2

Aluffi III.6.8 suggests proving that, for $R$ a ring and $A$ any set, $H = \text{Hom}_R(R^{\oplus A}, R)$ satisfies the universal property for the product of the familiy of $\{R_a\}_{a \in A}$, where $R_a \cong R$ for all $a$.

This question is already asked and partially answered here and here, but since I tried proving that myself before looking on math.SE and got a slightly different proof (one that shows the universal property explicitly, perhaps analogously to the second linked question, but I cannot see this easily), I'm curious if my version is correct.

So, define projection $\pi_a : H \rightarrow R_a$ by $\pi_a(\varphi) = \varphi(I_a)$, where $I_a \in R^{\oplus A}$ is the indicator function ($I_a(x) = 1$ if $x = a$ and $0$ otherwise)1. It's easy to verify that $\pi_a$ is in fact an $R$-module homomorphism.

Then, take arbitrary $R$-module $N$ and a collection of homomorphisms $\varphi_a : N \rightarrow R_a$. We need to define $\sigma : N \rightarrow H$ and show that it's unique.

Commutativity of the corresponding diagram forces $\forall a : A. \forall n : N. \pi_a (\sigma_n) = \varphi_a(n)$, or, expanding the definition for projections, $\sigma_n(I_a) = \varphi_a(n)$. Now, any morphism in $R^{\oplus A}$ can be defined as a finite sum of $I_a$, and the homomorphism condition forces $\sigma_n$ to be defined accordingly, so it exists and is unique, as desired.

Does this look reasonable?

1 Why this specific argument to $\varphi$? Frankly I don't know, it just feels like a good candidate based on the structure of the proofs I've encountered and built so far — it's a simplest nontrivial function useful as a building block of sorts.

0xd34df00d
  • 1,801
  • Any $R$-module homomorphism $R\times R\to R$ is determined by its restrictions to $R\times 0$ and $0\times R$, and any $R$-module homomorphism out of $R$ is determined by where $1$ is sent. In other words, a morphism $R\times R\to R$ is determined by its value on $(1,0)$ and $(0,1)$, which can be anything. This is like saying a linear map is determined by where it sends basis vectors. Viewing a tuple as a function on an index set, these basis vectors are indicator functions, so the $I_a$s are the basis vectors in $R^{\oplus A}$ in general. Does that answer your question? – anon Aug 16 '19 at 02:21
  • I think my remark is rather a rambling about one having to know that any function in $R^{\oplus A}$ is expressible as a finite sum of $I_a$s, otherwise I don't see a way to choose $I_a$s over anything else without this knowledge. But, ultimately, my question is about whether my proof as a whole is reasonable. – 0xd34df00d Aug 16 '19 at 02:24

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