Field axioms: Why do we have $ 1 \neq 0$?

Question

In the definitions of a field, we have $ 1 \neq 0$.

I know that in regular multiplication $0 \times 1=0$ but for reciprocal we don't have inverse of $0$.

But all the spaces and different definitions of multiplications that are satisfying field axioms, why do we need $ 1 \neq 0$?

Please do not use too technical of terminology. I am reading Baby Rudin right now.

This does not follow from the other axioms. But we do not want to count the ring ${0}$ as a field. So add that extra axiom to rule it out. — GEdgar, Feb 28 '15 at 18:44
Hi and welcome on Math.SE! Does it feel better to say that identity elements for multiplication and addition must be distinct? — MattAllegro, Feb 28 '15 at 18:45

score 8 · Answer 1 · answered Feb 28 '15 at 19:00

8

As pointed out already in another answer, a ring in which $1=0$ just consists of a single element as for example $$a=1a=0a=(0+0)a = (1+1)a= a+a$$ shows $0=a$ for each $a$.

And, one does not want this structure, just $\{0\}$, to be a field, since this would be inconvenient, since then on would one would write all the time let $K$ be a field other than the trivial field, instead of just writing let $K$ be a field.

For example, there is no reasonable notion of a (non-trivial) vectorspace over that "field" so what is $K^2,K^3$ and so on in that case? Again the trivial vectorspace, but then the dimension of $K^n$ is not $n$ anymore for this "field".

Polynomials over that ring make not much sense either and one can continue in this way.

answered Feb 28 '15 at 19:00

quid

42,135

1

so distributive law binds them all – aileia Feb 28 '15 at 19:19
so that is alreay implied, why to mention $1 \neq 0$ in axioms – aileia Feb 28 '15 at 19:24
so you are saying in trivial field, when everything is equal to each other, there is no discrimination among elements, 1≠0 can be relaxed – aileia Feb 28 '15 at 19:38
Yes mainly it is the distributive law, and then also the fact that it is a group under addition. What I mean to say is that if one drops the axiom $1 \neq 0$, then one gets only one more structure that is a field. Every field $K$ with $|K| \ge 2$ will automatically have $0 \neq 1$. The $1 \neq 0$ is really just another way to say $|K|\ge 2$. – quid Feb 28 '15 at 19:53
i thought it would be ideal, simple to show a few characteristics of the field, simple yet to grasp the meaning. A good example – aileia Mar 01 '15 at 07:12

score 3 · Answer 2 · answered Feb 28 '15 at 18:44

3

If $1 = 0$, the field is only $\Bbb K = \{0\}$. $$x = x \cdot 1 = x \cdot 0 = 0, \quad \forall\, x \in \Bbb K$$

answered Feb 28 '15 at 18:44

Ivo Terek

77,665

if 1=0 then this holds, this does not necessarily excludes 1=0. zero do not necessaril set 0 all elements – aileia Feb 28 '15 at 18:48
I'm not sure I follow your doubt now. Can you elaborate? – Ivo Terek Feb 28 '15 at 18:50
in axioms, there is no x.0=0 rule, so why do above argument requires $1 \neq 0$ – aileia Feb 28 '15 at 18:52
1

@aileia You can prove that $x\cdot 0=0$ in any ring by $$x\cdot 0=x\cdot (1-1)=x-x=0.$$ – J126 Feb 28 '15 at 18:53
That's what Joe said. And if you want to be $100%$ rigorous, you'll want to prove that $(-1)\cdot x = -x$ in any ring, too. – Ivo Terek Feb 28 '15 at 18:57
so you are saying, x.0 =0 is implied in axioms by distributive law – aileia Feb 28 '15 at 18:59
Yes, you can prove it from the axioms, like Joe did in his last comment. – Ivo Terek Feb 28 '15 at 19:02
what if -1 cannot interact x, according to my multiplication definition, but only with 1, it wil not equal to 0? – aileia Feb 28 '15 at 19:07
From the axioms, you can prove that $ (-1) \cdot x = - x $. Any field element interacts with $ x $ via the scalar multiplication operation. – Berrick Caleb Fillmore Feb 28 '15 at 19:19
I think i can do that:) by distributive law.:) (1+(-1))x=0 implies -1x=-x – aileia Mar 01 '15 at 07:30

score 0 · Answer 3 · answered Feb 28 '15 at 19:15

If 1=0, by definition of these two elements, one would have, for every element $a$ of this "field," $a=1a=0a=0$ as @quid has shown, in other words the ground set for this field would simply be a singleton. So there would be not much excitement or usefulness in this field. It is like lack of excitement in considering log base 1 -- it is no fun, or for that matter considering 1 to be a prime number.

Field axioms: Why do we have $ 1 \neq 0$?

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