Non-examples The following rings are not integral domains. The zero ring (the ring in which ). The quotient ring when m is a composite number. Indeed, choose a proper factorization (meaning that and are not equal to or ). Then and , but . A product of two nonzero commutative rings. In such a product , one has . The quotient ring for any . The images of and are nonzero, while their product is 0 in this ring. The ring of n × n matrices over any nonzero ring when n ≥ 2. If and are matrices such that the image of is contained in the kernel of , then . For example, this happens for . The quotient ring for any field and any non-constant polynomials . The images of and in this quotient ring are nonzero elements whose product is 0. This argument shows, equivalently, that is not a prime ideal. The geometric interpretation of this result is that the zeros of 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 is a power of an irreducible polynomial, which defines the same algebraic set. The ring of continuous functions on the unit interval. Consider the functions Neither nor is everywhere zero, but is. The tensor product . This ring has two non-trivial idempotents, and . They are orthogonal, meaning that , and hence is not a domain. In fact, there is an isomorphism defined by . Its inverse is defined by . This example shows that a fiber product of irreducible affine schemes need not be irreducible.
non-examples the following rings are not integral domains the zero ring (the ring in which ) the quotient ring when m is a composite number indeed choose a proper factorization (meaning that and are not equal to or ) then and  but  a product of two nonzero commutative rings in such a product  one has  the quotient ring for any  the images of and are nonzero while their product is 0 in this ring the ring of n × n matrices over any nonzero ring when n ≥ 2 if and are matrices such that the image of is contained in the kernel of  then  for example this happens for  the quotient ring for any field and any non-constant polynomials  the images of and in this quotient ring are nonzero elements whose product is 0 this argument shows equivalently that is not a prime ideal the geometric interpretation of this result is that the zeros of 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 is a power of an irreducible polynomial which defines the same algebraic set the ring of continuous functions on the unit interval consider the functions neither nor is everywhere zero but is the tensor product  this ring has two non-trivial idempotents and  they are orthogonal meaning that  and hence is not a domain in fact there is an isomorphism defined by  its inverse is defined by  this example shows that a fiber product of irreducible affine schemes need not be irreducible