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Let $K\subset L:=K(\alpha)$ be a primitive field extension of degree $n$ and we define $c_i\in L$ as \begin{align*} \sum_{i=0}^{n-1}c_i x^i=\frac{f^{\alpha}_K}{(x-\alpha)}\in L[x]\quad(1) \end{align*} where $f^{\alpha}_K$ is the minimal polynomial of $\alpha$ over $K$. The goal is to prove that $\{c_i\}_{i\in\{0,\ldots,n-1\}}$ is a $K$-basis for $L$. The book (considering introductions to Galois theory) does not define what a $K$-basis for $L$ is but I suppose it is defined as follows: for a (primitive) field extension $K\subset L$ of degree $n$, a $K$-basis for $L$ is defined as a set $\{c_i\}_{i\in\{0,\ldots,n-1\}}$ such that every element $l\in L$ can be written as a unique (because basis elements are independent) combination $k_0c_0+k_1c_1+\ldots+k_{n-1}c_{n-1}$ with $k_i\in K$. Is this a plausible definition?

Now let's look at how to manipulate the equation at (1). This can be written as $$(x-\alpha)=\frac{f^{\alpha}_K}{\sum_{i=0}^{n-1}c_i x^i}.$$ Since the linear term $(x-\alpha)$ is definitely in $L[x]=K(\alpha)[x]$, the RHS must also be. We make the remark that $\deg f^{\alpha}_K=n$ because $K\subset L$ is a field extension of degree $n$ and also that the degree of the polynomial in $L[x]$ in the denominator is at most $n-1$. Hence, at most a quantity of $n-1$ basis elements $c_i$ is needed. Whether these elements are independent, I wouldn't know; maybe it is shown by contradiction, or by long division arguments. Can someone pull me into the right direction? Thanks for the time!

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Yes, your definition of basis is correct. Here, "basis" is just the usual linear algebra definition of basis, which applies here because $L$ is indeed a $K$-vector space.

Your long division idea is the right direction to head towards. Setting $f^\alpha_K(x) = x^n + d_{n-1}x^{n-1} + \ldots + d_0$, the first few terms of $f^\alpha_K / (x - \alpha)$ are

$$ \frac{f^\alpha_K}{x-\alpha} = x^{n-1} + (d_{n-1} + \alpha)x^{n-2} + (\alpha(d_{n-1} + \alpha) + d_{n-2})x^{n-3} + \ldots$$

Note that the coefficients are increasing degree polynomials in $\alpha$ with $K$-coefficients. Thus, it's easy to construct $1, \alpha, \alpha^2, \ldots, \alpha^{n-1}$ as $K$-linear combinations of them. So we conclude that they are a $K$-basis for $L$.

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  • $\begingroup$ Do these coefficient on the RHS necessarily need to be independent of each other? $\endgroup$ – Algebear Mar 5 at 7:01
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    $\begingroup$ Yes, we can show that they are thanks to some finite dimensional linear algebra. There are $n$ coefficients, and they generate $1, \alpha, \alpha^2, \ldots, \alpha^{n-1}$, which is a basis for $L$. Thus, they generate all of $L$, and $L$ is $n$-dimensional, and there are $n$ of them, so they must be a basis for $L$. $\endgroup$ – Alex G. Mar 8 at 19:32

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