Fields Construction

Adjoining Roots

Proposition

Any ring homomorphism between two fields is injective.

Proof Suppose that is a ring homomorphism between two fields and such that for some . Then, we have , which implies that since is a field. Thus, , and hence is injective.

Lemma

Let be a field, and let be an irreducible polynomial in . Then the ring is an extension field of , and the residue of is a root of in .

Proof Since, the principle ideal in is maximal if and only if is irreducible, we have that is a field. The residue of in is the equivalence class of modulo . By definition, is a root of in .

Complete Splitting

A polynomial is said to be completely split over a field if it can be factored into linear factors in .

Proposition

Let be a field, and let be a monic polynomial in of positive degree. There exists a field extension of such that splits completely in .

Proof We use induction on the degree of . If , then is linear and has a root in . Thus, we can take . Now, suppose that the statement holds for all polynomials of degree less than , and let be a monic polynomial of degree . By the lemma, there exists a field extension of such that has a root in . Suppose that can be factored as , where is a irreducible polynomial with . By the induction hypothesis, there exists a field extension such that splits completely in , and a field extension of such that splits completely in . Then, the polynomial splits completely in the field extension .

Theorem

Let and be polynomials with coefficients in a field , with , and let be an extension field of . Then the following statements hold:

1. Division with remainder of by gives the same answer, whether carried out in or in .
2. divides in if and only if divides in .
3. The (monic) greatest common divisor of and is the same, whether computed in or in .
4. If and have a common root in , they are not relatively prime in . If and are not relatively prime in , there exists an extension field in which they have a common root.
5. If is an irreducible element of and if and have a common root in , then divides in .

Proof We prove each statement one by one:

1. In , suppose we have , this is also true in since . Meanwhile, division with remainder is unique, so we have in as well. Thus, the statement holds.
2. Directly follows from (1).
3. Let and be the (monic) greatest common divisor of and in and respectively. Then, by definition, . Note that there exist polynomials and in such that , and divides both and in . Thus, . So and are associates in . Since is monic, we have .
4. Suppose that and are not relatively prime in . Then, there exists a non-zero greatest common divisor such that and . By the lemma, there exists a field extension in which has a root, this will also be a root of and . Conversely, if and have a common root in , then is a common divisor of and in , so the greatest common divisor of and is not a unit in . Thus, and are not relatively prime in .
5. If is irreducible, its only monic divisors in are and . (4) tells us that the greatest common divisor of and g in isn’t . Therefore it is .

Multiple Roots of a Polynomial

Theorem

Let be a polynomial with coefficients in a field . An element in an extension field of is a multiple (double) root, meaning that divides , if and only if it is a root of and also a root of .

Proof Suppose that is a root of , say . Then by the product rule of differentiation, we have hence if and only if .

Corollary

Let be a polynomial with coefficients in . There exists a field extension of in which has a multiple root if and only if and are not relatively prime.

Proof If has a multiple root in , then and have a common root in , so they are not relatively prime in or . Conversely, if and are not relatively prime, then they have a common root in some field extension , hence has a multiple root there.

Proposition

Let be an irreducible polynomial in , where is a field. Then has no multiple root in any field extension of unless the derivative is the zero polynomial.

Proof We shall show that and are relatively prime unless is the zero polynomial. Since is irreducible, and are not relatively prime if and only if . And unless , should have degree no less than , however, . Thus, must be the zero polynomial.

Corollary

Let be an irreducible polynomial in , where is a field of characteristic zero, then has no multiple root in any field extension of .

Proof In a field of characteristic zero, the derivative of a polynomial is never the zero polynomial.

Primitive Elements

Primitive Element

Let be a field extension of . An element that generates , i.e. , is called a primitive element of over .

Lemma

Let be a field of characteristic , and let be an extension field that is generated over by two elements and . For all but finitely many , the element is a primitive element of over .

Proof See Artin’s Lemma 15.8.2.

Primitive Element Theorem

Let be a finite extension of a field of characteristic . Then contains a primitive element such that . In other words, is a simple extension of .

Proof Since is a finite extension, is generated by a finite set. In particular, a basis will generate over , say . We apply induction on . There is nothing to prove when . For any , assume the theorem holds for , that is the extension field . So is generated by a primitive element , . Now, we have . By the lemma, there exists a primitive element such that . Thus, the theorem holds.

The Fundamental Theorem of Algebra

Algebraically Closed Field

A field is said to be algebraically closed if every non-constant polynomial with coefficients in has a root in .

Fundamental Theorem of Algebra

is an algebraically closed field.

Proof

Remark

There are other proofs as well. See proof using Rouché’s theorem, proof using algebraic topology.

Proposition

Let be a finite set and . Then is maximal iff for some .

Hilbert’s Nullstellensatz

The maximal ideals in are in bijection with points in .

Theorem

Let be an ideal of generated by , let and let vector space .Then points in are in bijection with maximal ideals in .