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 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. 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.

Let $R$ be a Euclidean domain, and say $\varphi :R\setminus \left\{0\right\}\to N^{\ge 0}$ is a Euclidean function. That is,

• if $a$ divides $b$ then $\varphi \left(a\right)\le \varphi \left(b\right)$;
• for every $a$, and every $b$ not dividing $a$, we can find $q$ and $r$ such that $a=qb+r$ and $\varphi \left(r\right)<\varphi \left(b\right)$.

We need to show that every ideal is principal, so take an ideal $I\subseteq R$. If $I=\left\{0\right\}$ 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 $\varphi$. We claim that the ideal is in fact generated by $r$.

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

• If $r$ doesn't divide $i$, then we can find $\alpha$ and $\beta$ such that $i=\alpha r+\beta$ with $\varphi \left(\beta \right)<\varphi \left(r\right)$, since $\varphi$ is a Euclidean function. But then $i$ is in $I$, and $\alpha r$ is in $I$ (because $r$ is in $I$); so $i-\alpha r$ must be in $I$, and hence $\beta$ is in $r$. $\beta$ has smaller $\varphi$ 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 $i\in ⟨r⟩$ (because $i$ is a multiple of $r$).

Therefore if $i\in I$ then $i\in ⟨r⟩$.

Conversely, if $i\in ⟨r⟩$ 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.

The converse is false

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

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.