Does any (noetherian) integral domain have a "UFD closure"? Let $R$ be a (possibly noetherian if that helps) commutative unital integral domain.
Does there exist a UFD $\overline{R}$ such that $R$ embeds in $\overline{R}$ (via some map $\psi$) and such that $\overline{R}$ is universal with this property (i.e. such that for any UFD $S$ with an embedding $\varphi: R\to S$ there is a factorisation $\varphi = \rho\psi$ for some $\rho: \overline{R}\to S$.
One thing that seems like a natural requirement (and which it is not clear, at least to me, will follow from the above) is that if $x\in R$ is irreducible, then $\psi(x)$ is not a unit.
This question came up while talking to someone else about the "usual" proof that $\mathbb{Z}[i\sqrt{2}]$ is a UFD, which shows that the usual norm in $\mathbb{C}^2$ makes the ring euclidean using that any unit circle in $\mathbb{C}^2$ will contain an element from the ring in its interior. We then noted that the argument precisely fails to work for $\mathbb{Z}[i\sqrt{3}]$, and indeed looking at the way it fails gave a good idea of how to find elements where unique factorization failed to hold (the element $1+i\sqrt{3}$). But it also suggested that if one was to further add $\frac{1+i\sqrt{3}}{2}$ then one would have added the divisor needed to get a UFD (and indeed, $\mathbb{Z}[i\sqrt{3}][\frac{1+i\sqrt{3}}{2}] = \mathbb{Z}[\frac{1+i\sqrt{3}}{2}]$ is a UFD by the above mentioned argument.
This then lead to the above question of whether one can always formally add extra divisors of elements of a non-UFD to get a UFD (in a universal way, since otherwise one could just take the field of fractions to get a UFD).
We considered a construction along the lines of (for each pair of irreducible elements $p$ and $q$ such that $\langle p,q\rangle$ was not the entire ring) taking a quotient $R[k,s,t]/I$ where $I$ is generated by $ks - p$ and $kt - q$ (i.e. we are adding the missing common divisor $k$ of $p$ and $q$ along with the complementary divisors). But it did not seem like this would generally yield a UFD, much less a universal one.
 A: In general no, it doesn't.
Step 1: UFD closure is stable under localization.
Let us suppose that $\bar{R}$ is a UFD closure of $R$. Then, for every prime $p\subseteq R$, $R_p\otimes\bar{R}=\bar{R}_p$ is the UFD closure of $R_p$. A localization of an UFD is an UFD too,  we only need to check the universal property. In fact, if we have an homomorphism $R_p\to U$, with $U$ UFD, we get a unique homomorphism $\bar{R}\to U$ thanks to the universal property of $\bar{R}$, and hence a unique homomorpshism $\bar{R}_p\to U$ thanks to the universal property of the tensor product.
Step 2:
A Dedekind domain which is not a PID has no UFD closure.
Let $R$ be a Dedekind domain which is not a PID, and suppose it has an UFD closure $R\to\bar{R}$. A Dedekind domain is a PID if and only if it is an UFD, hence $R\neq\bar{R}$. For every prime $p$ of $R$, the localization morphism $R_p\to\bar{R}_p$ is still an UFD closure thanks to step 1. But $R_p$ is a PID, and hence an UFD, so $R_p\to\bar{R}_p$ is an isomorphism for every $p$. This implies that $R\to\bar{R}$ is an isomorphism too, absurd.
What is true is that you have a universal normal closure, i.e. a normalization. If the normalization is also an UFD (for example, $k[t^2,t^3]\subset k[t]$), then the normalization is also, in particular, an UFD closure.
Edit I added the answer to the question: when $R$ has an UFD closure $\psi:R\to\bar{R}$, is it true that $\psi(x)$ is not invertible for every irreducible $x$?
Yes, it is true. In fact, more generally the following is always true: if $x$ is a non invertible element of a noetherian domain $R$, there exists a UFD $U$ and a morphism $R\to U$ such that the image of $x$ is non invertible. When this is proved true, the universal property of the UFD closure immediately answers your question.
Step 1: We may suppose $R$ normal. In fact, if $f:R\to R'$ is the normalization of $R$, $R'$ is integral over $R$, and hence $f(x)$ is non invertible (for example, I may take a prime $p\subset R$ containing $x$ and pull it up to a prime in $R'$ containing $f(x)$).
Step 2: We may suppose $R$ local of dimension one. In fact, let $p$ be a minimal prime containing $x$: thanks to Krull's principal ideal theorem it has height one. But then, the image of $x$ in $R_p$ is non invertible, and $R_p$ is local of dimension one.
Step 3: A normal, local domain of dimension one is an UFD, so we conclude.
