Download Extension fields I

Extension fields I
Sergei Silvestrov
Spring term 2011, Lecture 12
Abstract
Contents of the lecture.
+ Introduction to extension fields.
+ Vector spaces.
Extension fields and Kronecker’s theorem
Definition 1. If E is a field containing F as a subfield, then E is called a extension field of F .
Theorem 1 (Kronecker). if F is a field and f (x) ∈ F[x] is a nonconstant polynomial, then there exist
an extension field E of F and an α ∈ E with f (α) = 0.
Proof. If the degree of f is 1, then f (x) is linear and we can choose E = F . If the degree of f is
greater than 1, write f (x) = p(x)g(x), where p(x) is irreducible. The quotient ring E = F[x]/⟨p(x)⟩
is a field. The natural map ϕ(a) : E 7→ F defined by ϕ(a) = a + ⟨p(x)⟩, is an isomorphism from F
to the subfield F ′ = { a + ⟨p(x)⟩ : a ∈ F } of E .
Put α = x + ⟨p(x)⟩ ∈ E . Let p(x) = a0 + a1 x + ⋅ ⋅ ⋅ + ad−1 xd−1 + xd , where ai ∈ F for all i. In
E = F[x]/⟨p(x)⟩, we have
p(α) = (a0 + ⟨p(x)⟩) + (a1 + ⟨p(x)⟩)α + ⋅ ⋅ ⋅ + (1 + ⟨p(x)⟩)α d
= (a0 + ⟨p(x)⟩) + (a1 + ⟨p(x)⟩)(x + ⟨p(x)⟩) + ⋅ ⋅ ⋅ + (1 + ⟨p(x)⟩)(x + ⟨p(x)⟩)d
= (a0 + ⟨p(x)⟩) + (a1 x + ⟨p(x)⟩) + ⋅ ⋅ ⋅ + (1xd + ⟨p(x)⟩)
= a0 + a1 x + ⋅ ⋅ ⋅ + xd + ⟨p(x)⟩
= p(x) + ⟨p(x)⟩ = ⟨p(x)⟩,
because p(x) ∈ ⟨p(x)⟩. But ⟨p(x)⟩ = 0 + ⟨p(x)⟩ is the zero element of E = F[x]/⟨p(x)⟩, and so α
is a root of p(x).
– Typeset by FoilTEX –
Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Example 1. The polynomial x2 + 1 ∈ ℝ[x] is irreducible, and so K = ℝ[x]/⟨x2 + 1⟩ is a field extension. If α is a root of x2 + 1, then α 2 = −1; moreover, every element of K has a unique expression
of the form a + bα , where a, b ∈ ℝ. Clearly, this is another construction of ℂ.
Algebraic and transcendental elements
Definition 2. Let E be an extension field of F . An element α ∈ E is algebraic over F if there is
some nonzero polynomial f (x) ∈ F[x] having α as a root; otherwise, α is transcendental over F .
Example 2. i ∈ ℂ is algebraic over ℚ and over ℝ. It is a nontrivial fact that π , e ∈ ℝ are transcendental over ℚ.
Theorem 2. Let E be an extension field of F . An element α ∈ E is transcendental over F if and
only if the evaluation homomorphism ϕα : F[x] 7→ E is a one-to-one map.
Proof.
2.
1. The element α is transcendental over F ⇔
f (α) ∕= 0 for all nonzero f (x) ∈ F[x] ⇔
3. ϕα ( f (x)) ∕= 0 for all nonzero f (x) ∈ F[x] ⇔
4. Ker(ϕα ) = {0} ⇔
5. ϕα is one-to-one.
The irreducible polynomial for α over F
Theorem 3. Let E be an extension field of F and let α ∈ E be algebraic over F . There exist a
unique irreducible monic polynomial f (x) ∈ F[x] such that
1.
f (α) = 0.
2. If g(x) ∈ F[x] and g(α) = 0, then f divides g.
Definition 3. Let E be an extension field of F and let α ∈ E be algebraic over F . The polynomial
described in Theorem 3 is called the irreducible polynomial for α over F and is denoted by
irr(α, F). The degree of irr(α, F) is the degree of α over F , denoted by deg(α, F).
