In this post, we consider generalizations of the real and complex number systems.
Definition 1. A field is a set equipped with two binary operations and such that the following properties hold:
(FA1) addition is commutative, viz., for all ;
(FA2) multiplication is commutative, viz., for all ;
(FA3) addition is associative, viz., for all ;
(FA4) multiplication is associative, viz., for all ;
(FA5) addition and multiplication are distributive, viz., and for all ;
(FA6) an additive identity exists, so that for all ;
(FA7) a multiplicative identity exists, so that for all ;
(FA8) addition is invertible, viz., each admits such that ;
(FA9) multiplication is invertible, viz., each admits such that .
The set of real numbers with usual addition and multiplication is a field. Similarly, the set of complex numbers and the set of rational numbers with usual addition and multiplication are also fields.
Note that a field is an ordered quintuplet of a set , two binary operations and on , and two identity elements and such that and are abelian groups, and that the dstributive property (FA5) holds.
If there is no danger of confusion, we simply write 0 and 1 for and , respectively. Moreover, given and , we write to denote the -fold sum .
Remark 2. is the unique additive identity, and is the unique multiplicative identity. Indeed, if is another additive identity, then . Similarly, if is another multiplicative identity, then .
Remark 3. Additive inverses and multiplicative inverses are unique. To this end, we fix . If and are additive inverses of , then
Similarly, if and if and are multiplicative inverses of , then
Remark 4. . Indeed, (FA7) implies that . (FA9), in conjunction with Remark 3, shows that must be an element of .
Remark 5. for all . indeed,
by (FA5), and subtracting from each side yields . Similarly, .
Remark 6. for all . Observe that
by (FA5) and Remark 5. Similarly, and it follows from Remark 3 that . An analogous argument shows that .
Exercise 7. A ring is a set with two binary operations and on such that (FA1), (FA3), (FA4), (FA5), (FA6), (FA7), and (FA8) hold true. Show that the properties established in Remark 2 through Remark 6 remain valid for rings.
Note that , , amd all share the same addition and multiplication operations. In what sense is a “substructure” of , or a “superstructure” of ? We formalize these notions below.
Definition 8. A subfield of a field is a subset of such that
(2) and whenever ;
(3) and whenever .
Definition 9. A field extension of a field is a field that contains as a subfield.
Example 10. has no proper subfield. Indeed, every subfield of must contain all the integers and their multiplicative inverses. Therefore, for all . Moreover, must be contained in , whence . It follows that .
Example 11. is a field extension of .
is quite a special one, in two respects. Firstly, with the usual less-than-or-equal-to ordering is the order-theoretic completion of . Indeed, is the smallest order-superstructure of such that the least-upper-bound property holds. A typical construction is given by the method of Dedekind cuts.
We remark that the method of Cauchy completion works for non-standard metrics on as well. See the Wikipedia article on -adic numbers for an important collection of fields we can obtain from completing certain non-standard metrics on .
Example 12. is a field extension of . In fact, is the algebraic closure of , i.e., the smallest field extension of such that every polynomial with coefficients in has a root in . That is an algebraically closed field is precisely the content of the fundamental theorem of algebra. To see that is the smallest such field, we consider the following general fact.
Let be a field extension of . Since is closed under addition and whenever and , we see that is a vector space over . Therefore, we can talk about the -dimension of . For example, the -dimension of is 2. Now, if is any field extension of that is also a subfield of , then must be a -vector subspace of . Therefore, the -dimension of is either 1 or 2. If , then . If , then . It follows that is the smallest proper field extension of . This, in particular, shows that is the algebraic closure of .
Example 13. A field extension of such that (see Example 12) is referred to as an algebraic number field. It can be shown that all such extensions of -dimension 2, called quadratic fields, are algebraic number fields of the form
where is a squarefree integer, viz., , , and the prime factorization of does not repeat any prime. If , then is said to be a real quadratic field; if , then is said to be an imaginary quadratic field.
Example 14 (Modular arthmetic). So far, every example of a field we have considered was infinite. Let us now construct finite fields. For each , we define the following equivalence relation on the set of integers : if and only if is divisible by . We denote by the quotient set and define the addition and multiplication operations by setting and for all .
Let us check that these operations are well-defined, viz., they do not depend on the choice of representatives. To this end, we assume that and . We can then find such that and . Observe that
and so . Therefore, . Similarly,
and so . It follows that .
