Question

Let \(R\) be an arbitrary ring, let \(a_{1}, \ldots, a_{\ell} \in R,\) and let $$ f:=\left(X-a_{1}\right)\left(X-a_{2}\right) \cdots\left(X-a_{\ell}\right) \in R[X] $$ For \(j \geq 0,\) define the "power sum" $$ s_{j}:=\sum_{i=1}^{\ell} a_{i}^{j} $$ Show that in the ring \(R\left(\left(X^{-1}\right)\right)\), we have $$ \frac{\mathbf{D}(f)}{f}=\sum_{i=1}^{\ell} \frac{1}{\left(X-a_{i}\right)}=\sum_{j=1}^{\infty} s_{j-1} X^{-j} $$ where \(\mathbf{D}(f)\) is the formal derivative of \(f\).

Short answer

Question: Prove that in the ring \(R\left(\left(X^{-1}\right)\right)\), the equation \(\frac{\mathbf{D}(f)}{f} = \sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)} = \sum_{j=1}^{\infty} s_{j-1} X^{-j}\) holds, where \(R\) is an arbitrary ring, \(a_1, \ldots, a_\ell\) are elements in \(R\), \(f \in R[X]\), \(f(X) = \prod_{i=1}^{\ell} (X - a_i)\), \(\mathbf{D}(f)\) is the formal derivative of \(f\), and \(s_j = \sum_{i=1}^{\ell} a_i^j\). Solution: We have proved that \(\frac{\mathbf{D}(f)}{f} = \sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)} = \sum_{j=1}^{\infty} s_{j-1} X^{-j}\) in the ring \(R\left(\left(X^{-1}\right)\right)\) by following these steps: 1. Found the formal derivative of \(f\): \(\mathbf{D}(f) = \ell\). 2. Computed \(\frac{\mathbf{D}(f)}{f}\): \(\frac{\mathbf{D}(f)}{f} = \frac{\ell}{\prod_{i=1}^{\ell} (X - a_i)}\). 3. Showed that \(\frac{\mathbf{D}(f)}{f} = \sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)}\). 4. Proved that \(\sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)} = \sum_{j=1}^{\infty} s_{j-1} X^{-j}\).

Step-by-step solution

Find the formal derivative of \(f\)

By definition of formal derivative, $$\mathbf{D}(f) = \sum_{k=1}^{\ell} \mathbf{D}\left(\prod_{i=1, i\neq k}^{\ell} (X - a_i)\right).$$ Now since differentiation is a linear operation, we can apply the product rule of differentiation to find the derivative of \(\prod_{i=1, i\neq k}^{\ell} (X - a_i)\). Notice that for each \(k = 1,\ldots, \ell\), the derivative of \(\prod_{i=1, i\neq k}^{\ell} (X - a_i)\) with respect to \(X\) is equal to \(\prod_{i=1, i\neq k}^{\ell} (1)\), which is just \(1\). Thus, the formal derivative of \(f\) is given by $$\mathbf{D}(f) = \sum_{k=1}^{\ell} 1 = \ell.$$

Compute \(\frac{\mathbf{D}(f)}{f}\)

Next, we need to compute the quotient \(\frac{\mathbf{D}(f)}{f}\). Recall that \(f = \prod_{i=1}^{\ell} (X - a_i)\) and \(\mathbf{D}(f) = \ell\). Therefore, $$\frac{\mathbf{D}(f)}{f} = \frac{\ell}{\prod_{i=1}^{\ell} (X - a_i)}.$$

Show that \(\frac{\mathbf{D}(f)}{f} = \sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)}\)

We will now show that \(\frac{\mathbf{D}(f)}{f} = \sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)}\). Observe that for \(i = 1,\ldots, \ell\), the partial fraction decomposition of \(\frac{\ell}{\prod_{i=1}^{\ell} (X - a_i)}\) is given by $$\frac{\ell}{\prod_{i=1}^{\ell} (X - a_i)} = \sum_{i=1}^{\ell} \frac{A_i}{X - a_i},$$ where \(A_i \in R\) for all \(i = 1,\ldots, \ell\). Then for each \(i\), we have that \(A_i(X - a_i) = \ell\), so \(A_i = \frac{\ell}{X-a_i}\). It follows that $$\frac{\mathbf{D}(f)}{f} = \sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)}.$$

