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 \(\displaystyle N\) of \(\displaystyle M''\) is the homomorphic image of its inverse image

\(\displaystyle \beta^{-1} N = \{ m \in M \ | \ \beta m \in M \}\)

in \(\displaystyle M\).

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


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

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

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

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

where \(\displaystyle M, M', M''\) are modules over the ring \(\displaystyle R\).

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

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

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

\(\displaystyle \Longrightarrow 0_M \in \beta^{-1} N\)

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

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

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

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

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

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

But \(\displaystyle N\) is a submodule of \(\displaystyle M''\) ... ... so \(\displaystyle nr \in N\)

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

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

\(\displaystyle \Longrightarrow mr \in \beta^{-1} N\)

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

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

So \(\displaystyle \beta^{-1} N\) is finitely generated

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

Now we have to show that ...

\(\displaystyle \beta^{-1} N\) is finitely generated \(\displaystyle \Longrightarrow N\) is finitely generated

Proceed as follows ... ...

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

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

Consider an arbitrary element \(\displaystyle y \in N\) ...

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

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

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

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

Now we have that

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

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

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

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