Question

Let \(R\) be a ring of characteristic \(m>0,\) and let \(n\) be an integer. Show that: (a) if \(\operatorname{gcd}(n, m)=1,\) then \(n \cdot 1_{R}\) is a unit; (b) if \(1<\operatorname{gcd}(n, m)

Short answer

A zero divisor? Equal to $0$? Answer: - If gcd$(n,m)=1$, then $n\cdot 1_R$ is a unit. - If $1<$ gcd$(n,m)<m$, then $n\cdot 1_R$ is a zero divisor. - If gcd$(n,m)=m$ or $m|n$, then $n\cdot 1_R=0$.

Step-by-step solution

Case (a): gcd\((n, m) = 1\)#

In this case, since gcd\((n, m) = 1\), there exist integers \(x\) and \(y\) such that \(nx + my = 1\). We want to show that \(n \cdot 1_R\) is a unit. Consider the element \(x \cdot 1_R\) in the ring \(R\). We have: $$(n \cdot 1_R)(x \cdot 1_R) = (nx) \cdot 1_R = ((1-my) \cdot 1_R) = 1_R - my \cdot 1_R.$$ Now, recall that the characteristic of \(R\) is \(m\). Therefore, \(m \cdot 1_R = 0\), and so, $$(n \cdot 1_R)(x \cdot 1_R) = 1_R - my \cdot 1_R = 1_R - 0 = 1_R.$$ Since \(n \cdot 1_R\) has an inverse, namely \(x \cdot 1_R\), we conclude that \(n \cdot 1_R\) is a unit.

Case (b): \(1

In this case, we have \(1<\) gcd\((n, m) < m\). Let \(d = \)gcd\((n, m)\), and write \(n = da\) and \(m = db\) for some integers \(a\), \(b\). Note that gcd\((a, b) = 1\). We want to show that \(n \cdot 1_R\) is a zero divisor. Since the characteristic of \(R\) is \(m\), we know that \(m \cdot 1_R = 0\). So, $$n \cdot 1_R = (da) \cdot 1_R = d(a \cdot 1_R) = (db) \cdot (a \cdot 1_R).$$ Now, since \(d > 1\), we have \((db) \cdot (a \cdot 1_R) \neq 0\), but $$n \cdot 1_R = (db) \cdot (a \cdot 1_R) = d(ba \cdot 1_R) = d(m \cdot 1_R) = d \cdot 0 = 0.$$ Thus, we have a nontrivial product \((db) \cdot (a \cdot 1_R) = 0\), and so \(n \cdot 1_R = 0\) is indeed a zero divisor.

Case (c): Otherwise: gcd\((n, m) = m\), or \(m|n\)#

In this case, we have gcd\((n, m) = m\), i.e., \(m|n\). We want to show that \(n \cdot 1_R = 0\). Since \(m|n\), there exists an integer \(k\) such that \(n = km\). Now, recall that the characteristic of \(R\) is \(m\). Therefore, \(m \cdot 1_R = 0\), and we have: $$n \cdot 1_R = (km) \cdot 1_R = k(m \cdot 1_R) = k \cdot 0 = 0,$$ as desired.

Key concepts

Characteristic of a Ring

In the fascinating world of ring theory, the characteristic of a ring is an essential concept. Imagine a set equipped with two operations: addition and multiplication, known as a ring. Now, one might wonder about how repetitive addition, like multiplying a unit by itself several times, behaves within this set.

• The characteristic of a ring is the smallest number of times you need to add the ring's multiplicative identity to itself to get zero.
• This number is denoted by \(m\), and it's always either a positive integer or zero, which means infinite times would be needed to reach zero.
• For example, if \(1_R + 1_R + \, ... \, + 1_R = 0\) after \(m\) times, then the characteristic is \(m\).

Understanding this property helps clarify other ring behaviors, especially in the provided problem, where it connects deeply with concepts like units and zero divisors.

Unit Element

A unit in a ring is an element that has an inverse concerning multiplication. If you multiply a unit by a particular element, the result will be the identity element of the ring.

• In mathematical terms, an element \(n \cdot 1_R\) is a unit if there exists another element \(y\) such that \((n \cdot 1_R)(y) = 1_R\).
• This idea is akin to multiplying a number by its reciprocal to get one, but we're doing it within the framework of a ring.
• In our exercise, when the greatest common divisor pr gcd of \(n\) and the characteristic \(m\) is 1, \(n\cdot 1_R\) turns out to be a unit.

By learning about units, students get to see how ring structures incorporate familiar concepts from basic arithmetic but add fascinating nuances.

Zero Divisor

In the realm of ring theory, zero divisors play a critical role. A zero divisor is a non-zero element that becomes zero when multiplied by another non-zero element within the ring.

• For the element \(n \cdot 1_R\) to be a zero divisor, it should satisfy \((n \cdot 1_R)(z) = 0\) for some \(z eq 0\).
• In our specific exercise, when \(1<\operatorname{gcd}(n, m)
• Zero divisors highlight interesting properties of ring elements, especially in rings that are not fields, where multiplication usually doesn't have full flexibility.

By grasping what zero divisors are, students understand how elements can interact strangely in non-intuitive ways.