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 July 6th, 2018, 02:40 PM #1 Senior Member   Joined: Nov 2011 Posts: 250 Thanks: 3 field... In field: Why the multiplicative identity (1) different from the zero-element (0)?
 July 6th, 2018, 04:05 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,458 Thanks: 1339 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
 July 7th, 2018, 02:33 AM #3 Senior Member   Joined: Nov 2011 Posts: 250 Thanks: 3 I don't understand I don't understand nothing. Can you explain to my in simple words what you post.
July 7th, 2018, 05:37 AM   #4
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 Originally Posted by shaharhada 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.

 July 7th, 2018, 07:05 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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
 July 7th, 2018, 12:38 PM #6 Global Moderator   Joined: May 2007 Posts: 6,764 Thanks: 697 $x+0=x$ $x\times 1=x$ $x+1\ne x$ $x\times 0=0$ Obviously 1 and 0 are different.
July 7th, 2018, 01:11 PM   #7
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 Originally Posted by mathman $x+0=x$ $x\times 1=x$ $x+1\ne x$ $x\times 0=0$ 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.

July 7th, 2018, 01:12 PM   #8
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 Originally Posted by Country Boy 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.

July 7th, 2018, 01:48 PM   #9
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 Originally Posted by Maschke 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.

July 7th, 2018, 03:01 PM   #10
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 Originally Posted by romsek 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.

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