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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.]

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    $\begingroup$ Every (associative unital) commutative ring is the union of its finitely generated subrings, which are noetherian, and hence is an filtering inductive limit of noetherian rings with injective homomorphisms. This maybe answers Question 4, but you haven't defined what you mean by limit. $\endgroup$ – YCor Feb 13 at 13:03
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    $\begingroup$ @YCor - Thanks! The question is about limits, not colimits. In question 1 I wrote "Is there a functor from a small category to $\mathsf{Noeth}$ whose limit in $\mathsf{CRing}$ is $A$?". Do you think it's not a clear enough definition of what is meant by "limit"? $\endgroup$ – Pierre-Yves Gaillard Feb 13 at 13:10
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    $\begingroup$ @FrançoisBrunault $\mathbb{Z}$ is Noetherian, so the ideal $\{0\}$ is such $\mathfrak{a}$ and so $I$ has an initial object, and so the limit is just $\mathbb{Z}$, no? $\endgroup$ – David Roberts Feb 13 at 14:20
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    $\begingroup$ Then take $A=$ the ring of all algebraic integers. Its only noetherian quotients are finite products of residue fields of maximal ideals. So the limit is the product of all those residue fields. $\endgroup$ – Laurent Moret-Bailly Feb 13 at 14:43
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    $\begingroup$ Sorry yes, "colimits" should be "limits" in my second sentence of my previous comment. $\endgroup$ – YCor Feb 13 at 16:02
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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!

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    $\begingroup$ Thanks for your answer and your comment! If I'm not mistaken your post also gives a negative answer to Question 1. Don't you think so? [In your answer I think $x=a$.] $\endgroup$ – Pierre-Yves Gaillard Feb 13 at 15:13
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    $\begingroup$ Yes, and in fact also to Question 3. I have edited. $\endgroup$ – Laurent Moret-Bailly Feb 13 at 19:57
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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.

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