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barbiemathgurl
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let alpha be algebraic over F of degree n, show that there exists at most n isomorphisms mapping F(alpha) onto a subfield of bar F (this means the algebraic closure).
thanx
thanx
Please, reread your questions. This is not the first time you made an error. I was lucky to understand what you asked.barbiemathgurl said:let alpha be algebraic over F of degree n, show that there exists at most n isomorphisms mapping F(alpha) onto a subfield of bar F (this means the algebraic closure).
thanx
An algebraic alpha is a symbol that represents a complex number in a field extension. It is usually denoted by the Greek letter alpha (α) and is a root of a polynomial equation.
This means that the polynomial equation in which the algebraic alpha is a root has a degree of n, and the field extension F is the smallest field in which the polynomial can be factored into linear factors.
This can be shown by using the fundamental theorem of algebra, which states that a polynomial of degree n has at most n distinct roots. Since an algebraic alpha is a root of a polynomial equation of degree n, it can have at most n distinct conjugates.
Having a finite number of conjugates is important because it allows us to fully characterize the field extension in which the algebraic alpha exists. It also helps us to understand the properties and behavior of the algebraic alpha within the field.
This can be proven by showing that the polynomial equation in which the algebraic alpha is a root has no more than n distinct roots over the given field. This can be done by using various techniques such as the rational root theorem, Descartes' rule of signs, or the Eisenstein's criterion.