Assume that you have an integral of the form , where are polynomials. It is well-known that the best way to solve this is to do a partial fraction decomposition of , which allows you to find an antiderivative of ; this can then be used to evaluate . Partial fraction decomposition of allows you to write as a sum of a polynomial with a linear combination of terms of the form , where the range over the (complex) roots of and lies between 1 and the multiplicity of the root of . Now,

whence finding an antiderivative of this is easy.

In case one has , one has that is a root of as well, and one can combine the terms and to a term , multiplied by a polynomial in of degree . This allows to give a partial fraction decomposition of purely in term of real polynomials. Note that is the minimal polynomial of over .

More generally, one can consider polynomials over an arbitrary field . Now will in general not split into linear (or at most quadratic) factors over . But nonetheless, one can still do a partial fraction decomposition. Moreover, one can continue this to another level of abstractness by working with arbitrary principal ideal domains.

The main ingredient which we need is the Chinese Remainder Theorem:

Let be a principal ideal domain and let . Write , where are pairwise coprime elements. Then

is an isomorphism. Moreover, there exist such that divides , that the inverse of the above isomorphism is given by

In case is Euclidean, one can choose the such that .

Consider , . We first show that . For that, note that

Now the are pairwise coprime, whence this is equivalent to dividing . Hence, . Therefore, induces a monomorphism . We have to show that is surjective, or alternatively, that it admits an inverse.

Let . We have that is coprime to ; hence, we have a Bézout identity, i.e. we can write with ; in case is Euclidean, we can also assume that , (see my post on the Extended Euclidean Algorithm). But this means that is zero in for all , in . Set ; therefore, , whence , as we had to show.

We need another simple lemma, which works for a certain class of Euclidean rings such as with for some fixed , and with . The two properties we need are the following:

We will say that an Euclidean domain is *admissible* if is multiplicative and for all , one can write with such that and .

Let be an admissible Euclidean domain and . Any can be written in the form with , , where .

We show this by induction on . For , , whence . Hence, is of the required form. Now, assume that . Write with , and . Hence, and , whence by induction, we can write with suitable , . If we set , we obtain with as required.

Now partial fraction decomposition is a direct corollary of these results:

Let be a principal ideal domain with field of fractions , and let be two elements. Assume that , where , are pairwise coprime polynomials and . Then there exist elements such that

In case is Euclidean with a multiplicative degree function, we can assume that for , .

As an example, we want to consider the Partial Fraction Decomposition for , i.e. for , , . Now and , whence

is the partial fraction decomposition of in .

We can assume that by replacing by .

By the Chinese Remainder Theorem, the map , is an isomorphism. The inverse is given by for some ; in case is Euclidean, can be chosen such that . Therefore, there exist (where in case is Euclidean, ) and such that

Dividing by , we get

This shows the claim in case is not Euclidean.

In case is Euclidean, we can write with ; by adding to , we can therefore assume that satisfies . By the previous lemma, we can now write with and ; as , we get . Therefore, we obtain with , which gives

and by reindexing the second sum we obtain the formula from the statement of the theorem.

Now assume that we have a rational function written as

where are pairwise coprime, with . Then

In case is prime, this can be considered an element in the localization of at the prime ideal . Let be the completion of . Note that both and have the residue field .

In the special case that , we see that . Hence, since is a prime element in , we obtain an isomorphism with . In , the element corresponds to

In particular, we can use the Hasse derivative of respectively (the definition can be translated to power series rings as well) to obtain

In the case , this is Heaviside's formula.

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