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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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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

1 Answer 1

up vote 3 down vote accepted

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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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
    
I think that what you did is right. But the thing is that it is unnecessarily involved, and that probably threw of the marker (it threw me off too). You want to prove an implication, then consider the contrapositive, and then prove the contrapositive by contradiction. Feels like too complicated; you probably shouldn't have lost marks on it. –  Martin Argerami Dec 8 '12 at 5:06
    
I will keep these comments in mind. Thanks. –  providence Dec 8 '12 at 5:19

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