Module Over a Division Ring - Blyth Theorem 1.1, Part 4

[Math Amateur]

I am reading T. S. Blyth's book "Module Theory: An Approach to Linear Algebra" ... ... and am currently focussed on Chapter 1: Modules, Vector Spaces and Algebras ... ...

I need help with an aspect of Theorem 1.1 part 4 ...

Theorem 1.1 in Blyth reads as follows:View attachment 5838In the above text, in part 4 of the Theorem we read:" ... ... when \(\displaystyle R\) is a division ring

(4) \(\displaystyle \lambda x = 0_M\) implies \(\displaystyle \lambda = 0_R\) or \(\displaystyle x = 0_M\) ... ... "


Blyth proves that if \(\displaystyle R\) is a division ring and \(\displaystyle \lambda x = 0_M\) with \(\displaystyle \lambda \neq 0_R\) then we have that \(\displaystyle x = 0_M\) ... ...But ... ... Blyth does not show that if \(\displaystyle R\) is a division ring and \(\displaystyle \lambda x = 0_M\) with \(\displaystyle x \neq 0_M\) then we have that \(\displaystyle \lambda = 0_R\) ... ...Can someone please help me to prove this ...

Peter

 

[steenis]

This is simple logic, Peter.
You want to prove: $s1: p\to (a\vee b)$
For some reason you have proven: $s2: (p\wedge \neg a)\to b$
And you want to know if $s1$ and $s2$ are equivalent: $s1\Leftrightarrow s2$
If so, you have proven $s1$, and you are ready

You know this rule: $r1: (x\to y)\Leftrightarrow (\neg x \vee y)$, we are going to use this

Ok, you have proven $s2$:
$$(p\wedge \neg a)\to b$$
by $r1$, this is equivalent with:
$$\neg(p\wedge \neg a) \vee b$$
which is equivalent with
$$\neg p \vee a \vee b$$
which is equivalent with
$$\neg p \vee (a \vee b)$$
by $r1$, this is equivalent with:
$$p\to (a\vee b)$$
And you are ready.

Therefore, you don't have to prove the second implication.

 

[Math Amateur]

This is simple logic, Peter.
You want to prove: $s1: p\to (a\vee b)$
For some reason you have proven: $s2: (p\wedge \neg a)\to b$
And you want to know if $s1$ and $s2$ are equivalent: $s1\Leftrightarrow s2$
If so, you have proven $s1$, and you are ready

You know this rule: $r1: (x\to y)\Leftrightarrow (\neg x \vee y)$, we are going to use this

Ok, you have proven $s2$:
$$(p\wedge \neg a)\to b$$
by $r1$, this is equivalent with:
$$\neg(p\wedge \neg a) \vee b$$
which is equivalent with
$$\neg p \vee a \vee b$$
which is equivalent with
$$\neg p \vee (a \vee b)$$
by $r1$, this is equivalent with:
$$p\to (a\vee b)$$
And you are ready.

Therefore, you don't have to prove the second implication.

Thanks Steenis ... but it is going on towards 2.00am here in Tasmania ... truly is the edge of the world ... ...

... will work through what you have said in the morning ...

... but thank you for your assistance ...

Peter