Showing equality of dimensions

[said]

Homework Statement

[/B]
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 dim_ℚℚ(α) is at most n.

Homework Equations

The Attempt at a Solution

Here[/B] 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 dim_ℚℚ(α) is at most n, we must show that 1,α,α²,..,α^n-1 is a basis of ℚ(α).
To show it is a basis,
i) 1,α,α²,..,α^n-1 must span ℚ(α)
ii) 1,α,α²,..,α^n-1 must be linearly independent.For span:I would say that since 1,α,... spans ℚ(α) then 1,α,...,α^n-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 + . . . + a_n-1α^n-1 = 0
Perhaps it helps showing linear independence since we want 1,α,...,α^n-1 to be written as
a₀ + a₁α + a₂α² + . . . + a_n-1α^n-1 = 0 where a₀ = a₁ = . . . = a_n-1 = 0

 

[Fredrik]

Homework Statement

[/B]
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,...>.

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.

Since ℚ(α) = <1,α,α²,...> we know that 1,α,α²,... span ℚ(α).

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.

To show that dim_ℚℚ(α) is at most n, we must show that 1,α,α²,..,α^n-1 is a basis of ℚ(α).

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

As for P(α) = 0 I am not quite sure what relevance it has.

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.

 

[said]

Is that last thing your notation for the space spanned by {αn|n∈N}\{\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.

My professor uses the notation <1,α,α²,...> but yes it is a spanning set.

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

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?

Recall that a set {x1,…,xr}\{x_1,\dots,x_r\} with all the xix_i distinct is said to be linearly independent if the following implication holds for all scalars a1,…,ara_1,\dots,a_r:

∑i=1raixi=0 ⇒ a1=⋯=ar=0.​

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.

 

[Fredrik]

So, by proving this, does it automatically prove that the subset of cardinality n is linearly dependent?

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).

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.

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.