4
$\begingroup$

Let $R$ be a commutative ring and let $F$ be a functor $\mathbf{Mod}_R\rightarrow \mathbf{Mod}_R$. Then for a module $M$ the split mono $M\rightarrow M\oplus R$ gives a split mono $F(M)\rightarrow F(M\oplus R)$. Define $F'(M)=\frac{F(M\oplus R)}{F(M)}$. Then $F'$ extends to a functor $\mathbf{Mod}_R\rightarrow \mathbf{Mod}_R$ (since given $\phi:M\rightarrow M'$ one can easily check that the induced morphism $\frac{F(M\oplus R)}{F(M)}\rightarrow \frac{F(M'\oplus R)}{F(M')}$ is well defined).

I'm looking for functors $F$ such that $F$ is naturally isomorphic to $F'$. The exterior algebra (with it's algebra structure forgotten) is such a functor, since

$$\Lambda(M\oplus R)\cong \Lambda(M)\otimes \Lambda(R) \cong \Lambda(M)\otimes (R\oplus R) \cong \Lambda(M) \oplus \Lambda(M)$$

and it can be checked that the canonical morphism $\Lambda(M)\rightarrow \Lambda(M\oplus R)$ does indeed hit the first $\Lambda(M)$.

We then also notice that $M\otimes\Lambda(-)$ is a solution for any $M$. Are these all the solutions?


The motivation for this question is that $F'$ is like the derivative (or first difference) of $F$ and so the equation $F\cong F'$ is like the differential equation with solutions $\mathrm{m\times exp(-)}$. So I am asking if $\Lambda$ is like $\mathrm{exp}$.

$\endgroup$
  • $\begingroup$ Also, does the tensor algebra $T$ satisfy $T'=T^2$? $\endgroup$ – Oscar Cunningham Jan 19 '16 at 15:39

Your Answer

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

Browse other questions tagged or ask your own question.