I would like to see that $\texttt{Nil}_n$ is isomorphic to $h^{\mathbb{Z}[x]/(x^n)}$ as categories.

$\texttt{Nil}_n: \texttt{Rings} \longrightarrow \texttt{Sets}$ is the functor that sends a ring $R$ to $\{x \in R | x^n = 0\}$.

$h^{\mathbb{Z}[x]/(x^n)}: \texttt{Mod}_R \longrightarrow \texttt{Mod}_R$ is the functor where $\mathbb{Z}[x]/(x^n)$ is an $R$-module and it sends an $R$-module $N$ to $Hom(\mathbb{Z}[x]/(x^n),N)$.

I know that two categories are isomorphic if when there exists two functors as above, they satisfy also the above conditions.

I want to see that if there exists two functors $F: \texttt{Nil}_n \longrightarrow h^{\mathbb{Z}[x]/(x^n)}$ and $G: h^{\mathbb{Z}[x]/(x^n)} \longrightarrow \texttt{Nil}_n$, then $FG=id_{h^{\mathbb{Z}[x]/(x^n)}}$ and $GF=id_{\texttt{Nil}_n}$.

But my problem comes when I have to find / see which are these functors. Does $F$ and $G$ even exist? How can I find them? Are they known functors?

I want to see that in order to see that $\texttt{Nil}_n$ is represented by the ring $h^{\mathbb{Z}[x]/(x^n)}$ and the natural equivalence

$\tau: h^{\mathbb{Z}[x]/(x^n)} \longrightarrow \texttt{Nil}_n$

given (for each ring $R$) by

$\tau_R: h^{\mathbb{Z}[x]/(x^n)}(R) \longrightarrow \texttt{Nil}_n(R): f ↦ f(\bar{x})$.

I was thinking on first showing that $\texttt{Nil}_n$ and $h^{\mathbb{Z}[x]/(x^n)}$ are isomorphic and then showing that there exists a natural iso $\mu: h^{\mathbb{Z}[x]/(x^n)} \longrightarrow \texttt{Nil}_n$ determined by $\mu_{\mathbb{Z}[x]/(x^n)}(Id_{\mathbb{Z}[x]/(x^n)})$ and it is equal to $\tau$.

  • 2
    $\begingroup$ You are asking for two functors to be isomorphic as categories, which doesn't make sense. They're not categories! Furthermore, you definitely don't want to consider $R$-module maps out of $Z[x]/x^n$ here-those have nothing to do with nilpotents. You should probably be showing that the representables associates to $Z[x]/x^n$ and $Nil_n$ are naturally isomorphic functors from rings to sets. $\endgroup$ – Kevin Arlin Dec 22 '18 at 16:03
  • $\begingroup$ And how can I check if two functors are isomorphic? $\endgroup$ – idriskameni Dec 22 '18 at 18:53
  • $\begingroup$ You've already given the natural map $\tau$. Now you just need to show $\tau_R$ is an isomorphism for every $R$. $\endgroup$ – Kevin Arlin Dec 22 '18 at 19:29

Ok, several problems. As Kevin Carlson points out in the comments, $\newcommand\Nil{\operatorname{Nil}}\Nil_n$ and (if we define $A=\Bbb{Z}[x]/(x^n)$) $h^A$ are functors, not categories.

Thus we need to show that they are isomorphic as functors.

Isomorphisms of functors

Recall, then, the definition of natural transformation, which is the notion of a morphism between functors. Let $F,G:C\to D$ be functors from a category $C$ to a category $D$. Then a natural transformation $\theta: F\to G$ is a family of maps, $\theta_x: F(x)\to G(x)$ for each $x\in C$ satisfying the following commutative diagram for every morphism $f:x\to y$ in $C$: $$\require{AMScd}\begin{CD}F(x) @>F(f)>> F(y) \\ @V\theta_xVV @V\theta_yVV\\G(x) @>>G(f)> G(y) \end{CD}$$

An isomorphism of functors (usually called a natural isomorphism) then is a natural transformation all of whose components are isomorphisms. (Check that this is equivalent to the natural transformation having an inverse natural transformation)

How to apply this to your question

First of all, you have the domains and codomains of your functors wrong. Well $\Nil_n$ is correct. It is a functor from $\newcommand\Ring{\mathbf{Ring}}\newcommand\Set{\mathbf{Set}}\Ring$ to $\Set$, but $h^A(R)=\newcommand\Hom{\operatorname{Hom}}\Hom_\Ring(A,R)$ is actually also a functor from $\Ring$ to $\Set$, not from right $R$-modules to right $R$-modules.

