2$. Hence, we have still that the sequence

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2$

is exact. In other words, we have a sum decomposition $$\ker(f_1 \otimes f_2) = \alpha(\ker(f_1) \otimes V_2) + \beta(V_1 \otimes \ker(f_2)),$$ where $\alpha : \ker(f_1) \otimes V_2 \to V_1 \otimes V_2$ and $\beta : V_1 \otimes \ker(f_2) \to V_1 \otimes V_2$ are the canonical morphisms. In general, these are not monic! However, this is the case, by definition, when $V_1$ and $V_2$ are flat. So in this case we can safely treat $\alpha$ and $\beta$ as inclusions and write $$\ker(f_1 \otimes f_2) = V_1 \otimes \ker(f_2) + \ker(f_1) \otimes V_2.$$

Answer 2

When $f_1$ and $f_2$ are surjective

Constructing an isomorphism

Suppose first that $f_1$ and $f_2$ are surjective, and that we are working over an arbitrary commutative ring $$. We may assume that $f_i$ is the canonical quotient map from $V_i$ to $V_i / U_i$ for a submodule $U_i$ of $V_i$. By abuse of notation, we denote by $U_1 ⊗ V_2 + V_1 ⊗ U_2$ the resulting submodule of $V_1 ⊗ V_2$, even if the homomorphisms from $U_1 ⊗ V_2$ and $V_1 ⊗ U_2$ to $V_1 ⊗ V_2$ are not injective. The submodule $U_1 ⊗ V_2 + V_1 ⊗ U_2$ of $V_1 ⊗ V_2$ lies in the kernel of $f_1 ⊗ f_2$, whence we get an induced homomorphism of $$-modules $$ φ \colon (V_1 ⊗ V_2) / (U_1 ⊗ V_2 + V_1 ⊗ U_2) \to (V_1 / U_1) ⊗ (V_2 / U_2) \,, \quad [v_1 ⊗ v_2] \mapsto [v_1] ⊗ [v_2] \,. $$ We want to show that $φ$ is an isomorphism. This can be done in multiple ways.

1. We can construct an inverse $ψ$ to $φ$ as follows. For every element $v_2$ of $V_2$, the composite $$ λ_{v_1} \colon V_1 \to V_1 ⊗ V_2 \to (V_1 ⊗ V_2) / (U_1 ⊗ V_2 + V_1 ⊗ U_2) \,, \quad v_1 \mapsto v_1 ⊗ v_2 \mapsto [v_1 ⊗ v_2] $$ is $$-linear and contains $U_2$ in its kernel. We get therefore an induced $$-linear map. $$ λ'_{v_1} \colon V_1 / U_1 \to (V_1 ⊗ V_2) / (U_1 ⊗ V_2 + V_1 ⊗ U_2) \,, \quad [v_1] \mapsto [v_1 ⊗ v_2] \,. $$ For every element $[v_1]$ of $V / U_1$ we have thus the well-defined map $$ ρ_{[v_1]} \colon V_2 \to (V_1 ⊗ V_2) / (U_1 ⊗ V_2 + V_1 ⊗ U_2) \,, \quad v_2 \mapsto λ'_{[v_1]}(v_2) = [v_1 ⊗ v_2] \,. $$ This map is $$-linear and contains $U_2$ in its kernel, and therefore induces a well-defined $$-linear map $$ ρ'_{[v_1]} \colon V_2 / U_2 \to (V_1 ⊗ V_2) / (U_1 ⊗ V_2 + V_1 ⊗ U_2) \,, \quad [v_2] \mapsto [v_1 ⊗ v_2] \,. $$ We have altogether a well-defined map \begin{align*} (V_1 / U_1) × (V_2 / U_2) &\to (V_1 ⊗ V_2) / (U_1 ⊗ V_2 + V_1 ⊗ U_2) \,, \\ ([v_1], [v_2]) &\mapsto ρ'_{[v_1]}([v_2]) = [v_1 ⊗ v_2] \,. \end{align*} This map is $$-bilinear, and therefore induces the desired $$-linear map $$ ψ \colon (V_1 / U_1) ⊗ (V_2 / U_2) \to (V_1 ⊗ V_2) / (U_1 ⊗ V_2 + V_1 ⊗ U_2) \,, \quad [v_1] ⊗ [v_2] \mapsto [v_1 ⊗ v_2] \,. $$

2. Let $T$ be another $$-module. Any $$-bilinear map from $β$ from $(V_1 / U_1) × (V_2 / U_2)$ to $T$ can be pulled back to a $$-bilinear map from $α$ from $V_1 × V_2$ to $T$, given by $α(v_1, v_2) = β([v_1], [v_2])$. This bilinear map $α$ satisfies $α(v_1, v_2) = 0$ whenever $v_1$ is contained in $U_1$ or $v_2$ is contained in $U_2$.

