# 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.)

# Examples

• $$\mathbb{Z}$$ 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 $$\mathbb{Z}_n$$.

# Properties

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>

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