Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am confused about what exactly I have to check or how to check when a problem asks that there is a natural isomorphism.

My confusion arises in the following problem from Lang's Algebra (Chapter XVI Ex 4):

Let $\phi:A\rightarrow B$ be a commutative ring homomorphism. Let $E$ be an $A$-module and $F$ a $B$-module. Let $F_A$ be the $A$-module obtained from $F$ via the operation of $A$ on $F$ through $\phi$, that is for $y\in F_A$ and $A\in A$ this operation is given by $$(a,y)\rightarrow \phi(a)y$$ Show there is a natural isomorphism $$\text{Hom}_B(B\otimes_A E,F)\cong \text{Hom}_A(E,F_A)$$

I am confused firstly about when an isomorphism means in this context because the left hand side is a $B$ module and the right hand side an $A$ module. Secondly I know the definition of natural transformation has to do with functors, but I do not see how this applies or what I have to check to show that an isomorphism is natural in this context.

Thank you all.

share|cite|improve this question
Just to clarify. I am not asking how to do this problem. Just basically what it means and what sort of function I have to construct and what I have to check with it. – user45150 Oct 23 '12 at 16:53
There are two functors here: one sends a pair $(E, F)$ to the LHS of your isomorphism, and one sends $(E, F)$ to the RHS. You can show that the isomorphism is actually in the category of $A$-modules, because every $B$-module is also an $A$-module via the homomorphism $\phi : A \to B$. – Zhen Lin Oct 23 '12 at 17:28
Would I have to check then that it is natural in both those components? – user45150 Oct 24 '12 at 20:09
Sure, because that's the definition. – Zhen Lin Oct 24 '12 at 21:11
That's what I thought. Just making sure. Thanks – user45150 Oct 24 '12 at 21:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.