# What is an easier way to think about tensor products of modules?

I'm having trouble understanding the the tensor product of modules. Sure I get the "it's quotient of the free abelian group by the subgroup generated by certain elements" construction, but this offers me no intuition for what it actually is, or how to compute one.

Can someone offer a real no BS way to see the tensor product, and perhaps provide an elementary example? Thank you very much.

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Strictly speaking, the definition is the only real `no BS' way to see it. –  treble Feb 13 '13 at 0:39
The same question has been asked (and answered) on math.SE at least 5 times. Have you used the search function? –  Martin Brandenburg Feb 13 '13 at 0:45

## 2 Answers

Think of tensor product as usual product. An element $a \otimes b$ of $\rm A \otimes \rm B$ can be thought as $ab$.

The idea is the same as for the direct sum, and the symbols are really well-chosen.

A very enlightening example : $$\mathbb R^2 \oplus \mathbb R^3 = \mathbb R^5$$

$$\mathbb R^2 \otimes \mathbb R^3 = \mathbb R^6$$

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A kind of 'abstract product'. –  Berci Feb 13 '13 at 0:05
This makes me wonder a how tensor product is not just all possible sums $a \otimes b$ where $a \in A, b \in B$. –  user62069 Feb 13 '13 at 0:07
Yes, it is. But some of them ought to be equal. For example, it should be like a multiplication: $a\otimes(b+c)=a\otimes b+a\otimes c$. This is not satisfied in the free Abelian group.. –  Berci Feb 13 '13 at 0:18
But! If you fix bases, you basically get a unique form of possible sums $e_i\otimes f_j$ where $e_i$ and $f_j$ are basis elements. –  Berci Feb 13 '13 at 0:20

Tensor product is a kind of cartesian products: we just form pairs.

I prefer the tensor product of modules (acts) over a semigroup, forgetting the additive structure, because it has similar effects, but much easier to define.

Let $S$ be a semigroup and $Y$ a set. A left action of $S$ on $Y$ is just a function $S\times Y\to Y$ (the image of $\langle s,y\rangle$ is written as $sy\in Y$) that satisfies $s_1(s_2y)=(s_1s_2)y$ for all $s_1,s_2,y$. Right action is defined likewise. Modules can also be thought as an action of a ring on an Abelian group.

Now if $X_S$ is a right act and ${}_SY$ is a left act, then their 'tensor product' is defined as $X\otimes_S Y:= X\times Y/\sim$ where $\sim$ is generated by all the $$\langle xs,y\rangle \sim \langle x,sy\rangle$$ pairs, i.e. they will be regarded as equal in the $X\otimes_S Y$ 'tensor set'.

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