Let $\alpha \in \mathbb{C}$ be a complex number. Let $V = \mathbb{Q}(\alpha)$ be the rational vector space spanned by powers of $\alpha$. That is $V = <1,\alpha,\alpha^2,\ldots>$.
If $P(t)$ is a polynomial of degree $n$ such that $P(\alpha) = 0$, show that $dim_{\mathbb{Q}}V$ is at most $n$.
Here is my take on this question. Please give me some feedback/corrections.
Since $\mathbb{Q}(\alpha) = <1,\alpha,\alpha^2,\ldots>$ we know that $1,\alpha,\alpha^2,\ldots>$ span $\mathbb{Q}(\alpha)$.
To show that $dim_{\mathbb{Q}}\mathbb{Q}(\alpha) $ is at most $n$, we must show that $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ is a bsis of $\mathbb{Q}(\alpha) $. To show it is a basis,
- $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ must span $\mathbb{Q}(\alpha) $
- $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ must be linearly independent.
For span:I would say that since $1,\alpha,\ldots$ spans $\mathbb{Q}(\alpha)$ then $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ spans $\mathbb{Q}(\alpha) $ because its elements are in the set $1,\alpha,\ldots$
For linear independence: I was thinking of using induction but I'm not sure how I should go about it.
As for $P(\alpha) = 0$ I am not quite sure what relevance it has. It tells us that $P(\alpha) = a_0 + a_1\alpha + a_2\alpha^2 + . . . + \alpha_{n-1}α^{n-1} = 0 $ Perhaps it helps showing linear independence since we want $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ to be written as $a_0 + a_1\alpha + a_2\alpha^2 + . . . + a_{n-1}α^{n-1} = 0$ where $a_0 = a_1 = . . . = a_{n-1} = 0$
You have to assume that $P$ has rational coefficients.
There is a theorem that states that any vector space which is spanned by $n$ elements, necessarily has dimension less than or equal to $n$.
Thus you do not need to show linear independence.
All you need to do is find $n$ elements which span $V$. I claim that such a spanning set is $\{ 1, \alpha,..., \alpha^{n-1} \}$. Your proof of this from above is incomplete.
To fully prove spanning, you need to prove that $\alpha^m$ can be written as a linear combination of $1, \alpha, \dots, \alpha^{n-1}$ for all $m \geq n$. To prove this, write out $P(x)=a_0 + \dots + a_nx^n$, and then use the fact that $\alpha^n=-\frac{a_0}{a_n} - \frac{a_1}{a_n}\alpha-\dots - \frac{a_{n-1}}{a_n} \alpha^{n-1}$.