I am working with the the tensor product $-\otimes_R -$ over some noncommutative ring $R$. Is the tensor product always commutative if $R$ is commutative? (That is: Is it true that $M \otimes_R N \cong N \otimes_R M$ for all $M$ and $N$?) If so, can the tensor product be commutative if $R$ is noncommutative?
Abstract Algebra – When Is the Tensor Product Commutative?
abstract-algebranoncommutative-algebratensor-products
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The tensor product of two Noetherian modules is indeed Noetherian. Even better, if $L$ is finitely generated and $N$ is Noetherian, then $L \otimes_R N$ is Noetherian.
Because $L$ is finitely generated, there is an exact sequence $R^m \to L \to 0$ for some integern $m \geq 0$. Applying the right exact functor $-\otimes_R N$ yields an exact sequence $R^m \otimes_R N \to L \otimes_R N \to 0$. But $R^m \otimes_R N \cong N^m$ is Noetherian, so its homomorphic image $L \otimes_R N$ must also be Noetherian.
(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.
Best Answer
The tensor product's commutativity depends on the commutativity of the elements. If the ring is commutative, the tensor product is as well. If the ring R is non-commutative, the tensor product will only be commutative over the commutative sub-ring of R. There will always be tensors over the ring that will not commute if R is non-commutative.