# checking the functor $\texttt{Nil}_n$ is represented by $(\mathbb{Z}[x]/(x^n), \tau_R)$

This is the continuation of another question I did some days ago. Here.

I have been working on it and I would like to know if my try to prove it is right or not. I would appreciate a lot any feedback I can get.

I want to prove that $$\texttt{Nil}_n$$ is represented by $$(\mathbb{Z}[x]/(x^n), \tau_R)$$. Meaning that there exists a family of functions for every ring $$R$$, $$\tau_R: h^{\mathbb{Z}[x]/(x^n)}(R) \longrightarrow \texttt{Nil}_n(R): f \mapsto f(\overline{x})$$ that is an isomorphism (for each $$R$$)(bijective application, since it is a function between sets). And $$\tau_R$$ is natural, i.e. the following map commutes for every map $$g:A\longrightarrow B$$ of rings:

$$\require{AMScd}\begin{CD}h^{\mathbb{Z}[x]/(x^n)}(A) @>g \circ - >> h^{\mathbb{Z}[x]/(x^n)}(B) \\ @V\tau_AVV @V\tau_BVV\\\texttt{Nil}_n(A) @>>\texttt{Nil}_n(g)> \texttt{Nil}_n(B) \end{CD}$$

Proof:

• NATURALITY.

This means that we have to check $$\tau_S \circ h^{\mathbb{Z}[x]/(x^n)}(f)= \texttt{Nil}_n (f) \circ \tau_R$$.

Let $$h\in h^{\mathbb{Z}[x]/(x^n)}$$ be an arbitrary element.

On the one hand, we have $$(\tau_S \circ h^{\mathbb{Z}[x]/(x^n)}(f))(h)= \tau_S(h^{\mathbb{Z}[x]/(x^n)}(f)(h)) = \tau_S (f\circ h) = (f\circ h)(\overline{x})\in \texttt{Nil}_n(S)$$.

On the other hand, we have $$(\texttt{Nil}_n (f) \circ \tau_R)(h)=\texttt{Nil}_n (f)(\tau_R(h))= \texttt{Nil}_n (f)(h(\overline{x}))= f(h(\overline{x}))= (f\circ h)(\overline{x}) \in \texttt{Nil}_n(S)$$.

Hence, $$\tau_S \circ h^{\mathbb{Z}[x]/(x^n)}(f)= \texttt{Nil}_n (f) \circ \tau_R$$.

• INJECTIVITY.

Let $$a,a'\in h^{\mathbb{Z}[x]/(x^n)}$$ be two different elements.

$$a: h^{\mathbb{Z}[x]/(x^n)} \longrightarrow R: \overline{x} \mapsto a(\overline{x})$$

$$a': h^{\mathbb{Z}[x]/(x^n)} \longrightarrow R: \overline{x} \mapsto a'(\overline{x})$$

Since $$\texttt{Nil}_n(R) \subseteq R$$, $$\tau_R(a)$$ or $$\tau_R(a')$$ is nothing else than the restriction of $$a,a'$$ to the subset $$\texttt{Nil}_n(R)$$. Hence, $$\tau_R(a)\neq \tau_R(a')$$ when $$a\neq a'$$.

• SURJECTIVITY.

Let $$b\in \texttt{Nil}_n(R)$$. Since it is a nilpotent element, we know that $$b^n=0$$. We know that there exists an homomorphism in $$h^{\mathbb{Z}[x]/(x^n)}$$, namely $$h$$, such that $$h(0)=0$$. That means $$b^n=0=h(0)$$. But since $$\overline{0}=\overline{x^n}$$, $$b^n=0=h(0)=h(\overline{x^n})=h(\overline{x})^n$$.

Hence $$b=h(\overline{x})$$.

Are Naturality, Injectivity and Surjectivity well-proven?

Thank you.

Naturality looks good to me.

Injectivity and surjectivity both look problematic to me though.

Again, let $$A=\Bbb{Z}[x]/(x^n)$$ for notational brevity.

Injectivity:

You say $$\tau_R(a)$$ is the restriction of $$a$$ to $$\newcommand\Nil{\operatorname{Nil}}\Nil_n(R)$$, which isn't right. $$\tau_R(a)=a(\bar{x})$$. Thus you need to prove that if $$a\ne a'$$, then $$a(\bar{x})\ne a'(\bar{x})$$. However, this is immediate, since $$a$$ and $$a'$$ are maps from $$A$$ to $$R$$, which is generated over $$\Bbb{Z}$$ by $$\bar{x}$$, so if $$a(\bar{x})=a'(\bar{x})$$, then $$a(p(\bar{x}))=a'(p(\bar{x}))$$ for any polynomial $$p\in \Bbb{Z}[x]$$, and any element of $$A$$ can be expressed as $$p(\bar{x})$$ for some polynomial, so we must have $$a=a'$$.

Note I am being much more explicit above than I usually would be, since I'm trying to explain it. It should suffice to write:

If $$\tau_R(a)=\tau_R(a')$$, then $$a(\bar{x})=a'(\bar{x})$$, so $$a=a'$$, since $$A$$ is generated over $$\Bbb{Z}$$ by $$\bar{x}$$.

Surjectivity:

Your argument for surjectivity doesn't make sense. You're saying there exists $$h\in h^A$$ such that $$h(0)=0$$, and then concluding that therefore $$h(\bar{x})=b$$. That doesn't make any sense, since every element $$h\in h^A$$ satisfies $$h(0)=0$$, but we just showed in the injectivity section that if $$h(\bar{x})=h'(\bar{x})$$, then $$h=h'$$, so they can't possibly all have $$h(\bar{x})=b$$.

Define $$\phi : \Bbb{Z}[x]\to R$$ by $$x\mapsto b$$. Since $$b^n=0$$, $$x^n$$ is in the kernel of $$\phi$$, so $$\phi$$ induces a map $$\tilde{\phi}:A\to R$$ with $$\tilde{\phi}(\bar{x})=b$$. $$\tilde{\phi}$$ is thus the desired map.
• Here, in surjectivity, why do you know $b^n=0$ if we only know that $b\in R$? Should $R$ be $Nil_n(R)$? @jgon – idriskameni Jan 2 at 15:01
• @idriskameni, yes $b$ should be in $\Nil_n(R)$, sorry for not being more explicit there. – jgon Jan 2 at 15:32