How can 0=1 in a ring?!
grocery_stocker
And I quote
"As I wrote in the other post, a field is a ring such that not(0=1)
and
every nonzero element has a multiplicative inverse.
In the econtext of vector spaces, this is where your scalars come
from.
If you keep the axioms for a vector space but allow your scalars to
come from a ring without insisting that the ring must be a field,
you obtain the notion of a module. "
Can someone give me a concrete example of a ring where 0=1. Thanks in
advance.
quasi
The zero ring.
R = {0} with the obvious addition and multiplication tables.
The zero ring plays the same role for rings, for those texts that
allow it (most don't), as the empty set does for sets.
quasi
Dave Seaman
> On Fri, 18 Apr 2008 19:57:11 -0700 (PDT), grocery_stocker
><cda...@gmail.com> wrote:
>>The question stems from the following url
>>
>>http://groups.google.com/group/sci.math/browse_thread/thread/4e3ece135c6b503a/cdb850fed66209d3?hl=en&lnk=gst&q=abelian+group+and+vector+spaces#cdb850fed66209d3
>>
>>And I quote
>>
>>"As I wrote in the other post, a field is a ring such that not(0=1)
>>and
>>every nonzero element has a multiplicative inverse.
>>In the econtext of vector spaces, this is where your scalars come
>>from.
>>If you keep the axioms for a vector space but allow your scalars to
>>come from a ring without insisting that the ring must be a field,
>>you obtain the notion of a module. "
>>
>>Can someone give me a concrete example of a ring where 0=1. Thanks in
>>advance.
> The zero ring.
> R = {0} with the obvious addition and multiplication tables.
And to expand on that, the reason we say 0=1 in that ring is that 1 is
used to denote the unity element of a ring R, having the property that
1*x = x*1 = x, for every x in R.
In the case R = {0}, we indeed find that 0*x = x*0 = x for every x in R
(the only possibility being x = 0), and therefore 0 is the unity element
of R = {0}, hence 0 = 1.
--
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>
Bill Dubuque
>grocery_stocker <cda...@gmail.com> wrote:
>
>> The question stems from the following url
>>
>>http://groups.google.com/group/sci.math/browse_thread/thread/4e3ece135c6b503a/cdb850fed66209d3?hl=en&lnk=gst&q=abelian+group+and+vector+spaces#cdb850fed66209d3
>>
>>And I quote
>>
>> "As I wrote in the other post, a field is a ring such that not(0=1)
>>
>> Can someone give me a concrete example of a ring where 0=1.
>
>
> The zero ring plays the same role for rings, for those texts
> that allow it (most don't), as the empty set does for sets.
The usual _convention_ is that {0} is a Ring, but not a Domain/Field,
for consistency with R/I is a Domain/Field iff I is Prime/Maximal
and the convention that 1 isn't Prime/Maximal (among other theorems).
Hence the ubiquitous proof by contradiction (0=1) in Domains/Fields.
Of course {0} does actually satisfy all the axioms for each structure.
The analogy with the empty set is not a good one, since the question
of whether or not empty sets (carriers) are permitted as algebras
is a rather subtle one. They are especially handy in the case of
many-sorted algebras, but one has to be careful with the associated
logic in order to maintain completeness -- see my prior posts
http://google.com/groups?threadm=y8zlodxnayi.fsf_-_%40martigny.ai.mit.edu
--Bill Dubuque
Rotwang
> On Fri, 18 Apr 2008 19:57:11 -0700 (PDT), grocery_stocker
>
> >advance.
>
> The zero ring.
>
> R = {0} with the obvious addition and multiplication tables.
Note also that this is the only such ring, since if 0 = 1 in R then
for any x in R we have x = x*1 = x*0 = 0.
> The zero ring plays the same role for rings, for those texts that
> allow it (most don't), as the empty set does for sets.
Namely it is an initial object, i.e. there is exactly one homomorphism
from {0} to any ring. It's also a terminal object, i.e. there is
exactly one homomorphism from any ring to {0}.
Arturo Magidin
quasi <qu...@null.set> wrote:
[...]
>The zero ring plays the same role for rings, for those texts that
>allow it (most don't), as the empty set does for sets.
I would not go so far as to say "most"; "many", certainly, but I'm not
sure "most" is accurate.
--
======================================================================
"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
grocery_stocker
> quasi <qu...@null.set> wrote:> >>http://groups.google.com/group/sci.math/browse_thread/thread/4e3ece13...
>
> >>And I quote
>
> >> "As I wrote in the other post, a field is aringsuch that not(0=1)
> >> and every nonzero element has a multiplicative inverse. ..."
>
> >> Can someone give me a concrete example of aringwhere 0=1.
