Euclidean domains are principal ideal domains

A com­mon theme in ring the­ory is the idea that we iden­tify a prop­erty of the in­te­gers, and work out what that prop­erty means in a more gen­eral set­ting. The idea of the eu­clidean do­main cap­tures the fact that in $$\mathbb{Z}$$, we may perform the di­vi­sion al­gorithm (which can then be used to work out great­est com­mon di­vi­sors and other such nice things from $$\mathbb{Z}$$). Here, we will prove that this sim­ple prop­erty ac­tu­ally im­poses a lot of struc­ture on a ring: it forces the ring to be a prin­ci­pal ideal do­main, so that ev­ery ideal has just one gen­er­a­tor.

In turn, this forces the ring to have unique fac­tori­sa­tion (proof), so in some sense the Fun­da­men­tal The­o­rem of Arith­metic (i.e. the state­ment that $$\mathbb{Z}$$ is a unique fac­tori­sa­tion do­main) is true en­tirely be­cause the di­vi­sion al­gorithm works in $$\mathbb{Z}$$.

This re­sult is es­sen­tially why we care about Eu­clidean do­mains: be­cause if we know a Eu­clidean func­tion for an in­te­gral do­main, we have a very easy way of recog­nis­ing that the ring is a prin­ci­pal ideal do­main.

For­mal statement

Let $$R$$ be a eu­clidean do­main. Then $$R$$ is a prin­ci­pal ideal do­main.

Proof

This proof es­sen­tially mir­rors the first proof one might find in the con­crete case of the in­te­gers, if one sat down to dis­cover an in­te­ger-spe­cific proof; but we cast it into slightly differ­ent lan­guage us­ing an equiv­a­lent defi­ni­tion of “ideal”, be­cause it is a bit cleaner that way. It is a very use­ful ex­er­cise to work through the proof, us­ing $$\mathbb{Z}$$ in­stead of the gen­eral ring $$R$$ and us­ing “size” noteThat is, if $$n > 0$$ then the size is $$n$$; if $$n < 0$$ then the size is $$-n$$. We just throw away the sign. as the Eu­clidean func­tion.

Let $$R$$ be a Eu­clidean do­main, and say $$\phi: \mathbb{R} \setminus \{ 0 \} \to \mathbb{N}^{\geq 0}$$ is a Eu­clidean func­tion. That is,

• if $$a$$ di­vides $$b$$ then $$\phi(a) \leq \phi(b)$$;

• for ev­ery $$a$$, and ev­ery $$b$$ not di­vid­ing $$a$$, we can find $$q$$ and $$r$$ such that $$a = qb+r$$ and $$\phi(r) < \phi(b)$$.

We need to show that ev­ery ideal is prin­ci­pal, so take an ideal $$I \subseteq R$$. We’ll view $$I$$ as the ker­nel of a ho­mo­mor­phism $$\alpha: R \to S$$; re­call that this is the proper way to think of ideals. (Proof of the equiv­alence.) Then we need to show that there is some $$r \in R$$ such that $$\alpha(x) = 0$$ if and only if $$x$$ is a mul­ti­ple of $$r$$.

If $$\alpha$$ only sends $$0$$ to $$0$$ (that is, ev­ery­thing else doesn’t get sent to $$0$$), then we’re im­me­di­ately done: just let $$r = 0$$.

Other­wise, $$\alpha$$ sends some­thing nonzero to $$0$$; choose $$r$$ to be nonzero with min­i­mal $$\phi$$. We claim that this $$r$$ works.

In­deed, let $$x$$ be a mul­ti­ple of $$r$$, so we can write it as $$ar$$, say. Then $$\alpha(ar) = \alpha(a) \alpha(r) = \alpha(a) \times 0 = 0$$. There­fore mul­ti­ples of $$r$$ are sent by $$\alpha$$ to $$0$$.

Con­versely, if $$x$$ is not a mul­ti­ple of $$r$$, then we can write $$x = ar+b$$ where $$\phi(b) < \phi(r)$$ and $$b$$ is nonzero. noteThe fact that we can do this is part of the defi­ni­tion of the Eu­clidean func­tion $$\phi$$. Then $$\alpha(x) = \alpha(ar)+\alpha(b)$$; we already have $$\alpha(r) = 0$$, so $$\alpha(x) = \alpha(b)$$. But $$b$$ has a smaller $$\phi$$-value than $$r$$ does, and we picked $$r$$ to have the small­est $$\phi$$-value among ev­ery­thing that $$\alpha$$ sent to $$0$$; so $$\alpha(b)$$ can­not be $$0$$, and hence nor can $$\alpha(x)$$.

So we have shown that $$\alpha(x) = 0$$ if and only if $$x$$ is a mul­ti­ple of $$r$$, as re­quired.

The con­verse is false

There do ex­ist prin­ci­pal ideal do­mains which are not Eu­clidean do­mains: $$\mathbb{Z}[\frac{1}{2} (1+\sqrt{-19})]$$ is an ex­am­ple. (Proof.)

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