Examples in Category theory
Jose Capco
I am looking for examples of subcategory of Sets that have no...
products, coproducts
equalizer, coequalizer
And then maybe also examples by which it has one and doesnt have the
other- Any nice examples would be appreciated.
Sincerely,
Jose Capco
PS: Of course I can arbitrary construct a category with finite objects
by which this happens, but I want an example of a category that is
commonly worked with (e.g. groups, divisible groups, rings..etc.).
Arturo Magidin
Jose Capco <cliom...@kriocoucke.mailexpire.com> wrote:
>Dear NG,
>
>I am looking for examples of subcategory of Sets that have no...
>
>products, coproducts
Does not have ->any<- products or coproducts, or just does not have
arbitrary ones?
For the latter, look at FinSet, the full subcategory of finite sets;
it does not have arbitrary products or coproducts, because those
require infinite sets.
>equalizer, coequalizer
>
>And then maybe also examples by which it has one and doesnt have the
>other- Any nice examples would be appreciated.
>PS: Of course I can arbitrary construct a category with finite objects
>by which this happens, but I want an example of a category that is
>commonly worked with (e.g. groups, divisible groups, rings..etc.).
Hmm... I would not call groups a "subcategory of sets" (though, I
guess, technically it is); rather, I would call it a "concrete
Set-based category" (objects are sets, and arrows are functions of the
underlying sets). But your mileage may vary.
Fields does not have all products or all coproducts. I think it may
also fail to have coequalizers.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Gonçalo Rodrigues
<cliom...@kriocoucke.mailexpire.com> fed this fish to the penguins:
>Dear NG,
>
>I am looking for examples of subcategory of Sets that have no...
>
>products, coproducts
>equalizer, coequalizer
>
>And then maybe also examples by which it has one and doesnt have the
>other- Any nice examples would be appreciated.
>
>Sincerely,
>Jose Capco
>
In areas other than algebra is where you will find such examples:
1. The category of Banach spaces and bounded linear maps is finitely
complete and finitely cocomplete but neither complete nor cocomplete.
2. The category of differentiable manifolds (finite dimension) and
differentiable maps has products but may fail to have some equalizers
(the set {x:f(x) = g(x)} can exhibit all sorts of pathologies
depending on how f and g intersect). If you restrict to some fixed
finite dimension then it does not have products.
Off the top of my head, I cannot recall any "natural" example of a
category with equalizers but not having products.
Hope it helps, best regards,
G. Rodrigues
Jose Capco
> In areas other than algebra is where you will find such examples:
> 1. The category of Banach spaces and bounded linear maps is finitely
> complete and finitely cocomplete but neither complete nor cocomplete.
We could simply say metric spaces instead couldn't we?
Sincerely
Jose Capco
Jose Capco
> Hmm... I would not call groups a "subcategory of sets" (though, I
> guess, technically it is); rather, I would call it a "concrete
> Set-based category" (objects are sets, and arrows are functions of the
> underlying sets). But your mileage may vary.
Well yes, I guess you are right. I meant of course something that has
a forgetful functor to the sets (concrete small category perhaps? Im
not very into the names).. Because I would define a subcategory as
some category whose classes of objects is a subclass of the objects of
the over-category and I have to use some kind of a forgetful functor
here. Well, no matter, you did understood what I meant :)
>
> Fields does not have all products or all coproducts. I think it may
> also fail to have coequalizers.
Fields trivially have coequalizers
Let K,L and R be fields and let g,h: K->L be parallel field morphisms,
then if there is a field morphism f: L->R such that fg=fh we must
necessarily have g=h because all field morphisms are monomorphisms.
Gonçalo Rodrigues
<cliom...@kriocoucke.mailexpire.com> fed this fish to the penguins:
>On May 9, 4:17 pm, Gonçalo Rodrigues <op73...@mail.telepac.pt> wrote:
Which morphisms? There are several candidates for the morphisms
between metric spaces: continuous maps, contractions, lipschitz maps,
uniformly continuous maps, etc.
For continuous maps, the category is equivalent to metrizable
topological spaces which has countable, but not arbitrary products.
The same, if you take uniformly continuous maps. For contractions, it
has finite products but not infinite ones (hint: think of a sequence
of metric spaces with diameter going to infinity), etc., etc.
