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1 Rings
Theorem 1.1 (Substitution Principle). Let ϕ : R → R′^ be a ring homomor- phism
(a) Given an element α ∈ R′^ there is a unique homomorphism Φ : R[x] → R′^ which agrees with the map ϕ on constant polynomials and sends x → α.
(b) Given elements α 1 , · · · , αn ∈ R′^ there is a unique ring homomorphism Φ : R[x 1 ,... , xn] such that Φ|R = φ and Φ[xi] = αi.
Lemma 1.2. For every ring R there is a unique ring homomorphism Z → R.
Lemma 1.3. If R is a ring and a ∈ R then {ra : r ∈ R} = (a) is an ideal.
Theorem 1.4. A ring R is a field if and only if it has exactly two ideals.
Corollary 1.5. Let F be a field and R a non-zero ring. Then every homo- morphism ϕ : F → R is injective.
Lemma 1.6. Every ideal in Z is principle.
Theorem 1.7. Let g(x) be a monic polynomial in R[x] and let α be an element of R such that g(α) = 0. Then x − α divides g(x).
Theorem 1.8. If F is a field then every ideal of F [x] is principle.
Corollary 1.9. Let F be a field and let f, g ∈ F [x] which are both non-zero. Then there is a unique monic d(x) ∈ F [x] called the greatest common divisor of f, g such that
(a) d generates the ideal (f, g) of F [x] generated by f, g.
(b) d divides f and g
(c) If h is any divisor of f and g then h divides d.
(d) There are p, q ∈ F [x] such that d = pf + qg
Theorem 1.10. Let I be an ideal of a ring R.
(a) There is a unique ring structure on the set R/I such that the canonical map π : R → R/I sending a → a + I is a homomorphism.
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(b) The kernel of π is I.
Theorem 1.11 (Mapping Property of Quotient Rings). Let f : R → R′^ be a ring homomorphism with kernel I and let J be an ideal which is contained in I. Denote R/J by R.
(a) There is a unique homomorphism f : R → R′^ such that f π = f
(b) First Isomorphism Theorem: If J = I then f maps R isomorphically to the image of f.
Theorem 1.12 (Correspondence Theorem). Let R = R/J and let π denote the canonical map R → R
(a) There is a bijective correspondence between the set of ideals of R which contain J and the set of all ideals of R given by
I → π(I) and I → π−^1 (I)
(b) (Third Isomorphism Theorem) If I ⊂ R corresponds to I ⊂ R then R/I and R/I are isomorphic rings.
Definition 1.13. Let R′^ be a ring extension of R and let α ∈ R′. WE define
R[α] = {Σriαi^ : ri ∈ R}
Theorem 1.14. Let R ⊂ R′^ and let α ∈ R′. Then there is a unique map
ϕ : R[x] → R′
such that ϕ is the identity on R and takes x → α. Further F [α] = im[α]
Definition 1.15. C = R[i]
Theorem 1.16. Let R be a ring and let f (x) be a monic polynomial of positive degree with coefficients in R. Let R[α] be the ring obtained by ad- joining an element satisfying f (α) = 0. The elements of R[α] are in bijective correspondence with vectors (r 0 , · · · , rn− 1 ) ∈ Rn^ via a map μ where
μ(r 0 , · · · , rn− 1 ) = r 0 + r 1 α + r 2 α^2 + · · · + rn− 1 αn−^1
Theorem 1.17. Let R be a ring and let a, b ∈ R such that ab = 0. Further let c ∈ R[α] be such that ψ(a)c = 1 where ψ : R → R[α]. Then if R[α] 6 = 0 we must have ψ(b) = 0.
Definition 1.18. We say b ∈ R is a Zero Divisor if there is a non-zero a ∈ R such that ab = 0
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Theorem 1.30 (Hilbert’s Nullstellensatz). The maximal ideals of the poly- nomial ring C[x 1 ,... , xn] are in a bijective correspondence with points of com- plex n-dimensional space. a = 〈a 1 , · · · , an〉 → Ma = (x 1 −a 1 , x 2 −a 2 ,... , xn− an).
