Is 1, α, α², ..., αⁿ⁻¹ a Basis for ℚ(α)?

[said]

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.Here is my attempt to solve 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

 

[caffeinemachine]

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.Here is my attempt to solve 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). Mistake here.
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,... This reasoning is not correct.[/color]

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

Hello smoha. Welcome to MHB.

It is strange that you questioned had not been answered earlier. Probably because you hadn't used LaTeX. Check out the LaTeX forum of our site. It doesn't take much time to get started.

Now.

As for the first mistake, note that a spanning set for a vector space is not necessarily a basis.

The second one is a bit more serious. What you have essentially argued is that a subset of a spanning set of a vector space is a basis.
This is clearly not true.

I can walk you through the proof once you understand these mistakes.

 

[said]

I simply do not see the mistakes that you pointed out.

As for the first one, I don't see where I implied that a spanning set of a vector space is a basis.

For the second one, I only said that we must show that this subspace is a basis. I did not say it is a basis.

 

[caffeinemachine]

I simply do not see the mistakes that you pointed out.

As for the first one, I don't see where I implied that a spanning set of a vector space is a basis.

For the second one, I only said that we must show that this subspace is a basis. I did not say it is a basis.

The line where I pointed out the first mistake reads:

To show that $\dim_{\mathbb Q}\mathbb Q(\alpha)$ is atmost $n$, we need to show that $1, \alpha, \ldots, \alpha^{n-1}$ is basis.

The reason why I called this a mistake is that: To show that $\dim_{\mathbb Q}\mathbb Q(\alpha)$ is atmost $n$, we only need to show that $1, \alpha, \ldots, \alpha^{n-1}$ spans $\mathbb Q(\alpha)$.

May be calling this a mistake was a bit extreme. Forgive me for that.

Now for the second mistake.

You argue that since $1, \alpha, \alpha^2,\ldots$ spans $\mathbb Q(\alpha)$, so does $1,\alpha,\alpha^2,\ldots, \alpha^{n-1}$.

This to me sounds like: Since $S\subseteq V$ spans a vector space $V$, therefore $T\subseteq S$ also spans $V$.

Can you please elaborate on your reasoning so that I can understand your question better.