MHB Modules - Northott: Proposition 1

[Math Amateur]

In D. G. Northcott's book: Lessons on Rings, Modules and Multiplicities, Proposition 1 reads as follows:View attachment 3453The first line of the above proof reads as follows:

"Since $$0_R + 0_R = 0_R$$, the definition of an R-module shows that

$$0_Rx = (0_R + 0_R)x = 0_Rx + 0_Rx,$$

whence $$0_Rx = 0_M$$, because $$M$$ is a group. ... ... "Now it seems highly plausible that

$$0_Rx = 0_Rx + 0_Rx$$

in the group $$M$$ leads to the conclusion that $$0_Rx = 0_M$$ ... ... BUT ... how do we know (prove) this? ...

Can someone help?

[NOTE:

Northcott seems to be saying that in a group M, for an element $$a$$:

$$a = a + a$$

$$\Longrightarrow a$$ is the identity

$$\Longrightarrow x + a = a + x = x$$ for all $$x \in M$$

... BUT ... how do we prove this? ]

 

[Math Amateur]

In D. G. Northcott's book: Lessons on Rings, Modules and Multiplicities, Proposition 1 reads as follows:View attachment 3453The first line of the above proof reads as follows:

"Since $$0_R + 0_R = 0_R$$, the definition of an R-module shows that

$$0_Rx = (0_R + 0_R)x = 0_Rx + 0_Rx,$$

whence $$0_Rx = 0_M$$, because $$M$$ is a group. ... ... "Now it seems highly plausible that

$$0_Rx = 0_Rx + 0_Rx$$

in the group $$M$$ leads to the conclusion that $$0_Rx = 0_M$$ ... ... BUT ... how do we know (prove) this? ...

Can someone help?

[NOTE:

Northcott seems to be saying that in a group M, for an element $$a$$:

$$a = a + a$$

$$\Longrightarrow a$$ is the identity

$$\Longrightarrow x + a = a + x = x$$ for all $$x \in M$$

... BUT ... how do we prove this? ]

... ... just been thinking about the above question ...

Now have another question ...

How do we know that $$0_Rx = 0_Ry$$ for $$x,y \in M$$

Help with the above question and the question in the first post would be appreciated ...

Peter

 

[Euge]

Fix $x\in M$. Since $M$ is a group under addition, $0_R x$ has an additive inverse, $-(0_R x)$. Add $-(0_R x)$ to both sides of the equation $0_R x = 0_R x + 0_R x$. The left hand side will be $0_M$, and the right hand side will be $(0_R x + 0_R x) + [-(0_R x)]$, which is the same as $0_R x + (0_R x + [-(0_R x)])$, by associativity of addition. This reduces to $0_R x + 0_M$, which equals $0_R x$. Thus, $0_R x = 0_M$. Since this holds for every $x\in M$, $0_R x = 0_R y = 0_M$ for all $x, y\in M$ (answering your second question).