# Some questions of field of quotients

1. Jun 5, 2010

### logarithmic

I have a few questions on these topics.

1) There's a theorem which says that the cancellation law holds in a ring iff the ring has 0 divisors. Does this mean that there are examples of rings where the cancellation law holds, but don't have multiplicative inverses? i.e. cancellation is more general than multiplying both sides of ab = ac by a^-1, which is how I thought of it in group theory.

2) If D is an integral domain, and it's field of quotients is Quot(D). Then D is a subring of Quot(D), however I don't see how this is true. The elements of Quot(D) are equivalence classes, so I don't see how any element of D can be in Quot(D). I know that every a in D, corresponds to a/1 in Quot(D) by an isomorphism, but how does this imply D is a subset of Quot(D), when the elements are still different (D isn't made up of equivalence classes)?

3) Can someone explain heuristically why Quot(D) is the smallest field containing D?

4) Since Quot(D) is a field, so multiplicative inverses exist for nonzero elements, can we take a/b in Quot(D) to mean a^-1 * b = a * b^-1?

Thanks for any help.

2. Jun 5, 2010

### chingkui

For 1), think about a simple example of integral domain, does cancellation law apply? Does inverse exist?
2) the statement should be understood as "D is isomorphic to a subring in Q(D)". What matter is isomorphism.
3) The smallest field F(D) containing D must contains all 1/a, for any non-zero a in D, and therefore must contains all b/a for any a,b in D.

3. Jun 5, 2010

### Hurkyl

Staff Emeritus
1) (you meant "does not have zero divisors"). Most elements of the ring of integers don't have multiplicative inverses....

2) The construction of Quot(D) comes equipped with a canonical function D -> Quot(D). For most purposes, it is... unpleasant to use an overly strict notion of "subring" that requires the map to be a set theoretic inclusion in addition to being an injective homomorphism.

3) What elements must be in a field containing D?

4) What is the defining property of b-1? Does 1/b have that property? (I assume that was a typo in the equation you wrote)

4. Jun 6, 2010

### logarithmic

Thanks for the help so far.

So the domain D isn't really a subset of Quot(D). If we define a/b := [(a, b)] as the equivalence class of all quotients (a', b') that are equivalent to (a, b) in the sense that a'b = ab'. Then technically D is only isomorphic to D' = { [(a, 1)] | a in D} which is a subring of Quot(D). If we then redefine D so any of its elements a in D is identified with [(a, 1)], i.e. a := [(a, 1)] = a/1, then D really is a subring of Quot(D).

Furthermore, if E is a field that contains (as a subset) D, then E must contain every a/1 in D, therefore E must also contain its multiplicative inverse 1/a, because E is a field. Thus, for any b in D, E must also contain all b*(1/a) = b/a. So E contains everything in Quot(D). Which is why Quot(D) is the smallest field containing D (the redefined D whose elements are a/1).

Also, a/b which is an element of Quot(D), is equal to (a/1) * (b/1)^-1, by field properties of Quot(D), where a/1 and b/1 are elements of the redefined domain D, but not the original D, because b may not have an multiplicative inverse in the original D, which was just an arbitrary domain.

Is that correct?

5. Jun 6, 2010

### Hurkyl

Staff Emeritus
It sounds like you have the idea.

I will point out that most fields that set-theoretically contain D do not set-theoretically contain Quot(D). It's another example of why describing things in terms of homomorphisms is more useful than describing things in set-theoretic terms. Specifically:

Theorem: If $D \to F$ is an injective homomorphism, it has a unique factorization of the form $D \to \text{Quot}(D) \to F$, where the first map is the one that makes Quot(D) the fraction field of D. Furthermore, the map $\text{Quot}(D) \to F$ is monic.​

6. Jun 6, 2010

### Landau

Yes, we should actually say that D can be embedded in Quot(D). More generally, an embedding from (of) a ring R to (into) another ring S is an injective homomorphism
$$f:R\to S.$$
Now the image $f(R)\subset S$ is a subring of S, and R is identified with f(R). You probably know the first isomorphism theorem:
$$R/\ker f\cong f(R).$$
Since f is injective, the kernel is trivial, so $R=R/\ker f\cong f(R).$

In other words, R is isomorphic to f(R), which is a subring of S. So S contains an isomorphic copy of R, and by abuse of language we just say that R "is" a subring of S.

In particular, D is embedded in Quot(D) via the embedding

$f:D\to Quot(D)$ given by $d\mapsto [(d,1)].$