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Proof

(related to Proposition: In a Field, $0$ Is Unequal $1$)

By hypothesis, $(F,+,\cdot)$ is a field.
Suppose, $0=1$.
- Take $x\in F$ with $x\neq 0$¹.
- Because $0$ is the neutral element of addition in $F$, $1$ is its additive inverse, i.e. $1-0=0.$
- From the distributivity laws in $F,$ we get $x\cdot 0=x\cdot (1-0)=x\cdot 1-x\cdot 0.$
- Because $1$ is the neutral element of multiplication in $F$, it follows $x\cdot 0=x-x\cdot 0,$ or $x=x\cdot 0+x\cdot 0.$
- From the distributivity laws, it follows $x=x\cdot (0+0),$ and, since $0$ is neutral with respect to the an addition, $x=x\cdot 0.$
- Altogether, it follows $x=x-x,$ or $x=0$, in contradiction to $x\neq 0.$
The assumption $0=1$ must be false, i.e. $0\neq 1.$

∎

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References

Bibliography

Knauer Ulrich: "Diskrete Strukturen - kurz gefasst", Spektrum Akademischer Verlag, 2001

Footnotes

Note that from the definition of a field it follows that it is, in particular, a ring $R$ which is not the zero ring and that "every $x\neq 0$ in $F$ has an inverse element". Thus, such elements, for which $x\neq 0$, must exist in $F$. ↩