Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    15
    ReputationRep:
    I ran into this problem while solving a larger one. Can anyone see where I've gone wrong?

    Suppose that L:K is a finite extension, and let \{a_1,...,a_n\} be a basis for L over K.

    [L:K(a_1)]=[K(a_1,...,a_n):K(a_1)]

    =[K(a_1,...,a_n):K(a_1,...,a_{n-1})][K(a_1,...,a_{n-1}):K(a_1)]

    =2[K(a_1,...,a_{n-1}):K(a_1)]

    and by induction on n we get


    [L:K(a_1)]=2^{n-1}

    which is certainly wrong.
    Offline

    3
    ReputationRep:
    (Original post by james22)
    I ran into this problem while solving a larger one. Can anyone see where I've gone wrong?

    Suppose that L:K is a finite extension, and let \{a_1,...,a_n\} be a basis for L over K.

    [L:K(a_1)]=[K(a_1,...,a_n):K(a_1)]

    =<b>[K(a_1,...,a_n):K(a_1,...,a_{n-1})]</b>[K(a_1,...,a_{n-1}):K(a_1)]

    =2[K(a_1,...,a_{n-1}):K(a_1)]

    and by induction on n we get


    [L:K(a_1)]=2^{n-1}

    which is certainly wrong.
    There's no way of knowing if this is 2 or not, without some characterisation of the a_i. This will be equal to the degree of the minimal polynomial of a_n over K(a_1,...,a_{n-1}).

    To expand, consider the splitting field of X^3-2 (and splitting fields are always finite extensions). If we take \alpha = 2^{\frac{1}{3}} \in \mathbb{R}, which has of course has minimal polynomial X^3-2 over \mathbb{Q}, then we get [\mathbb{Q}(\alpha):\mathbb{Q}] = 3, with basis \{1,2^{\frac{2}{3}},2^{\frac{1}{  3}}\}. This isn't yet the splitting field, but hereafter is irrelevant - whatever happens next, by the tower law the degree of the extension isn't going to be a power of 2!
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by FireGarden)
    There's no way of knowing if this is 2 or not, without some characterisation of the a_i. This will be equal to the degree of the minimal polynomial of a_n over K(a_1,...,a_{n-1}).

    To expand, consider the splitting field of X^3-2 (and splitting fields are always finite extensions). If we take \alpha = 2^{\frac{1}{3}} \in \mathbb{R}, which has of course has minimal polynomial X^3-2 over \mathbb{Q}, then we get [\mathbb{Q}(\alpha):\mathbb{Q}] = 3, with basis \{1,2^{\frac{2}{3}},2^{\frac{1}{  3}}\}. This isn't yet the splitting field, but hereafter is irrelevant - whatever happens next, by the tower law the degree of the extension isn't going to be a power of 2!
    OK, I see that it may not be 2, but surely it cannot be 1?

    This would mean that [L:K(a_1)] is at least 2^(n-1)? This would mean that [L:K] is at least 2^n, but 2^n is bigger than n, which is a contradiction.
    Offline

    3
    ReputationRep:
    (Original post by james22)
    OK, I see that it may not be 2, but surely it cannot be 1?

    This would mean that [L:K(a_1)] is at least 2^(n-1)? This would mean that [L:K] is at least 2^n, but 2^n is bigger than n, which is a contradiction.
    Well, one of the extensions is going to have degree 1. If \{a_1 \ldots a_n\} is the basis over K, then one of the a_i is going to be 1\in K and then that extension is of course of degree 1; the field won't change.

    I think an issue is you're starting with the basis of the big field to construct it from the base field.

