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Lecture 1: Field Extension:


Def 1: F and k are fields. If , then k is called a field extension of F and F is called a subfield of k.


Example: (Simple extension)

Given a field k, let k() = smallest field containing k and . Then k() is a field extension of k.

We claim that: k()=k[]={:; n is natural number}

There are two possibilities:

1.      1, ,,… are linearly independent. Then

      k[x]       has kernel 0; So, k[]k[x]. k[] is thus a field.

     ( p(x) )

      Remark: Field of fraction of k[]field of all rational function.

2.      Otherwise, there exist such that: Let f(x)= minimal polynomial of . Then:

       has kernel= ideal generated by f(x)= (f(x)), which is a maximal ideal.

So, . Since is a field, k[] is also a field.


Finally, since k()and k[] is a field, k[]=k().


Def 2: In the second case, we say is algebraic over k.


Example: k=R, F=C, =I, then f(x)= ; C

Example: Let k=Q=rational numbers. If  is algebraic over Q, we say  is an algebraic number.


Thm1.1: Let f(x)= irreducible polynomial in k[x] with deg f. Then there is an extension F/k and such that f()=0.

Proof: Consider E=k[x]/(f(x)).

The map:

 is an embedding of k in E. Identify k with its image in E. Let ==image of x in E. Then: f()==0.



Def 3: We say F/k is an algebraic extension if every is algebraic over k. Any extension F of k is a vector space over k. Furthermore,

          Let: [F:k]=dim of F as v.s. over k= degree of F/k. If [F:k]<  , then we say F/k is a finite extension.

Example: If  is algebraic over k, then k()=k[] is a finite extension of k, with [k():k]=degree of minimal polynomial of  over k.

Example: If k=Q={all rational numbers}, then the finite extension of Q is called number fields.


Lemma1.2: “Algebraic extension are finite in nature”

           Proof:  Say F/k is finite. Let . Then, 1, , can’t all be linearly independent which implies  is algebraic.

    Proof: Just write down the bases for each extension.


Def 4: (Compositum) Let E,F be contained in some larger field L. The compositum of E,F denoted by EF is the smallest subfield of L containing both E and F. In other words, EF=