A cheap, but not "natural" way to get examples with colimits is to go
through the list and take the duals (or opposites) of those
categories.
Best regards,
G. Rodrigues
Jose Capco
> >We could simply say metric spaces instead couldn't we?
>
> Which morphisms? There are several candidates for the morphisms
> between metric spaces: continuous maps, contractions, lipschitz maps,
> uniformly continuous maps, etc.
Well, I always thought that I can consider it to be a subcategory of
the category of topological spaces.. and it is a a standard practice
to take the continuous maps as the morphisms in the category of
topological spaces (we want the isomorphisms to turn out to be -just-
homeomorphism) .. but maybe in the metric spaces we want the
isomorphism to be not only homeomorphism but also have compatible
metrics, ie. an isometric homeomorphism.. thus probably for metric
spaces it would be most logical to take the morphisms as isometric
continuous functions.
Sincerely,
Jose Capco
Marc Olschok
This would be too restrictive. Instead one takes the Lipschitz-1 maps
as morphism between metric spaces, i.e. those maps f: (X,d) --> (X',d')
that satisfy d'(fx,fy) <= d(x,y).
Then the isomorphisms are exactly the isometries.
Marc
Jose Capco
> Jose Capco <cliomse...@kriocoucke.mailexpire.com> wrote:
> This would be too restrictive. Instead one takes the Lipschitz-1 maps
> as morphism between metric spaces, i.e. those maps f: (X,d) --> (X',d')
> that satisfy d'(fx,fy) <= d(x,y).
> Then the isomorphisms are exactly the isometries.
>
> Marc
I was actually thinking about that :) .. but I didnt know if anyone
would object.. yeah this is probably the natural way of doing it. But
still I don't suppose infinite products do exist here.
Mike
>
> Fields trivially have coequalizers
>
> Let K,L and R be fields and let g,h: K->L be parallel field morphisms,
> then if there is a field morphism f: L->R such that fg=fh we must
> necessarily have g=h because all field morphisms are monomorphisms.
I think you said the exact opposite of what you meant to say.
Your proved that fields trivially do NOT have coequalizers. If g,h:K-
>L then the coequalizer of g and h exists only in the trivial case
that g = h.
Marc Olschok
> On May 10, 6:00 pm, Marc Olschok <nob...@nowhere.invalid> wrote:
>
> > as morphism between metric spaces, i.e. those maps f: (X,d) --> (X',d')
> > that satisfy d'(fx,fy) <= d(x,y).
> > Then the isomorphisms are exactly the isometries.
> >
>
> would object.. yeah this is probably the natural way of doing it. But
> still I don't suppose infinite products do exist here.
There is a counterexample (I just made it up, but I am sure that I must
have seen something like this in a book before):
Consider for all natural k>0 the spaces (X_k,d_k) where
X_k = {0,1} and d_k(0,1) = k.
Suppose, you had a product (X,d) with projections p_k: (X,d) --> (X_k,d_k).
Let 1 be the one-point space {*} with the trivial metric.
The two families of constant maps
x_k: 1 --> X_k with x_k(*) = 0
and
y_k: 1 --> X_k with y_k(*) = 1
will produce two elements x,y in X with p_k(x) = 0 and p_k(y) = 1.
But then one has k = d_k(0,1) <= d(x,y) for all k, which is impossible.
Marc
Jose Capco
Oh, right .. sorry :)
Well, there was a question going around in this thread about whether
we could find some workable categories that have equalizers but no
products. I was suggested yesterday, that the rings with 1 but by
which the ring homomorphisms do not necessarily bring 1 to 1 will
work. Indeed it is possible that the equalizer in this case do not
have a 1.. I will try to think of a concrete example here.
Sincerely,
Jose Capco
Gonçalo Rodrigues
<cliom...@kriocoucke.mailexpire.com> fed this fish to the penguins:
>On May 10, 6:00 pm, Marc Olschok <nob...@nowhere.invalid> wrote:
No. Denote by p_i the would-be projections and the elements p_ix by
x_i.
Since all p_i are contractions
d(x_i, y_i) <= d(x, y) => sup_i {d(x_i, y_i)} <= d(x, y)
Now pick a sequence of metric spaces with diameter going to infinity.
