Extension of bodies

In Algebra, field extensions are the fundamental problem of Field Theory. A field is a set in which the operations sum and product are defined and "work well". When building an extension of a field, you are looking for a larger set in which the addition and product operations still work well and in addition you can solve polynomial equations.

Definition.

Let (K, +, ·) be a field. A body L is an extension of K if K is a subfield of L, that is, if (L,+,·) is a field and (K,+,·) is a field with the restriction to K of the operations + and · on L. If L is an extension over K, denote L:K or L/K.

Extension over a field as a vector space over the field

• Yeah. L is an extension of K, then L It's a vector space over K.

In effect, the addition of K also serves as addition in vector space, and the multiplication of an element of K by one of L defines the dot product of the vector space:

By definition of body, (L,+){displaystyle (L,+)} is abelian group, and we can consider the product by climbers ⋅ ⋅ :K× × LΔ Δ L{displaystyle cdot:Ktimes Llongrightarrow L} as a restriction to K× × L{displaystyle Ktimes L} of the product in ⋅ ⋅ :L× × LΔ Δ L{displaystyle cdot:Ltimes Llongrightarrow L}. In this way it is immediately fulfilled that:

• a⋅ ⋅ (α α +β β )=(a⋅ ⋅ α α )+(a⋅ ⋅ β β ){displaystyle acdot (alpha +beta)=(acdot alpha)+(acdot beta)},

• (a+b)⋅ ⋅ α α =(a⋅ ⋅ α α )+(b⋅ ⋅ α α ){displaystyle (a+b)cdot alpha =(acdot alpha)+(bcdot alpha)},

• (a⋅ ⋅ (b⋅ ⋅ α α ))=(a⋅ ⋅ b)⋅ ⋅ α α {displaystyle (acdot (bcdot alpha))=(acdot b)cdot alpha },

• 1⋅ ⋅ α α =α α {displaystyle 1cdot alpha =alpha },

Whatever. a,b한 한 K{displaystyle a,bin K} and α α ,β β 한 한 L{displaystyle alphabeta in L}. The first two properties are due to the distribution of the product with respect to the sum in L{displaystyle L} and to K L{displaystyle Ksubset L}, the third is because the product is associative L{displaystyle L}and the fourth is because K{displaystyle K} It's subbody. L{displaystyle L}, so the unit element L{displaystyle L} is the unit element K{displaystyle K}.

Simple extension

The whole K(α α ):={f(α α )g(α α ):f,g한 한 K[chuckles]x],g(α α )I was. I was. 0!{displaystyle K(alpha):=left{{frac {alpha)}{g(alpha)}}:f,gin K[x],g(alpha)neq 0right}}}}}. This set is a body, it is extension of K{displaystyle K}It's subbody. L{displaystyle L}, and in fact is the least extension K{displaystyle K} containing a α α {displaystyle alpha }. It's called extension generated by α on K{displaystyle K}.

Algebraic and transcendental extensions

Kronecker's theorem.

Sea K{displaystyle K} a body and p한 한 K[chuckles]x]{displaystyle pin K[x]} an irreducible polynomial, then there is some extension L:K{displaystyle L:K} so that p{displaystyle p} has some root. L{displaystyle L}.

Homomorphism evaluation

Function β β :K[chuckles]x]Δ Δ K(α α ){displaystyle beta:K[x]longrightarrow K(alpha)} every polynomial p(x)한 한 K[chuckles]x]{displaystyle p(x)in K[x]} to match your evaluation in α α {displaystyle alpha }, i.e., β β (p)=p(α α ){displaystyle beta (p)=p(alpha)}. This application is in fact an isomorphism of commutative and unitary rings, and is called homomorphism evaluation.

Algebraic Extension

An extension L:K{displaystyle L:K} is said to be algebraic if all element α α 한 한 L{displaystyle alpha in L} is algebraic about K{displaystyle K}.

Algebraic Elements

Suppose there's a polynomial p한 한 K[chuckles]x]{displaystyle pin K[x]} that has α α {displaystyle alpha } by root.

In this situation (ker (β β )I was. I was. {0!{displaystyle ker(beta)neq {0}}or equivalent, there is some p한 한 K[chuckles]x]{displaystyle pin K[x]} irreducible with K[chuckles]x](p) K(α α ){displaystyle {frac {K[x]}{(p)}}cong K(alpha)}}) it is said that α α {displaystyle alpha } is algebraic about K{displaystyle K}.

An element is then algebraic over a field if and only if it is the root of some polynomial with coefficients in said field.

Irreducible monic polynomial

Yeah. α α {displaystyle alpha } is an algebraic element on the body K{displaystyle K} so that α α K{displaystyle alpha notin K}, polynomial p{displaystyle p} which generates the core of the evaluation application (i.e., ker β β =(p){displaystyle ker beta =(p)}...is irreducible. Splitting p{displaystyle p} by its main coefficient (those scaling that multiplies the greatest power of the variable x{displaystyle x}) is obtained a monic polynomial (i.e., so its main coefficient is the unit), which is denoted by mα α K{displaystyle m_{alpha }^{K}} and called monic polynomial irreducible of α α {displaystyle alpha } concerning K{displaystyle K}.

