Question

Let k C F = kla) be a simple algebraic extension. Prove that F is normal over k if and only if for every algebraic extension FC K and every o E Autk(K),(F) = F.

Answer 1

ut ke F = k ) be a simple Algebrouc extension Then F is normal if and only if for every algebrail Ć extension FC K and every of Aut(K), (F) = F. Proof het F = x(x) é normal extension. 3 F is splitting field of minimal polynomial mare) of d over k[x] Claim. For any algebraic extension F Çk and every Te Aut(K)! we know o permutes the roots of irreducible polynomial For ß. for every Bek. l.e. for each BEK. 0(B) is again a root of Irreducible folynomial of ß over K[x]. why so easy to show suppose ß has irreducible polynomial P(x) = x + an, ahlp... taintage K [x] since ß is root of this polynomial. Þlp)-0. ß" + ani pale ... & aß ta s0.-('). - Now apply of this Equation. o on both sides Since sin field isomorphism it is lheart multiplicative So an applying o to equation (1) we get (B))" + an (693) **... + 2y = (B) + 20 = 0. o(B) is again a root of plug. So s permuter the roots of irr, polynomial of ß. & BEK. Prover the claim. This

for Te Aut (k) so alseady fixes k. –@) and o(a) will be again a root of minimal polynomial Man) (By Above claim.). Since F = K[w] is spuittin field of malas. So it contains all the roots of malas. ) EF -(3) 7 From (2) and (31 we get o(F)=F. mensely wt for every of Aut (k) ; (f)= F. in particular o(ade F Thet if possible to be splitting field of we are required to prove that F=F f wt FICK be any algebraic et a . In order to prove that F is normal we are requireal to prove that F is splittig field of d. se every root of males wes in F. if we vory o over Aut (k) we will get old) roots of males (since by claim o permuter the roots of irr, bolyrma } As given olarEF Not Aut (k) Hence all roots of malas lies in Fi Hence F is Normal.