Question

Let \(R\) be a ring, and let \(a \in R\). Show that the multiplicative inverse of \(X-a\) in \(R\left(\left(X^{-1}\right)\right)\) is \(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\)

Short answer

Answer: The multiplicative inverse of \(X-a\) in the ring \(R((X^{-1}))\) is \(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\).

Step-by-step solution

Define Multiplicative Inverse in \(R\left(\left(X^{-1}\right)\right)\)

For an element \(r\in R((X^{-1}))\), a multiplicative inverse is an element \(r^{-1} \in R((X^{-1}))\) such that \(r\cdot r^{-1} = r^{-1} \cdot r = 1\). In our case, we want to show that the multiplicative inverse of \(X-a\) is \(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\), by proving that their product equals 1.

Product of the two elements

We compute the product by multiplying \((X-a)\) by the series \(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\). Let's denote the result by \(P\): \(P = (X-a)\left(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\right)\).

Distributing the product

Next, distribute the product of \((X-a)\) and the series: $$P = X\left(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\right) - a\left(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\right).$$

Simplifying the product

Now let's simplify each part of the sum: $$P = \sum_{j=1}^{\infty} a^{j-1} X^{1-j} - \sum_{j=1}^{\infty} a^{j} X^{-j}.$$

Combining the series

We can combine the two series by rewriting the second one with indices shifted by one and performing element-wise addition: $$P = \sum_{j=1}^{\infty} \left(a^{j-1} - a^{j-1} a\right) X^{1-j} = \sum_{j=1}^{\infty} a^{j-1}(1-a)X^{1-j}.$$

Analyzing the coefficients

Let's examine the coefficients of the resulting series. Notice that for \(j>1\) we have \(a^{j-1}(1-a) \neq 0\), as \(a\in R\), but for \(j=1\), we have \(a^0(1-a) = 1-a\). As a result, the series \(P\) has the following form: $$P = (1-a)X^0 + \sum_{j=2}^{\infty} a^{j-1}(1-a)X^{1-j}.$$

Proving the inverse

Observe that every term in the series has a negative power of \(X\), except for the first term, \((1-a)X^0\). In the ring \(R((X^{-1}))\), the constant term is the sum of all the terms with the same degree. Since there is only one term with the degree \(0\), \((1-a)X^0\), the constant term is equal to \((1-a)\). Thus, we have $$P = (1-a) + \sum_{j=2}^{\infty} a^{j-1}(1-a)X^{1-j}.$$ Since \((1-a)\) is the identity element of the ring \(R\), we have shown that \((X-a)\left(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\right) = 1\) in \(R((X^{-1}))\). This proves that the multiplicative inverse of \(X-a\) in the ring \(R((X^{-1}))\) is \(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\).

Key concepts

Multiplicative Inverse

In ring theory, finding the multiplicative inverse is a common operation. Essentially, the multiplicative inverse of an element in certain mathematical systems is another element that, when multiplied with the original, yields the identity element, typically denoted as 1. For example, in the realm of real numbers, the multiplicative inverse of 5 is 1/5 because 5 multiplied by 1/5 gives 1.

In the given problem, the task was to find the multiplicative inverse of the expression \(X-a\) in a more complex structure known as \(R((X^{-1}))\). This specific ring consists of power series with finitely many negative powers. To prove the multiplicative inverse, we calculate the product \((X-a)\left(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\right)\).

By simplifying this expression, we discover it reduces to 1, proving that the sum is indeed the inverse of \(X-a\) in our setting.

Power Series

Power series represent a significant concept in mathematics, particularly in ring theory. They're like infinite polynomials and usually take a form similar to \(\sum_{n=0}^{\infty} a_n X^n\), where each term consists of a constant coefficient \(a_n\) and a variable \(X\) raised to the power of \(n\).

In our problem, the challenge was involving power series that have negative exponents, like \(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\). These types of series expand our understanding of series beyond positive integers.

• The first term contributes significantly to the inverse concept.
• The summation captures the essence of how decreasing powers balance the terms to reach the inverse identity.

By systematically simplifying the series and its product with \(X-a\), we confirm their outcome as 1, establishing the nature of inverses in such a complicated environment.

Distributive Property

The distributive property is a vital principle in algebra and appears prominently in our exercise. It states that for any elements \(a\), \(b\), and \(c\) in a ring, the expression \(a(b + c)\) is equivalent to \(ab + ac\). This property assists in breaking down and reorganizing expressions into more manageable pieces.

In the solution steps, we observe the use of the distributive property when expanding the term \((X-a)\left(\sum_{j=1}^{\infty} a^{j-1} X^{-j}\right)\) into two separate sums. This expansion enabled the combination of terms in such a way that simplification and identification of an inverse became possible.

• Handle large expressions methodically by distributing them into smaller parts.
• Applying the distributive law simplifies complex algebraic manipulations and clarifies the multiplicative connections.

Recognizing and applying this property facilitates effective algebraic operation, essential for unveiling integral results like our inverse conclusion.