Prove that \(\sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)} = \sum_{j=1}^{\infty} s_{j-1} X^{-j}\)

To show that \(\sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)} = \sum_{j=1}^{\infty} s_{j-1} X^{-j}\), we will express the left side of the equation in terms of the power sum \(s_{j-1}\). Notice that for each \(i = 1,\ldots, \ell\), the term \(\frac{1}{\left(X-a_i\right)}\) can be expanded as a geometric series: $$\frac{1}{\left(X-a_i\right)} = \frac{1}{X}\frac{1}{1 - \frac{a_i}{X}} = \frac{1}{X}\left(1 + \frac{a_i}{X} + \frac{a_i^2}{X^2} + \frac{a_i^3}{X^3} + \cdots\right).$$ Then, summing over all \(i = 1,\ldots, \ell\), we get $$\sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)} = \sum_{i=1}^{\ell} \left(\frac{1}{X} + \frac{a_i}{X^2} + \frac{a_i^2}{X^3} + \cdots\right) = \sum_{j=1}^{\infty} s_{j-1} X^{-j},$$ where \(s_{j-1} = \sum_{i=1}^{\ell} a_i^{j-1}\). Therefore, we have shown that \(\frac{\mathbf{D}(f)}{f} = \sum_{i=1}^{\ell} \frac{1}{\left(X-a_i\right)} = \sum_{j=1}^{\infty} s_{j-1} X^{-j}\) in the ring \(R\left(\left(X^{-1}\right)\right)\).

Key concepts

Ring Theory

When delving into the world of abstract algebra, ring theory emerges as a fundamental concept. It studies structures known as rings, which are sets equipped with two binary operations that we typically think of as addition and multiplication. A ring must satisfy several properties, notably, it should be closed under both operations, associative, and have an additive identity. Moreover, the distributive property must connect the two operations.

Think of the ring of polynomials with coefficients in an arbitrary ring R, often denoted as R[X]. This particular ring serves as the playground for our exercise, dealing with polynomials and the operations on them like the formal derivative, which we will later explore during this article.

Power Sum

A power sum is a specific type of series that holds great significance in number theory and combinatorics. In general, it takes the form s_j = a₁^j + a₂^j + ... + a_ℓ^j for some fixed natural number j. In our given problem, the power sum appears in a polynomial context, represented within a ring.

The calculation of power sums has a broad range of applications, such as in symmetric function theory and even in solving polynomial equations. The intrigue behind power sums stems from their capability to be utilized in describing the coefficients of polynomials and in the process of symmetric polynomial decomposition.

Formal Derivative

The concept of a formal derivative may sound sophisticated, but it's analogous to the familiar derivative from calculus, with an abstraction twist. Here, we strip away the notion of 'limits' and focus on the symbolic manipulation of polynomials. Like in calculus, the formal derivative of a polynomial is obtained by applying the power rule to each term.

For instance, given a polynomial f in the ring R[X], its formal derivative D(f) is computed algebraically, applying the derivative rules term-wise. This process doesn't consider convergence or continuity; it's purely formal, hence the name. The solution involves taking the formal derivative of a product of terms in the form (X - a_i), where the a_i's are elements of the ring.

Partial Fraction Decomposition

When faced with a complex rational function, partial fraction decomposition is the tool we reach for in order to simplify it into a sum of simpler fractions. It’s especially useful for integration and solving certain equations in calculus, but its utility extends to other areas like ring theory.

Applied to a polynomial in the form of a quotient, as seen in our exercise, partial fraction decomposition breaks down a single complex fraction into several simpler fractions, such that each denominator corresponds to a factor of the original denominator. In the context of ring theory and polynomials over rings, this technique allows us to express complicated algebraic relationships in a more manageable and insightful manner. It plays a central role in demonstrating the relation between the formal derivative of a polynomial and its factors.