Note that modular addition is always invertible, as is the additive inverse of . Most integers, however, do not have multiplicative inverses, and so invertibility of modular multiplication is a trickier matter.
We claim that is not a field if is not a prime. To see this, we suppose that is a positive prime number strictly less than that divides . We can then find such that . If for some , then for some , so that . But , and so does not admit a multiplicative inverse in . This is absurd, and we conclude that has no multiplicative inverse in .
Let us now show that is a field whenever is prime. In order to verify this claim, we shall make use of the Euclidean algorithm. Fix a nonzero integer . The Euclidean algorithm furnishes integers and such that . Since is assumed to be prime, . We now observe that
whence is the multiplicative inverse of .
Exercise 15. An integral domain is a ring (see Exercise 7) such that (FA2) holds (see Definition 1) and that, for all , implies either or . Show that is an integral domain but not a field. Show also that is an integral domain if and only if it is a field, i.e., when is prime. If you feel ambitious, prove that every finite integral domain is a field.
Note that in , even though there is no integer such that in . This difference is captured by the following notion:
Definition 16. A field is of characteristic in case is the smallest positive integer such that . If no such positive integer exists, then we say that is of characteristic zero. We write to denote the characteristic of .
Note that is of characteristic zero, and that is of characteristic . These are representative of all fields, as far as the computation of characteristics is concerned.
Proposition 17. Every field of positive characteristic must be of prime characteristic.
Proof. Let . We suppose for a contradiction that for some integers . Since and , we can find elements and of such that and , respectively. Now,
which is absurd. We conclude that is of prime characteristic.
Proposition 17, in tandem, with the argument given in Example 14, shows that the subset
of forms a subfield of . In other words, a field of positive characteristic contains a copy of . We formalize this notion below.
Definition 18. A field homomorphism is a function between two fields and that satisfies the following properties:
(1) for all ; (2) for all ; (3) ; (4) .
A field isomorphism is a bijective field homomorphism. We say that is isomorphic to if there exists a field isomorphism . In this case, we write .
We shall return to the study of the subfield at the end of this post. For now, we show that field homomorphisms are structure-preserving.
Proposition 19. The image of a field homomorphism is a subfield of the codomain field.
Proof. Let be a field homomorphism. Evidently, and are in . We fix and find such that and . Observe that
Similarly, . We also note that
whence . Similarly, . We conclude that is a subfield of .
Unlike group homomorphisms, field homomorphisms are quite rigid.
Proposition 20. Every field homomorphism is injective.
Proof. Let be a field homomorphism. We define the kernel of to be the set
We first show that is injective if and only if . If is injective, then the kernel is clearly trivial. Conversely, we suppose that the kernel is trivial and fix such that . Note that , so that . It follows that .
Let us now define an ideal of to be a subset of such that and imply , and that implies . We show that has only two ideals: and . Indeed, if is a nonzero element of an ideal , then , and so for all .
Observe that is an ideal of . Indeed, if and , then
so that . Moreover, if , then
and so . Now, , and so . Since is an ideal, it follows that , whence is injective.
As a simple application, we establish the following property of characteristics.
Corollary 21. If there exists a field homomorphism , then .
Proof. By Proposition 20, is an injective field homomorphism. If , then for all . Since is injective,
is never for any . We conclude that .
If is of positive characteristic, then an analogous reasoning implies that
is nonzero for all and is zero when . It follows that .
What the above corollary tells us is that it is necessaryto have matching characteristics in order for field homomorphisms to exist. We now show that, if is either or , then it is sufficient to have matching characteristics to ensure the existence of field homomorphisms. To lay out the proper context for this result, we introduce the notion of the prime subfield, or the subfield generated by .
Definition 22. Let be a field and let be a subset of . The subfield of generated by is the smallest subfield of that contains , viz., the intersection of all subfields of that contains . If , then the subfield generated by is referred to as the prime subfield of .
We conclude the post by establishing the promised sufficiency result:
Proposition 23. Let be a field. If , then the prime subfield of is isomorphic to . If for some prime , then the prime subfield of is isomorphic to .
Since we have shown in Proposition 17 that the characteristic of a field must be either 0 or prime, this covers all possible cases.
Proof. Let be a field and let be the prime subfield of . If , then the mapping is a field homomorphism from into . Since , we see that is a field extension of . Now, by Proposition 20. Since contains no proper subfield (Example 10), it follows that .
If , then the mapping is a field homomorphism from into . Analogously as above, we can conclude that .