This page covers section 3.6 (“The Field of Quotients of an Integral Domain”). Throughout, is an integral domain and is its field of fractions, defined in “topics covered”.
The section focuses on generalizing the relationship between and (field of fractions) to integral domains in general.
Definition: Ring can be imbedded in ring if there exists an injective ring homomorphism . If both and are unital then we also require the imbedding to map to . If imbeds in then is an extension of .
Theorem 3.6.1: Every integral domain may be imbedded in a field.
The theorem is proven by constructing the field of fractions where if and only if ; here and . This is motivated by equivalence of fractions if and only if . then imbeds in in the natural way, for any non-zero . The notation is shorthand for , the equivalence class under of the element .
The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.
Let with and such that and in . In the rational numbers, this would look like and .
We have and . The two products are the same if . Of course, we have that and so that and the multiplication is well-defined (recall that is commutative because it is an integral domain).
Let . We have
The two forms are equal if and are equal. Multiplying each out, we see that they are both
so that the distributive law holds.
We do not require that be unital, but rather we define for any non-zero . This is well-defined because if is another non-zero element of , then because .
Let with . The map respects ring addition because
where we see by the comments above that . It also respects ring multiplication because
Finally, the map is injective: means that , or . This can be rearranged as . Now, so we must conclude that because is an integral domain.
In , every element has a multiplicative inverse which may or may not exist in . We consider the map given by
We must show that this map is well-defined. If then . Then we see that
which has us conclude that , which is the desired statement. The map is also a homomorphism, because
By now we have exhibited that the field of fractions is homomorphic to a subset of . Last of all, the map is injective because means . Rewriting this as , we see that it’s the condition that and are the same equivalence class.
Therefore any field containing also contains the field of fractions of . In that sense, is the smallest field containing .
(a) Let and .
is reflexive: We have because any element satisfies .
is symmetric: Suppose so there exists with . As is commutative, we also have , which shows that .
is transitive: Suppose and . Then there exist with and . This is slightly tricky. We want to combine the two equations in such a way that we get terms and , and the dependence cancels out. One way to do that is to multiply the first equation by and the second equation by and then add the two equations together. This gives
which shows that .
(b) Let and . We define multiplication on by and see that is closed under it because . Suppose and . Then
but we can show that . If are such that and , then multiplying the first equation by and the second equation by and adding gives the desired result. Therefore multiplication is well-defined.
We define addition on by and again see that is closed under it because . Suppose again that and . Then
and again we can show that with the same technique. Therefore addition is well-defined.
The only other ring axiom to verify is the distributive property, and the proof is identical to that in problem 3.6.2. Thus is a ring as defined.
(c) This part deserves some commentary. First, we note that this problem (3.6.5) is concerned with generalizing the field of fractions. Before, we took and had an equivalence relation if . Now we let be more general and employ a new equivalence relation which makes sense in the context. The other big change is that may have zero divisors. In fact, if we choose to have no zero divisors, then the new, more complicated-looking equivalence relation immediately reduces to the old equivalence relation. This problem is foundational for ring localizations.
The question is now whether imbeds in . It can but need not. If contains no zero divisors, then the map for fixed is an imbedding (i.e. is one-to-one). This is true because means there exists such that . As and has no zero divisors, this implies so the map is injective.
On the other hand, if contains a zero divisor then it may not imbed. We can build some intuition by studying a particular example, such as with . In that case, ; all other elements fall into these equivalence classes. We see that has elements – it is smaller than , so clearly does not imbed in ! The zero divisors in cause some of the fractions to “reduce” in ways that we would not expect.
Judging by the content of the subsequent parts of the problem, it doesn’t seem that a proof of the general criterion for imbedding is called for. It appears to suffice to show that there are situations where imbeds in but also situations where it does not.
(d) Let and fix . We have . Now, because . Also,
Hence is a homomorphism from to . The kernel is the set of elements with . That is, those elements for which there exists with . Recalling the notation of previous sections, we see that
(e) If , then there exists with by the comments of part (d). On the other hand, we know that is closed under multiplication and , so if were in then . Hence is empty.
(f) The multiplicative identity element in is for any . We see this is true because since . As is some sort of generalization of the rational number , the natural thing to consider is
But and, as we have just seen, this is the identity element! Hence any element with is invertible in .
In this way, we have created something like the field of fractions. However, with the looser constraints on and , we end up having only some subset of the construction being invertible.
The subject of the section is the field of fractions of an integral domain, which is a big hint. We know that coprime implies that there exist such that . The natural thing to do is say that . If and are both positive then this is fine, but they cannot be as we will see momentarily. Negative powers are tricky in because we do not assume elements to have multiplicative inverses.
If or is then the result is trivially true, so we will consider which implies that (1) neither nor is zero, and (2) one of or is positive and the other is negative. We may assume without loss of generality that and . Then we prefer to write
In the field of fractions of , we may state the fact that
which follows directly from the hypotheses of the problem. This implies that
Now use that and subtract to find
Because is an integral domain, we have either or . The latter case is easily shown to imply that . Therefore, .
We have no notion of field of fractions or even ring localization (problem 3.6.5) for an arbitrary ring. However, the argument presented above was only suggested by the field of fractions and didn’t actually rely on the construction. Therefore we again take as before, with and and . We have
As before, we use that and subtract to find
which again forces us to conclude that .