Best regards,
G. Rodrigues
Marc Olschok
>[...]
> we could find some workable categories that have equalizers but no
> products.
Suppose that you have some category C that has equalizers and in which
there exist at least one pair of maps f,g: a --> b with f =/= g.
Now consider category C_m defined via
Objects of C_m := Objects of C
Morphisms of C_m := monomorphisms of C
Then
(*) C_m has equalizers but the product of a and b cannot exist in C_m.
Moreover, (*) holds also for any full subcategory of C_m that is
closed under taking equalizers in C_m.
Immediate examples are C=Rings and D=Fields.
> I was suggested yesterday, that the rings with 1 but by
> which the ring homomorphisms do not necessarily bring 1 to 1 will
> work.
Are you sure? I think, they should have all products (inherited from Rng).
> Indeed it is possible that the equalizer in this case do not
> have a 1.. I will try to think of a concrete example here.
What do you mean here?
Marc
Jose Capco
> Jose Capco <cliomse...@kriocoucke.mailexpire.com> wrote:
> >[...]
> > Well, there was a question going around in this thread about whether
> > we could find some workable categories that have equalizers but no
> > products.
>
I made an error there, I meant categories that dont have equalizers
but have products.
> > I was suggested yesterday, that the rings with 1 but by
> > which the ring homomorphisms do not necessarily bring 1 to 1 will
> > work.
>
> Are you sure? I think, they should have all products (inherited from Rng).
See above :)
Just a few days ago my supervisor suggested an easy example in algebra
having no equalizer. He regarded the sets that are totally ordered and
have a least element as our category, and the morphism between them
are the functions that preserve ordering although least elements dont
need necessarily be mapped to the least element.
Take the real numbers >=1 as set A and the reals >=0 as set B
consider the the parallel maps f,g : A-> B with f be the canonical
injections and g(x)=f(x) for all x>1 and g(1)=0
then these two maps dont have an equalizer in our category (their
equalizer in the category of sets in the reals > 1, which has no least
element).
They dont have products though, but if we consider partially ordered
sets instead of totally ordered ones, then I suppose they have
products.
Sincerely,
Jose Capco
Jose Capco
>
> Just a few days ago my supervisor suggested an easy example in algebra
> having no equalizer.
The example I just wrote.. probably I shouldn't say "in algebra" :)
Marc Olschok
>[...]
> there exist at least one pair of maps f,g: a --> b with f =/= g.
>
> Now consider category C_m defined via
> Objects of C_m := Objects of C
> Morphisms of C_m := monomorphisms of C
>
> Then
>
> (*) C_m has equalizers but the product of a and b cannot exist in C_m.
Sorry, the above nonsense is halfway between meaningless and wrong,
because I mixed two different parts of my reasoning.
Instead:
(1) In a category where all morphism are monomorphism,
if f,g: A --> B are two maps with f =/= g then the
product of A and B cannot exist.
( this is actually the standard argument in proving that two fields
may not have a product, even if they have the same characteristic )
For a category K, consider K_m with
objects of K_m := objects of K
morphisms of K_m := the monomorphisms of K
(2) If there are two different monomorphism f,g: A --> B in K,
then A and B do not have a product in K_m. Moreover, they cannot
have a product in any full subcategory of K_m which includes A and B.
(3) If there are two different f,g: A --> B in K and if the product
A x B exist in K, then one can apply (2) to the two different
monomorphisms (A,f): A --> A x B and (A,g): A --> A x B.
Consequently A and A x B do not have a product in K_m.
(4) K_m is closed under taking equalizers in K.
Hence, for every category K with equalizers and binary products
which is not a preorder, the category K_m has equalizers but
not all products.
Marc
Marc Olschok
> having no equalizer. He regarded the sets that are totally ordered and
> have a least element as our category, and the morphism between them
> are the functions that preserve ordering although least elements dont
> need necessarily be mapped to the least element.
>
> Take the real numbers >=1 as set A and the reals >=0 as set B
>
> consider the the parallel maps f,g : A-> B with f be the canonical
> injections and g(x)=f(x) for all x>1 and g(1)=0
> then these two maps dont have an equalizer in our category (their
> equalizer in the category of sets in the reals > 1, which has no least
> element).
Nice example. Of course one needs that the underlying functor to Sets
is representable.
