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Extension fields

by ehrenfest
Tags: extension, fields, solved
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ehrenfest
#1
Mar1-08, 09:22 AM
P: 1,996
1. The problem statement, all variables and given/known data
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]


2. Relevant equations



3. 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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morphism
#2
Mar1-08, 01:00 PM
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Is it [itex]\mathbb{Q}(\sqrt{2})[/itex] or [itex]\mathbb{Q}(\sqrt[3]{2})[/itex]?
ehrenfest
#3
Mar1-08, 02:49 PM
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Sorry. See the EDIT.

morphism
#4
Mar1-08, 03:36 PM
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Extension fields

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.
Hurkyl
#5
Mar1-08, 04:08 PM
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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 x-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]?
ehrenfest
#6
Mar1-08, 04:21 PM
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Quote Quote by morphism View Post
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?
morphism
#7
Mar1-08, 04:35 PM
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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.)
ehrenfest
#8
Mar1-08, 08:06 PM
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Quote Quote by morphism View Post
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.
Hurkyl
#9
Mar2-08, 05:02 AM
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Quote Quote by ehrenfest View Post
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?
ehrenfest
#10
Mar2-08, 12:23 PM
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Quote Quote by Hurkyl View Post
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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