Theorem 1.31. If a, b ∈ Z have no factor in common other than ± 1 then there are c, d such that ac + bd = 1.
Theorem 1.32. Let p be a prime integer and let a, b be integers. Then if p divides ab p divides a or p divides b.
Theorem 1.33 (Fundamental Theorem of Arithmetic). Every integer a 6 = 0 can be written as a product
a = cp 1 · · · pk
where c is ± 1 and each pi is prime. And further, up to the ordering this product is unique.
Theorem 1.34. Let F be a field
(a) If two polynomials f, g ∈ F [x] have no common non-constant factors then there are polynomials r, s ∈ F [x] such that rf + sg = 1
(b) If an irreducible polynomial p ∈ F [x] divides a product f g then p divides one of the factors.
(c) Every nonzero polynomial f ∈ F [x] can be written as a product
cp 1 · · · pn
where c ∈ F [x] and the pi are monic irreducible polynomials and n ≥ 0. This factorization is unique except for the ordering of terms.
Theorem 1.35. Let F be a field and let f (x) be a polynomial of degree n with coefficients in F. Then f has at most n roots in F.
Definition 1.36. Let R be an integral domain. If a, b ∈ R we say a divides b if (∃r ∈ R)ar = b
We say that a is a proper divisor of b if b = qa for some q ∈ R and nei- ther q ∈ R and neither q nor a is a unit.
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We say a non-zero element a ∈ R is irreducible if it is not a unit and if it has no proper divisor.
We say that a, a′^ ∈ R are associates if a divides a′^ and a′^ divides a. It is easy to show that if a, a′^ are associates then a = ua′^ for some unit u ∈ R.
Theorem 1.37. Let R be an integral domain.
u is a unit ⇔ (u) = (1)
a, a are associates ⇔ (a) = (a′) a divides b ⇔ (a) ⊃ (b) a is a proper divisor of b ⇔ (1) ) (a) ) (b)
Theorem 1.38. Let R be an integral domain. Then the following are equiv- alent.
(a) For every a ∈ R, a 6 = 0 if a is not a unit then
a = b 1 · · · bn
where each bi is irreducible.
(b) R does not contain an infinite increasing chain of principle ideals
(a 1 ) ( (a 2 ) ( (a 3 ) (
Definition 1.39. We say that an integral domain R is a Unique Factorization Domain (UFD) if
(i) Existence of factors is true for R
(ii) If a ∈ R and a = p 1 · · · pn and a = q 1 · · · qm where pi, qj are irreducible. Then m = n and after reordering pi, qi are associates for each i.
Definition 1.40. Let R be an integral domian. p ∈ R is prime if p 6 = 0 and (∀a, b ∈ R) if p divides ab then p divides a or p divides b.
Theorem 1.41. Let R be an integral domain such that existence of factor- ization holds. Then R is a UFD if and only if ever irreducible element is prime.
Lemma 2.2. Let F ⊂ K be fields with α ∈ K. Then if
ϕα : F [x] → K f (x) √ f (α)
we have α is transcendental if and only if ϕ is injective. Or more specifically if the kernel of ϕ is 0.
Definition 2.3. Let F ⊂ K be fields with α ∈ K. Further let
ϕα : F [x] → K f (x) √ f (α)
We then know that ker(ϕα) is principle as F [x] is a principle ideal domain. So in particular it is generated by a single element fα(x) ∈ F [x].
But because K is a field we must have fα(x) is irreducible (because otherwise K would have a zero divisor) Hence fα(x) is the only irreducible polynomial in (fα(x)) (because every element of the ideal is a multiple of fα(x)) and we call fα the Irreducible Polynomial for α over F.
Definition 2.4. Let F (α) be the smallest field containing both α and F. Similarly let F (α 1 ,... , αn) be the smallest field containing α 1 ,... , αn and F.
Lemma 2.5. Recall that F [α] is the ring
{Σanαn^ : an ∈ F }
and is the smallest ring containing both F and α. We then have F (α) is isomorphic to the field of fractions of F [α]
In particular we have that if α is transcendental then F [x] → F [α] is an isomorphism and hence F (α) is isomorphic to the field F (x) of rational func- tions.
