Field of Fractions
Throughout this article, A denotes an integral domain (which may not be a UFD).
The field of fractions of A is an embedding of A into a field K,
such that every element of K can be expressed as , where and .
Prove that the field of fractions of A is unique up to isomorphism.
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.
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.
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 .
Any can be uniquely written as
where , and . We call this the prime factorization of .
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 .
In we have
Here is our theorem of the day.
If is a UFD, so is its ring of polynomials .
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. ♦
Complete the proof by showing that satisfies a.c.c. on principal ideals.
While proving the above theorem, we also showed
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.
Every non-zero prime ideal of a PID is maximal.
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.
Prove that is not a UFD.
[Hint: define some norm function to an easier ring. ]