We will mostly adopt the Bourbaki definitions for the preliminary material. An ideal in a ring is *prime* if is an integral domain. Two ideals are *relatively prime* if A *multiplicative subset* of a set is a submagma of In a ring, we assume it is a submagma with respect to multiplication of course. We then have the canonical extension of by formal inverses of written (although I will use the notation to reinforce the conceptualization of a ring extension).

Now suppose is an -module. A natural question to ask is “can we induce some -module”? The natural choice will be the -module where we define

for

**Proposition 1.** Let be a ring and be a multiplicative subset. Then there exists a ring and homomorphism such that

- Elements of are invertible in
- For every homomorphism such that elements in are invertible in there exists a unique homomorphism such that

It naturally turns out that and for some Similarly we have:

**Proposition 2.** Let be a ring, be a multiplicative subset, and be an -module. Then there exists an -module and homomorphism such that

- For every the map defined by is bijective.
- For every -module such that the map is bijective for every and if we have a homomorphism there exists a unique homomorphism such that

Here we can show that and is called the **ring of fractions of ** by while often denoted for short, is called the **module of fractions of** by

**Definition 3.** Let be an ideal in We define the **radical** of in as

We call the **nilradical** of is a **reduced ring** if

Let be an -algebra and be a multiplicative subset of Then since is an -module, we can make into an -module

But we can also turn it into an -algebra. We need a homomorphism

But the domain collapses to and then to under the product of So we have the canonical product we need.

[1] Bourbaki, N. *Commutative Algebra, Ch 1-7*. Springer-Verlag. 1989.