How to prove the field extension is algebraically closed

In summary: I doubt it since the trick focuses on extensions K/F which can be put in the form K=F(\alpha) , for some \alpha\in K , and one achieves this using separability, but also: why? Separability shouldn't, imo, be a problem for someone trying to tackle what you're trying...I think it's because separability is a consequence of algebraic closure.
  • #1
lugita15
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Suppose that E is a field extension of F, and every polynomial f(x) in F[x] has a root in E. Then E is algebraically closed, i.e. every polynomial f(x) in E[x] has a root in E.

I've been told that this result is really difficult to prove, but it seems really intuitive so I find that surprising. Where can I find a proof of this result?

Any help would be greatly appreciated.

Thank You in Advance.
 
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  • #2
lugita15 said:
Suppose that E is a field extension of F, and every polynomial f(x) in F[x] has a root in E. Then E is algebraically closed, i.e. every polynomial f(x) in E[x] has a root in E.

I've been told that this result is really difficult to prove, but it seems really intuitive so I find that surprising. Where can I find a proof of this result?

Any help would be greatly appreciated.

Thank You in Advance.



No wonder you've been told that: it is false. Let [itex]E:= ℂ(t)[/itex] be the field of rational functions on the (transcendental) variable t. Then E is a field extension of [itex]ℂ[/itex] in which every pol. of [itex]ℂ[x][/itex] has a root, but E certainly is not alg. closed as, for example, the pol. [itex]x^2-t\in ℂ(t)[x][/itex] has no root in it...

Now, the claim is true if one adds the condition that E/F is an algebraic extension.

DonAntonio
 
  • #3
DonAntonio said:
Now, the claim is true if one adds the condition that E/F is an algebraic extension.
OK, so adding that condition where can I find the proof of the result?
 
  • #4
lugita15 said:
OK, so adding that condition where can I find the proof of the result?



It's theorem 2 in http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/algclosure.pdf

The proof isn't THAT easy but it isn't THAT hard, either.

Enjoy, and for this read the whole paper. The first part is just the well-known and "standard" proof by Artin.

DonAntonio
 
  • #5
my free algebra notes 844.I.1, do this in detail, in the first 12 pages.

http://www.math.uga.edu/%7Eroy/844-1.pdf
 
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  • #6
mathwonk said:
my free algebra notes 844.I.1, do this in detail, in the first 12 pages.

http://www.math.uga.edu/%7Eroy/844-1.pdf



The question wasn't about the existence of alg. closures of fields, which Artin's proof, that also appears in your notes, does, but the fact that after the first construction of an extension E/F s.t. every non-constant pol. in E[x] has a root we can ALREADY stop there since F is alg. closed. I couldn't find this point in your notes, though perhaps I oversaw it.

DonAntonio
 
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  • #7
oops, my mistake, thank you.
 
  • #8
DonAntonio said:
It's theorem 2 in http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/algclosure.pdf

The proof isn't THAT easy but it isn't THAT hard, either.

Enjoy, and for this read the whole paper. The first part is just the well-known and "standard" proof by Artin.
Is it possible to prove the result without invoking the notion of perfect fields?
 
  • #9
lugita15 said:
Is it possible to prove the result without invoking the notion of perfect fields?



I doubt it since the trick focuses on extensions K/F which can be put in the form [itex]K=F(\alpha)[/itex] , for some [itex]\alpha\in K[/itex] , and one

achieves this using separability, but also: why? Separability shouldn't, imo, be a problem for someone trying to tackle what you're trying...

DonAntonio
 

FAQ: How to prove the field extension is algebraically closed

How do you prove that a field extension is algebraically closed?

The most commonly accepted method of proving that a field extension is algebraically closed is by showing that every polynomial with coefficients in the field has at least one root in the field. This can be done by using the fundamental theorem of algebra, which states that every polynomial of degree n has n complex roots. If all the coefficients of the polynomial are in the field, then at least one of the roots must also be in the field, making it algebraically closed.

2. What does it mean for a field extension to be algebraically closed?

A field extension is considered algebraically closed if every polynomial with coefficients in the field has at least one root in the field. This means that every equation in the form of ax^n + bx^{n-1} + ... + c = 0, where a, b, c, etc. are elements of the field, has a solution within the field.

3. Can any field extension be proven to be algebraically closed?

No, not every field extension can be proven to be algebraically closed. It is possible to construct a field extension that is not algebraically closed by adding elements to the field that do not have a solution in the field for certain polynomials. In order to prove that a field extension is algebraically closed, it must be shown that every polynomial with coefficients in the field has a solution within the field.

4. What are some examples of algebraically closed field extensions?

The most well-known example of an algebraically closed field extension is the field of complex numbers, denoted by C. This field contains all real and imaginary numbers, and every polynomial with coefficients in C has at least one solution within C. Other examples of algebraically closed field extensions include the field of algebraic numbers and the field of formal Laurent series.

5. Are there any practical applications for proving that a field extension is algebraically closed?

Proving that a field extension is algebraically closed can have important applications in various fields such as algebraic geometry, number theory, and theoretical physics. It can also be used in computer science for designing algorithms and cryptography. Additionally, the concept of algebraically closed field extensions is crucial in understanding and developing more advanced mathematical theories and structures.

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