July 6th, 2018, 03:40 PM  #1 
Senior Member Joined: Nov 2011 Posts: 239 Thanks: 2  field...
In field: Why the multiplicative identity (1) different from the zeroelement (0)? 
July 6th, 2018, 05:05 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,163 Thanks: 1135 
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 
July 7th, 2018, 03:33 AM  #3 
Senior Member Joined: Nov 2011 Posts: 239 Thanks: 2  I don't understand
I don't understand nothing. Can you explain to my in simple words what you post. 
July 7th, 2018, 06:37 AM  #4  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 282 Thanks: 85 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
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.  
July 7th, 2018, 08:05 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. 
July 7th, 2018, 01:38 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,625 Thanks: 622 
$x+0=x$ $x\times 1=x$ $x+1\ne x$ $x\times 0=0$ Obviously 1 and 0 are different. 
July 7th, 2018, 02:11 PM  #7 
Senior Member Joined: Aug 2012 Posts: 2,075 Thanks: 593  
July 7th, 2018, 02:12 PM  #8  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  Quote:
 
July 7th, 2018, 02:48 PM  #9  
Senior Member Joined: Sep 2015 From: USA Posts: 2,163 Thanks: 1135  Quote:
But as CountryBoy noted this isn't a very interesting field.  
July 7th, 2018, 04:01 PM  #10 
Senior Member Joined: Aug 2012 Posts: 2,075 Thanks: 593  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.


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