Theorem 2.6. (a) Suppose that α is algebraic over F and let f (x) be its irreducible polynomial over F. The map F [x]/(f ) → F [α] is an iso- morphism and F [α] is a field. Thus F [α] = F (α)
(b) More generally let α 1 ,... , αn be algebraic elements of a field extension K of F. Then F [α 1 ,... , αn] = F (α 1 ,... , αn).
Theorem 2.7. Let α be an algebraic over F and let f (x) be its irreducible polynomial. Suppose f (x) has degree n. Then (1, α,... , αn−^1 ) is a basis for F [α] as a vector space over F
Theorem 2.8. Let α ∈ K and β ∈ L be algebraic elements of two extensions of F. There is an isomorphism of fields
σ : F (α) → F (β)
which is the identity on F and which sends α √ β if and only the irreducible polynomials for α and β over F are equal.
Definition 2.9. Let K, K′^ be field extensions of F. An isomorphism
ϕ : K → K′
which restricts to the identity on F is called an Isomorphism of field extensions of an F -isomorphism
Theorem 2.10. Let ϕ : K → K′^ be an isomorphism of field extensions of F and let f (x) be a polynomial with coefficients in F. Let α be a root of f in K and let α′^ = ϕ(α) be its image in K′. Then α′^ is also a root of f.
Definition 2.11. Let K be a field extension of a field F. We can always regard K as a vector space over F where addition is field addition and mul- tiplication by F is simply multiplication.
We say that the degree of K as an extension of F is the dimension of the vector space (denoted [K : F ]).
Extensions of degree 2 are called quadratic, of degree are called cubic, ect.
The term degree comes from the case when K = F (α) for an algebraic α over F and so (1, α,... , αn−^1 ) form a basis for the vector space (where n is the degree of the irreducible polynomial).
In this case we also call the degree the degree of α over F
Theorem 2.12. If α is algebraic over F then [F (α) : F ] is the degree of the irreducible polynomial of α.
Theorem 2.13. Let F ⊂ K ⊂ L be fields. Then [L : F ] = [L : K][K : F ]. These are called towers of field extensions.
Corollary 2.14. Let K be an extension of F of finite degree n and let α ∈ K. Then α is algebraic over F and it’s degree divides n.
then we define
f ′(x) = nanxn−^1 + (n − 1)an− 1 xn−^2 · · · a 1
Where we interpret n as the image of n ∈ Z under the unique ring homo- morphism Z → F.
It can be shown that things like the product rule hold for these formal deriva- tives.
Lemma 2.24. Let F be a field and let f (x) ∈ F [x]. Let α ∈ F be a root of f (x). Then α is a multiple root, i.e. that (x − α)^2 divides f (x) if and only if α is a root of f (x) and of f ′(x).
Theorem 2.25. Let f (x) ∈ F [x] where F is a field. Then there exists a field extension K of F in which f has a multiple root if and only if f and f ′^ are not relatively prime.
Theorem 2.26. Let f be an irreducible polynomial in F [x]. Then f has no multiple roots in any field extension of F unless the derivative f ′^ is the zero polynomial. In particular if F has characteristic 0, then f has no multiple root.
Definition 2.27. Let K be a field extension of F. Let α 1 ,... , αn be a se- quence of elements of K. We say that α 1 ,... , αn are Algebraically Dependent if there is a polynomial f ∈ F [x 1 ,... , xn] such that f (α 1 ,... , αn) = 0. We say α 1 ,... , αn are algebraically independent otherwise.
Lemma 2.28. Let F ⊆ K. α 1 ,... , αn are algebraically independent if and only if the substitution map ϕ : F [x 1 ,... , xn] → K which takes f (x 1 ,... , xn) to f (α 1 ,... , αn) has ker(ϕ) = 0.