>
> They dont have products though, but if we consider partially ordered
> sets instead of totally ordered ones, then I suppose they have
> products.
Yes, inherited from Ordered sets.
Marc
Jose Capco
that coproduct exists in the category of commutative unitary rings..
but I mean, coproducts are direct sums in the category of rings.. but
then the direct sum of commutative unitary rings need not be unitary..
Im not sure if Im making a mistake here somewhere, but this just got
me really confused... and what would be the injection in the
coproduct? say rings R_i and the coproduct (+)R_i, then I suppose
the injection would be having
x |---> mu(x) ... with mu(x)_i = x and mu(x)_j = 0 for j =/= i
Sincerely,
Jose Capco
Mike
wrote:
The coproduct in the category of commutative unitary rings is the
tensor product. As you point out in the category of all commutative
rings the coproduct of two unitary rings is not unitary.
Marc Olschok
> I am getting a little confused here. I was always somehow convinced
> that coproduct exists in the category of commutative unitary rings..
> but I mean, coproducts are direct sums in the category of rings..
As already pointed out, this is wrong; the tensor product (over Z)
provides the coproduct in CRing. You probably confused rings and modules.
> but
> then the direct sum of commutative unitary rings need not be unitary..
> Im not sure if Im making a mistake here somewhere, but this just got
> me really confused... and what would be the injection in the
> coproduct?[...]
(See above for the error :-) With the tensor product, you will see
easily, what the coproduct injections should be.
You have correctly observed that the forgetful functor
U: CRing --> Ab does not preserve coproducts.
Marc
Dom
> On 8 May 2007 07:27:22 -0700, Jose Capco
>
> >Dear NG,
>
> >I am looking for examples of subcategory of Sets that have no...
>
> >products, coproducts
> >equalizer, coequalizer
>
> >And then maybe also examples by which it has one and doesnt have the
> >other- Any nice examples would be appreciated.
>
> >Sincerely,
> >Jose Capco
>
> In areas other than algebra is where you will find such examples:
>
> 1. The category of Banach spaces and bounded linear maps is finitely
> complete and finitely cocomplete but neither complete nor cocomplete.
Although the above category is not closed under Cartesian products, it
has products--if I remember correctly the subspace of the Cartesian
product consisting of all bounded tuples.
Jose Capco
> The coproduct in the category of commutative unitary rings is the
> tensor product. As you point out in the category of all commutative
> rings the coproduct of two unitary rings is not unitary.
Oh right, and because they are Z-algebras we can make tensor product
with respect to Z (ring of whole numbers)... so I suppose this a
wellknown fact that the coproduct is the tensor product. I think we
discussed this a year ago in this group
http://groups.google.com/group/sci.math/browse_thread/thread/eaf5f2f867f2eb7a/
Marc ended the thread by saying he isn't sure why this is true for
infinite collections of commutative rings (but I hypothesize that its
true there too)
By the by.. a little off topic here.. a while back I posted on some
German usage of words in category theory, and a year ago in another of
my post there was a hot debate on the term "directed colimit" for what
half of the mathematicians would call "direct limit".. I think many of
us concluded that the former terminology was better for the purpose of
understanding and that its widely used by mathematicians to accept it
as standard:
see http://groups.google.com/group/sci.math/browse_thread/thread/eaf5f2f867f2eb7a/
Now going back to the German language, I guess theres no mend to
"gerichtete Limes".. Even if we find a term its not in standard use so
we have to stick to "gerichteter Limes" or does anyone know a term
that is in standard use that is also equivalent to "gerichteter
Limes"? Oh my bad, maybe I should have just posted this in
de.sci.mathematik
Sincerely,
Jose Capco
Marc Olschok
> On May 15, 12:16 pm, Mike <mat...@hofstra.edu> wrote:
> German usage of words in category theory, and a year ago in another of
> my post there was a hot debate on the term "directed colimit" for what
> half of the mathematicians would call "direct limit".. I think many of
> us concluded that the former terminology was better for the purpose of
> understanding and that its widely used by mathematicians to accept it
> as standard:
>
> "gerichtete Limes".. Even if we find a term its not in standard use so
> we have to stick to "gerichteter Limes" or does anyone know a term
> that is in standard use that is also equivalent to "gerichteter
> Limes"? Oh my bad, maybe I should have just posted this in
> de.sci.mathematik
Good idea! (F'up set accordingly)
They translate as follows:
(1) "colimit" --> "Kolimes".