   Suppose conversely that $α$ is any $$-bilinear map from $V_1 × V_2$ to $T$ such that $α(v_1, v_2) = 0$ whenever $v_1$ is contained in $U_1$ or $v_2$ is contained in $U_2$. For all $v_1, v'_1 ∈ V_1$ and $v_2, v'_2 ∈ V_2$ with $v_1 - v'_1 ∈ U_1$ and $v_2 - v'_2 ∈ U_2$ we then have \begin{align*} α(v_1, v_2) &= α(v'_1 + v_1 - v'_1, v'_2 + v_2 - v'_2) \\ &= α(v'_1, v'_2) + \underbrace{ α(v'_1, v_2 - v'_2) }_{= 0} + \underbrace{ α(v_1 - v'_1, v'_2) }_{= 0} + \underbrace{ α(v_1 - v'_1, v_2 - v'_2) }_{= 0} \\ &= α(v'_1, v'_2) \,. \end{align*} This shows that $α$ descends to a well-defined map $β$ from $(V_1 / U_1) × (V_2 / U_2)$ to $T$, which is then again bilinear. (We could also prove this by using the universal properties of $V_1 / U_1$ and $V_2 / U_2$, similar to how we have constructed the inverse $ψ$ before.) We have thus constructed an isomorphism \begin{align*} {}& \mathrm{Bilin}(V_1 / U_1 , V_2 / U_2 ;\, T) \\ ≅{}& \{ α ∈ \mathrm{Bilin}(V_1, V_2;\, T) \mid \text{$α(v_1, v_2) = 0$ whenever $v_1 ∈ U_1$ or $v_2 ∈ U_2$} \} \,. \end{align*} It follows that \begin{align*} {}& \mathrm{Hom}((V_1 / U_1) ⊗ (V_2 / U_2), T) \\ ≅{}& \mathrm{Bilin}(V_1 / U_1 , V_2 / U_2 ;\, T) \\ ≅{}& \{ α ∈ \mathrm{Bilin}(V_1, V_2;\, T) \mid \text{$α(v_1, v_2) = 0$ whenever $v_1 ∈ U_1$ or $v_2 ∈ U_2$} \} \\ ≅{}& \{ f ∈ \mathrm{Hom}(V_1 ⊗ V_2, T) \mid \text{$f(v_1 ⊗ v_2) = 0$ whenever $v_1 ∈ U_1$ or $v_2 ∈ U_2$} \} \\ ≅{}& \{ f ∈ \mathrm{Hom}(V_1 ⊗ V_2, T) \mid \text{$f(x) = 0$ whenever $x ∈ U_1 ⊗ V_2$ or $x ∈ V_1 ⊗ U_2$} \} \\ ≅{}& \{ f ∈ \mathrm{Hom}(V_1 ⊗ V_2, T) \mid \text{$f(x) = 0$ for all $x ∈ U_1 ⊗ V_2 + V_1 ⊗ U_2$} \} \\ ≅{}& \mathrm{Hom}((V_1 ⊗ V_2) / (U_1 ⊗ V_2 + V_1 ⊗ U_2), T) \,. \end{align*} By writing out these isomorphisms, we see that it is given (from top to bottom) by the induced homomorphism $φ^*$. It follows from Yoneda’s lemma (or more precisely from the fully faithfulness of the Yoneda embedding) that $φ$ is an isomorphism.

3. We could also combine both approaches: we use the isomorphism \begin{align*} {}& \mathrm{Hom}((V_1 / U_1) ⊗ (V_2 / U_2), T) \\ ≅{}& \{ α ∈ \mathrm{Bilin}(V_1, V_2;\, T) \mid \text{$α(v_1, v_2) = 0$ whenever $v_1 ∈ U_1$ or $v_2 ∈ U_2$} \} \end{align*} for $T = (V_1 ⊗ V_2) / (U_1 ⊗ V_1 + V_1 ⊗ U_2)$ to easily construct $ψ$ as the homomorphism corresponding to the bilinear map $(v_1, v_2) \mapsto [v_1 ⊗ v_2]$.

