Why is the tensor product of modules unique (up to isomorphism)?

Question

I read the definition of tensor products of modules which can also be found here. I was wondering why exactly it is unique up to isomorphism.

Let's say you have found a tensor product of two modules $M\otimes_RN$ and a balanced product $\otimes:M\times N \to M\otimes_RN$. Why can't you expand $M\otimes_RN$ to a bigger abelian group $G$ with the balanced product $\otimes:M\times N \to M\otimes_RN\hookrightarrow G$ (using the natural inclusion of $M\otimes_RN$ into $G$)? Since apparently the balanced product doesn't need to be surjective. Everything in the definition of the tensor product seems to only rely on $\text{Im}(\otimes)$ and not on $M\otimes_RN$.

Answer 1

The key part of the definition is that for any balanced product $f\colon M\times N\to H$ there is unique group homomorphism $\bar f\colon M\otimes N\to H$ such that $\bar f \circ \otimes = f$.

So, let's assume that $G$ has that universal property. Denote $i\colon M\otimes N\hookrightarrow G$ and $\otimes' = i\circ \otimes$. Now, $\otimes\colon M\times N\to M\otimes N$ is a balanced product so by the universal property there is unique group homomorphism $j\colon G\to M\otimes N$ such that $j\circ \otimes' = \otimes$.

Note that $\otimes'\colon M\times N\to G$ is a balanced product so there exists unique group homomorphism $f\colon G\to G$ such that $f\circ \otimes' = \otimes'$ by the universal property. However, we have $(i\circ j)\circ \otimes' = i\circ\otimes =\otimes'$ and $\mathrm{id}_G\circ \otimes' =\otimes'$, so by the uniqueness, $i\circ j = \mathrm{id}_G$.

Similarly, $\otimes\colon M\times N\to M\otimes N$ is a balanced product, so like above, we have $(j\circ i)\circ \otimes = j\circ \otimes' = \otimes$ and $\mathrm{id}_{M\otimes N} \circ \otimes = \otimes$, and by uniqueness we have $j\circ i = \mathrm{id}_{M\otimes N}$.

Therefore, $G\cong M\otimes N$.

Answer 2

Let's just quickly recap the universal property of this kind of tensor product.

1. We have a balanced product $\otimes : M \times N \to M \otimes_R N$ which means that it preserves addition in each variable and $\otimes(x \cdot r, y) = \otimes(x, r \cdot y)$,
2. For every abelian group $G$ and every balanced product $f : M \times N \to G$, there is a unique $\hat f : M \otimes_R N \to G$ such that $\hat f \circ \otimes = f$.

What you suggest is that if $M \otimes_R N$ is contained as a subgroup in a larger abelian group $G$, then this larger group also satisfies the universal property. Let's check.

1. A balanced product $\odot : M \times N \to G$. You define this to be $\otimes$ followed by the inclusion (we'll call it $i$) into $G$. This should work because addition is the same in this larger group $G$.

2. Let $H$ be any abelian group and $f : M \times N \to H$ be a balanced product. We want there to be a unique $g : G \to H$ such that $g \circ \odot = f$.

We can define $g$ on the elements of the form $\otimes(x, y)$, since those have to agree with $f$: $g(\otimes(x, y)) = f(x, y)$. Moreover, we can use the linearity of g to extend it to sums of such basic elements. But remember that $G$ doesn't consist of just those elements: we wanted it to be larger than just $M \otimes_R N$. So can we always extend $g$ to all of $G$ in a unique way?

The answer is no. Let $H$ be $G / (M \otimes_R N)$ and $f$ be $q \circ \odot$, where $q$ is the quotient map $G \to G / (M \otimes_R N)$. You can check that this $f$ is a balanced product if $\odot$ is.

Then both $0$ and $q$ are both maps satisfying (2), but they aren't equal if $G \neq M \otimes_R N$.

$q$ trivially satisfies the equality $q \circ \odot = q \circ \odot$. That we have $0 \circ \odot = q \circ \odot$ is less trivial, but follows from the definition of $\odot$. $q \circ \odot(x, y) = q \circ i \circ \otimes(x, y) = 0$ since $\otimes(x, y)$ is in $M \otimes_R N$. Thus, $q \circ \odot$ is the zero map, and so is equal to $0 \circ \odot$.

The contrapositive of this argument is essentially the one given for the proposition at the bottom of the Definition section in your link.