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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 $(-1_K)$?

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1 Answer 1

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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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  • $\begingroup$ What's $\mathbb{Z}/3\mathbb{Z}$? $\endgroup$
    – asdasdf
    Jan 1, 2017 at 17:49
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    $\begingroup$ Integers (mod 3). $\endgroup$ Jan 1, 2017 at 17:49
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    $\begingroup$ Yes, I believe that is true, since then $x+1>x$ for any $x$. $\endgroup$ Jan 1, 2017 at 18:09
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    $\begingroup$ $1>0$ is a consequence of the ordered field axioms btw. $\endgroup$ Jan 1, 2017 at 18:22
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    $\begingroup$ Consequence of $x^2\ge 0$ for all $x$. $\endgroup$ Jan 1, 2017 at 18:30

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