We find in either case that $φ$ is an isomorphism because both $(V_1 ⊗ V_2) / (U_1 ⊗ V_1 + U_2 ⊗ V_1)$ and $(V_1 / U_1) ⊗ (V_2 / U_2)$ have suitable universal properties.

Diagram chase

We know that tensoring is right exact. We get therefore the following commutative diagram with exact rows and exact columns: $$ \require{AMScd} \begin{CD} U_1 ⊗ U_2 @>>> V_1 ⊗ U_2 @>>> (V_1 / U_1) ⊗ U_2 @>>> 0 \\ @VVV @VVV @VVV \\ U_1 ⊗ V_2 @>>> V_1 ⊗ V_2 @>>> (V_1 / U_1) ⊗ V_2 @>>> 0 \\ @VVV @VVV @VVV \\ U_1 ⊗ (V_2 / U_2) @>>> V_1 ⊗ (V_2 / U_2) @>>> (V_1 / U_1) ⊗ (V_2 / U_2) @>>> 0 \\ @VVV @VVV @VVV \\ 0 @. 0 @. 0 @. \end{CD} $$ So let’s use the following result:

Lemma. Let $R$ be a ring and consider the following commutative diagram of $R$-modules with exact rows and exact columns: $$ \begin{CD} {} @. A @>β>> B @>>> 0 \\ @. @Vδ'VV @VVε'V @. \\ C @>δ>> D @>ε>> E @. {} \\ @. @VVV @VVγV @. \\ {} @. F @>>> G @. {} \end{CD} $$ The induced sequence $A ⊕ C \xrightarrow{f} D \xrightarrow{g} G$ is again exact.

Proof. It follows the exactness of the middle row and left column that $g ∘ f = 0$. Let $d ∈ \ker(f)$. We have $0 = g(d) = γε(d)$, whence $ε(d) ∈ \ker(γ) = \mathrm{im}(ε')$. There hence exists $b ∈ B$ with $ε(d) = ε'(b)$. By the surjectivity of $β$, there exists $a ∈ A$ with $b = β(a)$. Let $d' := d - δ'(a)$. We have $$ ε(d') = ε(d) - εδ'(a) = ε(d) - ε'β(a) = ε(d) - ε'(b) = ε(d) - ε(d) = 0 \,. $$ There hence exists $c ∈ C$ with $d' = δ(c)$. We have now $d - δ'(a) = d' = δ(c)$, and therefore $d = δ'(a) + δ(c) = f( (a, c) )$. This shows that the kernel of $g$ is contained in the image of $f$.

In our specific example, we find that the sequence $$ (U_1 ⊗ V_2) ⊕ (V_1 ⊗ U_2) \to V_1 ⊗ V_2 \to (V_1 / U_1) ⊗ (V_2 / U_2) $$ is exact. In other words, the kernel of $V_1 ⊗ V_2 \to (V_1 / U_1) ⊗ (V_2 / U_2)$ is given by $U_1 ⊗ V_2 + V_1 ⊗ U_2$.

When $f_1$ and $f_2$ are not both surjective

If $f_1$ and $f_2$ are not both surjective then we can proceed as already explained in Martin Brandenburg’s answer: By using the factorization $$ V_1 ⊗ V_2 \to \mathrm{im}(f_1) ⊗ \mathrm{im}(f_2) \to W_1 ⊗ W_2 \,, $$ we see that the desired equality holds if and only if the homomorphism $\mathrm{im}(f_1) ⊗ \mathrm{im}(f_2) \to W_1 ⊗ W_2$ is injective. This is, for example, the case if $\mathrm{im}(f_1)$ and $W_2$ are flat, or if $W_1$ and $\mathrm{im}(f_2)$ are flat. This is in particular the case when $$ is a field, since then every $$-module is free and thus flat.

Answer 3

The isomorphism $$ \ker (f_1\otimes f_2)=V_1\otimes \ker f_2 + \ker f_1 \otimes V_2 $$ is equivalent to the isomorphism $$ \operatorname{coker} (f_1\otimes f_2)=\operatorname{coker} f_1\otimes \operatorname{coker} f_2, $$ since for any ideals $I,J\subset R$ we have $$ R/(I+J)=R/I\otimes R/J $$