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?
Yes. The tensor product satisfies many algebraic identities, such as
$$ A \otimes (B \oplus C) \cong (A \otimes B) \oplus (A \otimes C) $$ $$ A \otimes (B \otimes C) \cong (A \otimes B) \otimes C$$ $$ R \otimes_R A \cong A$$ $$ (R/I) \otimes_R A \cong A / (IA)$$ $$ A \otimes \operatorname{colim}_j (B_j) \cong \operatorname{colim}_j (A \otimes B_j)$$ $$ \hom(A \otimes B, C) \cong \hom(A, \underline{\hom}(B, C)) $$
and furthermore, each of these isomorphisms is not merely "there exists an isomorphism", but the relevant theorems give a specific isomorphism.
With some basic calculations in your toolbox, you can use algebra to compute different isomorphic representations of tensor products.
If you've not seen colimits, an example of what that isomorphism means is that taking the tensor product with $A$ is right exact.