*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.*

* *

*Remark:*

*In fact, any nonzero ring homoeomorphism between two fields is an embedding fields.**Also, because: p(x)=q(x)f(x)+r(x) and so .*

*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”*

*Finite extensions are algebraic.*

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

*If , then: [E:k]=[E:F][F:k]*

* 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=*