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Let $M$ be a left $R$-module, and consider $\mathrm{Ann}(x)$ where $x \in M$. To show that $\mathrm{Ann}(x)$ must be a left ideal, one must first show that it is a subgroup (of the underlying additive group $R$), however it's not clear to me that if $r \in \mathrm{Ann}(x)$, then $-r \in \mathrm{Ann}(x)$. I'm not assuming $R$ has $1$. Does anyone know a way around this?

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2 Answers 2

Recall that $(r+s)x = rx + sx$ for $r,s \in R$ and $x \in M$. Then

$$0 = 0x = (r+(-r))x = rx + (-rx) = 0 + (-rx) \Longrightarrow -rx = 0$$

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Start from the very definition of $-r$, that is, $r + (-r) = 0$. Now $0 = 0 x = \dots$

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