(2) "direct limit" --> "direkter Limes".
(3) "directed colimit" --> "gerichteter Kolimes".
at least, the difference between "direkt" and "gerichtet" avoids
the guess wether the property of the diagram or the decision between
limit and colimit is indicated. But it cannot avoid possible confusion
in case (2) might be used as synonym for (1).
Hence I think, you should probably use (1) and (3) in your course.
Marc
galathaea
wrote:
> On May 15, 12:16 pm, Mike <mat...@hofstra.edu> wrote:
>
> > The coproduct in the category of commutative unitary rings is the
> > tensor product. As you point out in the category of all commutative
> > rings the coproduct of two unitary rings is not unitary.
>
> Oh right, and because they are Z-algebras we can make tensor product
> with respect to Z (ring of whole numbers)... so I suppose this a
its always hard to judge how well known some data is
but in crowds studying monoidal categories
the fact that the coproduct is the tensor product
is used extensively
it is general to all monoidal categories
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
Jose Capco
> Jose Capco <cliomse...@kriocoucke.mailexpire.com> wrote:
> > I am getting a little confused here. I was always somehow convinced
> > that coproduct exists in the category of commutative unitary rings..
> > but I mean, coproducts are direct sums in the category of rings..
>
> As already pointed out, this is wrong; the tensor product (over Z)
> provides the coproduct in CRing. You probably confused rings and modules.
>
Well I dont know if CRing is some standard notation, but I assume you
mean commutative unitary rings... well you could say Im confusing
rings and CRings.. I have seen direct sums used on the category of
just rings (no unit assumed, no commutativity assumed or ..etc.), I
think I have seen it in wikipedia or some source which I cannot be
much sure of.
Gonçalo Rodrigues
this fish to the penguins:
>> > The coproduct in the category of commutative unitary rings is the
>> > tensor product. As you point out in the category of all commutative
>> > rings the coproduct of two unitary rings is not unitary.
>>
>> Oh right, and because they are Z-algebras we can make tensor product
>> with respect to Z (ring of whole numbers)... so I suppose this a
>> well known fact that the coproduct is the tensor product.
>
>its always hard to judge how well known some data is
> but in crowds studying monoidal categories
> the fact that the coproduct is the tensor product
>is used extensively
>
>it is general to all monoidal categories
>
I'm having difficulty parsing what you wrote (and not just because of
your breaking-line or all-small-caps rules).
The way I am reading is that, in the first paragraph. you mean that a
category with all finite coproducts is monoidal (and yes, it is true).
But then that this fact is "general to all monoidal categories" makes
no sense.
Best regards,
G. Rodrigues
Jose Capco
disturbing me a bit now..
We know of the "empty" category.. the class of objects is empty and so
are the morphisms.. question is, functors taking values in the empty
category to any category.. how are they even visualized?
I question this because I cant imagine what empty products would be?
Is it the terminal object of the category? what if the category has no
terminal object.. then does this mean that the empty product does not
exist? and what is all the logic behind this?
Sincerely,
Jose Capco
Marc Olschok
> Ok here's another food for thought, which may be pedantic but its
> disturbing me a bit now..
>
> We know of the "empty" category.. the class of objects is empty and so
> are the morphisms.. question is, functors taking values in the empty
> category to any category.. how are they even visualized?
Pretty much the same way you would "visualise" the function
from the empty set to some other set. It is empty, so you would
do nothing.
>
> I question this because I cant imagine what empty products would be?
> Is it the terminal object of the category?
Yes.
> what if the category has no
> terminal object.. then does this mean that the empty product does not
> exist?
Yes.
> and what is all the logic behind this?
First recall that any class I can be viewed as a category with
no morphisms where the objects are the elements of I.
From this point of view, a family of objects A_i (i in I) in a category K
is the same as a functor A: I --> C with A(i) = A_i and a product
of such a family is the same as a limit of the corresponding functor.
Hence it makes sense that an empty family of objects corresponds
to a functor from the empty category and that an empty product
is a limit of such a funtor.
