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Question: Let $F_4 = \{0, 1, a, b\}$ be the field containing $4$ elements. Assume that $1 + 1 = 0$. Prove that $b = a^{−1} = a^2 = a + 1$.

Proof: Consider $ab \in F_4$. There are four possibilities for what $ab$ is equal to. (Note that $a$ and $b$ are fixed elements of the field; they are not arbitrary elements of it.) If $ab = 0$, then $a = 0$ or $b = 0$, a contradiction (i.e. fields do not have zero divisors. If you have not proved this already, then it is worth doing). If $ab = a$, then $b = 1$, a contradiction; likewise, $ab= b$. Hence, $ab$ must be equal to $1$. In other words, $b = a^{−1}$ ......... (irrelevant part continued).

My question is: Why is it a contradiction that "if $ab=0$, $a=0$ or $b=0$ (fields do not have zero divisors."? But why are we allowed to make the statement that "$0*0 = 0$ holds"?

Thank you.

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    It is a contradiction because you are assuming the four elements of the field are $0$, $1$, $a$, and $b$ (ie you have *assumed* that $a \neq 0$ and $b \neq 0$). – 2017-01-23
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    because you already stated $a,b$ are NOT $0$, ergo their productcannot be $0$ – 2017-01-23

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A field is a commutative ring whose nonzero elements are invertible.

In any ring $(A,+,\times)$ (and in particular in any field), we have for all $x\in A$, $0x=0$. That's because $0x=(0+0)x=0x+0x$ and every element in a groupe (here the groupe $(A,+)$) is regular (you can simplify any equality $r+u=s+u$ into $r=s$).

Let $(F,+,\times)$ be a field and let $x,y\in F$ such that $xy=0$. If $x\neq 0$ then $x$ is invertible and we have $y=1y=(x^{-1}x)y=x^{-1}(xy)=x^{-1}0=0$. This proves that if $xy=0$ then either $x=0$ or $y=0$.