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Suppose that $F$ is a field and prove that $0\ne1$

According to the definition of a field I know that the zero element is different from the one element, but is there a scientific proof for that?

SuperNova
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    Dear SuperNova: you can't prove a definition (scientifically or otherwise). –  Jun 23 '14 at 09:54
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    @AsalBeagDubh You cannot prove a definition that creates a conservative extension of your logic. But in practice, very few restrict themselves to conservative extensions. And the axioms of field theory are certainly not conservative extensions. It isn't strange to wonder if $0 = 1$ can be proven from field axioms, and to call it a "definition" is very misleading. – DanielV Jun 23 '14 at 10:21
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    @AsalBeagDubh There's a trivial sense in which you can: "$A$ (by definition), $A$ if and only if $A$, therefore $A$." – Shaun Jun 23 '14 at 10:22
  • @DanielV In my opinion, it is strange to wonder whether a formula, in this case $0\neq1$, can be proved from the field axioms (also called the definition of "field") when one knows that the formula is one of the axioms (as the OP says (s)he does). Given some axioms, it's very easy to prove any one of those axioms; a one-line "proof" suffices. Also, I agree with Asal Beag Dubh that you can't prove a definition; conservativity has nothing to do with it. – Andreas Blass Jun 26 '14 at 16:15

4 Answers4

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In a field, $0\neq 1$ by definition. This is to exclude the zero ring from being a field.

This leads to a natural question which is an easy but illuminating exercise: If we are in a ring with $0=1$, show that every element equals $0$.

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Some field axioms do not include $\,0 \ne 1\,$ but, rather, deduce it from other axioms. For example, the axioms may state that $\, F\setminus0\,$ is a monoid, hence the neutral element $1$ of the monoid is distinct from the neutral element $\,0\,$ of the additive group of $\,F.\,$ Therefore, in this axiomatization of a field, $\,0\ne 1\,$ is not an axiom but, rather, a derived theorem (albeit a trivial one). Thus it may well be that the exercise in your question was posed in the context of this or a similar axiom system.

Remark $\ $ This thread discusses the rationale for excluding the one-element ring from the class of fields, so it may be of interest. Beware that there are occasional (inconsequential) errors when using axioms involving monoids and unit groups, as the linked comment shows.

Community
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Bill Dubuque
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For every $x \in F $ we have $$0 \cdot x = 0 $$ because $$0\cdot x = (0 + 0 ) \cdot x = 0\cdot x + 0 \cdot x$$ by the axioms of field.

But also $$1 \cdot x = x \ \ \forall x \in F$$ and so if $0 = 1 $ we have $$x = 1\cdot x = 0 \cdot x = 0 \ \ \forall x \in F$$

and then $F = 0 $

WLOG
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  • But this hasn't really proven anything: it just passes the buck to a different question "for a field $F$, how can I prove $F\neq {0}$? For an algebraic structure satisfying all field axioms except "$0\neq 1$" the structure is nonzero iff $0\neq 1$. The $0\neq 1$ axiom just can't be derived from the other axioms (unless your system contains an axiom that says "nonzero ring," and then you're fine. I'm just not familiar with that being a common thing :) ) – rschwieb Jun 23 '14 at 15:17
  • Yes I supposed that $F \neq 0 $ – WLOG Jun 23 '14 at 15:50
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There is a principle in proof theory that if you can find a set of variables such that axioms hold but a conclusion does not, then it is impossible to prove the conclusion from the axioms. For example, suppose you know the following axioms:

$$x > y \tag{A1}$$ $$x > z \tag{A2}$$

Call $P$ a proposition to be proven as $y > z$. Well, if we let $x = 3, y = 1, z = 2$. Note that under this assignment, A1 and A2 are true, but $P$ is false. Therefore you cannot prove $P$ from A1 and A2.

There are 10 or so axioms of field theory. The one element field satisfies the axioms of field theory. The one element field also satisfies $0_F = 1_F$, therefore it is impossible to prove $0_F \ne 1_F$ from just field axioms.

DanielV
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    What one element field? – Tobias Kildetoft Jun 23 '14 at 10:14
  • http://en.wikipedia.org/wiki/Field_(mathematics) observe that there is no axiom that $0 \ne 1$. This leads to http://en.wikipedia.org/wiki/Field_with_one_element . Note that the field with one element is only inconsistent if you add the axiom $0 \ne 1$ to the other less-forgotten axioms of fields. – DanielV Jun 23 '14 at 10:26
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    It actually does if you read it properly. And the other page linked is not about a "field" with just one element. – Tobias Kildetoft Jun 23 '14 at 10:31