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Bypassing some of the background on fields we begin with the idea of separability. Separability is a relationship between two fields in an extension. Given a field extension ${K \subset{F}}$ , we may ask if elements of $F$ are separable over $K$ , ie: is it true given an element $u \in F$ , that the minimal polynomial $f(x) \in K[x]$ splits into distinct linear factors over some splitting field of $f(x)$ . If every element of $F$ is separable over $K$ , we say that the extension ${K \subset{F}}$ is separable, or that $F$ is a separable extension of $K$ .

Let’s let $K$ be of characteristic $0$ , and $F$ be an algebraic extension of $K$ , ie: every element of $F$ is the root of some polynomial in $K$ . Under what conditions, we might ask, is the extension separable? Calculus answers the question in part. Letting $u \in F$ with minimal polynomial (monic) $f(x)$ we can test for multiple roots by computing the derivative of ${f(x)= a_n x^n + \cdots + a_1 x + a_0}$ given by: ${f'(x)= n a_n x^{n - 1} + \cdots + 2 a_2 x + a_1}$ . Letting $f(x)$ factor in the splitting field as:
$f(x)=(x-u_1)^{n_1}(x-u_2)^{n_2}...(x-u_j)^{n_j}$
Then the derivative is given by: $f'(x)=\sum_{k=1}^{j}(x-u_k)^{n_k-1}\prod_{i \neq k} (x-u_i)^{n_i}$ where each $n_i$ is understood to be at least one. So, let’s consider $(f,f')$ , the GCD of our minimal polynomial and f. Unsurprisingly, if the GCD is not one then a linear factor of f must divide both $f$ and $f'$ so that the degree of that factor must be at least two, ie: $f$ has at least a double root. Further, since $1 \neq 0$ in a field and so by $char(K)=0$ , $n 1 \neq 0$ for any positive integer and hence $f'$ is not equal to zero.
So, since the derivative is also in $K[x]$ we know that if $f$ has u as a root with multiplicity then $f'(u)=0$ , with $f'$ non-zero, ⇒⇐ ( $f$ is minimal). Also if the other roots have multiplicity in $f$ then the minimal polynomial for these roots divides $f$ . These must have simple roots (not multiple), but $f$ is irreducible and so the minimal polynomial must be $f$ , ⇒⇐. So $f$ is separable over K.
Now what can we say if characteristic of $K$ is non-zero, one might ask? Since $char(K)=p$ in this case, we might consider whether it is possible for an irreducible polynomial’s derivative to be zero, in contrast to the case when $char(K)=0$ . Starting with the simple example of a field of prime order $|K|=p$ , Fermat’s little theorem implies that if a polynomial has terms $x^n$ with $p|n$ , these terms may be replaced with $x^m$ , for some power $p \nmid m$ , as $x^p \equiv x$ $\mod{p}$ . So an irreducible polynomial over these fields has non-zero derivative and the same argument applied above shows that $f$ is separable.

In Chapter two of Euler’s “Intro to the Analysis of the Infinites”, there is a technique to aid for quickening the rote calculation of partial fractions that a student might appreciate when confronted with a series of these type of problems. This technique is first introduced in section 41 continuing to 45 of chapter two of the text.

Supposing one starts with a fraction with the degree of the numerator M, less than the denominator N, and N is a product of factors one of which being at least “simple” as Euler puts it (linear, ie: ${pq-z}$ )

${\frac{M}{N}=\frac{A}{(p-qz) }+ \frac{P}{S}}$

${ \frac{M}{(p-qz)S}=\frac{A}{(p-qz) }+ \frac{P}{S}}$

${\frac{(M-AS)}{(p-qz)S}=\frac{P}{S}}$

${(p-qz) \vert {(M-AS)}}$

${z=p/q \Rightarrow M-AS=0}$

${A=M(\frac{p}{q})/S(\frac{p}{q})}$

(Note: Euler’s “Introduction to the Analysis of the Infinites”, courtesy of Ian Bruce, is available on his webpage along with a lot of other classic works by the “master of us all”, as Pierre-Simon Laplace would say. ) http://www.17centurymaths.com/contents/euler/introductiontoanalysisvolone/ch2vol1.pdf

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Field theory: Separability

Euler Partial Fractions Method