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Given two submodules $U,V \subseteq M$ over a (commutative) ring $R$, and a flat $R$-module $A$, I can interpret $U \otimes_R A$ and $V \otimes_R A$ as submodules of $M \otimes_R A$. Is it necessarily true that $$(U \cap V) \otimes_R A \cong (U \otimes_R A) \cap (V \otimes_R A) ?$$

I think it should be true in many cases, with intuition coming from $\mathbb{Z}$-modules and $A = \mathbb{Q}$, but I'm unsure about what happens in general.

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(My original answer contained a significant error, which is what the comments below refer to.)

Yes, it is true. However, it is not true if the collection of modules is infinite. User26857 gives an example in the comments.

Lemma. Let $M_i$ , $i \in I$, be a finite collection of modules. Then the natural map $$\phi : (\prod M_i) \otimes N \to \prod ( M_i \otimes N)$$ is an isomorphism. (The natural map is the one induced by the universal property of $\prod (M_i \otimes N)$, and $\_ \otimes N$ applied to the maps $\prod M_i \to M_i$.)

Proof. This is because finite products are finite coproducts, and tensor products commute with coproducts.

If $I$ is infinite, then this map is neither injective nor surjective in general.

We can use user26857s example to produce an example when $\phi$ is not injective: $(\prod_{n \geq 0} K[x]/x^n) \otimes K(x) \to \prod_{n \geq 0} (K[x]/x^n \otimes K(x)) = 0$. The LHS is not $0$ because the map $K[x] \to \prod_{n \geq 0} K[x]/x^n$ is injective, so because $K(x)$ is flat, $K(x)$ embeds as a submodule of $(\prod_{n \geq 0} K[x]/x^n) \otimes K(x)$.

An example when $\phi$ is not surjective is $(\prod_{\mathbb{N}} \mathbb{Z}) \otimes \mathbb{Q} \to \prod_{\mathbb{N}} \mathbb{Q}$. The element $(1,1/2,1/4,1/8, \ldots)$ is not in the image.

Proposition. Lei $I$ be finite again. Suppose $M_i$ are submodules of $M$. We identify each $M_i \otimes N$ with a submodule of $M \otimes N$, using that $N$ is flat. We also identify $(\bigcap_{i \in I} M_i) \otimes N$ with a submodule of $N$ in the same way. Then $\bigcap_{i \in I} (M_i \otimes N) = (\bigcap_{i \in I} M_i) \otimes N$, as submodules of $M$.

Proof.

Consider the exact sequence:

$$0 \to \bigcap_{i \in I} M_i \to M \to \prod_{i \in I} M / M_i.$$

Using flatness of $N$, we get another exact sequence:

$$0 \to (\bigcap_{i \in I} M_i) \otimes N \to M \otimes N \to (\prod_{i \in I} M / M_i) \otimes N\qquad (**)$$

We study the map $f: M \otimes N \to (\prod_{i \in I} M / M_i) \otimes N$ by composing it with $\phi: (\prod_{i \in I} M / M_i) \otimes N \to \prod_{i \in I} ((M / M_i) \otimes N)$ from the lemma above.

Claims:

1) The composition $\phi \circ f$ is the natural projection map. Hence the kernel of $\phi \circ f$ is $\bigcap_{i \in I} (M_i \otimes N)$. (Lazy justification -- all maps are induced canonically...)

2) $\ker(\phi \circ f) = \ker(f)$. This is because $\phi$ is injective, as we saw in the lemma.

Now it follows from exactness of the sequence $(**)$ above that, as submodules, $(\bigcap_{i \in I} M_i) \otimes N$ and $\ker f$ agree. This proves the claim.

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    $\begingroup$ You are wrong: the tensor product (with a flat module) doesn't commute with arbitrary intersections. If $R=K[X]$, $N$ is the field of fractions of $R$, and $M_n=(X^n)$, then $\bigcap_nM_n=0$, so $(\bigcap_nM_n)\otimes N=0$ while $M_n\otimes N=N$ for all $n$. (In fact, this example shows that the localization doesn't commute with arbitrary intersections.) $\endgroup$ – user26857 Sep 28 '17 at 9:36
  • $\begingroup$ @user26857 Thanks! Good example. $\endgroup$ – Lorenzo Najt Sep 28 '17 at 14:23
  • $\begingroup$ @user26857 I found my mistake. I had written a sum to denote a general element of the products. Thanks :) $\endgroup$ – Lorenzo Najt Sep 28 '17 at 20:35
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Finite intersections are pullbacks, so in particular are finite limits, and a module is flat iff tensoring with it commutes with all finite limits (exercise).

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