Separable Element: Minimal Polynomial Question

[Pratibha]

if a\inK be seperable element over the field F,let f(x) be minimal plynomial of a over F.let degree of f(x) be n, We know f(x) will be seperable over F, BUT is it necessary that f(x) will be MINIMAL POLYNOMIAL of other n-1 roots. if so, what is the reason?

 

[jvicens]

Hello,

Thank you for your post. I can provide some insight into your question.

Firstly, it is important to define what we mean by a "separable element" over a field. An element a is said to be separable over a field F if its minimal polynomial, denoted by f(x), has distinct roots in an algebraic closure of F. This means that f(x) can be factored into linear factors over the algebraic closure of F.

Now, to answer your question, it is not necessary that f(x) will be the minimal polynomial of the other n-1 roots. This is because the minimal polynomial of an element a over a field F is not unique. In fact, there can be multiple minimal polynomials of an element a over F.

The reason for this is that the minimal polynomial of an element a over F is determined by the field extension F(a), where F(a) is the smallest field containing both F and a. Therefore, if there are other elements in F(a) that share the same minimal polynomial as a, then f(x) will not be the minimal polynomial of those elements.

For example, let's consider the field extension Q(sqrt(2)) over Q. The minimal polynomial of sqrt(2) over Q is x^2 - 2. However, the element -sqrt(2) also has the same minimal polynomial as sqrt(2). Therefore, x^2 - 2 is not the minimal polynomial of both sqrt(2) and -sqrt(2) over Q.

In summary, the reason why it is not necessary for f(x) to be the minimal polynomial of the other n-1 roots is because the minimal polynomial of an element depends on the field extension it generates, and there can be multiple elements in the same field extension with the same minimal polynomial.

I hope this helps to clarify your question. Let me know if you have any further inquiries.