Interpretation of $B\otimes M$ when $B\subset A$

Question

Let us fix the setting to avoid ambiguity. Let $R$ be a commutative, unital ring. All tensor products will be considered for modules over $R$.

Let $A$ be an $R$-module, $B$ a submodule. Now, is $A\otimes M\subset B\otimes M$ ?

The question to the answer is given succinctly here. Essentially, flatness will guarantee it, as you can think of the inclusion as an injective module homomorphism, which will remain an injective morphism after tensoring with a flat module. But that is not all.

My main question is, is it common to use $B\otimes M$ to represent the submodule of $A\otimes M$, generated by elementary tensors of the form $\{b\otimes m: b\in B\subset A, m\in M\}$?

(If it is not clear why this distinction is important, consider the module homomorphism $\mathbb{Z}\to \mathbb{Z}$ given by multiplication by two, then the image of this homomorphism is $2\mathbb{Z}$ as a $\mathbb{Z}$-module, which is the same as $\mathbb{Z}$, so $2\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}$ . However, the elementary tensors mentioned above will just be zero, as $2n\otimes \bar{m}=2(n\otimes\bar{m})=n\otimes\overline{2m}=0$, so the submodule generated is the zero module.)

Because I am quite confident that that is what is happening in the accepted answer of this post. This post has caused me a good amount of confusion precisely because of the example mentioned above.

Answer 1

I disagree with the linked post. To answer your original question first, you should always interpret $B\otimes M$ as the tensor product of $B$ with $M$, not as some misleading notation for a particular subgroup of $A\otimes M$. $B$ itself is a group, so we should not ambiguate the product $B\otimes M$ with some 'relative' notion like $\langle b\otimes m:b\in B,m\in M\rangle\subset A\otimes M$ as this will depend on $A$. However, wearing our categorical hats, what matters in this discussion - as what matters for the $\ker$ discussion - is the associated maps. It is true that the associated map $B\otimes M\to A\otimes M$ has image $\langle b\otimes m\rangle_{b,m}$. However, don't be fooled: the tensor relations for $\otimes$ are different in $B\otimes M$ to $A\otimes M$.

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To discuss the linked post:

Say $f:A\to B$. $\ker(f)\otimes M\not\subset\ker(f\otimes1)$ in general; not in any natural way. There is a natural map $\ker(f)\otimes M\to A\otimes M$ which is killed by $f\otimes1:A\otimes M\to B\otimes M$, which the linked post implicitly shows, but this does not mean $\ker(f)\otimes M\subset\ker(f\otimes1)$. Rather, it just means there is an induced natural map $\ker(f)\otimes M\to\ker(f\otimes1)$ but this map is certainly not injective in general, nor is it surjective!

$\ker(f)\otimes M\to\ker(f\otimes1)$ injects iff. $\ker(f)\otimes M\to A\otimes M$ injects; this holds when $\mathsf{Tor}(\mathrm{im}(f),M)=0$ but is potentially false otherwise. $\ker(f)\otimes M\to\ker(f\otimes1)$ surjects iff. $\mathrm{im}(f)\otimes M\to B\otimes M$ injects and this again need not hold; $\mathrm{im}(f)\otimes M\not\subset\mathrm{im}(f\otimes1)$ in general, in fact $\mathrm{im}(f\otimes1)$ is a quotient of $\mathrm{im}(f)\otimes M$.

Try the map associated to multiplication by $2$, $\Bbb Z/4\Bbb Z\to\Bbb Z/4\Bbb Z$; let's take this to be our $f:A\to B$. Let's also take $M=\Bbb Z/2\Bbb Z$.

$A\otimes M\cong\Bbb Z/2\Bbb Z\cong B\otimes M$ in such a way that $f\otimes 1$ is the zero map. $\ker(f)\otimes M\cong\Bbb Z/2\Bbb Z$ so you might be tempted to say $\ker(f)\otimes M=\ker(f\otimes1)$ but the associated map (and this is what matters; we don't really care if two groups are abstractly and unnaturally isomorphic, we usually want to know the maps that provide the isomorphism; also, categorically "$\ker(f\otimes1)$" carries information about a map, not just an object) $\ker(f)\otimes M\to A\otimes M=\ker(f\otimes1)$ is in fact the zero map; we do not get the natural inclusion. It is neither surjective nor injective! Similarly $\mathrm{im}(f)\otimes M\cong\Bbb Z/2\Bbb Z$ is not a subgroup of $\mathrm{im}(f\otimes1)=0$.