Dtyjlui

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 $id_Ain Hom_{Ring}(A,A)$. We will show that $ain texttt{Nil} (A)$ 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:

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

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 $(gcirc -)(id_A)=g$

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.

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 $g:Ato B$ 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 $g:Ato Bbb{Z}[x]/(x^{n+1})$ with $g(a)=x$, but then $0=g(a^n)=g(a)^n=x^nne 0$. 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.