# Tensor Products I - Extending scalars

# Restriction of scalars

Suppose ^{[1]} You can verify this gives **restriction of scalars** from

# Extension of scalars

In light of the above process, one might ask the following:

Is it possible to go the other way? In other words, if

Sadly, not in general, as the example below illustrates:

## Frustrating example

Consider the ring ^{[2]}

Can we try to do the next best thing, which is to embed the original

## Second frustrating example

Suppose

# Category theory insight

In light of the second example, let's widen our scope and instead consider *all* ^{[3]} We are ideally looking for a functor

This looks very similar to the universal property of the free module, and so it is perhaps no surprise that the construction of the mystery module

Before we move on to the actual construction, it's worthwhile to consider the choice we've made above, which is that we're specifically looking for a functor **co-extension of scalars**.

# The construction

Given our ^{[4]} In this abelian group

for

Although the free *does* use the information about their algebraic structures. The first bullet point uses the additive structure in

We denote the resulting quotient group by **tensor product of and over **. Note that by construction it "remembers" the left

It is common to let **simple tensors** (or sometimes **pure tensors**).

In this new notation, our construction of

## The -module structure on

We have constructed the abelian group

As with any action (or function) defined on a quotient group by a formula using a choice of coset representative, we should check that this action is well defined. In other words, we should really verify that for each

is a morphism of abelian groups with

# Properties of our construction

There is a (natural)

We'll add the details of this argument later, but for now we'll simply note that the construction follows from a universal property of the free

Here is a nice corollary of our universal property:

Let

Why is this true? First observe that by the First Isomorphism Theorem the quotient

This implies that

## Suggested next note

More generally, suppose

is any ring morphism. We can then view as an -module using the ring morphism and the action of on , by defining . ↩︎ In fact, the only possible

-action! ↩︎ This is really an example of a forgetful functor, since we are simply forgetting part of the

-action, hence our choice for the letter . ↩︎ If we're being pedantic, we mean the set

, where is the forgetful functor from . ↩︎