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!


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 $L$.

  • $\begingroup$ Do these coefficient on the RHS necessarily need to be independent of each other? $\endgroup$ – Algebear Mar 5 at 7:01
  • 1
    $\begingroup$ 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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