Can you find a domain where $ax+by=1$ has a solution for all $a$ and $b$ relatively prime, but which is not a PID?

Question

In Intro Number Theory a key lemma is that if $a$ and $b$ are relatively prime integers, then there exist integers $x$ and $y$ such that $ax+by=1$. In a more advanced course instead you would use the theorem that the integers are a PID, i.e. that all ideals are principal. Then the old lemma can be used to prove that "any ideal generated by two elements is actually principal." Induction then says that any finitely generated ideal is principal. But, what if all finitely generated ideals are principal but there are some ideals that aren't finitely generated? Can that happen?

Answer 1

If I'm not mistaken, the integral domain of holomorphic functions on a connected open set $U \subset \mathbb{C}$ works. It is a theorem (in Chapter 15 of Rudin's Real and Complex Analysis, and essentially a corollary of the Weierstrass factorization theorem), that every finitely generated ideal in this domain is principal. This implies that if $a,b$ have no common factor, they generate the unit ideal. However, for instance, the ideal of holomorphic functions in the unit disk that vanish on all but finitely many of ${1-\frac{1}{n}}$ is nonprincipal.

Answer 2

How about this construction:

Define a domain $R_0$ as follows. Take a field $K$, adjoin an indeterminate $x_0$, and localize at $(x_0)$ (that is, adjoin inverses to everything not a multiple of $x_0$).

$R_0$ has all its ideals principal and linearly ordered: $(x_0)$ contains $(x_0^2)$ contains $(x_0^3)$...

Now given $R_i$, define $R_{i+1}$ inductively: Adjoin an indeterminate $x_{i+1}$, so we have $R_i[x_{i+1}]$. Quotient by $(x_{i+1}^2 - x_i)$. Finally, localize at the prime ideal $(x_{i+1})$.

This effectively just gives us one more principal ideal containing all the principal ideals from $R_i : (x_{i+1})$ contains $(x_{i+1}^2)=(x_i)$ contains $(x_i^2)$...

Now let $R$ be the union of all the $R_i$, and it's obvious that any finitely generated ideal is principal, but there's a non-fg one generated by all the $x_i$.

Answer 3

See the Wikipedia article Bézout domain.

Answer 4

Note: for more on Bézout domains, see e.g.

Section 8.2 of http://alpha.math.uga.edu/~pete/factorization2010.pdf

or

Section 12.4 of http://alpha.math.uga.edu/~pete/integral.pdf

Answer 5

The easiest example I know is the ring of all algebraic integers (roots of monic polynomials with integer coefficients). As noted, it is a Bezout domain, so every finitely generated ideal is principal, and in particular for every two algebraic integers $a$ and $b$ there exist algebraic integers $\alpha$ and $\beta$ such that $\alpha a+\beta b = d$, where $d$ is a gcd for $a$ and $b$. However, the ideal $(2, 2^{1/2}, 2^{1/4}, 2^{1/8}, \ldots, 2^{1/2^{n}},\ldots)$ is not principal, so the ring is not a PID.