I'm trying to show that any field $F$ is an integral domain. So assume that there is some zero-divisor $a \in F$ meaning that for $a \neq 0$ there exists some $b \in F$ with $b \neq 0$ such that $ab = 0$. Can I arrive at a contradiction in the following way, or is there an issue? $$ \begin{aligned} ab+a=a\\ a(b+1)=a \\ b+1 = 1\\ b=0 \end{aligned} $$ Going from one to two holds because of distributivity, two to three holds from left cancellation. Thanks in advance for the clarification.
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2Yes, I don't see a flaw there. You can also get there from $ab=0$ implies that $ab=a0$. Now cancel the $a$'s. – Randall Nov 17 '21 at 15:54
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2You could also just multiply with $a^{-1}$ in the beginning. – Severin Schraven Nov 17 '21 at 15:56
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Left cancellation only works only if in $((ab=ac)\Rightarrow b=c)$, $a$ is not a zero-divisor.
There is a simpler proof, namely that units are not zero-divisors. Suppose $a,b\in F$ with $a\ne 0$ and $ab=0$.
Then $b = 1b = (a^{-1}a)b = a^{-1}(ab) = a^{-1}0 = 0$. The last step follows since $0$ is absorbing.
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