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Integral Domains

Table of Contents

1. Definition and Properties

An Integral Domain is a unital ring with no nonzero divisors, i.e., if \(ab = 0\) then either \(a=0\) or \(b=0\).

Constructive mathematicians would phrase this differently, as saying \(ab\neq0\) whenever \(a\neq0\) and \(b\neq0\).

The integers \(\ZZ\), rationals \(\QQ\), reals \(\RR\), and complex numbers are integral domains.

Let \(X\) be a non-empty set with at least 2 disjoint subsets \(X_{1}\), \(X_{2}\); and let \(R\) be any ring. Then \(\hom(X,R)\) is not an integral domain. To see this, consider

\begin{equation} f(x) = \begin{cases} 1 & x\in X_{0}\\ 0 & x\notin X_{0} \end{cases} \end{equation}

and

\begin{equation} g(x) = \begin{cases} 1 & x\in X_{1}\\ 0 & x\notin X_{1}. \end{cases} \end{equation}

Then we have

\begin{equation} (fg)(x) = f(x)g(x) = 0 \end{equation}

since \(X_{0}\cap X_{1}=\emptyset\). Although both \(f\neq0\) and \(g\neq0\), we have \(fg=0\).

The ring \(\ZZ[\sqrt{-5}]\) is an integral domain.

1.1. Factors

Let \(R\) be an integral domain, and \(x,y\in R\). We say that \(x\) Divides \(y\), and write \(x\mid y\), if \(y=xt\) for some \(t\in R\).

We call an element \(x\in R\) a Unit if \(x\mid 1\). The set of units of \(R\) is denoted \(R^{*}\).

Calling an element \(x\in R\) a unit is a fancy-pants way of saying it has a multiplicative inverse. Every integral domain has at least one unit (indeed, all unital rings have at least one unit): the multiplicative identity \(1\in R\) is its own multiplicative inverse.

We say \(z\in R\) is a Greatest Common Divisor of \(x_{1}\), …, \(x_{n}\) and write \(z=\gcd(x_{1},\dots, x_{n})\) if

  1. \(z\mid x_{j}\) for \(1\leq j\leq n\)
  2. for any \(t\in R\), \(t\mid x_{j}\) for \(1\leq j\leq n\) implies \(t\mid z\).

Further, stealing borrowing terminology from number theory, we say that \(x\) and \(y\) are Coprime if there exists a unit \(u\in R^{*}\) which is a gcd of \(x\) and \(y\).

Let \(R\) be an integral domain, and \(x,y\in R\). If \(z\) and \(t\) are both GCDs of \(x\) and \(y\), then there exists a \(u\in R\) such that \(z=ut\).

  • Assume \(z = \gcd(x,y)\), \(t=\gcd(x,y)\), and \(t\neq z\).
  • Then \(t\mid x\) and \(t\mid y\) by definition of gcd.
  • Then \(t\mid z\) by \(z=\gcd(x,y)\).
  • Thus \(z=ut\) for some \(u\in R\).

(Bronstein, 1.1.1) Let \(R\) be an integral domain, and \(x,y\in R\). If \(z\) and \(t\) are both GCDs of \(x\) and \(y\), then \(z=ut\) and \(t=vz\) for some units \(u,v\in R^{*}\).

  • Assume \(z = \gcd(x,y)\), \(t=\gcd(x,y)\), and \(t\neq z\).
  • Consider \(u\in R\) such that \(z=ut\) by the previous lemma.
  • Consider \(v\in R\) such that \(t=vz\) by the previous lemma.
  • Then \(z=ut=u(vz)\).
  • If \(z\neq0\), then \(u,v\in R^{*}\).
  • If \(z=0\), then \(t=1z\) and \(z=1t\).
  • Thus \(u,v\in R^{*}\).

In an integral domain \(R\), let \(a,b\in R\). If \(a\) divides \(b\), and \(b\) divides \(a\), then we call them Associates. Equivalently, if there is some unit \(u\in R^{*}\) such that \(a=ub\).

1.2. Prime, Irreducible Elements

Let \(R\) be an integral domain. A nonzero element \(p\in R\setminus R^{*}\) is called Prime if for any \(a,b\in R\), \(p\mid ab\) implies \(p\mid a\) or \(p\mid b\).

Let \(R\) be an integral domain. A nonzero element \(p\in R\setminus R^{\ast}\) is called Irreducible if for any \(a,b\in R\), \(p = ab\) implies \(a\in R^{\ast}\) or \(b\in R^{\ast}\).

Greater intuition may be derived by considering the contrapositive: \(p\) is irreducible if for any \(a,b\in R\), we have \(a\notin R^{\ast}\) and \(b\notin R^{\ast}\) implies \(p\neq ab\). That is, it can't be written as the product of two non-units.

Let \(R\) be an integral domain. Then every prime \(p\in R\) is irreducible.

  • Let \(p\in R\) be prime.
  • Assume \(a,b\in R\) satisfy \(p=ab\).
  • Per cases \(p\mid a\) or \(p\mid b\) by definition of prime.
  • Suppose \(p\mid a\);
    • Then \(a=q_{a}p\);
    • Thus \(1=q_{a}b\);
    • Thus \(b\mid 1\);
    • Thus \(b\in R^{\ast}\) is a unit.
  • Suppose \(p\mid b\); then \(a\in R^{\ast}\) is a unit (by the exact same reasoning).
  • Hence \(p\) is irreducible.

In any integral domain, a prime is always irreducible. The converse is not always true.

2. References

  • Manuel Bronstein, Symbolic Integration 1: Transcendental Functions. Springer, second ed., 1996.

Last Updated 2021-06-01 Tue 10:00.