Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Topic: Finite Rings
Replies: 28   Last Post: Feb 6, 2013 8:33 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 11,581
Registered: 12/4/04
Re: Finite Rings
Posted: Feb 5, 2013 11:04 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Tuesday, February 5, 2013 2:53:59 AM UTC-6, William Elliot wrote:
> On Mon, 4 Feb 2013, Arturo Magidin wrote:

> > On Monday, February 4, 2013 10:55:03 AM UTC-6, Arturo Magidin wrote:
> >
> > > The only two books I have that do not allow 1=0 are: Zariski and
> > > Samuel's "Commutative Algebra", which restricts the use of the term
> > > "identity" to rings that are not nullrings; and Lam's "A First Course
> > > in Noncommutative rings" and "Lectures on Rings and Modules", which
> > > specifies this explicitly in the introduction.
> >
> > Added: Note, however, that the zero ring does not qualify as a "ring"
> > under Lam's definitions, so it cannot be an example. So the only book I
> > have that both allows the 0 ring, and *does not* recognize the 0 ring as
> > a ring with identity, is Zariski-Samuel. It also happens to be by far
> > the oldest, having been published originally in 1958.
> Much ado about nothing.

Says the man who (i) was wrong; and (ii) goes on and on about how everyone must always present things in ways that *he* finds convenient.

What a hypocrit.

> One man's heresy is another's definition.

You were wrong, Elliot. That's is all there is to it. You tried to show off, but instead put your foot in your mouth, firmly. Now, it's all about how the grapes were sour anyway.

> Yet the problem still remains problematic.

No, it does not. Under the standard definitions, the zero ring is not a counterexample because either (i) it does contain an identity; or (ii) it is not considered a ring by those that require 1=/=0.

So, your grandstanding was based on your ignorance and your misunderstanding.

The only problem that still remains is that you continue to have delusions of adequacy.

Arturo Magidin

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2015. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.