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(Intuition: in the category of non-empty sets, every function that coequalizes all points in the domain factors through the terminal object. I would like to know if something analogous happens in certain category of `connected algebraic spaces'. I formulate the precise question in terms of commutative algebra.)

Let $\cal A$ be the category of finitely generated $\mathbb{C}$-algebras with exactly two idempotents.

Let ${f : A \rightarrow B}$ be a map in $\cal A$ such that, for every ${g, h : B \rightarrow \mathbb{C}}$, ${g f = h f : A \rightarrow \mathbb{C}}$.

Does $f$ factor (inside $\cal A$) as ${f = k l}$ with ${l : A \rightarrow L}$, ${k : L \rightarrow B}$ and $L$ local?

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  • $\begingroup$ I have two questions. Local $\mathbb{C}$-algebras are rarely finitely generated, so they are not in $\mathcal{A}$, so the second part is not clear. Secondly, have you considered the map $A=\mathbb{C}[t^2,t^3]\subset \mathbb{C}[t]$ for your $f$? $\endgroup$
    – Mohan
    Commented Mar 1, 2018 at 1:53
  • $\begingroup$ @Mohan: That's does not satisfy the condition $\forall g,h: B\to\mathbb{C}: g\circ f=h\circ f$. Polynomial rings won't work because the condition implies $im(f) \subseteq \mathbb{C} + rad(B)$. $\endgroup$
    – Johannes Hahn
    Commented Mar 1, 2018 at 2:21
  • $\begingroup$ You are right, but you did not my question about local rings. $\endgroup$
    – Mohan
    Commented Mar 1, 2018 at 2:58
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    $\begingroup$ @Mohan: Maybe your remark about local $\mathbb{C}$-algebras can be extended to an example showing that the answer to my question is 'no', but I think that the question is clear. $\endgroup$
    – Mendieta
    Commented Mar 1, 2018 at 3:46

1 Answer 1

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Yes it's true.

1) First assume that $B$ is reduced (non nonzero nilpotent element). Let $I$ be the set of maximal ideals of $B$. Since for a f.g. algebra over a field, the Jacobson radical equals the radical, we have $\bigcap_{M\in I}M=0$. By the Nullstellensatz, for every $M\in I$ there is a unique $\mathbf{C}$-algebra homomorphism $u=u_M:B\to\mathbf{C}$ such that $\mathrm{Ker}(u)=M$. Thus we have a canonical embedding $j=(u_M)_{M\in I}$ of $\mathbf{C}$-algebras $B\to\mathbf{C}^I$.

I claim that $j(f(A))$ is valued in the diagonal of $\mathbf{C}^I$ (the constant maps $I\to\mathbf{C}$). Indeed otherwise this means that $j(f(A))$ contains a nonconstant map, which means that there exist $M,M'\in I$ and $a\in A$ such that $u_M(f(a))\neq u_{M'}(f(a))$. So $u_M\circ f\neq u_{M'}\circ f$. This precisely contradicts the assumption.

So $j(f(A))$ is contained in the diagonal (i.e., in the scalars of $\mathbf{C}^I$). Since $j$ is injective, this means that $f(A)$ is contained in the scalars. Hence the conclusion holds (with $L=\mathbf{C}$).

2) In general, let $R$ be the nilradical of $B$. Then the composite map $A\to B/R$ clearly satisfies the assumption, so is valued in the scalars. This means that $f(A)=(f(A)\cap R)\oplus\mathbf{C}1_B$, where the direct sum is in terms of vector spaces. This implies that $f(A)$ has Krull dimension zero (since its nilradical is a hyperplane, see details in edit below) and actually that $L=f(A)$ is local artinian.

Note: the assumption on idempotents seems unnecessary.

Conclusion:

For every algebraically closed field $K$, and $f:A\to B$ homomorphism of finitely generated (commutative associative unital) $K$-algebra we have equivalences

  1. $g_1f=g_2f$ for all $g_1,g_2$ $K$-algebra homomorphisms $B\to K$
  2. $f(A)$ is local (artinian)
  3. $f$ factors through a local finitely generated $K$-algebra (i.e., a commutative $K$-algebra that is local and of finite dimension as $K$-module).

Edit (to answer the OP's request in the comment)

Proposition: let $K$ be a field and $D$ a commutative $K$-algebra. Suppose that the nilradical $N$ of $D$ is a hyperplane. Then $D$ is local artinian.

Proof: since on an arbitrary commutative ring every prime ideal contains the nilradical, it follows from the assumption that $N$ is the only prime ideal, and thus $D$ is artinian. Also it follows that $N$ is the unique maximal ideal, and hence $D$ is local.

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  • $\begingroup$ PS: I realize that Part (1) of the proof precisely amounts to proving the claim in @JohannesHahn's comment. $\endgroup$
    – YCor
    Commented Mar 1, 2018 at 13:05
  • $\begingroup$ This is exactly the thing I was looking for. I hope you don't mind me requesting some further (possibly naive) details of your proof. The argument shows that the composite ${A \rightarrow B \rightarrow B/R}$ factors as $\require{AMScd}$ \begin{CD} A @>>> \mathbb{C} \\ @V f V V= @VV V\\ B @>>> B/R \end{CD} Why is it that you can conclude that ${f(A)}$ is local and Artinian? (I realise that this may be a triviality but sometimes very simple commutative-algebraic facts ellude me. In this case, a pointer to the right place in Atiyah-MacDonald would be enough.) $\endgroup$
    – Mendieta
    Commented Mar 1, 2018 at 13:31
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    $\begingroup$ PS this answer is quoted in the paper "<i>Level $\epsilon</i>" by Francisco Marmolejo, Matías Menni, arxiv.org/abs/1909.12757 $\endgroup$
    – YCor
    Commented Jun 10, 2020 at 18:11
  • $\begingroup$ Interesting that this has some kind of relation with topos theory. Anyway, the published version of the paper is also freely downloadable from the journal's webpage. cahierstgdc.com/index.php/volume-lx $\endgroup$
    – Mendieta
    Commented Jun 16, 2020 at 14:56

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