Proof of Fourth or Lattice Isomorphism Theorem for Modules

[Math Amateur]

Dummit and Foote give the Fourth or Lattice Isomorphism Theorem for Modules on page 349.

I need some help with the proof of Fourth or Lattice Isomorphism Theorem for Modules ... hope someone will critique my attempted proof ...

(I had considerable help from the proof of the theorem for groups in Project Crazy Project ...)

The Theorem reads as follows:

View attachment 2982

My attempt at the proof follows. Note that there are several points where I am unsure of the manipulations/mechanics with cosets of quotient modules ...In the Theorem stated above we read:

" ... ... There is a bijection between the submodules of \(\displaystyle M\) which contain \(\displaystyle N\) and the submodules of \(\displaystyle M/N\). ... ... "

So we need to demonstrate that there exists a bijection between the sets:

\(\displaystyle \mathcal{A} = \{ A \ | \ A \text{ is a submodule of } M \text{ and } A \text{ contains the submodule } N \}\)

\(\displaystyle \mathcal{B} = \{ A/N \ | \ A/N \text{ is a submodule of } M/N \}\)That is, we need to show that there exists a bijection \(\displaystyle \phi \ : \ \mathcal{A} \to \mathcal{B}\)

where \(\displaystyle \phi (S) = S/N\)

Now we know that there exists a surjective homomorphism:

\(\displaystyle \pi \ : \ S \to S/N\)

where \(\displaystyle \pi (s) = \overline{s}\)

so essentially, we have that:

\(\displaystyle \phi (S) = \pi \)

where \(\displaystyle \pi = \{ \overline{s}_i \ | \ \pi (s_i) = \overline{s}_i \}
\)

[Question 1 - is this right ... it seems so because \(\displaystyle \pi \) is essentially the set of all cosets of \(\displaystyle S/N\)]
Proof that \(\displaystyle \phi \) is injective

Suppose that \(\displaystyle \phi (S) = \phi (T)\) ... ... then we need to show \(\displaystyle S = T\) ... ...Note that we have that:

\(\displaystyle \phi (S) = \phi (T) \Longrightarrow \pi = \phi [T] \)

since we have that

\(\displaystyle \phi (S) = \pi \)

Now let \(\displaystyle x \in S\)

Then we have:

\(\displaystyle x \in S \)

\(\displaystyle \Longrightarrow \pi (x) = \pi (y) \text{ for some } y \in T \text{ since } \pi = \pi [T] \)

\(\displaystyle \Longrightarrow \overline{x} = \overline{y}\)

\(\displaystyle \Longrightarrow x - y \in N \) (Question 2, is that correct?)

\(\displaystyle \Longrightarrow x \in y + N\) (Question 3, is that correct?)But then ... \(\displaystyle y + N \subset T \) (Question 4, is that correct?)

So then \(\displaystyle x \in T\)

A similar (symmetrical) argument, I think, would show \(\displaystyle x \in T \Longrightarrow x \in S\) ...

So we have \(\displaystyle S=T\) and so \(\displaystyle \phi\) is injective ...

I am still working on the surjectivity of \(\displaystyle \phi\) ...

Can someone please critique the proof and indicate whether it is OK?

(Apologies for the large number of possibly trivial questions regarding cosets ... just ensuring my reasoning is sound ...)

If anyone can demonstrate or point to the existence of a better proof - online or in a text - I would be really interested ...

Peter

 

[Euge]

Dummit and Foote give the Fourth or Lattice Isomorphism Theorem for Modules on page 349.

I need some help with the proof of Fourth or Lattice Isomorphism Theorem for Modules ... hope someone will critique my attempted proof ...

(I had considerable help from the proof of the theorem for groups in Project Crazy Project ...)

The Theorem reads as follows:

View attachment 2982

My attempt at the proof follows. Note that there are several points where I am unsure of the manipulations/mechanics with cosets of quotient modules ...In the Theorem stated above we read:

" ... ... There is a bijection between the submodules of \(\displaystyle M\) which contain \(\displaystyle N\) and the submodules of \(\displaystyle M/N\). ... ... "

So we need to demonstrate that there exists a bijection between the sets:

\(\displaystyle \mathcal{A} = \{ A \ | \ A \text{ is a submodule of } M \text{ and } A \text{ contains the submodule } N \}\)

\(\displaystyle \mathcal{B} = \{ A/N \ | \ A/N \text{ is a submodule of } M/N \}\)That is, we need to show that there exists a bijection \(\displaystyle \phi \ : \ \mathcal{A} \to \mathcal{B}\)

where \(\displaystyle \phi (S) = S/N\)

Now we know that there exists a surjective homomorphism:

\(\displaystyle \pi \ : \ S \to S/N\)

where \(\displaystyle \pi (s) = \overline{s}\)

so essentially, we have that:

\(\displaystyle \phi (S) = \pi \)

where \(\displaystyle \pi = \{ \overline{s}_i \ | \ \pi (s_i) = \overline{s}_i \}
\)

[Question 1 - is this right ... it seems so because \(\displaystyle \pi \) is essentially the set of all cosets of \(\displaystyle S/N\)]

It's almost right. In the definition of $\mathcal{B}$, the $ A/N$ should be $A$.

\(\displaystyle \Longrightarrow x - y \in N \) (Question 2, is that correct?)

Yes.

\(\displaystyle \Longrightarrow x \in y + N\) (Question 3, is that correct?)

Yes.

But then ... \(\displaystyle y + N \subset T \) (Question 4, is that correct?)

Yes.

A similar (symmetrical) argument, I think, would show \(\displaystyle x \in T \Longrightarrow x \in S\) ...

Yes, that's right.

I am still working on the surjectivity of \(\displaystyle \phi\) ...

Well, if you're still stuck, let me help you. Given $B\in\mathcal{B}$, $\pi^{-1}(B)$ is an element of $A$ that maps to $B$ under $\phi$. This proves surjectivity of $\phi$. Now can you show that in fact $\phi(\pi^{-1}(B)) = B$?

If anyone can demonstrate or point to the existence of a better proof - online or in a text - I would be really interested ...

It suffices to show that $\phi$ has an inverse. Define $\theta : \mathcal{B} \to \mathcal{A}$ such that $\theta(B) = \pi^{-1}(B)$. Then

$\displaystyle \theta\phi(A) = \theta(A/N) = A$

and

$\displaystyle\phi\theta(B) = \phi(\pi^{-1}(B)) = B$

for all $A\in \mathcal{A}$ and $B\in\mathcal{B}$. Thus, $\theta$ is the inverse of $\phi$.

 

[Math Amateur]

It's almost right. In the definition of $\mathcal{B}$, the $ A/N$ should be $A$.
Yes.
Yes.
Yes.
Yes, that's right.
Well, if you're still stuck, let me help you. Given $B\in\mathcal{B}$, $\pi^{-1}(B)$ is an element of $A$ that maps to $B$ under $\phi$. This proves surjectivity of $\phi$. Now can you show that in fact $\phi(\pi^{-1}(B)) = B$?
It suffices to show that $\phi$ has an inverse. Define $\theta : \mathcal{B} \to \mathcal{A}$ such that $\theta(B) = \pi^{-1}(B)$. Then

$\displaystyle \theta\phi(A) = \theta(A/N) = A$

and

$\displaystyle\phi\theta(B) = \phi(\pi^{-1}(B)) = B$

for all $A\in \mathcal{A}$ and $B\in\mathcal{B}$. Thus, $\theta$ is the inverse of $\phi$.

Thanks so much for your help Euge ...

Peter