Corollary 2.29. If α 1 ,... , αn are algebraically independent over F then F (α 1 ,... , αn) is isomorphic to F (x 1 ,... , xn) the field of rational functions in x 1 ,... , xn.
Definition 2.30. An extension o the form F (α 1 ,... , αn) where α 1 ,... , αn are algebraically independent is called a Pure Transcendental extension.
Definition 2.31. A transcendence basis for a field K of F is a set of elements α 1 ,... , αn such that K is algebraic over F (α 1 ,... , αn)
Theorem 2.32. Let (α 1 ,... , αm) and (β 1 ,... , βn) be elements in a field ex- tension K of F which are algebraically independent. If K is algebraic over F (β 1 ,... , βn) then m ≤ n and (α 1 ,... , αm) can be completed to a transcen- dence basis for K by adding n − m many of the βi.
Corollary 2.33. Any two transcendence basis for a field extension F ⊆ K have the same number of elements.
2.1 Finite Fields
Definition 2.34. We say that q = pr^ = |K| is the order of a field K. When dealing with finite fields p will always be a prime and q will be the order of the field we are talking about.
Fields with q = pr^ elements are often denoted Fq.
Theorem 2.35. Let p be a prime and let q = pr^ be a power of p with r ≥ 1. Let K be a field with order q.
(a) There exists a field of order q
(b) Any two fields of order q are isomorphic.
(c) Let K be a field of order q. The multiplicative group K×^ of nonzero elements of K is a cyclic group of order q − 1.
(d) The elements of K are roots of the polynomial xq^ − x. This polynomial has distinct roots and it factors into linear factors in K
(e) Every irreducible polynomial of degree r in Fp[x] is a factor of xq^ − x. The irreducible factors of xq^ − x in Fp[x] are precisely the irreducible polynomials in Fp[x] whose degree divides r.
(f ) A field K of order q contains a subfield of order q′^ = pk^ if and only if k divides r.
Corollary 2.36. Let K be a finite field. Then there is an element a ∈ K such that for all b ∈ K, b 6 = 0 there is an n ∈ ω such that an^ = b.
Definition 2.37. A generator for the cyclic group F p× is called an primitive element modulo p.
2.2 Field Extensions
Definition 2.48. If K is a field extension of F we say K/F.
Definition 2.49. An F -automorphism of K is an automorphism of K which is the identity on F.
Definition 2.50. The group of all F -automorphisms of K is called the Galois Group of the field extension (G(K/F ))
Theorem 2.51. For any finite extension K/F the order of |G(K/F )| divides the degree [K : F ] of the field extension.
Definition 2.52. A finite field extension K/F is called a Galois Extension if |G(K/F )| = [K : F ]
Definition 2.53. Let G be a group of automorphisms of K. The set of elements fixed by every element of G is called the fixed field of G
KG^ = {α ∈ K : ϕ(α) = α for all ϕ ∈ G}
Corollary 2.54. Let K/F be a Galois extension with Galois group G = G(K/F ). The fixed field of G is F.
Definition 2.55. Let f (x) ∈ F [x] be a nonconstant monic polynomial. A splitting field for f (x) over F is an extension K of F such that
(i) f (x) factors into linear factors in K : f (x) = (x − α 1 ) · · · (x − αn) with αi ∈ K
(ii) K is generated by the roots of f (x) : K = F (α 1 ,... , αn)
Theorem 2.56. If K is a splitting field of a polynomial f (x) over F then K is a Galois extension of F. Conversely, every Galois extension is a splitting field of some polynomial f (x) ∈ F [x].
Corollary 2.57. Every finite extension is contained in a Galois extension.
Corollary 2.58. Let K/F be a Galois extension and let L be an intermediate field: F ⊂ L ⊂ K. Then K/L is a Galois extension too.
Theorem 2.59. (a) Let K be an extension of a field F , let f (x) be a polynomial with coefficients in F and let σ be an F -automorphism of K. If α is a root of f (x) in K then σ(α) is also a root.
(b) Let K be a field extension generated over F by elements α 1 ,... , αr and let σ be an F -automorphism of K. If σ fixes each of the generators αi then σ is the identity automorphism.
