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Suppose $x$ is a transcendental over field $k$ and $k(x)$ is the field of fractions of $k[x]$. Can we explicitly express a basis of the $k$ vector space $k(x)$?

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marked as duplicate by Watson, ervx, Chill2Macht, Shailesh, user91500 Aug 16 '16 at 4:39

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Using the partial fraction decomposition of rational functions we can get an explicit basis, yes. If the field $k$ is not algebraically closed, the description will be dependent on a description of the irreducible polynomials in $k[x]$.

Suppose, then. $k$ is algebraically closed. Then the result about partial fraction decomposition tells us that every rational function can be written as a sum of a polynomial and scalar multiples of fractions of the form $\frac{1}{(x-\alpha)^r}$ with $\alpha\in k$ and $r\geq1$. Thus monomials $x^r$ with $r\geq0$ and these fractions generate $k(x)$ as a vector space. That these rational functions are linearly independent can be shown easily; in fact, this is usually done when one shows that the decomposition exists, to prove it is unique.

For exampe, let us show that the elements of $k(x)$ listed span $k(x)$. Suppose we have a function $f=p/q$, with $p$ and $q$ in $k[x]$ coprime and $q$ not constant. Suppose $q=\prod_{i=1}^n(x-\alpha_i)^{m_i}$, and let $$s_i=\prod_{\substack{1\leq j\leq n\\j\neq i}}(x-\alpha_j)^{m_j}.$$ The polynomials $s_1$, $\dots$, $s_n$ are clearly coprime, so there exist polynomials $p_1$, $\dots$, $p_n$ such that $p=\sum_{i=1}^np_is_i$, and then $$f=\frac{p}{q}=\sum_{i=1}^n\frac{p_is_i}{q}=\sum_{i=1}^n\frac{p_i}{(x-\alpha_i)^{m_i}}$$ We see that to show that out elements span $k(x)$ it is enough to show that they span all rational functions of the form $p/(x-\alpha)^r$ with $p\in k[x]$, $\alpha\in k$ and $r\geq1$.

So let $p\in k[x]$, $\alpha\in k$ and $r\geq1$. Let $y$ be a new variable. The polynomial $p(x+y)$ is an element of $k[x,y]$, and can be written in the form $$p(x+y)=\sum_{i=0}^np_i(x)y^i \tag{$\star$}$$ for some $n\geq0$ and $p_0$, $\dots$, $p_n\in k[x]$. It follows that $$p(x) = p(\alpha+(x-\alpha))=\sum_{i=0}^np_i(\alpha)(x-\alpha)^i$$ and then clearly $p(x)/(x-\alpha)^r$ is a linear combination of functions of the form $1/(x-\alpha)^i$ with $i\in\mathbb Z$.

You can show uniqueness :-)

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  • $\begingroup$ (The right hand side if ($\star$) is essentially the Taylor development of $p$ around $\alpha$, provided the characteristic of $k$ is zero —and in that case $p_i=p^{(i)}/i!$, of course— but in general that breaks down in interesting ways. The polynomials $p_i$ are called the Dieudonné derivatives of $p$) $\endgroup$ – Mariano Suárez-Álvarez Apr 7 '14 at 4:18

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