Now write down the definition of a product and check that for an
empty index-set it reduced to the definition of a terminal object.
Marc
Marc Olschok
By "direct sum" you really mean the sum of the underlying additive
abelian groups, equipped with some ring multiplication?
I do not see how this would work.
(a) for a start, you would need a noncommutative multiplication in the
coproduct of two rngs, even if they are both commutative: consider e.g.
the polynomial ring Z{x,y} over Z with two _noncommuting_ variables
and the to maps f: Z --> Z{x,y} and g: Z --> Z{x,y} with f(1)=x and g(1)=y.
(b) even for commutative rngs (= rings without identity) the usual
componentwise multiplication does not seem to work: take the
identity map Z --> Z twice and consider the induced map h: Z (+) Z --> Z.
Marc
Jose Capco
> By "direct sum" you really mean the sum of the underlying additive
> abelian groups, equipped with some ring multiplication?
>
> I do not see how this would work.
>
> (a) for a start, you would need a noncommutative multiplication in the
> coproduct of two rngs, even if they are both commutative: consider e.g.
> the polynomial ring Z{x,y} over Z with two _noncommuting_ variables
> and the to maps f: Z --> Z{x,y} and g: Z --> Z{x,y} with f(1)=x and g(1)=y.
>
> (b) even for commutative rngs (= rings without identity) the usual
> componentwise multiplication does not seem to work: take the
> identity map Z --> Z twice and consider the induced map h: Z (+) Z --> Z.
>
> Marc
Yeah, I think you are right. I made a booboo again. Although I believe
if the coproduct exist then it must be some sort of "free" ring ..
don't even know if coproduct always exist in this category.
Sincerely,
Jose Capco
Marc Olschok
>[...]
> Although I believe if the coproduct [ in Rng ] exist then it must be
> some sort of "free" ring .. don't even know if coproduct always
> exist in this category.
Actually it exists. The coproduct of R and S should be
R (+) S (+) R(x)S equipped with a suitable multiplication
and the obvious injections.
The argument runs like this:
Let Rng be the category of rings (possibly) without unity and
let Ring be the category of unitary rings and unitary homomorphisms.
(1) The inclusion of Ring into Rng has a left adjoint F: Rng --> Ring
where on objects F(R) = Z x R with componentwise addition and
multiplication (m,r)(n,s) = (mn , nr + ms + rs) and on maps
F(f)(m,r) = (m,fr).
(2) Suppose F: X --> A is left adjoint to G: A --> X and assume also,
that G has pullbacks. Then for any object z of X, the pair F and G induce
another pair of functors
F': X/z --> A/F(z) and G': A/F(z) --> X/z
such that F' is left adjoint to G'.
F' is defined via F'(u: x->z) = F(u): F(x) -> F(z)
and for v: a --> Fz, one defines G'v as the pullback
of Gv and the original unit X => GF evaluated at z.
p ---> Ga
| |
|G'v |Gv
v v
z ---> GFz
switching via adjoint gives
Fp ---> a
| |
|F'G'v | v
v v
Fz ==== Fz
which exhibits the counit F'G' => A/Fz of the new adjunction.
(3) Applying (2) to the (1) with z = 0 (and using Rng/0 = Rng) one
calculates ( I write "(+)" instead of "x" now )
F(0) = Z,
F'(R) = (Z (+) R --> Z) (the projection to Z)
G'(u: A --> Z) = ker(u)
(4) Both functors F': Rng --> Ring/Z and G': Ring/Z --> Rng
are full and faithful.
Hence one has an equivalence between Rng and Ring/Z.
Therefore the coproduct of R and S in Rng can be calculated by
(i) replacing R and S by Z(+)R --> Z and Z(+)S --> Z
(ii) calculation the coproduct of the above to objects in Ring/Z
p: (Z(+)R) (x) (Z(+)S) --> Z with p( (m+r) (x) (n+s) ) = mn
which can be rewritten as
p: Z (+) R (+) S (+) R(x)S with p(m + r + s + v) = m
(iii) going back to Rng, which gives the kernel of p as the
desired coproduct.
Marc
Jose Capco
> Jose Capco <cliomse...@kriocoucke.mailexpire.com> wrote:
> >[...]
