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.

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".

• Thank you very much. FWIW, this definition comes from the celebrated textbook by Dummit and Foote (2004, p. 34). Apparently, the condition “$(F\setminus\{0\},\cdot)$ is an Abelian group” should be complemented as “$(F\setminus\{0\},\cdot)$ is an Abelian group and $\cdot$ is commutative on the whole of $F$.” – triple_sec Apr 6 '16 at 17:43
• @triple_sec Gross! But I just found it's allready in the errata (of which there seem to be quite a few beyond typoes). - Back at school I complained that the textbook we had forgot to require $1\ne 0$ in its field axioms and my teacher didn't believe me it was important ... – Hagen von Eitzen Apr 6 '16 at 17:53
• Thank you for the link to the errata sheet. Indeed, if $1=0$, then $X=\{0\}$. However, many algebraists don’t regard this “trivial field” as a bona fide field. [On a related note: My pet peeve is how the important qualifier “non-trivial” is often left out of the statement “every vector space has a basis in ZFC.” Really? Please give me a basis for $\{0\}$ then...] – triple_sec Apr 6 '16 at 18:02
• @triple_sec: Here you go: $\{\}$. – Henning Makholm Apr 6 '16 at 18:06
• @triple_sec: No, $\operatorname{span}\varnothing$ is $\{0\}$. The span is the set of all linear combinations, including the empty linear combination (which in this case is the only one). – Henning Makholm Apr 6 '16 at 18:08

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.

• Thank you, I corrected the mistakes. That’s the whole point: that $x\cdot 0=0$ does not imply that $0\cdot x=0$, because $\cdot$ is commutative only on $F\setminus\{0\}$ according to the axioms I presented! @HagenvonEitzen pointed out my axioms are not right, and are compatible with multiplication not being commutative. – triple_sec Apr 6 '16 at 17:53