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.proof; but we approach it from a slightly different direction, because it is a bit cleaner that way. It is a very useful exercise to work through the proof, using Z instead of the general ring R and using "size" ^[1] as the Euclidean function.

We need to show that every ideal is principal, so take an ideal I⊆R. IfWe'll view I as the kernel of a homomorphism=α{:0R}→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∈R then we are immediately done: itsuch that α(x)=0 if and only if x is principal, being generated by the elementa multiple of r.

If α only sends 0.

So there is some nonzero element in to I0 (that is, everything else doesn't get sent to 0), then we're immediately done: just let r=0.

Otherwise, α sends something nonzero to 0; choose a nonzero element r to be nonzero with minimal ϕ. We claim that the ideal is in fact generated bythis r. works.

Indeed, given anylet i∈I, we need i∈⟨r⟩x (the ideal generated bybe a multiple of r: that is, the set of all, so we can write it as ar, say. Then α(ar)=α(a)α(r)=α(a)×0=0. Therefore multiples of r) are sent by α to 0.

• If

  Conversely, if x is not a multiple of r doesn't divide, then we can write i, then we can find α and β such that ix=αar+βb withwhere ϕ(βb)<ϕ(r), since and ϕb is a Euclidean function. But thennonzero. i^[2] is in I, andThen α(x)=α(ar)+α(b) is in; we already have Iα(r)=0 (because r is in I);, so iα−(x)=αr(b must be in I, and hence β is in r). But βb has a smaller ϕ-value than r does, and this contradicts the minimality ofwe picked r; so this bullet point can't happen after all.

• If to have the smallest rϕ divides i then we know instantly-value among everything that i∈⟨r⟩α (becausesent to i0; so α(b) cannot be 0, and hence nor can α(x).

  So we have shown that α(x)=0 if and only if x is a multiple of r).

Therefore if i∈I then i∈⟨r⟩.

Conversely, if i∈⟨r⟩ then i is a multiple of r. But r is in I already, so i must also be in i.

Hence I=⟨r⟩, so I is principal.as required.

1. ^{^︎}

   That is, if n>0 then the size is n; if n<0 then the size is −n. We just throw away the sign.

2. ^{^︎}

   The fact that we can do this is part of the definition of the Euclidean function ϕ.

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 Z, we may perform the division_algorithm (which can then be used to work out greatest common divisors and other such nice things from 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_ring_theory 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 Z is a unique factorisation domain) is true entirely because the division algorithm works in 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.

Let R be a Euclidean domain, and say ϕ:R∖{0}→N≥0 is a Euclidean function. That is, - if a divides b then ϕ(a)≤ϕ(b); - for every a, and every b not dividing a, we can find q and r such that a=qb+r and ϕ(r)<ϕ(b).

We need to show that every ideal is principal, so take an ideal I⊆R. If I={0} then we are immediately done: it is principal, being generated by the element 0.

So there is some nonzero element in I; choose a nonzero element r with minimal ϕ. We claim that the ideal is in fact generated by r.

Indeed, given any i∈I, we need i∈⟨r⟩ (the ideal generated by r: that is, the set of all multiples of r).

• If r doesn't divide i, then we can find α and β such that i=αr+β with ϕ(β)<ϕ(r), since ϕ is a Euclidean function. But then i is in I, and αr is in I (because r is in I); so i−αr must be in I, and hence β is in r. β has smaller ϕ than r does, and this contradicts the minimality of r; so this bullet point can't happen after all.
• If r divides i then we know instantly that i∈⟨r⟩ (because i is a multiple of r).

The converse is false

There do exist principal ideal domains which are not Euclidean domains: Z[12(1+√−19)] is an example. (Proof.)