– Typeset by FoilTEX –
1
Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
√
√
√
n
Example
3. irr( 2, ℚ) = x2 − 2 and deg( 2, ℚ)√= 2. Similarly, irr( 2, ℚ) = xn − 2 and
√
n
deg(√n 2, ℚ) = n √
for n ≥ 2. Over ℝ, the element 2 is of degree 1, with minimal polynomial
irr( n 2, ℝ) = x − n 2.
Simple extensions
Definition 4. Let E be an extension field of F and let α ∈ E . The smallest subfield of E containing
both F and α is called the simple extension of F and is denoted by F(α).
If α is algebraic over F , then F(α) = ϕα [F[x]]. If α is transcendental over F , then F(α) is
the quotient field of ϕα [F[x]].
Theorem 4. Let E be an extension field of F and let α ∈ E be algebraic over F . Let n = deg(α, F).
Then
F(α) = { a0 + ⋅ ⋅ ⋅ + an−1 α n−1 : a0 , . . . , an−1 ∈ F }.
Example 4. Let F = ℤ2 , let p(x) = x2 + x + 1. There exist a simple extension field ℤ2 (α) of ℤ2
containing a zero α of p(x). Then
ℤ2 (α) = { a0 + a1 α : a0 , a1 ∈ ℤ2 }
which is a new field containing 4 elements.
Vector spaces
Definition 5. Let F be a field. A vector space over F is an additive abelian group V equipped with
a scalar multiplication of each element α ∈ V by each element a ∈ F on the left, such that for all
a, b ∈ F and α , β ∈ V the following is true
V1 : αa ∈ V .
V2 : a(bα) = (ab)α .
V3 : (a + b)α = aα + bα .
V4 : a(α + β ) = aα + aβ .
V5 : 1α = α .
The elements of V are vectors and the elements of F are scalars.
– Typeset by FoilTEX –
2
Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Examples of vector spaces
Example 5. The Cartesian product F n is a vector space over F with scalar multiplication
a(a1 , . . . , an ) = (aa1 , . . . , aan ).
Example 6. Let E be an extension field of F
√. Then E is a vector space over F . In particular, ℝ is a
ℚ-vector space, ℂ is a ℝ-vector space, ℚ( 2) is a ℚ-vector space.
Linear independence
Definition 6. Let V be a vector space, and let S ⊂ V . The vectors of S span or generate V if for
any β ∈ V there exist n ∈ ℤ+ , scalars ai ∈ F and vectors αi ∈ S for 1 ≤ i ≤ n such that
n
β = ∑ ai αi .
i=1
In other words, β is a linear combination of the ai .
Definition 7. A vector space V is finite-dimensional if there is a finite subset S ⊂ V whose vectors
span V .
Example 7. The vectors (1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, 0, . . . , 1) span the finite-dimensional
space F n .
Example 8. Let E be an extension field of F and let α ∈ E be algebraic over F . Let n = deg(α, F).
Then the elements
1, α, . . . , α n−1
span F(α).
Definition 8. Let V be a vector space, and let S ⊂ V . The vectors in S are linearly independent,
if for any n ∈ ℤ+ , scalars ai ∈ F and distinct vectors αi ∈ S for 1 ≤ i ≤ n we have
n
∑ aiαi = 0 ⇔ a1 = a2 = ⋅ ⋅ ⋅ = an = 0.
i=1
Example 9. The vectors defined in Examples 5 and 6, are linearly independent.
– Typeset by FoilTEX –
3
Algebra course FMA190/FMA190F
2011, Spring term 2011, Sergei Silvestrov lectures
Definition 9. Let V be a vector space over a field F , and let B ⊂ V . The vectors in B form a basis
for V over F if they span V and are linearly independent.
Example 10. The vectors defined in Examples 5 and 6, form a basis.
Dimension
Theorem 5. Every finite-dimensional vector space has a basis.
dimensional vector space have the same number of elements.
Any two bases of a finite-
This theorem remains true without the assumption that the vector space is finite dimensional.
Definition 10. If V is a finite-dimensional vector space over F , then the number of elements in a
basis is called the dimension of V over F .
Example 11. F n is n-dimensional vector space over F .
Example 12. Let E be an extension field of F and let α ∈ E be algebraic over F . Then F(α) is
deg(α, F)-dimensional space over F .
– Typeset by FoilTEX –
4