>
>
> > The zeroringplays the same role for rings, for those texts
> > that allow it (most don't), as the empty set does for sets.
>
> for consistency with R/I is a Domain/Field iff I is Prime/Maximal
> and the convention that 1 isn't Prime/Maximal (among other theorems).
> Hence the ubiquitous proof by contradiction (0=1) in Domains/Fields.
> Of course {0} does actually satisfy all the axioms for each structure.
>
> The analogy with the empty set is not a good one, since the question
> of whether or not empty sets (carriers) are permitted as algebras
> is a rather subtle one. They are especially handy in the case of
> many-sorted algebras, but one has to be careful with the associated
>
> --Bill Dubuque
In the pictures of the two cubes, would 0 be the hollow cube or the
plane on the y z axis?
grocery_stocker
> On Apr 19, 7:32 am, Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>
>
>
>
>
> > quasi <qu...@null.set> wrote:
> > >grocery_stocker <cdal...@gmail.com> wrote:
>
> > >> The question stems from the following url
>
> > >>http://groups.google.com/group/sci.math/browse_thread/thread/4e3ece13...
>
> > >>And I quote
>
> > >> "As I wrote in the other post, a field is aringsuch that not(0=1)
> > >> and every nonzero element has a multiplicative inverse. ..."
>
> > >> Can someone give me a concrete example of aringwhere 0=1.
>
> > > The zeroring. R = {0} with obvious addition and multiplication tables.
>
> > > The zeroringplays the same role for rings, for those texts
> > > that allow it (most don't), as the empty set does for sets.
>
> > The usual _convention_ is that {0} is aRing, but not a Domain/Field,
> > for consistency with R/I is a Domain/Field iff I is Prime/Maximal
> > and the convention that 1 isn't Prime/Maximal (among other theorems).
> > Hence the ubiquitous proof by contradiction (0=1) in Domains/Fields.
> > Of course {0} does actually satisfy all the axioms for each structure.
>
> > The analogy with the empty set is not a good one, since the question
> > of whether or not empty sets (carriers) are permitted as algebras
> > many-sorted algebras, but one has to be careful with the associated
> > logic in order to maintain completeness -- see my prior postshttp://google.com/groups?threadm=y8zlodxnayi.fsf_-_%40martigny.ai.mit...
>
> > --Bill Dubuque
>
> In the pictures of the two cubes, would 0 be the hollow cube or the
>
> - Show quoted text -
Annd the pictures of the two cubs I'm asking about are
http://groups.google.com/group/sci.math/browse_frm/thread/3e8fac43081fcbf8
Jannick Asmus
> On 19 Apr, 04:29, quasi <qu...@null.set> wrote:
>> The zero ring plays the same role for rings, for those texts that
>> allow it (most don't), as the empty set does for sets.
>
> Namely it is an initial object, i.e. there is exactly one homomorphism
> from {0} to any ring.
... in the category C of rings (where the unity element is not
necessarily preserved by homomorphisms) only.
> It's also a terminal object, i.e. there is
> exactly one homomorphism from any ring to {0}.
... in C only.
Note that the category of commutative rings with unity and homomorphisms
preserving unity has an initial object, namely Z, and no terminal object.
J.
Mariano Suárez-Alvarez
Assuming you consider only rings in which 0
is different from 1, otherwise the zero ring
is terminal.
-- m
Jannick Asmus
Correct. Thanx.
--
Best wishes,
J.
grocery_stocker
>
> >>And I quote
>
> >> "As I wrote in the other post, a field is a ring such that not(0=1)
> >> and every nonzero element has a multiplicative inverse. ..."
>
> >> Can someone give me a concrete example of a ring where 0=1.
>
> > The zero ring. R = {0} with obvious addition and multiplication tables.
>
> > The zero ring plays the same role for rings, for those texts
> > that allow it (most don't), as the empty set does for sets.
>
> The usual _convention_ is that {0} is a Ring, but not a Domain/Field,
> for consistency with R/I is a Domain/Field iff I is Prime/Maximal
> and the convention that 1 isn't Prime/Maximal (among other theorems).
> Hence the ubiquitous proof by contradiction (0=1) in Domains/Fields.
> Of course {0} does actually satisfy all the axioms for each structure.
>
> The analogy with the empty set is not a good one, since the question
> of whether or not empty sets (carriers) are permitted as algebras
> is a rather subtle one. They are especially handy in the case of
> many-sorted algebras, but one has to be careful with the associated
>
> --Bill Dubuque
Okay, let me try asking this question again.
On the following URL
http://groups.google.com/group/sci.math/browse_frm/thread/3e8fac43081fcbf8
There is are two cubes to show how 1 + 1 = 1. Now, would 0 be the
hollow cube or the y-z axis (plane)?