By now you can build the quotient of a ring by an ideal, and you know that in \mathbb{Z} every ideal turned out to be n\mathbb{Z} — a single generator did all the work. That was not a lucky accident of the integers. It is a structural property with a name, and a whole family of rings share it. A ring in which every ideal is generated by one element is a principal ideal domain (PID), and PIDs are the rings where number theory still feels like number theory: gcds exist, Bézout's identity holds, and factorization behaves.

Where does that good behaviour come from? Almost always from a single humble tool you have used since primary school — division with remainder. The Euclidean algorithm works in \mathbb{Z} because you can always divide a by a nonzero b and get a remainder strictly smaller than b. Abstract exactly that ability — a "size" function that shrinks under division — and you get a Euclidean domain. This page pins down both notions, proves the theorem that ties them together (every Euclidean domain is a PID), and watches the same machinery run in three very different rings: the integers, polynomials over a field, and the Gaussian integers \mathbb{Z}[i].

Principal ideal domains

Recall that in a commutative ring the principal ideal generated by an element a is (a) = \{\, ra : r \in R \,\}, the set of all multiples of a. Some rings have ideals that need two or more generators, written (a, b) = \{\, ra + sb : r, s \in R \,\}. A PID is exactly the ring where that never happens — one generator is always enough.

You have already met the founding example. We showed that every ideal of \mathbb{Z} is n\mathbb{Z} = (n), so \mathbb{Z} is a PID. A second heavyweight is F[x], the polynomials over a field F: it too is a PID. Both facts are proved the same way — and that shared proof is what the rest of this page is about.

Euclidean domains: division with remainder, abstracted

What do \mathbb{Z} and F[x] have in common that forces every ideal to be principal? In each you can divide and get a smaller remainder. In \mathbb{Z} "smaller" means smaller absolute value; in F[x] it means lower degree. Distil that into one axiom and you have named a whole class of rings.

Three rings carry a natural Euclidean function, and they are the workhorses of the subject:

• \mathbb{Z} with N(n) = |n| — ordinary division with remainder, 0 \le r < |b|.
• F[x] (F a field) with N(p) = \deg p — polynomial long division, remainder of lower degree than the divisor.
• the Gaussian integers \mathbb{Z}[i] = \{\, a + bi : a, b \in \mathbb{Z} \,\} with the norm N(a + bi) = a^2 + b^2 — the squared distance from the origin in the complex plane.

The theorem that unifies them: Euclidean \Rightarrow PID

Here is the payoff. That one division axiom is powerful enough to force every ideal to collapse onto a single generator.

Proof (the Euclidean algorithm in disguise). The zero ideal is (0), done. Otherwise pick a nonzero d \in I whose N(d) is as small as possible (an actual minimum exists because norms are non-negative integers). Certainly (d) \subseteq I by absorption. For the reverse, take any a \in I and divide: a = dq + r with r = 0 or N(r) < N(d). But r = a - dq lies in I (both a and dq do), and d had the smallest norm of any nonzero element of I. So N(r) < N(d) is impossible — the only escape is r = 0, i.e. a = dq \in (d). Hence I = (d). \blacksquare

That single argument retroactively explains everything: \mathbb{Z}, F[x] and \mathbb{Z}[i] are all Euclidean, so all three are PIDs. And the generator d of the ideal (a, b) is precisely a greatest common divisor of a and b — which hands you Bézout for free.

Worked example 1: gcd and Bézout in \mathbb{Z}

Let us find \gcd(48, 18) and write it as an integer combination. Run division with remainder repeatedly, each remainder becoming the next divisor:

The last nonzero remainder is 6, so \gcd(48, 18) = 6 and the ideal (48, 18) = (6) = 6\mathbb{Z}. Now back-substitute to get Bézout:

So 6 = (-1)\cdot 48 + 3 \cdot 18 — check: -48 + 54 = 6. The Euclidean function N(n) = |n| strictly dropped at every line (18 > 12 > 6 > 0), which is why the process had to stop.

Worked example 2: dividing polynomials in \mathbb{Q}[x]

The same division-with-remainder runs in \mathbb{Q}[x], with N = \deg. Divide a = x^3 + 2x^2 - 1 by b = x^2 + 1:

• Leading terms: x^3 \div x^2 = x. Subtract x(x^2 + 1) = x^3 + x, leaving 2x^2 - x - 1.
• Leading terms again: 2x^2 \div x^2 = 2. Subtract 2(x^2 + 1) = 2x^2 + 2, leaving -x - 3.

The remainder -x - 3 has degree 1 < 2 = \deg b, so the algorithm halts — exactly the Euclidean condition N(r) < N(b). Crucially this needs F to be a field: you must divide by the leading coefficient of b. Over \mathbb{Z}, where you cannot, \mathbb{Z}[x] is not Euclidean (and, we will see, not even a PID).

Worked example 3: dividing Gaussian integers

The Gaussian integers \mathbb{Z}[i] are lattice points in the complex plane, and the norm N(a + bi) = a^2 + b^2 measures squared distance from the origin. Division with remainder is beautifully geometric: to divide, snap the exact complex quotient to its nearest lattice point. Divide 3 + 2i by 1 + i. First compute the true quotient by rationalising:

Round each coordinate to the nearest integer: the closest Gaussian integer is q = 2. Then the remainder is

Check the shrinking condition: N(r) = N(1) = 1, while N(b) = N(1 + i) = 1^2 + 1^2 = 2, so indeed N(r) < N(b). Why can we always succeed? The multiples of 1 + i form a coarser sublattice (below), and every point of the plane lies within distance \tfrac{\sqrt{2}}{2}\,|b| of some multiple of b — comfortably less than |b|. That geometric fact is the proof that \mathbb{Z}[i] is Euclidean, hence a PID.

The hierarchy so far

Each new condition is strictly stronger than the last, nesting these rings into a tower. Reading inward (most special first):

• Every field is Euclidean (take any constant norm; you can always divide exactly, r = 0).
• Every Euclidean domain is a PID — the theorem above.
• Every PID is a unique factorization domain (UFD) — factorization into primes exists and is unique, the subject of the next page.
• Every UFD is an integral domain, by definition.

All four inclusions are strict — there are PIDs that are not Euclidean, UFDs that are not PIDs, and domains with no unique factorization at all. Those separating examples are what the two vignettes below are for.