# 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?

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?

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This is one of my all-time favorite ring theory questions. It took us several days to get it the first time. – Noah Snyder Jul 23 '10 at 3:17
I think it's great that you are seeding the site with these sorts of questions. – Akhil Mathew Jul 23 '10 at 3:30
Good Question.. – Himadri Jul 23 '10 at 10:11
Is the question equivalent to the question posed in the subject, though? Clearly, if $R$ is a commutative ring with 1 in which every finitely generated ideal is principal, then for any $a$ and $b$ such that $(a,b)$ is not contained in any proper principal ideal you will have $x$ and $y$ for which $ax+by=1$. But if the latter condition holds, does it follow that every finitely generated ideal is principal? – Arturo Magidin Sep 7 '10 at 14:26

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.

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Cool, I didn't know that these had a name. – Noah Snyder Aug 3 '10 at 16:44
With apologies to M. Bézout, I took out the accent aigu to repair the link. – Pete L. Clark Sep 8 '10 at 8:52
@Peter: I copied the original link, with accent, from a wikipedia page. Perhaps they have "fixed" that. Thanks for fixing it here. – lhf Sep 8 '10 at 19:59

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 $x_i$.

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An easier way to give roughly this example is just to look at a field adjoin x^{1/2^n}] for all n. For any finite collection of elements they're contained in some polynomial ring, and so the ideal generated is principal. Your example is roughly the same but you use localization instead of the fact that polynomial rings over a field are PIDs – Noah Snyder Aug 2 '10 at 23:54

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.

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@Arturo: it is easy to see that this ring is not a PID, but it is not so easy to see that it is a Bezout domain (in fact a proof of that is not contained in my notes). For a more elementary example, I would recommend some (specific) non-discrete valuation ring, like the ring of Puiseux series over $\mathbb{C}$. – Pete L. Clark Sep 7 '10 at 14:02
The proof that it is a Bezout domain can be found in Dedekind's exposition of 1877. You can find it today in "Theory of Algebraic Integers" by Richard Dedekind, Cambridge Mathematical Library, 1996, ISBN 0-521-56518-9, translated by John Stillwell, announced in Section 14 and proven towards the end (p 151 in my edition). An excellent read, by the way, and highly recommended. Despite its age, you could reasonably use it as a textbook in a course today. – Arturo Magidin Sep 7 '10 at 14:12
@Pete: That said, yes, I agree it is not easy or immediate to show any finitely generated ideal is principal. If you've done some algebraic number theory it is fairly straightforward, but otherwise it would likely be a mystery. – Arturo Magidin Sep 7 '10 at 14:15
I like this example. The fact that the ring of entire functions is a Bezout domain is also not obvious. These are rings which everyone already knows about anyway and it is interesting that they happen to be Bezout domains. – George Lowther Sep 7 '10 at 23:48
@Arturo: recently on MO I asked a question having to do with the ring of all algebraic integers. Part of what I wondered was who first showed it is a Bezout domain. I had forgotten about your answer here: do you think that Dedekind's 1877 work is the first to prove this? ("Dedekind" at least feels like a fitting answer to this question.) – Pete L. Clark Dec 28 '10 at 15:12

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

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

or

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

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