TITLE: Is every commutative ring a limit of noetherian rings?
QUESTION [26 upvotes]: Edit of Feb. 14, 2019. After Laurent Moret-Bailly's accepted answer, only Questions 4 and 5 remain open. I don't care that much about Question 4, but I'm very curious about Question 5, which is

Do binary coproducts always exist in the category of noetherian commutative rings?

End of edit.
I asked the question on Mathematics Stackexchange but got no answer.
Let $\mathsf{Noeth}$ be the category of noetherian rings, viewed as a full subcategory of the category $\mathsf{CRing}$ of commutative rings with one.
Let $A$ be in $\mathsf{CRing}$.

Question 1. Is there a functor from a small category to $\mathsf{Noeth}$ whose limit in $\mathsf{CRing}$ is $A$?

Let $f:A\to B$ be a morphism in $\mathsf{CRing}$ such that the map
$$
\circ f:\text{Hom}_{\mathsf{CRing}}(B,C)\to\text{Hom}_{\mathsf{CRing}}(A,C)
$$ 
sending $g$ to $g\circ f$ is bijective for all $C$ in $\mathsf{Noeth}$.

Question 2. Does this imply that $f$ is an isomorphism?

Yes to Question 1 would imply yes to Question 2.

Question 3. Does the inclusion functor $\iota:\mathsf{Noeth}\to\mathsf{CRing}$ commute with colimits? That is, if $A\in\mathsf{Noeth}$ is the colimit of a functor $\alpha$ from a small category to $\mathsf{Noeth}$, is $A$ naturally isomorphic to the colimit of $\iota\circ\alpha$?

Yes to Question 2 would imply yes to Question 3, and yes to Question 3 would imply that many colimits, and in particular many binary coproducts, do not exist in $\mathsf{Noeth}$: see this answer of Martin Brandenburg.  
Here are two particular cases of the above questions:

Question 4. Is $\mathbb Z[x_1,x_2,\dots]$ a limit of noetherian rings?

(The $x_i$ are indeterminates.)

Question 5. Do binary coproducts exist in $\mathsf{Noeth}$?


One may try to attack the first question as follows:
Let $A$ be in $\mathsf{CRing}$ and $I$ the set of those ideals $\mathfrak a$ of $A$ such that $A/\mathfrak a$ is noetherian. Then $I$ is an ordered set, and thus can be viewed as a category. We can form the limit of the $A/\mathfrak a$ with $\mathfrak a\in I$, and we have a natural morphism from $A$ to this limit. I'd be interested in knowing if this morphism is bijective. [Edit. By a comment of Laurent Moret-Bailly the morphism in question is not always bijective.]

REPLY [13 votes]: Binary coproducts do not always exist in $\textrm{Noeth}$.
Assume that the coproduct $C=\overline{\mathbb{Q}} \sqcup \overline{\mathbb{Q}}$ exists in $\textrm{Noeth}$. Letting $A=\overline{\mathbb{Q}} \otimes_{\mathbb{Z}} \overline{\mathbb{Q}}$, we then have a canonical ring map $\varphi : A \to C$. Now the idea is to consider some kind of completion of $A$, similar to the one you suggest in the last paragraph of your question: for every $\sigma \in G=\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, there is a noetherian quotient $\mu_\sigma : A \to \overline{\mathbb{Q}}$ given by $\mu_\sigma(x \otimes y)=x \sigma(y)$. Moreover, the resulting morphism $i : A \to \prod_G \overline{\mathbb{Q}}$ is injective (this can be checked by restricting to $K \otimes K$ where $K$ is any finite Galois extension of $\mathbb{Q}$). Applying the coproduct property in $\textrm{Noeth}$, each $\mu_\sigma$ factors through $C$, so that $i$ factors through $\varphi$. In particular, the map $\varphi$ is injective, and we may identify $A$ with a subring of $C$.
Since $A$ is integral over $\overline{\mathbb{Q}}$, it has Krull dimension $0$. In particular, every maximal ideal $\mathfrak{m}_\sigma=\ker \mu_\sigma$ is a minimal prime ideal of $A$. By a result in Bourbaki, Commutative algebra (Chap. 2, Sect. 2.6, Prop. 16), the ideal $\mathfrak{m}_\sigma$ lies below a minimal prime ideal $\mathfrak{p}_\sigma$ of $C$. So we have constructed infinitely many minimal prime ideals of $C$, which is absurd because $C$ is noetherian.