In Atiyah Macdonald, for a commutative and unital ring $A$ the tensor product of $A$-modules, $M,N$ is defined by the usual universal property:

A pair $(T,g)$ with $T$ an $A$-module and $g$ an $A$-bilinear map $g\colon M \times N \to T$ such that for any $A$-module $P$ with an $A$-bilinear map $f\colon M \times N \to P$ there exists a unique $A$-module homomorphism $h\colon T \to P$ such that $h \circ g = f$.

And then, as usually goes with universal properties:

Moreover, if $(T,g), (T',g')$ are any two pairs satisfying this condition, then there is a unique isomorphism $j \colon T \to T'$ such that $j \circ g = g'$

My question is, in the second highlighted selection, is the converse true? If I have already established the existence of the usual tensor product, $M \otimes_A N$, via quotient of free module, and I encounter some other pair $(T', g')$ and a unique isomorphism $j: T \to T'$ such that $j \circ g = g'$, can I conclude that $T'$ also satisfies the universal property defining the tensor product, thus I can regard both $T$ and $T'$ as the tensor product of this $M$ and $N$?

I believe I proved that I can indeed do this, but I need reassurance because I have overlooked small details with tensor products before, and don't want to make that mistake again. My proof was as follows:

Suppose $(T,g)$ satisfies the universal property and $(T',g')$ is a pair where $T'$ is an $A$-mod and $g'$ an $A$-bilinear map $g' \colon M \times N \to T'$. Suppose $j$ is the unique isomorphism $j\colon T \to T'$ such that $j \circ g = g'$, or $g = j^{-1} \circ g'$. Suppose $P$ is any $A$-mod with an $A$-bilinear map $f\colon M \times N \to P$. We wish to show there exists a unique $A$-module homomorphism $\phi\colon T' \to P$ such that $\phi \circ g' = f$.

Since $(T,g)$ already satisfies the property by hypothesis, we know there exists a unique $A$-module homomorphism $\varphi \colon T \to P$ such that $\varphi \circ g = f$. So define the map $\phi \colon T' \to P$ by $\varphi \circ j^{-1}$. Then $\phi \circ g' = \varphi \circ j^{-1} \circ g' = \varphi \circ g = f$. So the diagram commutes as desired and the map $\phi$ is defined as the composition of two unique maps so is unique as well, thus $(T',g')$ satisfies the universal property.

Sorry that got a little (unnecessarily) complicated, I don't know how to draw diagrams on here, which really makes everything easier. As an aside, I feel like this probably is true and Atiyah & Macdonald didn't mention it in the text because they deemed it trivial.

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    $\begingroup$ I think this is the first time I've ever seen someone use $\phi$ and $\varphi$ as distinct variable names at the same time. $\endgroup$ – Eric Wofsey Mar 12 '18 at 6:39
  • $\begingroup$ I do it all the time! $\endgroup$ – Prince M Mar 12 '18 at 15:08

This is correct, although your argument for uniqueness of $\phi$ is not very clear: you chose to define it as a composition with certain maps that are unique, but how do you know you couldn't get a $\phi'$ that works and is not given by those compositions? What you want to say is that if $\phi'$ makes the diagram commute for $(T',g')$, then the map $\varphi'=\phi'\circ j$ would make the diagram commute for $(T,g)$ by just swapping the roles of $(T,g)$ and $(T',g')$ in your argument. Therefore by uniqueness of $\varphi$, $\varphi=\varphi'$, and so $\phi'=\varphi'\circ j^{-1}=\varphi\circ j^{-1}=\phi$.

Intuitively, what's going on here is that the isomorphism $j$ says that $(T,g)$ and $(T',g')$ have exactly the same structure (just the elements of $T$ have to be renamed via $j$ to get $T'$). So any property of $(T,g)$ (which is defined using only the module structure on $T$ together with the map $g$) will also be true of $(T',g')$. In particular, this applies to the universal property of the tensor product.

  • $\begingroup$ Thanks Eric! I must admit, I knew my uniqueness justification when I posted was inadequate, and your answer clears it up without question. In your experience, do you feel that this fact I ask about is ever useful? Or worth proving / observing? $\endgroup$ – Prince M Mar 12 '18 at 21:23
  • $\begingroup$ Yes, it's certainly useful. But it's not something you usually consciously make note of; usually you just think of it as an obvious consequence of how isomorphic modules are "the same", as in my second paragraph. $\endgroup$ – Eric Wofsey Mar 12 '18 at 21:38
  • $\begingroup$ Ok, thanks. One last question I had was, I just now wrote the whole argument out to ensure I comprehend the entirety of it and I cannot identify where the uniqueness (w/ respect to commuting diagrams) of the isomorphism j is invoked? I see plainly that this theorem fails if it is not unique, because there will be at least 2 distinct ways to construct the map $/Phi$, but I can’t seem to identify, where in the uniqueness argument, we actually invoke the assumption that j is unique? $\endgroup$ – Prince M Mar 12 '18 at 22:16
  • $\begingroup$ You don't need uniqueness of $j$, just its existence. If $j$ exists, it turns out to automatically be unique. $\endgroup$ – Eric Wofsey Mar 12 '18 at 22:17
  • $\begingroup$ Ah, yes. By the universal property of (T,g). Both j_1 and j_2 would satisfy the conditions so by uniqueness j_1 = j_2. $\endgroup$ – Prince M Mar 12 '18 at 22:36

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