Clearly, K(α α ) K[chuckles]x](mα α K){displaystyle K(alpha)cong {frac {K[x]}{(m_{alpha }^{K}}}}}}}}}}.

Transcendent Extension

An extension L:K{displaystyle L:K} it is said to be transcendent if there is any element α α 한 한 L{displaystyle alpha in L} that is transcendent about K{displaystyle K}.

Transcendent elements

If the ker(β β )={0!{displaystyle(beta)={0}}It will be β β {displaystyle beta } A monomorphism. In that case, K(x){displaystyle K(x)} isomorph to K(α α ){displaystyle K(alpha)}.

It will be said that the element α α {displaystyle alpha } It's transcendent about K{displaystyle K} and that K(α α ){displaystyle K(alpha)} is a transcendent extension on K{displaystyle K}. In addition, there will be no polynomial with coefficients in K{displaystyle K} root cause α α {displaystyle alpha } (i.e., yes. p한 한 K[chuckles]x]{displaystyle pin K[x]}, then p(α α )I was. I was. 0{displaystyle p(alpha)neq 0}).

Degree of an extension

As all vector space is based, we can calculate the dimension of L{displaystyle L} as vector space over K{displaystyle K}, denoted by dimK (L){displaystyle operatorname {dim} _{K}(L)}. It is called degree of extension L:K{displaystyle L:K} to the dimension of L{displaystyle L} Like K{displaystyle K}- vectorial space: [chuckles]L:K]=dimK (L){displaystyle [L:K]=operatorname {dim} _{K}(L)}.

Let's take several examples:

K = Q{displaystyle mathbb {Q} } the body of the rational and L = R{displaystyle mathbb {R} } the body of the real; R{displaystyle mathbb {R} } seen as vectorial space over Q{displaystyle mathbb {Q} }It is of infinite dimension, that is, [chuckles]R:Q]=∞ ∞ {displaystyle [mathbb {R}:mathbb {Q}]=infty }.

The result is not surprising if the cardinals of both sets are considered: If the dimension of R{displaystyle mathbb {R} } on Q{displaystyle mathbb {Q} } be fine. R{displaystyle mathbb {R} } would be isomorph to Qn,n한 한 N{displaystyle mathbb {Q} ^{n},nin mathbb {N} }What is not possible because <math alttext="{displaystyle |mathbb {Q} ^{n}|=|mathbb {Q} |=|mathbb {N} |日本語Qn日本語=日本語Q日本語=日本語N日本語.日本語R日本語{displaystyle Δmathbb {Q} ^{n}Student=sexmathbb {Q} cult=sexmathbb {N}.<img alt="{displaystyle |mathbb {Q} ^{n}|=|mathbb {Q} |=|mathbb {N} |.

Yeah. K = Q{displaystyle mathbb {Q} }the body of the rational and L = Q(2){displaystyle mathbb {Q} ({sqrt {2}}}}}, the lower body that contains at once Q{displaystyle mathbb {Q} } and √2, clearly Q(2){displaystyle mathbb {Q} ({sqrt {2}}}}} is an algebraic extension Q{displaystyle mathbb {Q} }Since 2{displaystyle {sqrt {2}}} root of polynomial x2− − 2{displaystyle x^{2}-2}.

At the same time:

Q(2) Q[chuckles]x]/(x2− − 2){displaystyle mathbb {Q} ({sqrt {2}}})cong mathbb {Q} [x]/(x^{2}-2)}}

since the ideal (x2− − 2){displaystyle (x^{2}-2}}} is the core of morphism β β :Q[chuckles]x]Δ Δ Q(2){displaystyle beta:mathbb {Q} [x]longrightarrow mathbb {Q} ({sqrt {2}}})}, clearly this is a suprajective morphism, followed by the first isomorphism theorem that are isomorphic fields.

Plus [chuckles]Q(2):Q]=2{displaystyle [mathbb {Q} ({sqrt {2}}}):mathbb {Q}=2}, that is, the dimension of Q(2){displaystyle mathbb {Q} ({sqrt {2}}}}} as vector space over Q{displaystyle mathbb {Q} } is 2, this is so since 2 is the degree of the monic and irreducible polynomial that has Ã2 as root: x2− − 2{displaystyle x^{2}-2}.

Overall:

[chuckles]K(α α ):K]=n{displaystyle [mathbb {K} (alpha):mathbb {K} ]=n} Yeah. n{displaystyle n} is the degree of myonic and irreducible polynomial K[chuckles]x]{displaystyle mathbb {K} [x]} that has α α {displaystyle alpha } as root, where K{displaystyle mathbb {K} } It's a body and K[chuckles]x]{displaystyle mathbb {K} [x]} are polynomials with coefficients in K{displaystyle mathbb {K} }.