A^2 algebraic over F -> a algebraic over F

[calvino]

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?

 

[matt grime]

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 a satisfies the poly x^2 - a^2 which is in F[x], so let's not make that assumption

then we simply write a^2=a *a, and since F is a field, then b_n * a is in F,

that is true if and only if 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?

a^2 is a root of some poly f(x) in F[x], right? I.e. f(a^2)=0. So what is the only g we can reasonably write down so that g(a)=0? HINT: if h is a polynomial over F[x], then so is the composite f(h(x)).