# Field of Fractions

Throughout this article, *A* denotes an integral domain (which may not be a UFD).

Definition.The

field of fractionsof A is an embedding of A into a field K,such that every element of K can be expressed as , where and .

**Exercise A**

Prove that the field of fractions of *A* is unique up to isomorphism.

**Examples**

1. is a field of fractions of .

2. is a field of fractions of .

3. A field of fractions of (*k* = field) is given by the set of all where and .

The third example strongly hints at a general construction.

Proposition 1.Every domain A has a field of fractions.

The trick is to “formally” invert elements. Although the proof is long, most of it is tedious mechanical work and not particularly illuminating.

**Proof**

We take the set of all pairs , together with the relation:

This is an equivalence relation: to check transitivity, if and where , then and since we have .

Write for the equivalence class of and let *K* be the set of equivalence classes. We define addition and product on *K* as follows.

Now show that these operations are well-defined, e.g. for addition if then , which follows from

We leave it to the reader to verify that *K*, with the above operations, forms a commutative ring with 1. It is a field since if , then so . ♦

# Field of Fractions of UFD

Let *A* be a UFD and *K* its field of fractions. We say that are **associates** if is a unit in *A*.

Now fix a set of primes modulo associates. For example, in ℤ, we can take .

Proposition 2.Any can be uniquely written as

where , and . We call this the

prime factorizationof .

**Proof**

Write where . Each of *a* and *b* can be written as where *u* is a unit, and . Expanding and removing terms with exponent 0 gives us an expression as above.

For uniqueness, suppose where . Without loss of generality assume . If , then . The LHS is a multiple of in *A* but the RHS is not; this gives a contradiction. Hence . This holds for the remaining terms as well. ♦

Now we can extend the following notions for . Let

be their prime factorizations.

**Divisibility**: we write if , equivalently for each .**Gcd**: write .**Lcm**: write .

As before if , then for each *i* if and only if . Similarly, if , then for each *i* if and only if .

**Example**

In we have

# Gauss’s Lemma

Here is our theorem of the day.

Theorem (Gauss).If is a UFD, so is its ring of polynomials .

**Proof**

We say is **primitive** if its coefficients have gcd 1.

**Step 0. Any prime is also prime in .**

Because is an integral domain.

**Step 1. The product of two primitive polynomials is primitive.**

Let and be primitive.

- Let be any prime; it suffices to show there is a coefficient of
*fg*not divisible by . - Since
*f*is primitive there is a maximum*d*such that ; similarly, there is a maximum*e*such that . - The coefficient of in
*fg*is not a multiple of since it is:

**Step 2. Extend A into a field.**

Let *K* be the field of fractions of *A*. For any , let be the gcd of the coefficients of and . Note that since divides every coefficient of . Also, the gcd of all coefficients of is 1 so we have:

This expression is clearly unique: if where and is primitive, we have and .

**Step 3. Prove that c and p are multiplicative.**

To show that and , write

Then . By step 1, is primitive; hence and .

**Step 4. Every irreducible non-constant f in A[X] is irreducible in K[X].**

Let be non-constant and irreducible. Note that it must be primitive since a prime element of is also prime in (step 0). If is reducible in we have where are not constant. Then gives a factorization in as well, which is a contradiction.

**Step 5. Every irreducible f in A[X] is prime.**

First suppose is non-constant and irreducible.

Suppose where . Then since is primitive. Hence we may assume are primitive. Since is a PID, one of is a multiple of in (by step 4). Without loss of generality write for some . But now so in fact is a multiple of in .

Finally if is constant and irreducible, it is irreducible in *A*. ♦

### Exercise B

Complete the proof by showing that satisfies a.c.c. on principal ideals.

# Consequences

While proving the above theorem, we also showed

Proposition 3.Let A be a UFD and K its field of fractions. Then is irreducible in if and only if either

- is a prime element of A, or
- , is primitive, and irreducible in .

Thus and are UFDs for any field *k*. Looking at the spectra of these rings, we see a striking difference between PIDs and general UFDs.

Proposition 4.Every non-zero prime ideal of a PID is maximal.

**Proof**

Suppose we have ideals of *A* with a non-zero prime. Since *A* is a PID we write and where . Then for some . Since is irreducible either or is a unit. In the former case, ; in the latter case . Hence is maximal. ♦

In particular, if *A* is a PID and not a field, its Krull dimension is 1.

In contrast, each is a UFD of Krull dimension at least *n*. Thus *UFDs are structurally much more complicated than PIDs*. They can be of arbitrarily high dimension.

### Exercise C

Prove that is not a UFD.

[Hint: define some norm function to an easier ring. ]

In the lemma of the section on consequences, it is mentioned that is primitive. I guess you mean is primitive and degree f >0.

Thanks. Corrected. 🙂

Can you please post the solution to Exercise C?

Here’s a hint. Use the fact that is a UFD. In the extension , define a norm map, similar to the case of .

Do we need A to be a PID in proposition 3? I suppose it is sufficient that A is a UFD.

Yup I think you’re right.