Editing Integral domain (section)

== Non-examples ==
The following rings are ''not'' integral domains.
* The [[zero ring]] (the ring in which <math>0=1</math>).
* The quotient ring <math>\Z/m\Z</math> when ''m'' is a [[composite number]]. To show this, choose a proper factorization <math>m = xy</math> (meaning that <math>x</math> and <math>y</math> are not equal to <math>1</math> or <math>m</math>). Then <math>x \not\equiv 0 \bmod{m}</math> and <math>y \not\equiv 0 \bmod{m}</math>, but <math>xy \equiv 0 \bmod{m}</math>.
* A [[product ring|product]] of two nonzero commutative rings.  In such a product <math>R \times S</math>, one has <math>(1,0) \cdot (0,1) = (0,0)</math>.
* The quotient ring <math>\Z[x]/(x^2 - n^2)</math> for any <math>n \in \mathbb{Z}</math>.  The images of <math>x+n</math> and <math>x-n</math> are nonzero, while their product is 0 in this ring.
* The [[matrix ring|ring]] of ''n'' × ''n'' [[Matrix (mathematics)|matrices]] over any [[zero ring|nonzero ring]] when ''n'' ≥ 2. If <math>M</math> and <math>N</math> are matrices such that the image of <math>N</math> is contained in the kernel of <math>M</math>, then <math>MN = 0</math>.  For example, this happens for <math>M = N = (\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix})</math>.
* The quotient ring <math>k[x_1,\ldots,x_n]/(fg)</math> for any field <math>k</math> and any non-constant polynomials <math>f,g \in k[x_1,\ldots,x_n]</math>. The images of {{math|''f''}} and {{math|''g''}} in this quotient ring are nonzero elements whose product is 0. This argument shows, equivalently, that <math>(fg)</math> is not a [[prime ideal]]. The geometric interpretation of this result is that the [[zero of a function|zeros]] of {{math|''fg''}} form an [[affine algebraic set]] that is not irreducible (that is, not an [[algebraic variety]]) in general. The only case where this algebraic set may be irreducible is when {{math|''fg''}} is a power of an [[irreducible polynomial]], which defines the same algebraic set.
* The ring of [[continuous function]]s on the [[unit interval]].  Consider the functions
*: <math> f(x) = \begin{cases} 1-2x & x \in \left [0, \tfrac{1}{2} \right ] \\ 0 & x \in \left [\tfrac{1}{2}, 1 \right ] \end{cases} \qquad g(x) = \begin{cases} 0 & x \in \left [0, \tfrac{1}{2} \right ] \\ 2x-1 & x \in \left [\tfrac{1}{2}, 1 \right ] \end{cases}</math>
: Neither <math>f</math> nor <math>g</math> is everywhere zero, but <math>fg</math> is.
* The [[tensor product of algebras|tensor product]] <math>\Complex \otimes_{\R} \Complex</math>. This ring has two non-trivial [[idempotent (ring theory)|idempotent]]s, <math>e_1 = \tfrac{1}{2}(1 \otimes 1) - \tfrac{1}{2}(i \otimes i)</math> and <math>e_2 = \tfrac{1}{2}(1 \otimes 1) + \tfrac{1}{2}(i \otimes i)</math>. They are orthogonal, meaning that <math>e_1e_2 = 0</math>, and hence <math>\Complex \otimes_{\R} \Complex</math> is not a domain. In fact, there is an isomorphism <math>\Complex \times \Complex \to \Complex \otimes_{\R} \Complex</math> defined by <math>(z, w) \mapsto z \cdot e_1 + w \cdot e_2</math>. Its inverse is defined by <math>z \otimes w \mapsto (zw, z\overline{w})</math>. This example shows that a [[fiber product of schemes|fiber product]] of irreducible affine schemes need not be irreducible.