(c) Let K be a splitting field of a polynomial f (x) over F. The Galois group G(K/F ) operates faithfully on the set {α 1 ,... , αr}.
Theorem 2.60 (The Main Theorem). Let K be a Galois extension of a field F and let G = G(K/F ) be its Galois group. The function
H √ KH
is a bijective map from the set of subgroups of G to the set of intermediate fields F ⊂ L ⊂ K. It’s inverse is
L √ G(K/L)
This correspondence has the property that if H = G(K/L) then
[K : L] = |H| hence [L : F ] = [G : H]
Theorem 2.61 (Existence of a primitive element). Let K be a finite exten- sion of a field F of characteristic 0. there is an element γ ∈ K such that K = F (γ).
Definition 2.62. We call an element γ ∈ K such that F (γ) = K a primitive element for K over F.
Theorem 2.63. Let G be a finite group of automorphisms of a field K and let F be its fixed field. Let {β 1 ,... , βr} be the orbit of an element β = β 1 ∈ K under the action of G. Then β is algebraic over F , it’s degree over F is r and its irreducible polynomial over F is g(x) = (x − β 1 ) · · · (x − βr). Further note that r divides |G|.
Corollary 2.64. Let K/F be a Galois extension. Let g(x) be an irreducible polynomial in F [x]. If g has one root in K then it factors into linear factors in K[x].
(i) K is a Galois extension of F.
(ii) K is the splitting field of an irreducible polynomial f (x) ∈ F [x].
(ii’) K is the splitting field of a polynomial f (x) ∈ F [x].
(iii) F is the fixed field for the action of the Galois group G(K/F ) on K.
(iii’) F is the fixed field for an action of a finite group of automorphisms of K.
Theorem 2.75 (Main Theorem). Let K be a Galois extension of a field F and let G = G(K/F ) be its Galois group. the function
H √ KH
is a bijective map form the set of subgroups of G to the set of intermediate fields F ⊂ L ⊂ K. Its inverse function is
L √ G(K/L)
This correspondence has the property that if H = G(K/L) then
[K : L] = |H| and [L : F ] = [G : H]
Theorem 2.76. Let K/F be a Galois extension and let L be an intermediate field. Let H = G(K/L) be the corresponding subgroup of G = G(K/F ). Then
(a) Let σ be an element of G. The subgroup of G which corresponds to the conjugate subfield σL is the conjugate subgroup σHσ−^1. In other words G(K/σL) = σHσ−^1.
(b) L is a Galois extension of F if and only if H is a normal subgroup of G. When this is so, then G(L/F ) is isomorphic to the quotient group G/H
Definition 2.77. Let F ⊆ C be a subfield of C which contains a primitive pth root of unity ζp = e^2 πi/p.
Lemma 2.78. If α is a root of f (x) = xp^ − a then α, ζpα, ζ p^2 α,... , ζ pp −^1 α are the roots of f (x). So the splitting field of xp^ − a is generated by a single root K = F [α]
Theorem 2.79. Let F ⊆ C and let F contain a pth root of unity. Further let a ∈ F be an element which is not a pth power in F. Then the splitting field of f (x) = xp^ − a has degree p over F and its Galois group is a cyclic group of order p.
Theorem 2.80. Let F be a subfield of C which contains a pth root of unity ζp and let K/F be a Galois extension of degree p. Then K is obtained by adjoining a pth root to F.
Theorem 2.81. Let p be a prime integer and let ζp = e^2 πi/p. For any subfield F of C the Galois group of F (ζp) over F is a cyclic group.
Lets consider the Galois group of a product of polynomials f (x)g(x) over F. Let K′^ be a splitting field of f g. Then K′^ contains a splitting field of K of f and F ′^ of g. So we have the following diagram.
K′ ∪ ∪ K F ′ ∪ ∪
Theorem 2.82. With the above notation, let G = G(K/F ) and H = G(F ′/F ) and G = G(K′/F ).
(i) G and H are quotients of G
(ii) G is isomorphic to a subgroup of the product G × H.
