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barbiemathgurl
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let E be an algebraic over F where F is perfect. Show that E is perfect.
Palindrom said:Just out of interest: What does the notation irr<a,E> mean?
A field F is considered perfect if every irreducible polynomial over F has distinct roots in its algebraic closure. This means that every polynomial equation over F has a solution in the algebraic closure of F.
To prove that E is perfect if F is perfect, we must show that every irreducible polynomial over E has distinct roots in its algebraic closure. Since E is an extension of F, every irreducible polynomial over E can be factored into irreducible polynomials over F. And since F is perfect, each of these factors has distinct roots in its algebraic closure. Therefore, E is also perfect.
Yes, an example of an algebraic field that is not perfect is the field of rational functions over a finite field. In this field, the polynomial x^p - t has no roots, where p is the characteristic of the field and t is an element that is not a pth power. This shows that not all irreducible polynomials have distinct roots in the algebraic closure, making it not perfect.
Algebraic closure and perfection are closely related concepts. A field F is perfect if every irreducible polynomial over F has distinct roots in its algebraic closure. This means that in order to prove that a field is perfect, we must consider its algebraic closure. On the other hand, if a field is not perfect, then there exist irreducible polynomials that do not have distinct roots in its algebraic closure.
The concept of perfection is important in many areas of mathematics, including algebraic geometry and number theory. For example, in algebraic geometry, the concept of perfection is used to study varieties and their points. In number theory, perfection is important in understanding the behavior of prime numbers in certain fields. Furthermore, the concept of perfection allows us to classify fields in terms of their properties, making it a useful tool in exploring the structure of fields.