Let $1$ be the multiplicative identity, so that $1\cdot a = a$ (where $a\in \mathbb{F})$. Let $0$ be the additive identity, so that $a+0=a$. Prove that $0\ne 1$. (Here we don't yet know that $0$ and $1$ must be unique, nor do we know that $0\cdot a = 0$).

My approach:

Suppose that $1=0$, then $a+1 = a \iff 1\cdot(a+1)=a\iff$ $1\cdot(a+1-a)=0\iff 1\cdot (a-a+1)=0 \iff 1+1 = 0\iff 1+ ... + 1 =0$.

I'm not sure what to do next because it doesn't look obvious that multiplicative identity must necessarily make other numbers in $\mathbb{F}$ by addition. I.e., if $1=-1$, as above, then $1$ is simply the additive inverse of itself.


closed as unclear what you're asking by user223391, Martin Sleziak, user91500, zhoraster, Watson Jan 6 '17 at 10:46

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    $\begingroup$ What is your definition of field? Most definitions require $0\neq 1$. $\endgroup$ – user223391 Jan 6 '17 at 4:50

If $1=0$, then $a=a\cdot 1=a\cdot 0=0$, i.e. every element is the zero element, which means that the ring is the zero ring. About why $a.0=0$, just note that $a\cdot 0=a\cdot (0+0)=a\cdot 0+a\cdot 0$, so $a\cdot 0=0$.

Actually the fact that $1\neq 0$ comes from the definition of integral domain. As Wikipedia says, "an integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero".

So since every field is an integral domain (*), then every field is in particular a nonzero commutative ring, in other words, in a field we must have $1\neq 0$.

About (*) we can prove it in the following way. Call our field $F$, we only need to show that $F$ doesn't have any non zero divisors of zero. Let's suppose that there is some divisor of zero $a$, then exits $b\in F$ such that $a\cdot b=0$. Now because we are in a field $b$ has an inverse $b^{-1}$, then multiplying by $b^{-1}$ we get $$(a\cdot b)\cdot b^{-1}=a\cdot (b\cdot b^{-1})=a\cdot 1=a=0\cdot b^{-1}=0.$$ Therefore $a=0$ and hence $F$ is an integral domain.

New edit: As a matter of fact, again by Wikipedia we have that a field "is a nonzero commutative division ring". So again, we have by definition that $1\neq 0$. The conclusion is that you don't have to prove that $1\neq 0$ in a field. It's just part of the definition of a field.

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    $\begingroup$ And why precisely do we exclude $\{0\}$ from being a field? $\endgroup$ – user223391 Jan 6 '17 at 4:50
  • $\begingroup$ @ZacharySelk math.stackexchange.com/questions/427078/is-0-a-field $\endgroup$ – Q the Platypus Jan 6 '17 at 4:52
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    $\begingroup$ @QthePlatypus And the answer is "we don't want $1=0$", which is tautological. But my point is that without knowing exactly what definition the OP is using, this question as it stands is slightly ill posed. $\endgroup$ – user223391 Jan 6 '17 at 4:53
  • $\begingroup$ @ZacharySelk I edited my answer. $\endgroup$ – Xam Jan 6 '17 at 4:57
  • $\begingroup$ "A field is a set $\mathbb{F}$ with at least two elements and two binary operations "+" and "$\cdot$"". $\endgroup$ – sequence Jan 6 '17 at 4:57

If $0=1$ and all else about the field axioms hold then the field only has one element. If we assume a field has at least two elements we can prove $0\ne 1$.

We start by proving $0*a=0$ for all $a$:

$0*a +0*a = (0+0)*a=0*a $

$0*a+0*a+(-(0*a))=0*a + (-(0*a)) $

$0*a = 0$

Now if $0=1$ then $a=1*a=0*a=0 $ for all $a $. so $a=0$ for all $a $. So the field only has one element; $0$.

So any field with at least two elements must have $0\ne 1$.


$0\ne 1$ sometimes appears as one of the field axiom, and this is equivalent to requiring the field to have at least two elements.

For the purpose of this exercise, let's assume that this not part of the field axioms, and consider any field with at least two elements, $0$ and $x$. Of course, the field also contains the element $1$, but it may be $x$ or (a priori) $0$.

We argue by contradiction and assume $0=1$. Then

$$x=1x=0x=(0+0)x = (1+1)x = 1x+1x = x+x.$$

(first equality from definition of $1$, second from contradiction assumption, third from definition of $0$, fourth from contradiction assumption, fifth from distributive property, sixth from definition of $1$).

Adding $-x$ (additive inverse of $x$) to both sides (and using associative property of addition on RHS) gives


a contradiction.


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