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:


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.

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

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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.

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