Theory 9.) Let K be an algebraically closed field. Show that every isomorphism ơ 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 to o 10. Let E be an algebraic extension of a field F. Show that every isomorphism of E onto a subfield of F leaving F fixed car be extended to an automorphism of F. E ond E of F and E,

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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,
Transcribed Image Text: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 σK 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 σbe an isomorphism of F onto the field F'. Let F¯' be an algebraic closure of F'. Then σ can be extended to an isomorphism 𝞽 of E onto a subfield of F'¯ such that 𝞽a=σa for all aF."

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