**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](https://math.stackexchange.com/q/3096154/660) 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 $A$?
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> 
> 


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](https://math.stackexchange.com/a/338437/660) 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.]



Is every commutative ring a limit of noetherian rings?
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!


