Understanding the isomorphisms and direct sums in Structure Theorem proof

Question

Background:

SNF Consequence: Given $M$, a submodule of $R^n$, there exists elements $d_1,...,d_r$ and $f_1,...,f_n$ where $f_i$ forms a basis for $R^n$ such that $d_1f_1,..,d_rf_r$ forms a basis for $M$ and $d_1|d_2|...|d_r.$

Structure Theorem: Let M be a finitely generated $R$- module then $\exists d_1,...,d_k$ such that $M\cong R^r\bigoplus\frac{R}{\langle d_1\rangle}\bigoplus...\bigoplus\frac{R}{\langle d_k\rangle}$ and $d_1|d_2|...|d_k$.

Proof: Suppose $M$ is generated by $\{x_1,...,x_n\}$ then there exists a surjective module homorphism such that $\phi:R^n\to M$. By First Isomorphism Theorem, we have $\frac{R^n}{\ker\phi}\cong M$. Now since $\ker\phi$ is a submodule of $M$ we can apply SNF Conseq. from above, $$M\cong\frac{\langle f_1,..,f_n\rangle}{\langle d_1f_1,...d_kf_k\rangle}\cong\frac{\langle f_1\rangle}{\langle d_1f_1\rangle}\bigoplus...\bigoplus \frac{\langle f_k\rangle}{\langle d_kf_k\rangle}\bigoplus\langle f_{k+1}\rangle\bigoplus...\bigoplus\langle f_n\rangle.$$Now $\frac{\langle f_i\rangle}{\langle d_if_i\rangle}\cong\frac{R}{\langle d_i\rangle}$ and ${\langle f_{k+1}\rangle}\cong R$ and similarly for other summands, we have the required result.

My Problem with this proof:

I wanted to know more details that go into this $$M\cong\frac{\langle f_1,..,f_n\rangle}{\langle d_1f_1,...d_kf_k\rangle}\cong\frac{\langle f_1\rangle}{\langle d_1f_1\rangle}\bigoplus...\bigoplus \frac{\langle f_k\rangle}{\langle d_kf_k\rangle}\bigoplus\langle f_{k+1}\rangle\bigoplus...\bigoplus\langle f_n\rangle.$$

Does this stand because $$M\cong\frac{\langle f_1,..,f_n\rangle}{\langle d_1f_1,...d_kf_k\rangle}\cong \frac{\langle f_1\rangle}{\langle d_1f_1,..d_kf_k\rangle}\bigoplus...\bigoplus \frac{\langle f_n\rangle}{\langle d_1f_1,...,d_kf_k\rangle}.$$ But how would this lead to the aforementioned result? (I have a feeling this will have something to do with the basis but I cannot seem to explain rigorously in my head)

Moreover, I am not convinced that ${\langle f_{k+1}\rangle}\cong R$ is necessarily true, why would this have to stand? (i.e., I am not convinced that if $n$ elements generate $R^n$ then each element must generate $R$).

Many thanks in advance and I really would appreciate any help!

Answer 1

Does this stand because $M≅\frac{⟨f_1,..,f_n⟩}{⟨d_1f_1,...,d_kf_k⟩}≅\frac{⟨f_1⟩}{⟨d_1f_1,...,d_kf_k⟩}⨁...⨁\frac{⟨f_n⟩}{⟨d_1f_1,...,d_kf_k⟩}$?

This is not quite correct as written, because quotients on the RHS cannot be taken as the denominators are not submodules of the numerators.

If you want an explicit isomorphism $$\frac{\langle f_1,...,f_n\rangle}{\langle d_1f_1,...,d_kf_k\rangle}\cong\frac{\langle f_1\rangle}{\langle d_1f_1\rangle}\bigoplus...\bigoplus \frac{\langle f_k\rangle}{\langle d_kf_k\rangle}\bigoplus\langle f_{k+1}\rangle\bigoplus...\bigoplus\langle f_n\rangle,$$ you should map $\lambda_1 f_1 + \cdots \lambda_n f_n \mapsto (\lambda_1 f_1, \cdots, \lambda_n f_n)$. This is well defined, because the $f_i$ form a basis of $M$.

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Now $\frac{⟨f_i⟩}{⟨d_if_i⟩}≅R⟨d_i⟩$ and $⟨d_{k+1}⟩≅R$ and similarly for other summands, we have the required result.

I think you wanted to say $⟨f_{k+1}⟩≅R$. This is true since the elements of $⟨f_{k+1}⟩$ are just scalar multiples of $f_{k+1}$.