Martín-Blas Pérez Pinilla suggests that "=" can be considered a logical symbol obeying logical axioms. While I agree that it fundamentally is so, I would like to note that it is possible to consider it an equivalence relation obeying 'internal' field axioms, because for example the rational numbers can be taken as equivalence classes of a certain set of pairs of integers, and so it is not quite right to consider the equality between these rationals as a logical equality. Also, Ittay made a mistake where he used an unstated axiom that allows substitution. What you need, either way, is something equivalent to the following for any field $F$:
$a=a$ for any $a \in F$ [reflexivity of =]
$a=b \Rightarrow b=a$ for any $a,b \in F$ [commutativity of =]
$a=b \wedge b=c \Rightarrow a=c$ for any $a,b,c \in F$ [transitivity of =]
(These describe "=" as an equivalence relation on $F$)
$a=b \Rightarrow P(a)=P(b)$ for any $a,b \in F$ and predicate $P$ [substitution]
(This describes substitution, which can be used to replace separate axioms governing how "=" and the field operations interact. Ittay used this in one of his steps.)
These allow us to "do the same thing to both sides", for example:
For any $a,b,c \in F$ such that $a=b$,
Let $d=a+c$ [closure under +]
$a+c=a+c$ [transitivity of =; $a+c=d=a+c$]
$a+c=b+c$ [substitution; where the predicate is given by $P(x) \equiv (a+c=x+c)$]
Note that to prove that something is a field, we will have to prove the substitution axiom, which boils down to proving the following equivalent set of axioms:
$a=b \Rightarrow a+c=b+c$ for any $a,b,c \in F$
$a=b \Rightarrow ac=bc$ for any $a,b,c \in F$
The original problem can then be proven as follows:
For any $a,b,c \in F$ such that $a+b=a+c$,
$b = 0+b = (-a+a)+b = (-a)+(a+b) = (-a)+(a+c) = (-a+a)+c = 0+c = c$
(1). The additive and multiplicative identities are (obviously) $0$ and $1$ and are (obviously) unique. And $0\not \equiv 1\pmod p$. (That is necessary to say, because in a field the additive and multiplicative identities are not allowed to be equal).
(2). Additive inverses:
(2A). Existence: If $0< x<p$ then $0<p-x<p$ with $x+(p-x)\equiv 0.$ And $0+0\equiv 0.$
(2B). Uniqueness: If $x+y\equiv x+y'\equiv 0$ then $y-y'\equiv 0.$ but if $0\leq y<p$ and $0\leq y'<p$ then $|y-y'|<p.$ And if $p$ divides $|y-y'|$ with $0\leq |y-y'|<p$ then $y-y'=0 .$
(3). Multiplicative inverses:
(3A). Existence: For $1\leq x<p$ let $M=\min \{z:0<z\land \exists y\;(xy\equiv z \pmod p\}.$
First, by considering the case $y=1$ we have $M\leq x<p.$
Second, we will show that if $xy\equiv k$ with $2\leq k<p$ then there exists $y'$ and $k'$ with $0< k'<k$ and $xy'\equiv k'.$ From this we will conclude that if $2\leq k<p$ then $k\ne M.$
Thirdly, from the first and second parts above, we will have $M=1$, so $xy\equiv 1$ for some $y$.
To prove the second part: Suppose $xy\equiv k$ with $2\leq k<p.$ Let $m=\max \{m'>0: m'k<p\}.$ Then $mk<p\leq (m+1)k.$ But $p$ is prime and $m+1>1<k$ so we cannot have $(m+1)k=p.$ So $mk<p<(m+1)k.$
Let $y'=(m+1)y$ and $k'=(m+1)k-p.$ We have $0<k'<k'-mk=k.$ Now $$xy'=xy(m+1)\equiv k(m+1)\equiv k(m+1)-p=k'.$$
(3B). Uniqueness of multiplicative inverse: $$xy\equiv xy'\equiv 1\implies$$ $$\implies y\equiv y(1)\equiv y(xy')\equiv (yx)y'\equiv (1)y'\equiv y'.$$
Best Answer
What do you think of this?
Lemma. In a field $a\cdot 0=0\cdot a=0$, for any $a\in\mathbb{F}$
proof
$a\cdot 0=a\cdot (1+(-1))=a+(-a)=0$
$0\cdot a=a\cdot 0=0$ for the commutativity of product
end proof
end lemma
main proof
$1=1\\ 1+0\cdot 0=1\\ 1+(1-1)(1-1)=1\\ 1+1\cdot 1+1(-1)-1(1)+(-1)(-1)\\ 1+1-1-1+(-1)(-1)=1\\ 0+0+(-1)(-1)=1\\ (-1)(-1)=1$