Thus you want to find a family of maps for every ring $R\in\Ring$, $\tau_R:h^A(R)\to \Nil_n(R)$ with $\tau_R$ an isomorphism of sets (i.e. a bijection) for every ring $R$. You should also check that $\tau_R$ is natural. I.e. that the following diagram commutes for every map $\phi :R\to S$ of rings: $$\begin{CD}h^A(R) @>h^A(\phi)>> h^A(S) \\ @V\tau_RVV @V\tau_SVV\\\Nil_n(R) @>>\Nil_n(\phi)> \Nil_n(S) \end{CD}$$

Note: I've used $\tau_R$, because you've already identified the correct family of maps $\tau_R$ in your question, you just have to prove it has the right properties.

Edit: In response to your comment asking how is $\tau_R$ defined exactly.

Well, as you mentioned, the elements of $h^A(R)$ are ring homomorphisms $f:A\to R$, and the elements of $\Nil_n(R)$ are elements $r$ of $R$ with $r^n=0$. Thus $\tau_R$ should take ring homomorphisms $f:A\to R$ and produce appropriate elements of $R$.

Well, $A=\Bbb{Z}[x]/(x^n)$, so let $\bar{x}$ be the image of $x\in \Bbb{Z}[x]$ in the quotient, $A$. Then if $f:A\to R$ is a ring homomorphism, $r=f(\bar{x})$ is an element of $R$, and moreover, $r^n=f(\bar{x})^n=f(\bar{x}^n)=f(0)=0$, so $r$ is in $\Nil_n(R)$.

| cite | improve this answer | |
  • 3
    $\begingroup$ To the downvoter: I don't particularly care about the rep, but if you think there's something wrong with my answer, please leave a comment, and I'll try to address it. $\endgroup$ – jgon Dec 23 '18 at 14:48
  • $\begingroup$ Thank you very much. I do understand what you say. But I have a hard problem trying to understand what $\tau_R$ does. Because I know that elements in $h^A(R)$ are ring homomorphisms from $A=\mathbb{Z}[x]/(x^n)$ to $R$. And elements in $Nil_n(R)$ are elements in $R$ that are zero when we take the $n$-th power. But how can you map $f \mapsto f(\overline{x})$?I do not understand what that means. Any help? $\endgroup$ – idriskameni Dec 28 '18 at 18:11
  • 1
    $\begingroup$ @idriskameni edited. $\endgroup$ – jgon Dec 28 '18 at 19:52
  • $\begingroup$ Could you tell me if this would be correct to check commutativity? Let $h\in h^A(R)$. Then $(\tau_S \circ h^A(f))(h)=\tau_S(h^A(f)(h))=\tau_S(f\circ h)=( f\circ h)(\overline{x})$. On the other hand, $(Nil_n(f) \circ \tau_R)(h)=Nil_n(f)(\tau_R(h))= Nil_n(f)(h(\overline{x}))= f(h(\overline{x}))=(f\circ h)(\overline{x})$. Hence, $\tau_S \circ h^A(f)=Nil_n(f) \circ \tau_R$. $\endgroup$ – idriskameni Dec 30 '18 at 16:14
  • 1
    $\begingroup$ @idriskameni Sorry, didn't see these till now, commutativity looks good, I'll edit to address bijectivity later, since I've got to head out now. $\endgroup$ – jgon Dec 31 '18 at 17:13

Hint: there is a universal property of $\mathbb Z[x]$ as the free ring on one generator. This means there is a functorial identifcation $$Hom_{Rings}(\mathbb Z[x], R) = Hom_{Sets}(\{x\},R) = \{x\in R\} = R$$

Now describe this equality in terms of an isomorphism of functors and adapt the proof to your situation. (Yes the functors natural transformations exist and are well known.)

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.