Question

Let \( K \) be an algebraically closed field containing a finite field \( F \). Let \( L \) be the subfield of \( K \) consisting of elements of \( K \) that are algebraic over \( F \). 
Consider the following statements: 
S1: \( L \) is algebraically closed. 
S2: \( L \) is infinite. 
Then, which one of the following is correct?

• A. S1 is TRUE and S2 is FALSE
• B. S2 is TRUE and S1 is FALSE
• C. both S1 and S2 are TRUE
• D. neither S1 nor S2 is TRUE

Hint:

When dealing with algebraically closed fields, remember that subfields consisting of algebraic elements over a finite field are typically infinite, and the field can be algebraically closed.

Answer 1

The Correct Option is C

To determine the correctness of the given statements about the subfield \( L \), we need to explore each statement.

1. Statement S1: \( L \) is algebraically closed.
   • If \( L \) consists of elements of \( K \) that are algebraic over \( F \), where \( F \) is finite, then every polynomial with coefficients in \( F \) that has solutions in \( K \) should also have solutions in \( L \). Since \( K \) is algebraically closed, it contains solutions to all polynomials over any subfield, including \( L \). It implies that the algebraic elements that generate solutions are already in \( L \).
   • Thus, \( L \) contains all roots of such polynomials and is algebraically closed.
2. Statement S2: \( L \) is infinite.
   • The field \( F \) is finite; however, any finite extension of a finite field is also finite. The algebraic closure of a finite field, \( K \), is infinite because it contains roots of all polynomials.
   • Considering \( L \) as the set of elements in \( K \) that are algebraic over \( F \), it includes all extensions derived from polynomials in \( F \).
   • Since polynomial extensions rapidly increase size, \( L \) effectively becomes infinite like \( K \) by including all these extensions.

Conclusion: Both statements S1 and S2 are correct because \( L \) is algebraically closed and infinite. The correct option is, therefore, "both S1 and S2 are TRUE".