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Like in the title, I'm trying to prove

that if a^2 is algebraic over F, then a is algebraic over F.

my idea- by assumption, a^2 is a solution to the polynomial equation

b_n x^n + b_(n-1) x^(n-1) + ... + b_0 = 0 , b_n's are in F

if a^2 is in F, then we simply write a^2=a *a, and since F is a field, then b_n * a is in F, for any n. further more, the solution works out almost trivially.

now if a^2 isn't in F, we consider a^2 to be in the simple extension E of F. I'm not sure how to examine from here. any help?

that if a^2 is algebraic over F, then a is algebraic over F.

my idea- by assumption, a^2 is a solution to the polynomial equation

b_n x^n + b_(n-1) x^(n-1) + ... + b_0 = 0 , b_n's are in F

if a^2 is in F, then we simply write a^2=a *a, and since F is a field, then b_n * a is in F, for any n. further more, the solution works out almost trivially.

now if a^2 isn't in F, we consider a^2 to be in the simple extension E of F. I'm not sure how to examine from here. any help?

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