# functor $\texttt{Nil}: Ring \longrightarrow Set$ is not representable

Consider the functor $$\texttt{Nil}: Ring \longrightarrow Set$$. I want to show that it is not representable.

I have been trying to adapt a proof from the pdf of Zach Norwood:

HOW TO PROVE THAT A NON-REPRESENTABLE FUNCTOR IS NOT REPRESENTABLE

But I do not know if I am doing good. I will let my try here:

Let $$\texttt{Nil}: Ring \longrightarrow Set$$ be the functor which sends a ring $$R$$ to its nilradical and ring homomorphisms to its restriction to nilradicals.

Suppose that $$\texttt{Nil} \cong h^A=Hom_{Ring}(A,-)$$ for some ring $$A$$. In particular, we have that $$\texttt{Nil}(A) \cong Hom_{Ring}(A,A)$$.

Consider an element $$a \in \texttt{Nil}(A)$$ corresponding via this isomorphism to $$id_A\in Hom_{Ring}(A,A)$$. We will show that $$a\in \texttt{Nil} (A)$$ has the following universal property:

$$\forall B \in Ring$$, and every $$b \in B$$ such that $$b^n=0$$ for some integer $$n$$, there exists a unique homomorphism $$A \longrightarrow B$$ sending $$a$$ to $$b$$.

Consider $$\tau: h^A \longrightarrow \texttt{Nil}$$ the natural transformation and the commutative diagram:

$$\require{AMScd}\begin{CD}h^A(A) @>g \circ - >> h^A(B) \\ @V\tau_AVV @V\tau_BVV\\\texttt{Nil}(A) @>>\texttt{Nil}(f)> \texttt{Nil}(B) \end{CD}$$

The map $$id_A \in Hom_{Ring}(A,A)$$ has the universal property:

$$\forall g \in h^A(B)$$, is the unique element in $$h^A(B)$$ such that $$(g\circ -)(id_A)=g$$

By naturality, $$a \in \texttt{Nil}(A)$$ has the universal property that for every $$y \in \texttt{Nil}(B)$$ such that $$y^n=0$$ for some $$n$$, there exist a unique homomorphism such that $$\texttt{Nil}(g)(a)=y$$, i.e. $$g(a)=y$$.

Let $$B=\mathbb{Z}[x]$$ and $$b=x$$. There is (supposedly) a unique $$g : A \longrightarrow B$$ such that $$g(a)=x$$. That means $$0=g(0)=g(a^n)=g(a)^n=x^n$$. Which is a contradiction.

• Continuation of this question. See also this question for representability. – Dietrich Burde Dec 28 '18 at 19:53
• Have already seen it. But not helpfull at all. That is why I have posted this, my try. – idriskameni Dec 28 '18 at 19:55

Qiaochu Yuan has given good advice in general, but I'd already started writing this answer, and it's more a review of what you've written.

You've almost got it. It's all correct up to your last paragraph. The problem is that you chose $$B=\Bbb{Z}[x]$$ and $$b=x$$, but the universal property of $$a$$ only guarantees that there exists a map $$g:A\to B$$ with $$g(a)=b$$ when $$x$$ is nilpotent. However in this case it is not, since $$\Bbb{Z}[x]$$ is a domain.

Instead, observe that if $$a^n=0$$ for some $$n$$, which must exist since $$a$$ is nilpotent, then consider $$B=\Bbb{Z}[x]/(x^{n+1})$$. Then there must exist $$g:A\to \Bbb{Z}[x]/(x^{n+1})$$ with $$g(a)=x$$, but then $$0=g(a^n)=g(a)^n=x^n\ne 0$$. Contradiction.

I assume that by "ring" you mean "commutative ring." Suppose $$\text{Nil}$$ is represented by some commutative ring $$N$$. Then $$\text{id}_N \in \text{Hom}(N, N) \cong \text{Nil}(N)$$ must be the "universal nilpotent" $$n \in N$$: that is, it is a nilpotent with the property that it maps to every other nilpotent in every other commutative ring $$R$$ under a (unique) homomorphism $$N \to R$$. (Every representable functor works this way: see universal element. This is $$a$$ in your work.)

But there can't be a universal nilpotent, because any nilpotent element must be nilpotent of some particular degree $$k$$. And if $$n^k = 0$$ then $$n$$ can't map to nilpotents of degree larger than $$k$$. For a bunch of variations on this argument see this blog post. In your work you attempt to map the universal nilpotent to something which is not a nilpotent at all.

Alternatively although similarly, you can argue that $$\text{Nil}$$ doesn't preserve infinite limits. (A representable covariant functor preserves all limits.) In fact it already fails to preserve infinite products, as follows: consider the product

$$R = \prod_{k \in \mathbb{N}} \mathbb{Z}[x]/x^k.$$

Then each $$x \in \mathbb{Z}[x]/x^k$$ is nilpotent but the product element $$\prod x$$ is not.