# Verifying a $K$-basis for a primitive extension $K\subset K(\alpha)$

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!

## 1 Answer

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 $$L$$.

• Do these coefficient on the RHS necessarily need to be independent of each other? – Algebear Mar 5 at 7:01
• 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 $L$. – Alex G. Mar 8 at 19:32