#16843: Zeromorphism
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Reporter: mkamalakshya | Owner: mkamalakshya
Type: defect | Status: needs_work
Priority: minor | Milestone: sage-6.4
Component: algebra | Resolution:
Keywords: days60 | Merged in:
Authors: Kamalakshya Mahatab | Reviewers:
Report Upstream: N/A | Work issues:
Branch: | Commit:
Dependencies: | Stopgaps:
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Comment (by tscrim):
Replying to [comment:9 pbruin]:
> No, it isn't. '''The zero map between two rings is a ring homomorphism
if and only if the codomain is the zero ring''' (note: codomain, not
image).
Warning for the next bit, I'm not a commutative algebraist or category
theorist, so please don't hate me if I'm spouting nonsense.
I agree that there would be problems if the 1,,R,, did not go to
1,,Im(f),, under a funtion `f` and that we cannot show this from the ring
axioms (unlike fields). I don't like that enlarging the codomain changes
whether the function is a morphism or not. Can you think of another
category where this happens or some property of the morphism that would
change?
> You are not correctly checking the definition of ring homomorphism. A
ring is a triple (''R'', +, .) satisfying the ring axioms, which say
(among other things) that there exists a two-sided multiplicative identity
1 in ''R''; this is automatically unique. A ring homomorphism from ''R''
to ''S'' is a homomorphism of additive groups that respects the
multiplication ''and'' sends the multiplicative identity element of ''R''
to that of ''S''. (One can rephrase this as saying that is compatible
with the empty product. See also [http://arxiv.org/abs/1404.0135].) This
implies in particular that any ring homomorphism from '''Q''' to itself
sends 1 to 1, and that the only ring homomorphism from '''Q''' to '''Q'''
is the identity.
That was an interesting note.
> The zero map is only a ''rng'' homomorphism (since rngs don't have 1, we
can't require rng homomorphisms to preserve 1). So someone who cares
about rngs might want the following to work (it currently doesn't):
> {{{
> sage: H = QQ.Hom(QQ, category=Rngs())
> sage: H([0])
> }}}
> The following is also problematic (in my opinion even more so):
> {{{
> sage: H = QQ.Hom(QQ, category=VectorSpaces(QQ))
> Traceback (most recent call last):
> ...
> ValueError: Rational Field is not in Category of vector spaces over
Rational Field
> }}}
> We should make sure that `QQ` is in `VectorSpaces(QQ)`. For the moment,
one has to construct '''Q''' as a 1-dimensional vector space over itself:
> {{{
> sage: H = Hom(QQ^1, QQ^1)
> sage: f = H([0])
> sage: f
> Vector space morphism represented by the matrix:
> [0]
> Domain: Vector space of dimension 1 over Rational Field
> Codomain: Vector space of dimension 1 over Rational Field
> }}}
Agreed. I believe there was only a wish/desire for letting rings know they
are also a free module/algebra over themselves; so no actual code yet. And
this ticket got a lot more involved than I was originally had thought.
--
Ticket URL: <http://trac.sagemath.org/ticket/16843#comment:10>
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