# Field basics: show $a \cdot b = 0 \iff a = 0 \vee b = 0$.

I was recently reviewing an old exam and noticed I lost marks for the following Q/A. I cannot for the life of me figure out why this was the case. If someone could highlight what I did incorrectly and suggest a correction, that would be greatly appreciated.

Question:

Prove $a \cdot b = 0 \iff a = 0 \vee b = 0$ where $a$ and $b$ are elements of a field $F$.

First we show $a = 0 \vee b = 0 \Longrightarrow a \cdot b = 0$.

Let $x \in F$ be arbitrary. Then $$0 \cdot x = 0 \cdot x + 0$$ But $x \in F$, so there exists an additive inverse $(-x)$ for $x$. Hence $$0 \cdot x + 0 = 0 \cdot x + (x + (-x))$$ Using the associativity of addition and the distributivity of multiplication over addition in $F$ we have $$0 \cdot x + (x + (-x)) = (0 \cdot x + x) + (-x) = x \cdot (0 + 1) + (-x)$$ Thus $$0 \cdot x = x \cdot 1 + (-x) = 0$$ Because $x$ was arbitrary, we can conclude that $0$ multiplied by any element of $F$ is $0$. Given that $a = 0 \vee b = 0$, it follows that $a \cdot b = 0$.

Next we show $a \cdot b = 0 \Longrightarrow a = 0 \vee b = 0$. If we assume the contrary, we have $$a \cdot b = 0 \wedge (a \ne 0 \wedge b \ne 0)$$ We know that there exists a multiplicative inverse $a^{-1}$ for $a$, so $$a^{-1} \cdot a \cdot b = a^{-1} \cdot 0$$ But (remembering that associativity holds) this reduces to $1 \cdot b = 0$, which is a contradiction. Hence it follows that $$a \cdot b = 0 \iff a = 0 \vee b = 0$$

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It looks pristine to me. Any reference to where points were lost? – andybenji Dec 8 '12 at 4:39
The first part is more complicated than needs be, and it looks like you have extra unneeded stuff in parts of it. For example, the line before "Thus", the $x+(-x)$ on the LHS is completely irrelevant to the proof.... BUt the proof looks right to me... – N. S. Dec 8 '12 at 4:43
P.S. I assume your field is commutative, that would make a difference. In some countries field is defined as not necesarily commutative. – N. S. Dec 8 '12 at 4:48
@N.S. It is commutative. I think you mean ring though? I'm fairly sure that all fields are commutative, regardless of country. – providence Dec 8 '12 at 4:49
@andybenji The points were lost because of what Martin Argerami describes following "in the last part..." in his answer. – providence Dec 8 '12 at 5:01

As mentioned in the comments, the first part can be shorter: $$0\cdot x=(0+0)\cdot x=0\cdot x+0\cdot x,$$ and then adding $-0\cdot x$ to both sides we get $0=0\cdot x$.
In the last part, the only thing I would comment on is that when you have $a\cdot b=0$, the usual reasoning is like this: "if $a=0$, we are done; otherwise, consider $a^{1-}$..."
The objection could be that if $a=0$ you cannot do your reasoning with $a^{-1}$ (you have to do it with $b^{-1}$).
Ah, now I recall, that your comments regarding the last part are exactly why I lost the mark. However, my question then (as it is now) is why is $a$ being $0$ a concern? I have assumed $a \ne 0 \wedge b \ne 0$. – providence Dec 8 '12 at 4:58