> > Although I believe if the coproduct [ in Rng ] exist then it must be
> > some sort of "free" ring .. don't even know if coproduct always
> > exist in this category.
>
> Actually it exists. The coproduct of R and S should be
> R (+) S (+) R(x)S equipped with a suitable multiplication
> and the obvious injections.
>
Well I'm confused here alone, what do you mean by (x)? a tensor
product?.. how can you make a tensor product of R and S? We don't
assume that R and or S are Z-algebras.. if 1 were not say in R, then R
can possibly be not a Z-algebra and so I don't know how you do the
tensor product.
> The argument runs like this:
>
> Let Rng be the category of rings (possibly) without unity and
> let Ring be the category of unitary rings and unitary homomorphisms.
>
I suppose first that here you assume Rng has also objects as rings
without units.
> (1) The inclusion of Ring into Rng has a left adjoint F: Rng --> Ring
> where on objects F(R) = Z x R with componentwise addition and
> multiplication (m,r)(n,s) = (mn , nr + ms + rs) and on maps
> F(f)(m,r) = (m,fr).
>
What gaurantees F(R) to be unitary? If R has no unit.. then ZxR has no
unit if its just the componentwise addition/multiplication.
Sincerely,
Jose Capco
Marc Olschok
> On May 19, 9:38 pm, Marc Olschok <nob...@nowhere.invalid> wrote:
> > Jose Capco <cliomse...@kriocoucke.mailexpire.com> wrote:
> > >[...]
> > > Although I believe if the coproduct [ in Rng ] exist then it must be
> > > some sort of "free" ring .. don't even know if coproduct always
> > > exist in this category.
> >
> > Actually it exists. The coproduct of R and S should be
> > R (+) S (+) R(x)S equipped with a suitable multiplication
> > and the obvious injections.
> >
>
> Well I'm confused here alone, what do you mean by (x)? a tensor
> product?.. how can you make a tensor product of R and S? We don't
> assume that R and or S are Z-algebras.. if 1 were not say in R, then R
> can possibly be not a Z-algebra and so I don't know how you do the
> tensor product.
Actually R and S _are_ Z-algebras --- in fact symmetric two-sided
Z-algebras (which is used in the construction of F(R)).
Of course they may not be _unitary_ algebras, but that is no problem.
R(x)S is just the tensor product of the abelian groups.
>
> > The argument runs like this:
> >
> > Let Rng be the category of rings (possibly) without unity and
> > let Ring be the category of unitary rings and unitary homomorphisms.
> >
>
> I suppose first that here you assume Rng has also objects as rings
> without units.
Well, my above "unity" is not good. I meant "Identity".
I use Rng as mnemonic for "Ring without Identity", so i drop
the "i" from the word "Ring".
>
> > (1) The inclusion of Ring into Rng has a left adjoint F: Rng --> Ring
> > where on objects F(R) = Z x R with componentwise addition and
> > multiplication (m,r)(n,s) = (mn , nr + ms + rs) and on maps
> > F(f)(m,r) = (m,fr).
> >
>
> What gaurantees F(R) to be unitary? If R has no unit.. then ZxR has no
> unit if its just the componentwise addition/multiplication.
______________________________________________^^^^^^^^^^^^^^
It is important, that the _multiplication_ on Z x R is _not_ componentwise.
One has (m,r)(n,s) = (mn , nr + ms + rs) , so the unit is (1,0).
One should think of (m,r) = (m,0) + (0,r) as a "sum" m + r,
then it is clear what the product (m+r)(n+s) should be.
Also, F(R) can be described as the set of matrices of the form
(m r)
(0 m)
with m in Z and r in R. Then the ordinary addition and multiplication
of matrices will give the ring structure of F(R).
Marc
Jose Capco
> Actually R and S _are_ Z-algebras --- in fact symmetric two-sided
> Z-algebras (which is used in the construction of F(R)).
> Of course they may not be _unitary_ algebras, but that is no problem.
> R(x)S is just the tensor product of the abelian groups.
If a ring has no identity, could you please show me or prove to me how
it could be a Z-algebra?
Could you define me the "tensor product of abelian group"? I only know
of tensor product of algebras and modules.
> It is important, that the _multiplication_ on Z x R is _not_ componentwise.
