suppose $K|F$ is a field extension & $\alpha \in K$ is such that $[F(\alpha):F]>=n$,
if $\lambda_1,...,\lambda_n$ are distinct scalars of $F$,prove that $\frac{1}{\alpha-\lambda_1},...,\frac{1}{\alpha-\lambda_n}$ are linear independent over $F$.
from $[F(\alpha):F]>=n$ i get that $\alpha$ is not the root of a polynomial over $F$ with degree less than $n$ and i did a lot of calculation and play with the equation,but i failed.
is there any hint?
thank you.
If you set an arbitrary $F$-linear combination of the $\frac{1}{\alpha-\lambda_i}$'s equal to zero and multiply by the product of the $\alpha-\lambda_i$'s, then you get an $n-1$ degree polynomial in $\alpha$ equal to zero. Thus the polynomial is zero. (The product of the $\alpha-\lambda_i$'s is not zero because $\alpha$ is not in $F$.)
edit:
Let $\sigma_{i,n}$ be the $i$th elementary symmetric polynomial on $n$ variables.
$ \begin{align*} 0 &=\sum_{k=1}^nA_k\prod_{1\leq j\leq n,j\neq k}(\alpha-\lambda_j)\\ &=\sum_{k=1}^nA_k\sum_{i=0}^{n-1}(-1)^{n-1-i}\sigma_{i,n-1}(\lambda_j;1\leq j\leq n,j\neq k)\alpha^i\\ &=\sum_{i=0}^{n-1}\left[\sum_{k=1}^nA_k(-1)^{n-1-i}\sigma_{i,n-1}(\lambda_j;1\leq j\leq n,j\neq k)\right]\alpha^i \end{align*} $
Since $[F(\alpha):F]\geq n$, we know that all the coefficients are $0$.
So $\sum_{k=1}^nA_k(-1)^{n-1-i}\sigma_{i,n-1}(\lambda_j;1\leq j\leq n,j\neq k)=0$ for all $1\leq i\leq n-1$. Let's apply Gaussian elimination to the matrix
$ M_n=\left[\begin{array}{ccccc} 1 & -\sum_{j\neq 1}\lambda_j & \sum_{1\leq j_1<j_2\leq n,j\neq 1}\lambda_{j_1}\lambda_{j_2} &\cdots & (-1)^{n-1}\prod_{j\neq 1}\lambda_{j}\\ 1 & -\sum_{j\neq 2}\lambda_j & \sum_{1\leq j_1<j_2\leq n,j\neq 2}\lambda_{j_1}\lambda_{j_2} &\cdots & (-1)^{n-1}\prod_{j\neq 2}\lambda_{j}\\ 1 & -\sum_{j\neq 3}\lambda_j & \sum_{1\leq j_1<j_2\leq n,j\neq 3}\lambda_{j_1}\lambda_{j_2} &\cdots & (-1)^{n-1}\prod_{j\neq 3}\lambda_{j}\\ 1 & -\sum_{j\neq 4}\lambda_j & \sum_{1\leq j_1<j_2\leq n,j\neq 4}\lambda_{j_1}\lambda_{j_2} &\cdots & (-1)^{n-1}\prod_{j\neq 4}\lambda_{j}\\ \vdots &\vdots &\vdots &\ddots\\ 1 & -\sum_{j\neq n}\lambda_j & \sum_{1\leq j_1<j_2\leq n,j\neq n}\lambda_{j_1}\lambda_{j_2} &\cdots & (-1)^{n-1}\prod_{j\neq n}\lambda_{j}\\ \end{array}\right] $
Subtracting the first row from all the others gives
