Section 105 Sarah Vilardi April 12 2011 Abstract Algebra II From Thursday Let F be a field A polynomial fx in Fx of degree n is said to be separable if fx has n distinct roots in every splitting field If K is an extension field of F then an element u in K is ID: 399001
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Slide1
Separability—Section 10.5
Sarah Vilardi
April 12, 2011
Abstract Algebra IISlide2
From Thursday…
Let F be a field. A polynomial f(x) in F[x] of degree n is said to be
separable
if f(x) has n distinct roots in every splitting field. If K is an extension field of F, then an element u in K is
separable
over F if u is algebraic over F and its minimal polynomial p(x) in F[x] is separable. The extension field K is a
separable extension
if every element of K is separable over F.
(
f+g
)’(x)=f’(x)+g’(x)
(
fg
)’(x)=f(x)g’(x)+f’(x)g(x)Slide3
Lemma 10.16
Let F be a field and f(x) be in F[x]. If f(x) and f’(x) are relatively prime in F[x], then f(x) is separable.Slide4
Definition
A field F is said to have
characteristic 0
if n1
F
≠ 0
F
for every positive n.Slide5
Theorem 10.17
Let F be a field of characteristic 0. then every irreducible polynomial in F[x] is separable, and every algebraic extension field K of F is a separable extension.Slide6
Theorem 10.18
If K is a finitely generated separable extension field of F, then K = F(u) for some u in K.
This proof is a beast…an outline will help us!Slide7
Theorem 10.18 Outline
By hypothesis, K = F(u
1
,…,u
n
). Proof is by induction on n.
Work with n = 2 case, where K = F(v, w).
Establish preliminary assumptions (min. polys, roots, splitting fields, etc.).
Claim: K = F(u). Prove that w is an element of F(u).
Let r(x) be the minimal polynomial of
w
over F(u).
Show that r(x) is linear.
Prove that K = F(v, w) = F(u).