Problem 5
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)
Step-by-Step Solution
Verified 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)
1Step 1: 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.
2Step 2: Case (b): \(1<\) gcd\((n, m) < m\)#
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.
3Step 3: 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 RingUnit ElementZero Divisor
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 \(m\).
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.
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.
Other exercises in this chapter
Problem 3
Let \(R\) be a ring, and let \(a, b \in R\) such that \(a b \neq 0 .\) Show that \(a b\) is a zero divisor if and only if \(a\) is a zero divisor or \(b\) is a
View solution Problem 4
Suppose that \(R\) is a non-trivial ring in which the cancellation law holds in general: for all \(a, b, c \in R,\) if \(a \neq 0\) and \(a b=a c,\) then \(b=c
View solution Problem 6
Let \(D\) be an integral domain, \(m \in \mathbb{Z},\) and \(a \in D .\) Show that \(m a=0\) if and only if \(m\) is a multiple of the characteristic of \(D\) o
View solution Problem 7
Show that for all \(n \geq 1,\) and for all \(a, b \in \mathbb{Z}_{n},\) if \(a \mid b\) and \(b \mid a\), then \(a r=b\) for some \(r \in \mathbb{Z}_{n}^{*}\).
View solution