Proving that $0\cdot x=0$ using field axioms
Consider the following axiomatic definition of a field:
A field is a set $F$ together with two binary operations $+$ and $\cdot$ on $F$ such that $(F,+)$ is an Abelian group with identity $0$ and $(F\setminus\{0\},\cdot)$ is an Abelian group with identity $1$, and the following left-distributive law holds: $$a\cdot(b+c)=(a\cdot b)+(a\cdot c)\quad\forall a,b,c\in F.$$
I want to show that $0\cdot x=0$ for any $x\in F$ using these, and only these, field axioms. I can prove that $x\cdot 0=0$ using left-distributivity, but multiplication with $0$ is not necessary commutative a priori [that $(F\setminus\{0\},\cdot)$ is an Abelian group does not say anything about multiplication with $0$].
Any hint would be appreciated.
To elaborate on my point, let me prove that $x\cdot 0=0$ for any $x\in F$: \begin{align*} 0+0=&\,0\\ \Downarrow&\,\\ x\cdot(0+0)=&\,x\cdot0\\ \Downarrow&\,\text{(left-distributivity)}\\ (x\cdot 0)+(x\cdot 0)=&\,x\cdot 0\\ \Downarrow&\,\\ [(x\cdot 0)+(x\cdot 0)]+[-(x\cdot 0)]=&\,x\cdot 0+[-(x\cdot 0)]\\ \Downarrow&\,\\ (x\cdot0)+\{(x\cdot0)+[-(x\cdot0)]\}=&\,0\\ \Downarrow&\,\\ (x\cdot0)+0=&\,0\\ \Downarrow&\,\\ x\cdot0=&\,0. \end{align*} My problem is I would need to exploit right-distributivity to show that $0\cdot x=0$, but right-distributivity does not follow immediately from the axioms.
$\endgroup$2 Answers
$\begingroup$You can't.
Let $F=\Bbb Q$, define addition as usual and $$x\cdot y =\begin{cases}xy&\text{if }x\ne 0\\y&\text{if }x=0\end{cases}$$ Then
- $(F,+)$ is an abelian group because $\Bbb Q$ really is a field;
- $(F\setminus\{0\},\cdot)$ is an abelian group because $\Bbb Q$ really is a field and $\cdot $ conincides with standard multiplication here
- Left distribution holds for $a\ne 0$ because it holds in the field $\Bbb Q$
- left distribution holds for $a=0$ by direct verification
In other words: your collection of axioms is "wrong".
$\endgroup$ 9 $\begingroup$What you are calling "commutativity" is "distributivity". While only "left-distributivity" is defined, multiplication in a field is commutative. Once you have shown that x0= 0, it follows immediately from the commutativity of multiplication that 0x= 0.
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