Is every commutative ring a limit of noetherian rings?

Question

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.]

Answer 1

The answer is no to all questions except 4.

Negative answers to 1,2 and 3: It is easy to construct a ring $A$ with an element $a$ satisfying:
(i) $a≠0$,
(ii) $a$ is nilpotent,
(iii) for each $n≥1$ there is $y_n\in A$ such that $a=y_n^n$.

For every morphism $\varphi:A\to C$, the image of $a$ inherits properties (ii) and (iii), hence it is zero if $C$ is noetherian (observe that the radical of $C$ is finitely generated, hence nilpotent). In other words, any morphism from $A$ to a noetherian ring $C$ factors through $B:=A/(a)$, and of course the same holds if $C$ is a limit of noetherian rings. This proves that $A$ is not such a limit (Question 1) and the natural morphism $A\to B$ provides a negative answer to Question 2.

For Question 3, consider the following special case: let $k$ be a nonzero noetherian ring and take $$A:=k[X_1,X_2,\dots]/(X_1, X_{mn}^m-X_n)_{m,n≥1}.$$ If $x_n$ denotes the class of $X_n$, we easily check that $x_n^n=0$ for all $n$, $x_n≠0$ if $n≥2$, and each $x_n$ is an $m$-th power for all $m$. If $\varphi:A\to C$ is a morphism with $C$ noetherian, the above argument (with $a=$any $x_n$) shows that $\varphi$ factors through $A/(x_n)_{n≥1}\cong k$. Now if $\Lambda$ is the ordered set of finitely generated $k$-subalgebras of $A$, then of course $A$ is the colimit of $\Lambda$ in CRing, but the above shows that there is a colimit in Noeth, which is $k$.

Positive answer to 4: $\mathbb{Z}[x_1,x_2,\dots]$ is a Krull domain (even a UFD), hence an intersection of discrete valuation rings inside its fraction field.

Negative answer to 5: Given a ring $R$, let us call a noetherian $R$-algebra $S$ a noetherian hull of $R$ if it is an initial object of the category of noetherian $R$-algebras.

Proposition. If $R$ has a noetherian hull $S$, the natural map $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ is surjective. In particular, $\mathrm{Spec}(R)$ is a noetherian space.
Proof: for each $p\in \mathrm{Spec}(R)$ its residue field $\kappa(p)$ (i.e. $\mathrm{Frac}(R/p)$) is a noetherian $R$-algebra, hence $R\to\kappa(p)$ factors through $S$. QED
(Remark: it is easy to see that $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ is in fact bijective, with trivial residue field extensions).

Now, if $A$ and $B$ are two noetherian rings, a coproduct of $A$ and $B$ in Noeth is the same thing as a noetherian hull of $A\otimes B$: this is clear from the universal properties. Thus, to answer Question 5 negatively, it suffices to find two noetherian rings $A$, $B$ such that $\mathrm{Spec}(A\otimes B)$ is not a noetherian space. There are plenty of examples, for instance $A=B=\overline{\mathbb{Q}}$.

Note: the same example was given by François Brunault in his answer, with a different argument. Since I have edited my answer several times, I am not sure who came first!

Answer 2

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.