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#1
Mar108, 09:22 AM

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1. The problem statement, all variables and given/known data
Prove that x^23 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? 


#2
Mar108, 01:00 PM

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Is it [itex]\mathbb{Q}(\sqrt{2})[/itex] or [itex]\mathbb{Q}(\sqrt[3]{2})[/itex]?



#3
Mar108, 02:49 PM

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Sorry. See the EDIT.



#4
Mar108, 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.



#5
Mar108, 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 x²3 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]? 


#6
Mar108, 04:21 PM

P: 1,996

[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? 


#7
Mar108, 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.) 


#8
Mar108, 08:06 PM

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#9
Mar208, 05:02 AM

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#10
Mar208, 12:23 PM

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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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