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

Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
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 $$R$$ is a division ring

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


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

Peter
 
Last edited:
Physics news on Phys.org
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.
 
steenis said:
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
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
Back
Top