My Math Forum  

Go Back   My Math Forum > College Math Forum > Abstract Algebra

Abstract Algebra Abstract Algebra Math Forum


Thanks Tree5Thanks
Reply
 
LinkBack Thread Tools Display Modes
July 6th, 2018, 02:40 PM   #1
Senior Member
 
Joined: Nov 2011

Posts: 224
Thanks: 2

field...

In field:
Why the multiplicative identity (1) different from the zero-element (0)?
shaharhada is offline  
 
July 6th, 2018, 04:05 PM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,105
Thanks: 1093

let both be represented by $\alpha$

let $x \in f$

$\alpha x = x$

$x = \alpha x = \displaystyle \left(\sum \limits_{k=1}^n \alpha \right) x= \sum \limits_{k=1}^n \alpha x = \sum \limits_{k=1}^n x = n x$

contradiction
Thanks from Maschke and topsquark
romsek is offline  
July 7th, 2018, 02:33 AM   #3
Senior Member
 
Joined: Nov 2011

Posts: 224
Thanks: 2

I don't understand

I don't understand nothing.
Can you explain to my in simple words what you post.
shaharhada is offline  
July 7th, 2018, 05:37 AM   #4
Senior Member
 
Joined: Aug 2017
From: United Kingdom

Posts: 234
Thanks: 78

Math Focus: Algebraic Number Theory, Arithmetic Geometry
Quote:
Originally Posted by shaharhada View Post
I don't understand nothing.
Can you explain to my in simple words what you post.
Suppose $0 = 1$ in your field and let $x$ be any element. He's shown that, no matter how many times you add $x$ to itself, you always end up back at $x$. In other words, $x = x + x = x + x + x = \dots$. Now, this is certainly an odd phenomenon, but I'm not sure exactly what the intended contradiction is.

The more accurate answer is that we just exclude the possibility that $0 = 1$ in the definition of a field - it's not something that can be proven from the other field axioms (at least in any definition of a field I've ever come across). Why might we want to do this? Well, firstly recall that the only ring in which $0 = 1$ is the ring with just one element, which we call the zero ring. So saying $0 \neq 1$ in a field is just saying that the zero ring is not a field.

One reason for wanting the zero ring not to be a field is comparable to wanting $1$ not to be a prime number: it makes it much more convenient to state definitions and theorems. Many basic results regarding prime and maximal ideals in rings don't hold for the zero ring. So rather than repeating "any field except the zero ring" or "any integral domain except the zero ring" at the start of loads of results, we just incorporate this exception into the definition and never need to think about it again. Another reason is that things would start to get very weird if we looked at vector spaces over the zero ring. I'd recommend having a think about this yourself.
Thanks from topsquark
cjem is online now  
July 7th, 2018, 07:05 AM   #5
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,261
Thanks: 894

Suppose we were to allow 0= 1 in a field. That is, the 'additive identity' is the same as the 'multiplicative identity'. In a field the "distributive law", a(b+ c)= ab+ ac, is true and, in particular, a(b+ 0)= ab+ a0. But since 0 is the additive identity b+ 0= b so ab= a(b+ 0)= ab+ b0. In a field, every element, in particular ab, has an additive identity. Adding the additive identity of ab to both sides 0= b0 for every element b. If 0= 1 then 0 is also the multiplicative identity so that b0= b. That is, for every element, b, of the field we have b= 0. That's not a very interesting field!

Some textbooks use, instead of the requirement that 0 is not equal to 1, the requirement that the field not be "trivial"- that it contain more than one element.
Thanks from topsquark and romsek
Country Boy is offline  
July 7th, 2018, 12:38 PM   #6
Global Moderator
 
Joined: May 2007

Posts: 6,589
Thanks: 612

$x+0=x$
$x\times 1=x$
$x+1\ne x$
$x\times 0=0$

Obviously 1 and 0 are different.
mathman is offline  
July 7th, 2018, 01:11 PM   #7
Senior Member
 
Joined: Aug 2012

Posts: 2,010
Thanks: 574

Quote:
Originally Posted by mathman View Post
$x+0=x$
$x\times 1=x$
$x+1\ne x$
$x\times 0=0$

Obviously 1 and 0 are different.
How do you get line three, $x+1\ne x$? Aren't you assuming the thing you're trying to prove? If $0 = 1$ then $x + 1 = x$ and there is no contradiction.
Maschke is offline  
July 7th, 2018, 01:12 PM   #8
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,261
Thanks: 894

Quote:
Originally Posted by Country Boy View Post
Suppose we were to allow 0= 1 in a field. That is, the 'additive identity' is the same as the 'multiplicative identity'. In a field the "distributive law", a(b+ c)= ab+ ac, is true and, in particular, a(b+ 0)= ab+ a0. But since 0 is the additive identity b+ 0= b so ab= a(b+ 0)= ab+ b0. In a field, every element, in particular ab, has an additive inverse. Adding the additive inverse of ab to both sides 0= b0 for every element b. If 0= 1 then 0 is also the multiplicative identity so that b0= b. That is, for every element, b, of the field we have b= 0. That's not a very interesting field!

Some textbooks use, instead of the requirement that 0 is not equal to 1, the requirement that the field not be "trivial"- that it contain more than one element.
I have replaced "additive identity" in two places with "additive inverse" which was intended.
Country Boy is offline  
July 7th, 2018, 01:48 PM   #9
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,105
Thanks: 1093

Quote:
Originally Posted by Maschke View Post
How do you get line three, $x+1\ne x$? Aren't you assuming the thing you're trying to prove? If $0 = 1$ then $x + 1 = x$ and there is no contradiction.
Yeah I believe that the singleton field is valid with the single element acting as both additive and multiplicative identity.

But as CountryBoy noted this isn't a very interesting field.
romsek is offline  
July 7th, 2018, 03:01 PM   #10
Senior Member
 
Joined: Aug 2012

Posts: 2,010
Thanks: 574

Lightbulb

Quote:
Originally Posted by romsek View Post
Yeah I believe that the singleton field is valid with the single element acting as both additive and multiplicative identity.

But as CountryBoy noted this isn't a very interesting field.
I did not say that at all. A field is required to have $0 \neq 1$. I merely pointed out that mathman's third line was unjustified.
Maschke is offline  
Reply

  My Math Forum > College Math Forum > Abstract Algebra

Tags
field



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
does cyclic field implies Galois field rayman Abstract Algebra 1 March 11th, 2014 03:27 PM
Fin. gen field extension -> intermediate field f.g. also? watson Abstract Algebra 1 September 20th, 2012 06:53 PM
Show R (comm. domain) over a field k is a field if dimR<inft watson Abstract Algebra 1 September 14th, 2012 09:07 PM
Every quadratic field is contained in a cyclotomic field brunojo Abstract Algebra 0 June 5th, 2009 06:25 PM
field goodfeeling Algebra 0 December 31st, 1969 04:00 PM





Copyright © 2018 My Math Forum. All rights reserved.