Question

Let \(p\) be a prime, and consider the ring \(\mathbb{Q}^{(p)}\) (see Example 7.26). Show that every non-zero ideal of \(\mathbb{Q}^{(p)}\) is of the form \(\left(p^{i}\right),\) for some uniquely determined integer \(i \geq 0\)

Short answer

In conclusion, we have shown that every non-zero ideal of the ring \(\mathbb{Q}^{(p)}\) is of the form \(\left(p^{i}\right)\), where \(i\) is a uniquely determined integer greater than or equal to 0. We achieved this by proving that \((p^i)\) is an ideal for each integer \(i\geq0\), demonstrating that any non-zero ideal must be of the form \((p^i)\), and confirming that the corresponding integer \(i\) is unique for each ideal. This result highlights the special structure of the ring of rational numbers with a denominator not divisible by a prime \(p\).

Step-by-step solution

Show \((p^i)\) is an ideal

First, let's show that for every integer \(i \geq 0\), \(\left(p^{i}\right)\) is indeed an ideal of \(\mathbb{Q}^{(p)}\). Consider any two elements \(a, b \in \left(p^{i}\right)\) and \(x \in \mathbb{Q}^{(p)}\). Then, we will check that the following conditions hold: 1. \(a + b \in \left(p^{i}\right)\) 2. \(ax \in \left(p^{i}\right)\) Since \(a, b \in \left(p^{i}\right)\), there exist integers \(m\) and \(n\) such that \(a = \frac{m}{p^{i}}\) and \(b = \frac{n}{p^{i}}\). Now, $$ a + b = \frac{m}{p^{i}} + \frac{n}{p^{i}} = \frac{m + n}{p^{i}}. $$ Since \(m + n\) is also an integer, \(a + b \in \left(p^{i}\right)\), fulfilling condition 1. Next, consider the product \(ax\). Since \(x\in \mathbb{Q}^{(p)}\), we have \(x = \frac{r}{s}\), where \(r, s \in \mathbb{Z}\) and \(p \nmid s\). Thus, $$ ax = \frac{m}{p^{i}} \cdot \frac{r}{s} = \frac{mr}{p^i s}. $$ Because \(p^i s\) is not divisible by \(p\), this fraction is still in \(\mathbb{Q}^{(p)}\). Moreover, since \(mr\) is an integer, \(ax \in \left(p^{i}\right)\), fulfilling condition 2. Since both conditions hold, we concluded that \((p^i)\) is an ideal for every integer \(i\geq0\).

Show every non-zero ideal is of the form \((p^i)\)

Now, we'll show that every non-zero ideal of \(\mathbb{Q}^{(p)}\) must be of the form \((p^i)\). Let \(I\) be a non-zero ideal of \(\mathbb{Q}^{(p)}\), and let \(x \in I\) be a non-zero element. Then, there exist integers \(a\) and \(b\) with \(p \nmid b\) such that \(x = \frac{a}{b}\). Let \(i\) be the highest power of \(p\) that divides \(a\), i.e., \(p^i \nmid a\) and \(p^{i+1} \mid a\). Then, we can write \(a = p^i c\) where \(c\) is an integer not divisible by \(p\). Thus, \(x = \frac{p^{i}c}{b}\). Now, let's multiply \(x\) by the integer \(b\), which is not divisible by \(p\): $$ x \cdot b = \frac{p^i c b}{b} = p^i c. $$ The element \(p^i c\) is in the ideal \(I\) since \(I\) is an ideal and closed under multiplication. Then, the ideal \(I\) contains an element that is a multiple of \(p^i\). Therefore, \(I \subseteq (p^i)\). Next, we show that \((p^i) \subseteq I\). Since \(p^i c \in I\), given that \(I\) is an ideal, we have that for any integer \(d\), \(p^i c \cdot \frac{d}{c} = p^i d \in I\). Therefore, \((p^i) \subseteq I\). By the double containment, we conclude that the ideal \(I\) must be of the form \((p^i)\).

Confirming uniqueness

Let's verify that the integer \(i\) is uniquely determined for any ideal \(I\). If two non-zero ideals \(I\) and \(J\) in \(\mathbb{Q}^{(p)}\) correspond to the same value of \(i\), then \(I = (p^i) = J\). In other words, if there is only one \(i\) that works for a given ideal, the integer that corresponds to the ideal is unique. Thus, every nonzero ideal of \(\mathbb{Q}^{(p)}\) is of the form \((p^i)\), where \(i\) is a uniquely determined integer greater than or equal to 0.

Key concepts

Non-zero ideals

In ring theory, an ideal is a special subset of a ring that maintains certain algebraic properties, such as being closed under addition and multiplication by any ring element. But what exactly is a "non-zero ideal"? Let's break it down.

A non-zero ideal is simply an ideal that contains more than just the zero element. Essentially, it's a subset of numbers in the ring that includes at least one non-zero element. This makes them useful since they can interact with other parts of the ring in interesting ways.

In the case of the ring \(\mathbb{Q}^{(p)}\), non-zero ideals are pivotal. For every non-zero ideal in this ring, we can represent it as \((p^i)\) for some integer \(i \geq 0\). This means each non-zero ideal is generated by some power of the prime number \(p\). The uniqueness comes from the property's power, helping ensure the structure and behavior of the ideals within \(\mathbb{Q}^{(p)}\).

When thinking about non-zero ideals, remember they are not just abstract; they are concrete subsets with well-defined behaviors in the ring. They help form the backbone of many computations and proofs in algebraic structures.

Prime numbers

Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves. They are the building blocks of number theory because every integer can be uniquely factored into a product of primes—a fundamental theorem known as the prime factorization theorem.

The role of primes in ring theory, particularly concerning ideals, is crucial. When considering \(\mathbb{Q}^{(p)}\), the prime numbers dictate the structure of its ideals. Specifically, the elements of these ideals \((p^i)\) involve powers of the prime \(p\).


Using prime numbers, these ideals help in understanding how numbers can be divided and multiplied, especially under constraints like denominator restrictions (as we'll explore soon). The property of being made up of a prime makes these ideals uniquely decomposable, meaning that an ideal generated by \(p^i\) is hardly reducible to other forms.

In the context of \(\mathbb{Q}^{(p)}\), the primes help enforce rules about which numbers can be considered part of these special structures, preserving purity of multiplication and addition within the ideal.

Rational numbers with restricted denominators

What happens when you want to work with fractions but with a twist? That's where rational numbers with restricted denominators come into play. This idea becomes particularly interesting in rings like \(\mathbb{Q}^{(p)}\).

Here, we're talking about fractions that look normal, \(\frac{a}{b}\), but with one crucial condition—the denominator \(b\) cannot be divisible by a particular prime \(p\). This "restriction" emphasizes how rational numbers behave differently under certain algebraic settings.

• This ensures that when operations involve ideals, especially those generated by primes \(p^i\), they remain well-behaved and consistent within the ring.
• By constraining the denominators, these rational numbers avoid divisibility by prime \(p\), helping maintain algebraic purity and facilitating smoother analysis and computations in ring theory.

These restrictions create a neat environment to study how elements interact in ring structures like \(\mathbb{Q}^{(p)}\), where the absence of \(p\) in denominators ensures that the multiplication and addition rules of ideals are preserved. In essence, it's about maintaining order and predictability within complex algebraic systems.