Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a commutative ring and let $M$ and $N$ be $R$-modules. Let $\sigma:R\rightarrow R$ be a ring automorphism.

Let $f: M\rightarrow N$ be a $\sigma$-semilinear map, i.e. a map of abelian groups satisfying $f(rm)=\sigma(r)f(m)$. Let $Hom_{R,\sigma}(M,N)$ denote the set of all $\sigma$-semilinear maps from $M$ to $N$.

The usual definition of the dual of $f$ does not give the right result, as it is a map $N^*=Hom_{R,id}(N,R)\rightarrow Hom_{R,\sigma}(M,R)\qquad n^* \mapsto n^* \circ f$. To force the target to be $Hom_{R,id}(M,R)$ I would like to postcompose with $\sigma^{-1}\in Hom_{R,\sigma^{-1}}(R,R)$. But of course the $\sigma$ is not uniquely determined by the map $f$. For example the zero map is $\sigma$-semilinear for any $\sigma$.

So my question is: Is this well defined, i.e. Given any $f\in Hom_{R,\sigma}(M,N)\cap Hom_{R,\sigma'}(M,N)$ and any $n^* \in N^*$. Do we have

$\sigma^{-1} \circ n^* \circ f= \sigma '^{-1}\circ n^* \circ f$?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

No.

Take $R=k[x,y]/(x^2,xy,y^2)$ and let $\sigma=id$ and $\sigma'$ swap $x$ and $y$.

Let $M=N=R/(x,y)$, then the identity map $M\to N$ is $\sigma$ and $\sigma'$ semilinear.

One element of $N^*$ sends the generator to $x$. But $\sigma^{-1}$ sends $x$ to itself and $\sigma'^{-1}$ sends $x$ to $y$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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