# In general, is there a “fast” way to check whether a tensor product is isomorphic to something?

I'm working on some problems in Dummit and Foote on tensor products and can't help but feel I'm missing something.

Often a question will ask to show whether a tensor product is isomorphic to something. Usually I just try to explicitly construct the isomorphism and see if it works. I find it's rather time-consuming to check that it's well-defined, and then if it preserves $+$ and $\cdot$, and then finally make sure that it's a bijection.

Is there a better way to do these sorts of questions, or am I just being lazy?

Edit: Here is what I know about the universal property, perhaps it will be of help:

If we have a bilinear, $R$-balanced map $\psi: M \times N \rightarrow L$, where $M,N$ are $R$-modules, and $L$ is some abelian group, then there is a unique homomorphism $\phi: M \otimes_R N \rightarrow L$ factoring through the inclusion map from $M\times N$ to $M \otimes_R N$. Conversely, if we have $\phi$, then $\psi := \phi(\text{inclusion map})$ is bilinear and $R$-balanced.

How does this help me establishing an isomorphism?

• Isomorphic as what? – edm Nov 2 '17 at 1:09
• Have you tried making use of the universal property? – Ben P. Nov 2 '17 at 1:12
• @edm As modules, I suppose. – Alex Nov 2 '17 at 1:12
• I think this question is overly broad without giving some examples of problems you are attempting. Properties of the tensor product, such as distributivity over direct sums, its behavior with respect to quotients, and base changing polynomial rings are often useful for finding isomorphisms. See exercises 16, 25, and 26 of section 10.4 in Dummit and Foote. – André 3000 Nov 2 '17 at 1:15
• As Ben P. indicated, once you've shown the map $(s, r(x)) \mapsto s r(x)$ is $R$-bilinear, the universal property gives that the induced map $S \otimes_R R[x] \to S[x]$ is a well-defined homomorphism for free. Then you can either construct the inverse explicitly, or you can now use the universal property of polynomial rings (see Lemma 21.3 here) to get a map going the other way. – André 3000 Nov 2 '17 at 1:38

$$A \otimes (B \oplus C) \cong (A \otimes B) \oplus (A \otimes C)$$ $$A \otimes (B \otimes C) \cong (A \otimes B) \otimes C$$ $$R \otimes_R A \cong A$$ $$(R/I) \otimes_R A \cong A / (IA)$$ $$A \otimes \operatorname{colim}_j (B_j) \cong \operatorname{colim}_j (A \otimes B_j)$$ $$\hom(A \otimes B, C) \cong \hom(A, \underline{\hom}(B, C))$$
If you've not seen colimits, an example of what that isomorphism means is that taking the tensor product with $A$ is right exact.