15

Could someone provide me a link to the proof of the adjointness of Hom and Tensor. I did an extensive google search but could not find anything self contained that presented the proof in full generality (or at least the generality I know). Let $R\to S$ be a ring homomorphism, let $M,N$ be $S$-modules and $Q$ an $R$-module. Then, we have $$\textrm{Hom}_R(M\otimes_S N,Q) \cong \textrm{Hom}_S(M,\textrm{Hom}_R(N,Q)$$

Stefan Hamcke
  • 27,733
Milo Kunis
  • 239

1 Answers1

14

Let $f \in \operatorname{Hom}_R(M\otimes_S N,Q)$. We define $g \in \operatorname{Hom}_S(M, \operatorname{Hom}_R(N,Q))$ by:

$$g(m)(n)=f(m \otimes n)$$

Similarly, if $g$ is defined, we can easily define $f$.

I'll leave it to you to prove that this map between $f$ and $g$ actually goes to the appropriate sets, but this is the basic argument. As mentioned by one of the comments, it does depend on how you define the tensor product.

Thomas Andrews
  • 177,126
  • Can you give me more details about how it depends of the tensor product definition? I consider that the rings are commutative. – pmtm Aug 28 '21 at 22:05
  • The definitions are so different, it requires some work to prove they are equivalent. One is in terms of universal properties, the other is a complicated technical definition in terms of a quotient of a free module on the set $M\times N$ . Proving things using these definitions can be very different. – Thomas Andrews Jan 12 '22 at 18:27
  • @ThomasAndrews defining the tensor product in terms of the quotient of a free module is like defining an even perfect number to be a number of the form $p+1\choose 2$ for a prime $p$ one less than a power of $2$. It's not logicaly incorrect, but it's in poor taste to call it a definition instead of a construction :-) – Vladimir Sotirov Jan 14 '22 at 20:54
  • 2
    Not really - using generators and relations is often how we define groups. The other definitions I know use universal properties, so you need to then prove the object exists. The free group modulo a subgroup is the easiest and most direct construction to get the properties we want. @VladimirSotirov – Thomas Andrews Jan 14 '22 at 21:03
  • @ThomasAndrews Using primes and their prescribed exponents is often how we define numbers too :-) More seriously, a definition in terms of generators and relations is also a definition using a universal property, namely the one resulting from combining that of a free group (or other free gadget) with that of a quotient. So good news: it turns out all definitions of the tensor product you know use universal properties. – Vladimir Sotirov Jan 14 '22 at 21:58
  • @VladimirSotirov not clear if you are being deliberately obtuse or just accidentally. (1) we most definitely define groups in terms of quotients of free groups. We have a whole notation for it. (2) while universal properties are cool, and I like them, they are sometimes a bit much for beginners. I knew about free abelian groups, subgroups, subgroups generated by a set, and quotients long before I learned anything about universal properties and the like. I like universal properties, and think of tensor products primarily in terms of the universal property. But is it the way to teach them? – Thomas Andrews Jan 14 '22 at 22:16
  • @ThomasAndrews if you're going to insult me, then I've nothing more to say to you, so I'll reiterate my points. There is only one definition of the tensor product: in terms of its universal property. The tensor product's construction as a quotient of a free module, while sometimes called a definition, is more accurately called a construction. That the result of the construction satisfies the univesal property of tensor products follows easilly from the univesal properties of free groups and quotients. – Vladimir Sotirov Jan 15 '22 at 00:06