Jtdylktuy

I want to show that $forall n geq 1$, there is a covariant functor $texttt{Nil}_n: Ring longrightarrow Set$ that sends a ring $R$ to the set ${xin R | x^n = 0}$.

I have being thinking that the functor sends ring homomorphisms $f:Xlongrightarrow Y$ to $texttt{Nil}_n(f):texttt{Nil}_n(X) longrightarrow texttt{Nil}_n(Y)$ with $texttt{Nil}_n(f)(x)= f(x)$ if $f(x) in Y$ and $texttt{Nil}_n(f)(x)=0$ if $f(x) notin Y$.

With that, I should verify if all the conditions in functor's definition are satisfied.

I have seen:

(1) $forall A in Ring$, it specifies a unique object $F(A)$ in $Set$ (Checked)

(2) $forall f in Hom_C (A, B)$, it specifies a morphism $F(f) in Hom_D(F(A), F(B))$. (Checked)

It should hold also:

(3) $forall A in Ring$, $F(Id_A)= Id_{F(A)}$ (Problem)

(4) $forall f: A longrightarrow B$, $g:B longrightarrow C$ then $F(gcirc f) = F(g) circ F(f)$. (Problem)

I have some problems verifying the last two conditions. Any help?

Is that suficient to show what I wrote in the first sentence?

If I want to do the same with the functor $texttt{Nil}: Ring longrightarrow Set$ that sends a ring $R$ to its nilradical. What should the functor do to the ring homomorphisms $f:X longrightarrow Y$?

Thank you.

Let $f : Rto S$ be a ring homomorphism.

Let's first examine $texttt{Nil}_n(f).$ You've written that $texttt{Nil}_n(f)(x) = f(x)$ if $f(x)in S$ and $0$ if $f(x)notin S.$ But $f(x)$ is always in $S,$ by definition - $f$ is a ring morphism with codomain $S$! So, the condition is unnecessary; you simply have
$$
texttt{Nil}_n(f)(x) = f(x)
$$
for all $xin texttt{Nil}_n(R).$

One should check that this is well-defined. Well, if $xintexttt{Nil}_n(R),$ then by definition $x^n = 0$ in $R.$ Hence, $f(x)^n = f(x^n) = f(0) = 0,$ so that $f(x)intexttt{Nil}_n(S),$ and we actually have a well-defined map of sets.

Let's check the identity condition. Let $S = R$ and $f = operatorname{id}_R,$ and let $xintexttt{Nil}_n(R).$ Then
$$
texttt{Nil}_n(operatorname{id}_R)(x) := operatorname{id}_R(x) = x,
$$
so that $texttt{Nil}_n(operatorname{id}_R)$ is the identity of $R$ restricted to the set $texttt{Nil}_n(R).$ This means exactly that $texttt{Nil}_n(operatorname{id}_R) = operatorname{id}_{texttt{Nil}_n(R)}.$

Notice that the above holds essentially because all the functor $texttt{Nil}_n$ does to morphisms is restrict them to the subset $texttt{Nil}_n(R)subseteq R.$ That is, if we have a ring homomorphism $f : Rto S,$ we may also say $texttt{Nil}_n(f) = left. fright|_{texttt{Nil}_n(R)}.$

Now, you could check element-wise that the assignment $fmapsto texttt{Nil}_n(f)$ behaves nicely with respect to composition, but you can also see this immediately via our new description of $texttt{Nil}_n(f).$ Let $f : Rto S$ and $g : Sto T$ be ring homomorphisms. Then
begin{align*}
texttt{Nil}_n(gcirc f) &= left. left(gcirc fright)right|_{texttt{Nil}_n(R)}\
&= left. gright|_{texttt{Nil}_n(S)}circ left.fright|_{texttt{Nil}_n(R)}quadtextrm{(because }left.fright|_{texttt{Nil}_n(R)}(texttt{Nil}_n(R))subseteq texttt{Nil}_n(S).)\
&= texttt{Nil}_n(g)circ texttt{Nil}_n(f).
end{align*}

Your functor $texttt{Nil}$ which assigns to $R$ the nilradical of $R$ should do essentially the same thing - simply restrict a morphism $f : Rto S$ to the nilradical. Notice also that your work with the $texttt{Nil}_n$ immediately implies this is well-defined, because $texttt{Nil}(R) = bigcup_{ngeq 1}texttt{Nil}_n(R).$