TITLE: How to understand "tensor" in commutative algebra?
QUESTION [12 upvotes]: Tensor is sure an important concept in commutative algebra, but the definition is kind of abstract, so is there any way to understand it which is easier?
Thanks advance!
The definition I see is the one defined by modules.

Proposition 2.12. Let $M, N$ be $A$-modules. Then there exists a pair $(T,g)$ consisting of an $A$-module $T$ and an $A$-bilinear mapping $g \colon M \times N \to T$, with the following property:
Given any $A$-module $P$ and any $A$-bilinear mapping $f \colon M \times N \to P$, there exists a unique $A$-linear mapping $f' \colon T \to P$ such that $f = f' \circ g$ (in other words, every bilinear function on $M \times N$ factors through $T$).
Moreover, if $(T,g)$ and $(T',g')$ are two pairs with this property, then there exists a unique isomorphism $j \colon T \to T'$ such that $j \circ g = g'$.

REPLY [5 votes]: Let's start with bilinear maps. In particular, choose a map $f:M\times N\to P$. Now consider the pairs $(u,\alpha v)$ and $(\alpha u,v)$ where $u\in M$, $v\in N$ and $\alpha$ is a scalar. We find that $f(u,\alpha v)=\alpha f(u,v) = f(\alpha u, v)$, and this does not depend on what bilinear map we have chosen. In other words, as far as bilinear maps are concerned, there's no difference between the pairs $(u,\alpha v)$ and $(\alpha u,v)$.
Therefore, a natural question arises: Can we write down an object that captures exactly the features of pairs of vectors that are essential for the bilinear map? That is, can we have a set $T$ and a function $g:M\times N\to T$ so that $g(u,v) = g(u',v')$ exactly if for any bilinear map $f$ we have $f(u,v)=f(u',v')$?
Obviously, if we can find such a function $g$, then we can write any bilinear map as $f = f'\circ g$ where $f'$ maps objects from $T$ to $P$. After all, by definition the objects in $T$ capture exactly those properties of the pairs of vectors which are relevant for $f$.
Now since we are doing linear algebra, we would additionally like $T$ to also be a linear space, and $f'$ to be linear. It is not hard to see that in this case, the function $g$ has to be bilinear.
The theorem you quoted now states that such a function $g$ not only always exists, but moreover is unique up to isomorphisms.