Number 9: Theory9.) Let K be an algebraically closed field. Show that every isomorphism o of K onto a subfield of itself such thatK is algebraic over o [K]is an automorphism of K, that is, is an onto map. [Hint: Apply Theorem 49.3 too".]10. Let E be an algebraic extension of a field F. Show that every isomorphism of E onto a subfield of F leavingS fixed can be extended to an automorphism of F.olgebraic closures F and E of F and E,

Name: Number 9: Theory9.) Let K be an algebraically closed field. Show that
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Description: Number 9: Theory9.) Let K be an algebraically closed field. Show that every isomorphism o of K onto a subfield of itself such thatK is algebraic over o [K]is an automorphism of K, that is, is an onto map. [Hint: Apply Theorem 49.3 too".]10. Let E be

Answered: Theory 9.) Let K be an algebraically…

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter7: Real And Complex Numbers

Section7.1: The Field Of Real Numbers

Problem 23E: If F is an ordered field, prove that F contains a subring that is isomorphic to . (Hint: See Theorem...

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Number 9:

Theory
9.) Let K be an algebraically closed field. Show that every isomorphism o of K onto a subfield of itself such that
K is algebraic over o [K]is an automorphism of K, that is, is an onto map. [Hint: Apply Theorem 49.3 too".]
10. Let E be an algebraic extension of a field F. Show that every isomorphism of E onto a subfield of F leaving
S fixed can be extended to an automorphism of F.
olgebraic closures F and E of F and E,

Expert Solution

Step 1

$(9)$ Let K be an algebraically closed field. We have to show that every isomorphism $σ$ of K onto a subfield of itself such that K is algebraic over $σ [K]$ is an automorphism of K that is, is an onto map.

To prove this we will use Isomorphism Extension Theorem,that says:

"Consider E be an algebraic extension of field F. Let $σ$ be an isomorphism of F onto the field F'. Let $\bar{F}'$ be an algebraic closure of F'. Then $σ$ can be extended to an isomorphism $𝞽$ of E onto a subfield of $\bar{F'}$ such that $𝞽 (a) = σ (a)$ for all $a \in F$ ."

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