> One has (m,r)(n,s) = (mn , nr + ms + rs) , so the unit is (1,0).
>
Oh, sorry.. I didnt noticed that.. I haven't look at the rest of your
construction, but if you dont mind I first want to make things clear
here before reading the rest :)
Sincerely,
Jose Capco
Marc Olschok
> On May 20, 8:59 pm, Marc Olschok <nob...@nowhere.invalid> wrote:
> > Actually R and S _are_ Z-algebras --- in fact symmetric two-sided
> > Z-algebras (which is used in the construction of F(R)).
> > Of course they may not be _unitary_ algebras, but that is no problem.
> > R(x)S is just the tensor product of the abelian groups.
>
> If a ring has no identity, could you please show me or prove to me how
> it could be a Z-algebra?
You could keep Lang's "Algebra" under your pillow at night.
The latest edition is big enough to make sleeping uncomfortable,
so when you lie awake, you can open page 121 (in III,1) and find
that for a commutative ring A, he defines an A-algebra as an A-modul E
together with an A-bilinear map E x E --> E. This map is then the
multiplication of the algebra.
(he says that in the rest of the book the multiplication of algebras
usually will be associative and have an identity ; but the example
from analysis and the Lie-multiplication make clear, that this
requirement is not always essential)
In the above situation, the ring multiplication provides such a map
when A = Z.
>
> Could you define me the "tensor product of abelian group"? I only know
> of tensor product of algebras and modules.
____________________________________^^^^^^^
Splendid! So you know in particular the tensor product of Z-modules
(also known as abelian groups :-).
Marc
Jose Capco
> You could keep Lang's "Algebra" under your pillow at night.
> The latest edition is big enough to make sleeping uncomfortable,
> so when you lie awake, you can open page 121 (in III,1) and find
> that for a commutative ring A, he defines an A-algebra as an A-modul E
> together with an A-bilinear map E x E --> E. This map is then the
> multiplication of the algebra.
Things that can make my night pleasant :) , I've not read Lang's work
yet .. yes, sorry I was thinking of something else. Yes, you are
right.. I was thinking of some ring morphism from Z to R, but thats
only one special case.
> (he says that in the rest of the book the multiplication of algebras
> usually will be associative and have an identity ; but the example
> from analysis and the Lie-multiplication make clear, that this
> requirement is not always essential)
Yes now I understand what you are trying to show.
> Splendid! So you know in particular the tensor product of Z-modules
> (also known as abelian groups :-).
>
Oh is that so? oh ok :) well I'll read the rest of your construction
carefully and soon write some comments =) .. thanks a lot :)
Sincerely,
Jose Capco
Jose Capco
> One should think of (m,r) = (m,0) + (0,r) as a "sum" m + r,
> then it is clear what the product (m+r)(n+s) should be.
>
> Also, F(R) can be described as the set of matrices of the form
>
> (m r)
> (0 m)
>
> with m in Z and r in R. Then the ordinary addition and multiplication
> of matrices will give the ring structure of F(R).
>
Btw not quite related to this topic, but where I can I read about
these direct sum of a ring with Z that you just describe. Apparently
this is a well known construction. I am in particular interested in
the prime ideals of F(R) for any commutative unitary ring.
Sincerely,
Jose Capco
Marc Olschok
Actually, the description of F(R) via matrices is wrong, unless the
original multiplication on R is trival (i.e. xy=0 for all x,y in R).
So there are in fact two different constructions:
(1) for a ring A and a two sided balanced A-Modul M
(i.e. a(mb) = (am)b for all a,b in A and m in M)
the set of all matrices of the form
(a x)
(0 a)
(with a in A and x in M) is a ring with the ordinary matrix operations,
which has an identity iff A has one.
There should be more about this in Rowen: Ring Theory.
(2) for a ring with identity A and a two sided balanced A-algebra R,
the sum (of abelian groups) A (+) R can be made into a ring with
multiplication (a+r)(b+s) = ab + rb + as + rs.
The construction is mentioned (for the special case A = Z)
in Bourbaki: Algebra (appendix of chapter II), and I am almost
sure it must be in Jacobson: Basic Algebra.
Since rings of the form F(R) are exactly those rings (with identity) A
that have a homomorphism A --> Z, you may also find more information
under the heading "augmented" rings.
Marc