College Algebra Exam Review 328

# College Algebra Exam Review 328 - n Corollary 7.5.3 Aut K.E...

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338 7. FIELD EXTENSIONS – FIRST LOOK This theorem asserts, in particular, that for any intermediate ﬁeld K ± M ± E , there are sufﬁciently many M –automorphisms of E so that Fix . Aut M .E// D M . So the ﬁrst task in proving the theorem is to see that there is an abundance of M –automorphisms of E . We will maintain the notation: K is a subﬁeld of C , f.x/ 2 KŒxŁ , and E is the splitting ﬁeld of f.x/ in C . Proposition 7.5.2. Suppose p.x/ 2 KŒxŁ is an irreducible factor of f.x/ and ˛ and ˛ 0 are two roots of p.x/ in E . Then there is a ± 2 Aut K .E/ such that ±.˛/ D ˛ 0 . Sketch of proof. By Proposition 7.4.2 , there is an isomorphism ² W K.˛/ ! K.˛ 0 / that ﬁxes K pointwise and sends ˛ to ˛ 0 . Using an inductive argu- ment based on the fact that E is a splitting ﬁeld and a variation on the theme of 7.4.2 , we show that the isomorphism ² can be extended to an automorphism of E . This gives an automorphism of E taking ˛ to ˛ 0 and ﬁxing K pointwise.
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Unformatted text preview: n Corollary 7.5.3. Aut K .E/ acts faithfully by permutations on the roots of f.x/ in E . The action is transitive on the roots of each irreducible factor of f.x/ . Proof. By Exercise 7.4.5 , Aut K .E/ acts faithfully by permutations on the roots of f.x/ and, by the previous corollary, this action is transitive on the roots of each irreducible factor. n Theorem 7.5.4. Suppose that K ± C is a ﬁeld, f.x/ 2 KŒxŁ , and E is the splitting ﬁeld of f.x/ in C . Then Fix . Aut K .E// D K . Sketch of proof. We have a priori that K ± Fix . Aut K .E// . We must show that if a 2 L n K , then there is an automorphism of E that leaves K ﬁxed pointwise but does not ﬁx a . n Corollary 7.5.5. If K ± M ± E is any intermediate ﬁeld, then Fix . Aut M .E// D M ....
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