2018-02-18 15:18:45 +0100 received badge ● Notable Question (source) 2017-12-02 18:40:08 +0100 received badge ● Popular Question (source) 2017-07-14 19:37:02 +0100 received badge ● Editor (source) 2017-07-14 19:27:34 +0100 asked a question Compute preimage of Ideals It's me again. I am wondering how to compute the preimage of an ideal $J\subseteq B$ in a ring along a ring homomorphism $f\colon A\to B$. For simplicity I would be fine with assuming $A$ and $B$ are polynomial rings of finitely many variables over a field. As far as I know, preimages can then be computed algorithmically using Gröbner bases so I think this should be implemented in Sage. However I did not find such a possibility when browsing through the documentation. I hope you can help me (and maybe also point me to where such issues are documented). Thank you! P.S.: The question of how to compute kernels of homorphisms of affine algebras over a field is of course equivalent. 2017-07-14 17:21:02 +0100 received badge ● Student (source) 2017-07-14 12:33:40 +0100 asked a question Declaring a vector space map of rings Hello sage community! I am new to sage and have a question which is originally motivated from my tries to implement Berlekamp's algorithm. One is working over a finite field $K=GF(q)$ and wants to factor a polynomial $f\in K[x]$. To do so one considers the ring $R=K[x]/f$ and the map $\beta\colon g\mapsto g^q-g$ from $R$ to itself and wants to compute its kernel. The problem for me is that the definition of $\beta$ uses the ring multiplication of of $R$ but is only a map of $K$-vector spaces, not a ring homomorphism. I am wondering how I can declare this map using sage. I know I can find a standard $K$-basis of $R$ and I know I can describe $\beta$ by describing its matrix corresponding to this base and I know how to compute the kernel of that matrix and how to translate its elements back to elements of $R$. But still I am interested in wether it is possible to declare a vector space homomorphism of rings in sage and how to work with it. Thank you!