Let $(K,+,*)$ be a field. Let $1_K$ be the field's multiplicative identity. Let $(-1_K)$ be its additive inverse. Let $0_K$ be the field's additive identity.
I'm trying to prove that for every $n>0_{\mathbb{N}}$, $f=\underbrace{1_K+...+1_K}_{n\, times\,+}$ isn't equal to $0_K$.
Since $+:K{\times}K{\rightarrow}K$, I know that I can write $f$ as $1_K+k$. I can also prove that for any two field elements $x$ and $y$, $x+y=0{\implies}x+y+(-x)=0+(-x){\implies}y=-x$ and so $f=0$ only iff $k=(-1_K)$.
And here we arrive at my original question: how to show that $k$ can't be equal to $(-1_K)$?
Answer
The statement is not true. Consider $\mathbb Z/3\mathbb Z$. This is a finite field, and $1+1+1=0$.
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