2.3: Divisibility in Integral Domains
In this section, we'll seek to answer the questions:
- What multiplicative properties can we generalize from \(\mathbb{Z}\) to any integral domain?
- What are the differences between a prime and irreducible element in a commutative ring?
When we introduced the notion of integral domain, we said that part of the reason for the definition was to capture some of the most essential properties of the integers. This is the heart of abstraction and generalization in mathematics: to distill the important properties of our objects of interest and explore the consequences of those properties. One such important property of \(\mathbb{Z}\) is cancellation .
Let \(R\) be a ring. Then \(R\) is a domain if and only if for all \(a,b,c\in R\) with \(c\ne 0\) and \(ac = bc\text{,}\) we have \(a = b\text{.}\)
We may read Theorem 2.3.1 as saying that the defining property of an integral domain is the ability to cancel common nonzero factors. Note that we have not divided ; division is not a binary operation, and nonzero elements of rings need not be units. However, as was the case in \(\mathbb{Z}\text{,}\) there are notions of divisibility and factorization in rings.
Let \(R\) be a commutative ring with identity, and let \(a,b\in R\text{.}\) We say \(a\) divides \(b\) and write \(a\mid b\) if there is a \(c\in R\) such that \(ac = b\text{.}\) We then say that \(a\) is a factor of \(b\text{.}\)
Find all factors of \(\overline{2}\) in the following rings:
- \(\displaystyle \mathbb{Z}_5\)
- \(\displaystyle \mathbb{Z}_6\)
- \(\displaystyle \mathbb{Z}_{10}\)
Our definition of prime also extends nicely to domains. Indeed, the desire to extend the familiar notion of prime from \(\mathbb{Z}\) to any ring is the reason for our less-familiar definition given in Definition: Prime .
Let \(R\) be a domain. We say a nonzero nonunit element \(a\in R\) is prime if whenever \(a\mid bc\) for some \(b,c\in R\text{,}\) either \(a|b\) or \(a|c\text{.}\)
A notion related to primality is irreducibility. In fact, one might reasonably say that irreducibility is the natural generalization of the typical definition of prime one encounters in school mathematics.
Let \(R\) be a domain. We say a nonzero nonunit element \(a\in R\) is irreducible if whenever \(a = bc\) for some \(b,c\in R\text{,}\) one of \(b\) or \(c\) is a unit. (Note that in some areas of the literature, the word atom is used interchangeably with irreducible.)
Find the units, primes, and irreducibles in the following rings.
- \(\displaystyle \mathbb{R}\)
- \(\displaystyle \mathbb{Z}\)
- \(\displaystyle \mathbb{Z}_{5}\)
- \(\displaystyle \mathbb{Z}_6\)
In domains, all primes are irreducible.
Let \(R\) be a domain. If \(a\in R\) is prime, then \(a\) is irreducible.
In familiar settings, the notion of prime and irreducible exactly coincide.
Every irreducible in \(\mathbb{Z}\) is prime.
Despite their overlap in familiar settings, primes and irreducibles are distinct types of elements. As the next exploration demonstrates, not all primes are irreducible. What is more, Exploration 2.3.3 will show that not all irreducibles are primes, even in domains!
Find an example of a ring \(R\) and prime \(p\in R\) such that \(p\) is not irreducible.
Consider the set \(R\) of all polynomials in \(\mathbb{Z}[x]\) for which the coefficient on the linear term is zero. That is,
\begin{equation*} R = \{a_0 + a_2 x^2 + \cdots + a_{n-1} x^{n-1} + a_n x^n: a_i\in \mathbb{Z},\ n\in\mathbb{N}_0 \}\text{.} \end{equation*}
(You should convince yourself that \(R\) is an integral domain, but do not need to prove it.) Then, find a polynomial of the form \(x^n\) in \(R\) that is irreducible, but not prime.
Our last straightforward generalization from the multiplicative structure of \(\mathbb{Z}\) is the notion of greatest common divisor. As our next definition again demonstrates, our careful work in the context of \(\mathbb{Z}\) generalizes nicely to all domains. Indeed, we intentionally did not appeal to \(\le\) to define the greatest common divisor in Definition: Greatest Common Divisor , as not all rings have a natural order relation like \(\mathbb{Z}\) does.
Let \(R\) be a domain, and let \(a,b\in R\text{.}\) A nonzero element \(d\in R\) is a greatest common divisor of \(a\) and \(b\) if
- \(d\mid a\) and \(d\mid b\) and,
- if \(e\in R\) with \(e\mid a\) and \(e\mid b\text{,}\) then \(e\mid d\text{.}\)
Let \(R\) be a domain and \(a,b\in R\) and suppose \(d\) is a greatest common divisor of \(a\) and \(b\text{.}\) Then any associate of \(d\) is also a greatest common divisor of \(a\) and \(b\text{.}\) (Recall Definition: Unit )
In most familiar domains, GCDs exist. However, they don't always! Find an example of elements in the ring from Exploration 2.3.3 which do not have a GCD. Justify your assertion.
Fill in the following blanks in order of increasing generality with the words ring , integral domain , field , and commutative ring .
__________ \(\Rightarrow\) __________ \(\Rightarrow\) __________ \(\Rightarrow\) __________