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Showing equality of dimensions

  1. Mar 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Let α ∈ ℂ be a complex number. Let V = ℚ(α) be the rational vector space spanned by powers of α. That is
    ℚ(α) = <1,α,α2,....>.
    1) If P(t) is a polynomial of degree n such that P(α) = 0, show that dimℚ(α) is at most n.


    2. Relevant equations


    3. The attempt at a solution

    Here
    is my take on this question. Please give me some feedback/corrections.

    Since ℚ(α) = <1,α,α2,....> we know that 1,α,α2,... span ℚ(α).
    To show that dimℚ(α) is at most n, we must show that 1,α,α2,..,αn-1 is a basis of ℚ(α).
    To show it is a basis,
    i) 1,α,α2,..,αn-1 must span ℚ(α)
    ii) 1,α,α2,..,αn-1 must be linearly independent.


    For span:I would say that since 1,α,... spans ℚ(α) then 1,α,...,αn-1 spans ℚ(α) because its elements are in the set 1,α,...

    For linear independence: I was thinking of using induction but I'm not sure how I should go about it.

    As for P(α) = 0 I am not quite sure what relevance it has. It tells us that
    P(α) = a0 + a1α + a2α2 + . . . + an-1αn-1 = 0
    Perhaps it helps showing linear independence since we want 1,α,...,αn-1 to be written as
    a0 + a1α + a2α2 + . . . + an-1αn-1 = 0 where a0 = a1 = . . . = an-1 = 0
     
  2. jcsd
  3. Mar 24, 2015 #2

    Fredrik

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    Is that last thing your notation for the space spanned by ##\{\alpha^n|n\in\mathbb N\}##? <x,y,z> usually denotes the ordered triple that most books write as (x,y,z), and since ordered n-tuples are finite sequences, my first instinct is to interpret your notation as representing a sequence, not a vector space or a subset of a vector space.

    We know that ##\{1,\alpha,\dots\}## is a spanning set for ##\mathbb Q(\alpha)## because ##\mathbb Q(\alpha)## is defined as the space spanned by ##\{1,\alpha,\dots\}##. So you don't need to provide an argument for it.

    This may be unnecessary. It's sufficient to prove that ##\mathbb Q(\alpha)## doesn't contain a linearly independent subset of cardinality ##n+1##.

    Recall that a set ##\{x_1,\dots,x_r\}## with all the ##x_i## distinct is said to be linearly independent if the following implication holds for all scalars ##a_1,\dots,a_r##:
    $$\sum_{i=1}^r a_i x_i=0\ \Rightarrow\ a_1=\dots=a_r=0.$$
    Edit: Is there anything in the problem statement that specifies that the polynomial function P has rational coefficients? I think we need it to have only rational coefficients.
     
    Last edited: Mar 24, 2015
  4. Mar 24, 2015 #3
    My professor uses the notation <1,α,α2,....> but yes it is a spanning set.

    So, by proving this, does it automatically prove that the subset of cardinality n is linearly dependent? If yes then what theorem/definition says this?

    But P(α) = 0 does not tell me anything about linear independence. Even though it is equal to zero, I do not know whether all of its coefficients are equal to zero.
     
  5. Mar 24, 2015 #4

    Fredrik

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    No, but that's not a problem here. To prove that the space is n-dimensional, you would have to prove that it has a linearly independent subset of cardinality n, and doesn't have a linearly independent subset of cardinality n+1. To prove that it's at most n-dimensional, you only need to prove that it doesn't have a linearly independent subset of cardinality n+1.

    Note that what exactly you need to prove, or what theorems you need to use, depends on what exactly your definitions are. I like the definition of "dimension" that says that the dimension of a non-trivial vector space X is the largest integer n such that X contains a linearly independent set of cardinality n. If this is the definition used in the book, then it's sufficient to prove that there's no linearly independent subset of cardinality n+1. If the book uses the definition that says that the dimension of X is the cardinality of a basis for X, it's sufficient to prove that there's no basis for this vector space of cardinality n+1 (and since a basis is linearly independent, you can do this by proving that there's no linearly independent subset of cardinality n+1).

    When n>0, you know that they're not all equal to zero, because if they were, we wouldn't say that the degree of P is n.
     
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