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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 $F$.

Answer:

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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    @providence No, I mean field. I went to school in Europe, and we used the same word for field and division ring. To differentiate we used the word commutative.... Sometimes things are lost (or gained) in translation...2012-12-08

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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 $b^{-1}$).

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    I will keep these comments in mind. Thanks.2019-02-09