    Take \mathbb{Q}(\zeta_5)/\mathbb{Q}, where \zeta_5 is a primitive 5th root of unity. This is clearly a simple extension, but the basis will be \{1,\zeta_5,\zeta_5^2,\zeta_5^3,  \zeta_5^4\}. Now if we tried to adjoin these basis elements individually to  \mathbb{Q} as it seems to appear in your inductive step, you'd get the degree of the extension being 5^4 if you didn't notice something crucial: After the first element, you're done. \zeta_5 and all its non-real powers are all primitive 5th roots; adjoining any one of them will give the same field. In your example, it could be like this case, where adjoining a_1 necessarily puts a_2\ldots a_n into your field to keep the operations closed. Adjoining the 'next element', as your inductive step is trying to, puts in an element that's now already there; the degree of that extension shall then be 1.

    I hope that was clear..!
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by FireGarden)
    Well, one of the extensions is going to have degree 1. If \{a_1 \ldots a_n\} is the basis over K, then one of the a_i is going to be 1\in K and then that extension is of course of degree 1; the field won't change.

    I think an issue is you're starting with the basis of the big field to construct it from the base field.

    Take \mathbb{Q}(\zeta_5)/\mathbb{Q}, where \zeta_5 is a primitive 5th root of unity. This is clearly a simple extension, but the basis will be \{1,\zeta_5,\zeta_5^2,\zeta_5^3,  \zeta_5^4\}. Now if we tried to adjoin these basis elements individually to  \mathbb{Q} as it seems to appear in your inductive step, you'd get the degree of the extension being 5^4 if you didn't notice something crucial: After the first element, you're done. \zeta_5 and all its non-real powers are all primitive 5th roots; adjoining any one of them will give the same field. In your example, it could be like this case, where adjoining a_1 necessarily puts a_2\ldots a_n into your field to keep the operations closed. Adjoining the 'next element', as your inductive step is trying to, puts in an element that's now already there; the degree of that extension shall then be 1.

    I hope that was clear..!
    I see, I was too caught up in the linear algebra bit to notice that. I think that what I have done is still enough to solve my problem though, as I don't thinkt the errors matter too much (this thing I was trying to prove was much strong than I needed). I'll post the whole question and result when I've done it or failed again.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by FireGarden)
    x
    Thanks for your help, could you please tell me if my overall solution is correct?

    Let L:K be a finite extension of degree m, and f be an irreducible polynomial over K. If deg(f)=n, if m and n are coprime, then f has no roots in L.

    Solution

    Suppose f does have a solution in L, f(a)=0.

    [L:K]=[L:K(a)][K(a):K]=n[L:K(a)]

    which contradicts n and m being coprime.

    I think this works. This was my first solution but I was put off because it only relies on n not dividing m, which is weaker than them being coprime.
    Offline

    1
    ReputationRep:
    (Original post by james22)
    Thanks for your help, could you please tell me if my overall solution is correct?

    Let L:K be a finite extension of degree m, and f be an irreducible polynomial over K. If deg(f)=n, if m and n are coprime, then f has no roots in L.

    Solution

    Suppose f does have a solution in L, f(a)=0.

    [L:K]=[L:K(a)][K(a):K]=n[L:K(a)]

    which contradicts n and m being coprime.

    I think this works. This was my first solution but I was put off because it only relies on n not dividing m, which is weaker than them being coprime.
    This is correct, however, the hypothesis is giving you more; namely, your polynomial is actually irreducible over the extension. Of course, under an obvious condition on its degree.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Mladenov)
    This is correct, however, the hypothesis is giving you more; namely, your polynomial is actually irreducible over the extension. Of course, under an obvious condition on its degree.
    I have used irreducibility to get that p is the minimum polynomial, not just any 0 polynomial.
    Offline

    1
    ReputationRep:
    (Original post by james22)
    I have used irreducibility to get that p is the minimum polynomial, not just any 0 polynomial.
    I am not sure I understand what you are saying.

    You are using irreducibility to get the degree of the root.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Mladenov)
    I am not sure I understand what you are saying.

    You are using irreducibility to get the degree of the root.
    Same thing. If p was reducible then the degree of the extension would not be n.
    Offline

    1
    ReputationRep:
    (Original post by james22)
    Same thing. If p was reducible then the degree of the extension would not be n.
    So yeah, that's fine. I just said that the polynomial is also irreducible over the extension of your base field.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.