MHB Help Needed: Analzying Berrick & Keating's Prop. 3.1.2 on Noetherian Rings

[Math Amateur]

I am reading the book "An Introduction to Rings and Modules with K-theory in View" by A.J. Berrick and M.E. Keating ... ...

I am currently focused on Chapter 3; Noetherian Rings and Polynomial Rings.

I need help with another aspect of the proof of Proposition 3.1.2.

The statement and proof of Proposition 3.1.2 reads as follows (pages 109-110):View attachment 4841
https://www.physicsforums.com/attachments/4842In the above passage from Berrick and Keating we read the following:

" ... ... A submodule $$N$$ of $$M''$$ is the homomorphic image of its inverse image

$$\beta^{-1} N = \{ m \in M \ | \ \beta m \in M \}$$

in $$M$$.

Since $$\beta^{-1} N$$ is finitely generated, so also is $$N$$. ... ... "


Now I am not sure I completely understand why the above statements are true ... so I will set down my thoughts ... and request that someone critique my analysis and let me know if it is correct and/or let me know the errors/shortcomings in it ...... ... ... ...

... now the homomorphic image of $$\beta^{-1} N$$ is $$N$$ since $$\beta$$ is surjective ... that is, we have $$\beta ( \beta^{-1} N ) = N$$ ... not quite sure why we need this result, but it is there anyway ... ...

Further B&K seem to assume that $$\beta^{-1} N$$ is a submodule of $$M$$ ... and then assume that the homomorphism $$\beta$$ between the finitely generated submodule $$\beta^{-1} N$$ of the Noetherian module $$M$$ and $$N$$ means that $$N$$ is finitely generated ... so that then $$M''$$ is Noetherian ...

So to show these assumptions are true, we first show that $$\beta^{-1} N$$ is a submodule of $$M$$ ... We need to show that

(i) $$0_M \in \beta^{-1} N$$
(ii) $$m_1, m_2 \in \beta^{-1} N \Longrightarrow (m_1 + m_2) \in \beta^{-1} N$$
(iii) $$m \in \beta^{-1} N \text{ and } r \in R \Longrightarrow mr \in \beta^{-1} N$$

where $$M, M', M''$$ are modules over the ring $$R$$.

-----------------------------------------------------------------------------------

(i) Show $$0_M \in \beta^{-1} N
$$$$\beta 0_M = 0_{M''}$$

$$\Longrightarrow \beta^{-1} 0_{M''} = \{ 0_M \text{ and possibly other elements of } \beta^{-1} N \}$$ ... ... since $$0_{M''} \in N$$

$$\Longrightarrow 0_M \in \beta^{-1} N$$

-------------------------------------------------------------------------------------(ii) Show $$m_1, m_2 \in \beta^{-1} N \Longrightarrow (m_1 + m_2) \in \beta^{-1} N$$$$m_1, m_2 \in \beta^{-1} N \Longrightarrow$$ there is (at least) elements $$n_1, n_2 \in N$$ such that $$\beta m_1 = n_1$$ and $$\beta m_2 = n_2$$

But ... $$N$$ is a submodule of $$M''$$ so $$n_1 + n_2 = \beta m_1 + \beta m_2 \in N$$

$$\Longrightarrow \beta ( m_1 + m_2 ) \in N $$

$$\Longrightarrow ( m_1 + m_2 ) \in \beta^{-1} N$$

---------------------------------------------------------------------------------------

(iii) Show $$m \in \beta^{-1} N$$ and $$r \in R \Longrightarrow mr \in \beta^{-1} N$$$$m \in \beta^{-1} N \Longrightarrow \exists \ $$ at least one $$n$$ such that $$\beta m = n $$

But $$N$$ is a submodule of $$M''$$ ... ... so $$nr \in N$$

$$\Longrightarrow ( \beta m ) r \in N $$

$$\Longrightarrow ( \beta ) m r \in N$$ since $$\beta$$ is a homomorphism ...

$$\Longrightarrow mr \in \beta^{-1} N$$

====================================================

Thus we have shown that $$\beta^{-1} N$$ is a submodule of $$M$$

So $$\beta^{-1} N$$ is finitely generated

====================================================

Now we have to show that ...

$$\beta^{-1} N$$ is finitely generated $$\Longrightarrow N$$ is finitely generated

Proceed as follows ... ...

----------------------------------------------------------------------------------------------

Assume $$\beta^{-1} N$$ is finitely generated by elements $$m_1, m_2, \ ... \ ... \ , m_k$$ ... ...

Consider an arbitrary element $$y \in N$$ ...

Then $$\exists$$ some element (at least one) $$x \in \beta^{-1} N$$ such that $$\beta x = y$$ ... ...

Now we have $$x = m_1 r_1 + m_2 r_2 + \ ... \ ... \ + m_k r_k$$ ... ... for some $$r_1, r_2, \ ... \ ... \ , r_k \in R$$

But ... ... since $$m_1, m_2, \ ... \ ... \ , m_k \in \beta^{-1} N$$ ... ...

... ... $$ \exists \ a_1, a_2, \ ... \ ... \ , a_k \in N$$ such that $$\beta m_1 = a_1, \beta m_2 = a_2, \ ... \ ... \ , \beta m_k = a_k$$ ... ...

Now we have that

$$\beta x = y = \beta ( m_1 r_1 + m_2 r_2 + \ ... \ ... \ + m_k r_k ) $$

$$= \beta ( m_1 r_1) + \beta (m_2 r_2 ) + \ ... \ ... \ + \beta ( m_k r_k ) $$ since $$\beta$$ is a homomorphism

$$= \beta ( m_1) r_1 + \beta (m_2) r_2 + \ ... \ ... \ + \beta ( m_k ) r_k
$$

$$= a_1 r_1 + a_2 r_2 + \ ... \ ... \ + a_k r_k$$ Thus we have shown that that an arbitrary element $$y \in N$$ is generated by $$a_1, a_2, \ ... \ ... \ , a_k \in N$$ ... that is an arbitrary submodule $$N \text{ of } M''$$ is finitely generated ... so that $$M''$$ is Noetherian ...

=====================================================Could someone please critique my analysis pointing out any errors or shortcomings ... ...

Hope someone can help ... ...

Peter

 

[Deveno]

That looks correct.

A simpler way to "package" this in your mind is the following:

Let $f: M \to N$ be any $R$-module homomorphism. Then:

1) If $M'$ is a submodule of $M$, $f(M')$ is a submodule of $N$ (in particular, $f(M)$ is a submodule of $N$).

2) If $N'$ is a submodule of $f(M)$, then $f^{-1}(N')$ is a submodule of $M$ containing $K = \text{ker }f = f^{-1}(0_N)$.

These two facts allow us to establish the analogue of the correspondence theorem for rings. In fact, if $R$ is a commutative ring, a submodule of $R$ considered as an $R$-module over itself, is just an ideal of $R$.

 

[Math Amateur]

That looks correct.

A simpler way to "package" this in your mind is the following:

Let $f: M \to N$ be any $R$-module homomorphism. Then:

1) If $M'$ is a submodule of $M$, $f(M')$ is a submodule of $N$ (in particular, $f(M)$ is a submodule of $N$).

2) If $N'$ is a submodule of $f(M)$, then $f^{-1}(N')$ is a submodule of $M$ containing $K = \text{ker }f = f^{-1}(0_N)$.

These two facts allow us to establish the analogue of the correspondence theorem for rings. In fact, if $R$ is a commutative ring, a submodule of $R$ considered as an $R$-module over itself, is just an ideal of $R$.

Thanks Deveno ... that is most helpful ...

Peter