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Extension fields

  • Thread starter ehrenfest
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  • #1
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[SOLVED] extension fields

Homework Statement


Prove that x^2-3 is irreducible over [tex]\mathbb{Q}(\sqrt{2})[/tex].

EDIT: it should be [tex]\mathbb{Q}(\sqrt[3]{2})[/tex]


Homework Equations





The Attempt at a Solution


So, we want to show that [itex]\sqrt{3}[/itex] and [itex]-\sqrt{3}[/itex] are not expressible as [itex]a +b(\sqrt[3]{2})^2 +c\sqrt[3]{2}[/itex]. We could take the sixth power of both sides, but then the equation would be just messy. What else can we do?
 
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Answers and Replies

  • #2
morphism
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Is it [itex]\mathbb{Q}(\sqrt{2})[/itex] or [itex]\mathbb{Q}(\sqrt[3]{2})[/itex]?
 
  • #3
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Sorry. See the EDIT.
 
  • #4
morphism
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If x^2 - 3 is reducible over [itex]\mathbb{Q}(\sqrt[3]{2})[/itex], then [itex]\mathbb{Q}(\sqrt[3]{2})[/itex] contains the roots [itex]\pm \alpha[/itex] of x^2 - 3. This in turn means that [itex]\mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt[3]{2})[/itex]. Try to think about why this cannot happen.
 
  • #5
Hurkyl
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morphism hints at what is probably the easiest approach.

Another one is the analog of what you would do to show it's irreducible over the rationals -- if x²-3 is reducible, it has a root. It's monic, so the root must be an algebraic integer... in fact, it must be an algebraic integer that divides 2.... how many divisors does 2 have in [itex]\mathbb{Q}(\sqrt[3]{2})[/itex]?
 
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  • #6
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If x^2 - 3 is reducible over [itex]\mathbb{Q}(\sqrt[3]{2})[/itex], then [itex]\mathbb{Q}(\sqrt[3]{2})[/itex] contains the roots [itex]\pm \alpha[/itex] of x^2 - 3. This in turn means that [itex]\mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt[3]{2})[/itex]. Try to think about why this cannot happen.
Why don't you just write [itex] \pm \sqrt{3}[/itex] instead of alpha?

[itex][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3[/itex] and [itex][\mathbb{Q}(\sqrt[2]{3}):\mathbb{Q}] = 2[/itex] and if [itex]\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\sqrt[3]{2})[/itex] it must be the case that [itex]\mathbb{Q}(\sqrt[3]{2})[/itex] is a finite extension of [itex]\mathbb{Q}(\sqrt[2]{3})[/itex].

and then we apply the theorem that says

[K:F] = [K:E][E:F]

which in our case is

3 = n*2, where n must be an integer, which is absurd.

Can you help me prove the statement that I underlined?
 
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  • #7
morphism
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I used alpha because it was faster to type. :p

If E is an intermediate field of the finite extension K/F, then [K:E]=[K:F]/[E:F], which is certainly finite because both [K:F] and [E:F] are. (The first is finite by assumption, and the second is finite because a subspace of a finite dimensional space is finite dimensional.)
 
  • #8
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If E is an intermediate field of the finite extension K/F, then [K:E]=[K:F]/[E:F]
I have not seen a proof of that. My book proves a theorem that says: If E is a finite extension field of a field F, and K is a finite extension of E, then K is a finite extension of F, and [K:F]=[K:E][E:F], but what you said is different.
 
  • #9
Hurkyl
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I have not seen a proof of that. My book proves a theorem that says: If E is a finite extension field of a field F, and K is a finite extension of E, then K is a finite extension of F, and [K:F]=[K:E][E:F], but what you said is different.
Have you thought at all about how this statement might imply what morphism said?
 
  • #10
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Have you thought at all about how this statement might imply what morphism said?
So, the only thing that you need to prove is that this intermediate field E is a finite extension of F and finitely extends to K. Let's do the former first.

We want to show that E is finite dimensional as a vector space over F. We know that E over F must be a subspace of K over F, which is finite dimensional. So, we need to find a finite set of vectors in E over F that span E over F. Now, we already have a finite number of vectors (call them B) in K over F that span K over F, which includes E over F. If we take the subset of B that is contained in E over F, is it true that those vectors will span E over F? It is definitely not true (consider the standard bases in R^2 and the line x=y; in fact, neither of the two basis vector are contained in that line). What we need to do is this. Let the basis B have size n. Find an arbitrary vector in E over F. Find another vector that is linearly independent with the first. Keep doing this until you get n vectors. If you cannot get n vectors, then you must have a set of less than n vectors that spans E over F. If you get n vectors, then use the fact any set of linearly independent vectors in K over F must be less than or equal to the dimension of the space, which is n. Thus the n vectors we have must span all K over F, which by assumption contains E over F.

Now the latter. Get a maximal set of linearly independent vectors of E over F. We just proved that this set must be finite. Then we can enlarge this set to a basis of K over F just be adding linearly independent vectors until there are none left. We even know how many we need to add because the dimension of K over F is n.

Please confirm this proof.
 

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