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Let $A$ be a commutative ring and let $M$ be a $A$-module. The following statement is taken from chapter two in Attiyah, McDonald, commutative algebra. If $a \subseteq Ann(M)$, we may regard $M$ as an $A/a$-module as follows, if $\overline{x}$ is an element of $A/a$ define $\overline{x}m=xm$ for $m\in M$ and $x$ representative of $\overline{x}$. Could you help me to show that this is welldefined, in the book it is stated as something obvious but I can't see why. If $x, x'$ is two different representatives of $\overline{x}$ then $xa=x'a$ or equivalently $x^{-1}x'\in a$, $\rightarrow$ $x^{-1}x'\in Ann(M)$, hence $x^{-1}x'm=0$ for all $m\in M$ how does this show that $x'm=xm$? I appoligies if this is obvious but I just got stuck and want some help.

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2 Answers 2

up vote 5 down vote accepted

If $\bar x = \bar y$ then $x-y\in \mathcal a\subseteq Ann(M)$. So $(x-y)m=0$ for all $m\in M$ and therefore, $xm=ym$ for all $m\in M$.

Note, there is no $x^{-1}$ in a general ring $A$. If $x,x'$ are two different representatives of $\bar x$, then $x+\mathcal a = x' + \mathcal a$. Not multiplication, addition.

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Keep in mind that $\mathfrak{a}$ is an ideal, so a coset in $A / \mathfrak{a}$ has the form $x + \mathfrak{a}$. If $x + \mathfrak{a} = x' + \mathfrak{a}$ then $x' = x + a$ with $a \in \mathfrak{a}$. Then $x'm = xm + am$, but $a \in \mathfrak{a} \subseteq \text{Ann}(M)$, so $am = 0$ and thus $x'm = xm$.

Hope that helps clear up your confusion!

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