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Is the following true:

If $R$ is a commutative unital ring with $\mathrm{Spec}(R)\!=\!\mathrm{Max}(R)\!\cup\!\{0\}$, then $R$ is a PID.

If yes, how can one prove it?

Since $0$ is a prime ideal, $R$ is a domain. Thus we must prove that every ideal is principal. I'm not sure if this link (first answer) helps.

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You are claiming that every 1-dimension domain is PID, which is not true. For example, take a curve which is not smooth, and localize at the non-smooth point (Say $\mathbb{C}[x,y]/(y^2-x^3)$ localizing at $(x,y)$) will give you a counterexample. – user27126 Nov 30 '12 at 11:26
@Sanchez : you needn't even localize to get a counter-example... – Georges Elencwajg Nov 30 '12 at 11:30
@GeorgesElencwajg, that's true. I guess I focused on the singularity part so I overlooked its unnecessity. Thanks! – user27126 Nov 30 '12 at 11:33
Thank you very much! I'll accept if you write it as an answer. – Leon Nov 30 '12 at 11:34
You are looking for an integral domain of Krull dimension $1$ which is not a PID. Well, there are thousands of examples, but let me mention only one which is not even GCD (I'm supposing that we are talking here about the arithmetical properties of this class of rings): $R=\mathbb{Z}[i\sqrt{3}]$. – user26857 Nov 30 '12 at 12:06
up vote 5 down vote accepted

You are claiming that every 1-dimension domain is PID, which is not true. For example, take a curve which is not smooth (Say $\mathbb{C}[x,y]/(y^2−x^3)$) will give you a counterexample, since it won't be a PID when you localize at $(x,y)$.

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Sanchez's example is from algebraic geometry. Here's an example from algebraic number theory.

The ring of integers of an algebraic number field is always a Dedekind domain; in particular, it is a 1-dimensional domain. However, there are (plenty of) ring of integers which are not PIDs. As you can see here, only a finite number of imaginary quadratic extensions of $\mathbb{Q}$ have a PID as ring of integers.

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As mentioned, there are easy counterxamples. However, it is true for UFDs since PIDs are precisely the $\rm UFDs$ of dimension $\le 1,\:$ i.e. such that prime ideals $\ne 0$ are maximal. Below is a sketch of a proof of this and closely related results.

Theorem $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$
$(1)\ \ $ prime ideals are maximal if nonzero
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1$
$(5)\ \ $ $\rm D$ is Bezout
$(6)\ \ $ $\rm D$ is a $\rm PID$

Proof $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1$)

$(1\Rightarrow 2)$ $\rm\ \ P\supset (p)\ \Rightarrow\ P = (p)$
$(2\Rightarrow 3)$ $\ \: $ Clear.
$(3\Rightarrow 4)$ $\ \ \rm (a,b) \subsetneq P = (p)\ $ so $\rm\ (a,b) = 1$
$(4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\ \Rightarrow\ (a,b) = c\ (a/c,b/c) = (c)$
$(5\Rightarrow 6)$ $\ \ \rm 0 \ne I \subset D\:$ Bezout is generated by an elt with the least number of prime factors
$(6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$

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