Another question on representability:

let $F$ be the covariant functor from the category CommutativeRing to Set which sends a ring to its nilradical. Is it representable? What if you restrict to the subcategory NoetherianCommutativeRing?

My intuitive guess is yes in the first case and no in the second case, but I'm not sure.

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    $\begingroup$ As Kevin's answer says, the functor is not representable, in either case. On the other hand, if instead of sending $R$ to its nilradical, you sent $R$ to $\{x \in R : x^n = 0\}$ for a fixed $n$ (which is not necessarily an ideal, but just a set), then it would be representable. $\endgroup$ – Ted Oct 4 '11 at 8:57
  • $\begingroup$ The functor Ted suggested is represented by the ring $\mathbb{Z}[x]/(x^n)$. This just says that to specify a ring morphism $\mathbb{Z}[x]/(x^n) \to R$ you just need to decide where $x$ goes and it can go to any element $y$ of $R$ such that $y^n=0$. $\endgroup$ – Omar Antolín-Camarena Oct 4 '11 at 14:02

No. Any representing ring would have to contain a universal nilpotent element (corresponding to the identity map on the representing ring), which clearly can not exist.

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    $\begingroup$ This is what Kevin means: Assume that the nilradical functor is represented by some ring $R$. The morphisms $R \to R$ should correspond to the nilpotent elements of $R$, let $r$ be the nilpotent corresponding to the identity morphism. By definition $r$ has the following property: for any nilpotent $a$ in some ring $A$, there is a unique ring morphism $R \to A$ mapping $r \mapsto a$. But if $r^n = 0$, clearly $r$ cannot be mapped to a nilpotent $a$ with $a^n \neq 0$. $\endgroup$ – Omar Antolín-Camarena Oct 4 '11 at 14:11

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