Integral domain

In keep­ing with ring the­ory as the at­tempt to iso­late each in­di­vi­d­ual prop­erty of \(\mathbb{Z}\) and work out how the prop­er­ties in­ter­play with each other, we define the no­tion of in­te­gral do­main to cap­ture the fact that if \(a \times b = 0\) then \(a=0\) or \(b=0\). That is, an in­te­gral do­main is one which has no “zero di­vi­sors”: \(0\) can­not be non­triv­ially ex­pressed as a product. (For un­in­ter­est­ing rea­sons, we also ex­clude the ring with one el­e­ment, in which \(0=1\), from be­ing an in­te­gral do­main.)


  • \(\mathbb{Z}\) is an in­te­gral do­main.

  • Any field is an in­te­gral do­main. (The proof is an ex­er­cise.)

Sup­pose \(ab = 0\), but \(a \not = 0\). We wish to show that \(b=0\).

Since we are work­ing in a field, \(a\) has an in­verse \(a^{-1}\); mul­ti­ply both sides by \(a^{-1}\) to ob­tain \(a^{-1} a b = 0 \times a^{-1}\). Sim­plify­ing, we ob­tain \(b = 0\). <div><div>

  • When \(p\) is a prime in­te­ger, the ring \(\mathbb{Z}_p\) of in­te­gers mod \(p\) is an in­te­gral do­main.

  • When \(n\) is a com­pos­ite in­te­ger, the ring \(\mathbb{Z}_n\) is not an in­te­gral do­main. In­deed, if \(n = r \times s\) with \(r, s\) pos­i­tive in­te­gers, then \(r s = n = 0\) in \(\mathbb{Z}_n\).


The rea­son we care about in­te­gral do­mains is be­cause they are pre­cisely the rings in which we may can­cel prod­ucts: if \(a \not = 0\) and \(ab = ac\) then \(b=c\).

In­deed, if \(ab = ac\) then \(ab-ac = 0\) so \(a(b-c) = 0\), and hence (in an in­te­gral do­main) \(a=0\) or \(b=c\).

More­over, if we are not in an in­te­gral do­main, say \(r s = 0\) but \(r, s \not = 0\). Then \(rs = r \times 0\), but \(s \not = 0\), so we can’t can­cel the \(r\) from both sides. <div><div>

Finite in­te­gral domains

If a ring \(R\) is both finite and an in­te­gral do­main, then it is a field. The proof is an ex­er­cise.

Given \(r \in R\), we wish to find a mul­ti­plica­tive in­verse.

Since there are only finitely many el­e­ments of the ring, con­sider \(S = \{ ar : a \in R\}\). This set is a sub­set of \(R\), be­cause the mul­ti­pli­ca­tion of \(R\) is closed. More­over, ev­ery el­e­ment is dis­tinct, be­cause if \(ar = br\) then we can can­cel the \(r\) (be­cause we are in an in­te­gral do­main), so \(a = b\).

Since there are \(|R|\)-many el­e­ments of the sub­set \(S\) (where \(| \cdot |\) refers to the car­di­nal­ity), and since \(R\) is finite, \(S\) must in fact be \(R\) it­self.

There­fore in par­tic­u­lar \(1 \in S\), so \(1 = ar\) for some \(a\). <div><div>