Exercise 3.1

Consider the set $5\mathbb{Z}$ of integer multiples of 5. We have seen that this is not a ring with the usual operations (cf. Example 3.5; it is at best a rng). However, consider the usual addition and the following different multiplication:

$$a\odot b:=\frac{ab}{5}$$

Verify that $5\mathbb{Z}$ is closed with respect to this operation, and that $(5\mathbb{Z},+,\odot )$ is a ring according to Definition 3.1.

Answer

We will first verify that $a\odot b$ is closed in $5\mathbb{Z}$. Let $a,b\in 5\mathbb{Z}$. Then, $a=5k$ and $b=5n$ for some $n,k\in \mathbb{Z}$. Therefore,

$$a\odot b=\frac{ab}{5}=\frac{25nk}{5}=5nk\in 5\mathbb{Z}.$$

Thus, it is closed.

Now, we will verify that it is associative. Let $a,b,c\in 5\mathbb{Z}$. Then,

$$(a\odot b)\odot c=\frac{\frac{ab}{5}c}{5}=\frac{abc}{25}=\frac{a\frac{bc}{5}}{5}=a\odot (b\odot c)$$

Thus, it is associative. Lastly, we will verify that distributivity holds. Let $a,b,c\in 5\mathbb{Z}$. Then,

$$a\odot (b+c)=\frac{a(b+c)}{5}=\frac{ab+ac}{5}=\frac{ab}{5}+\frac{ac}{5}=(a\odot b)+(a\odot c)$$

The proof for right distributivity is similar. Since the addition is derived, the property for addition holds automatically. Thus, $(5\mathbb{Z},+,\odot )$ is a ring.