Thanks Nils. Actually, I think that what I need is a conversion rather than
a coercion as `register_as_conversion()` lets me change the category.
On Wednesday 31 July 2024 at 2:49:16 am UTC+10 Nils Bruin wrote:
> On Monday 29 July 2024 at 22:13:27 UTC-7 Andrew wrote:
>
> [Not sure if this belongs here or in sage-dev...]
>
> I am trying to implement coercions between algebras that are related by
> base change. For example,consider
>
> A=CombinatorialFreeModule(ZZ['x'], ['1','2'])
> B=CombinatorialFreeModule(ZZ, ['1','2'])
> A.module_morphism(
> lambda a: B._from_dict({b: c.subtitute(x=1) for (b,c) in a}),
> codomain=B, category=A.category()
> ).register_as_coercion()
>
>
> Are you sure you want to register that as a *coercion*? Those are to be
> used in other coercion discoveries as well and can be used implicitly to
> resolve things like a+b, where a in A and b in B (and extensions of these!)
>
>
> This is entirely reasonable because I have not specified how Z is a
> Z[x]-module, but when I try to define this it seems I need slightly
> different syntax:
>
> Rx = ZZ['x]
> R = ZZ
> Rx.module_morphism(function=lambda f: f.substitute(x=1), codomain=R)
>
>
> These are bases, so I think you should define a ring homomorphism between
> them (I think a CombinatorialFreeModule has a ring as its base). And then
> you see how you'd get a problem if you insert a coercion from ZZ['x'] to
> ZZ: there's already one in the oppositie direction and sage really prefers
> its coercion graph to not have directed cycles.
>
> So I expect that your original map A -> B should really be one from a
> ZZ['x']-module to a ZZ-module, where ZZ is really ZZ['x']/(x-1).
>
> In fact, with
>
> P.<x>=ZZ[]
> R=P.quo(x-1)
> A=CombinatorialFreeModule(P, ['1','2'])
> B=CombinatorialFreeModule(R, ['1','2'])
>
> I get:
>
> sage: x*A('1')+B('2')
> B['1'] + B['2']
>
> so it seems to discover the coercion correctly. (I do get that A('1')
> prints as B['1'] so there is something fishy there. Are combinatorial
> modules always printing as `B`? or is that the default name for its
> "basis"?)
>
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