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

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In summary, the conversation discusses how to prove that if a^2 is algebraic over F, then a is algebraic over F. The idea is to assume that a^2 is a solution to a polynomial equation with coefficients in F. If a^2 is in F, then a satisfies the polynomial x^2 - a^2, which is also in F. This means that a is algebraic over F. However, if a^2 is not in F, then it belongs to a simple extension of F. In this case, we need to consider a polynomial that has a as a root. The conversation also provides a hint on how to approach this problem.
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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?
 
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calvino said:
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)).
 

1. What is the difference between A^2 algebraic over F and a algebraic over F?

The notation A^2 algebraic over F refers to a polynomial ring over a field F, where A is a variable. This means that any polynomial in A can be expressed as a linear combination of the monomials 1, A, A^2, A^3, etc. On the other hand, a algebraic over F refers to a specific element in this polynomial ring, which can be represented as a polynomial in A with coefficients from the field F.

2. Can you give an example of A^2 algebraic over F?

One example of A^2 algebraic over F is the polynomial ring F[A], where A is the indeterminate variable. The elements in this ring can be expressed as a linear combination of the monomials 1, A, A^2, A^3, etc. For instance, the element 2A^2 + 3A + 1 is an element in F[A].

3. How is A^2 algebraic over F related to vector spaces?

A^2 algebraic over F can be seen as a vector space over the field F, with the monomials 1, A, A^2, A^3, etc. as the basis. This means that any element in A^2 algebraic over F can be expressed as a linear combination of these basis elements. The degree of the polynomial in A determines the dimension of the vector space.

4. What does it mean for an element to be algebraic over a field?

An element a is said to be algebraic over a field F if there exists a polynomial with coefficients from F, such that a is a root of this polynomial. In other words, a is a solution to a polynomial equation with coefficients from F.

5. How is A^2 algebraic over F related to field extensions?

A^2 algebraic over F is an example of a field extension of F, which means it is a larger field that contains F as a subfield. In this case, A^2 algebraic over F contains all the elements of F, as well as the additional variable A and all the polynomials in A with coefficients from F.

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