$ \left[\begin{array}{ccccc} 1 & -\sum_{j\neq 1}\lambda_j & \sum_{1\leq j_1<j_2\leq n,j\neq 1}\lambda_{j_1}\lambda_{j_2} &\cdots & (-1)^{n-1}\prod_{j\neq 1}\lambda_{j}\\ 0 & \lambda_2-\lambda_1 & -(\lambda_2-\lambda_1)\sum_{j\neq 2}\lambda_j &\cdots & (-1)^{n-2}(\lambda_2-\lambda_1)\sum_{1\leq j_1<\ldots<j_{n-2}\leq n,j\neq 2}\prod_{1\leq i\leq n-2}\lambda_{j_i}\\ 0 & \lambda_3-\lambda_1 & -(\lambda_3-\lambda_1)\sum_{j\neq 3}\lambda_j &\cdots & (-1)^{n-2}(\lambda_3-\lambda_1)\sum_{1\leq j_1<\ldots<j_{n-2}\leq n,j\neq 3}\prod_{1\leq i\leq n-2}\lambda_{j_i}\\ 0 & \lambda_4-\lambda_1 & -(\lambda_4-\lambda_1)\sum_{j\neq 4}\lambda_j &\cdots & (-1)^{n-2}(\lambda_4-\lambda_1)\sum_{1\leq j_1<\ldots<j_{n-2}\leq n,j\neq 4}\prod_{1\leq i\leq n-2}\lambda_{j_i}\\ \vdots &\vdots &\vdots &\ddots\\ 0 & \lambda_n-\lambda_1 & -(\lambda_n-\lambda_1)\sum_{j\neq n}\lambda_j &\cdots & (-1)^{n-2}(\lambda_n-\lambda_1)\sum_{1\leq j_1<\ldots<j_{n-2}\leq n,j\neq n}\prod_{1\leq i\leq n-2}\lambda_{j_i}\\ \end{array}\right] $
Since the $\lambda$'s are all distinct, we can divide each row by $\lambda_k-\lambda_1$ without affecting the invertibility of the matrix. By induction, we see that this matrix is invertible. Thus all the $A_k$'s must be zero.
I think that this matrix is related to the inverse of the Vandermonde matrix, but I'm not sure how.
edit2: The induction proof. The $1\times 1$ case is the matrix $[1]$, which is invertible. Suppose that the $(n-1)\times(n-1)$ matrix $M_{n-1}$ is invertible. Then the $n\times n$ matrix $M_n$ given above has $\det M_n=\left(\prod_{k\neq 1}\lambda_1-\lambda_k)\right)\det M_{n-1}$ by the row reduction shown above. Since the $\lambda$'s are distinct and $\det M_{n-1}\neq 0$, we see that $\det M_n\neq 0$ as well.
edit3: I'll demonstrate the $2\times2$ and $3\times3$ cases:
$\det M_2=\left\vert\begin{array}{cc} 1 & -\lambda_2\\ 1 & -\lambda_1\\ \end{array}\right\vert=\left\vert\begin{array}{cc} 1 & -\lambda_2\\ 0 & \lambda_2-\lambda_1\\ \end{array}\right\vert=\lambda_2-\lambda_1 $
and
$\det M_3=\left\vert\begin{array}{ccc} 1 & -(\lambda_2+\lambda_3) & \lambda_2\lambda_3\\ 1 & -(\lambda_3+\lambda_1) & \lambda_3\lambda_1\\ 1 & -(\lambda_1+\lambda_2) & \lambda_1\lambda_2\\ \end{array}\right\vert=\left\vert\begin{array}{ccc} 1 & -(\lambda_2+\lambda_3) & \lambda_2\lambda_3\\ 0 & \lambda_2-\lambda_1 & -(\lambda_2-\lambda_1)\lambda_3\\ 0 & \lambda_3-\lambda_1 & -(\lambda_3-\lambda_1)\lambda_2\\ \end{array}\right\vert=(\lambda_2-\lambda_1)(\lambda_3-\lambda_1)\left\vert\begin{array}{ccc} 1 & -(\lambda_2+\lambda_3) & \lambda_2\lambda_3\\ 0 & 1 & -\lambda_3\\ 0 & 1 & -\lambda_2\\ \end{array}\right\vert=(\lambda_2-\lambda_1)(\lambda_3-\lambda_1)(\lambda_3-\lambda_2) $
Note that the $2\times 2$ minor of $M_3$ is a copy of $M_2$. So $\det M_3=(\lambda_2-\lambda_1)(\lambda_3-\lambda_1)\det M_2$