I'm struggling to understand this step in this proof in Rotman's advanced modern algebra. The related theorems are:
Let R be a euclidean ring, let F be a finitely generated free R-module, and let S be a submodule of F. Then there exists a basis $z_1, . . . , z_n$ of F and nonzero $σ_1, . . . , σ_q$ in R, where 0 ≤ q ≤ n, such that $σ_1 | · · · | σ_q$ and $σ_1z_1, . . . , σ_qz_q$ is a basis of S.
Let R be a euclidean ring, let Γ be the n × t presentation matrix associated to an R-map $λ: R^t → R^n$ relative to some choice of bases, and let M = cokerλ. If Γ is R-equivalent to a Smith normal form diag$(σ_1, . . . , σ_q)$ ⊕ 0, then those $σ_1, . . . , σ_q$ that are not units are the invariant factors of M.
The proof of the second theorem using the first is:
If diag$(σ_1, . . . , σ_q)$ ⊕ 0 is a Smith normal form of Γ, then there are bases $y_1, . . . , y_t$ of $R^t$ and $z_1, . . . , z_n$ of $R^n$ with $λ(y_1) = σ_1z_1, . . . , λ(y_q) = σ_qz_q$ and $λ(y_j)$ = 0 for all $y_j$ with j > q, if any. Now R/(0) $\cong$ R and R/(u) = {0} if u is a unit. If $σ_s$ is the first $σ_i$ that is not a unit, then M $\cong$ $R^{n−q}$ ⊕ $R/(σ_s)$ ⊕· · ·⊕ $R/(σ_q)$ a direct sum of cyclic modules for which $σ_s | · · · | σ_q$.
From the first line of the proof I gather that $σ_1z_1, . . . , σ_qz_q$ is a basis of the free submodule im(λ) so then M $\cong R^t/< σ_1z_1, . . . , σ_qz_q >$ ? What I don’t understand is how the bottom line (if this is correct) decomposes into $(σ_s)$ ⊕· · ·⊕ $(σ_q)$ i.e. where does z go and why is it not $(σ_sz_1)$ ⊕· · ·⊕ $(σ_qz_q)$ (or are these equivalent).
Any help would be appreciated.
Where does "$z$" go? $R/(\sigma z)$ doesn't mean anything as $z\notin R$. You can think of these quotients as coming from $Rz/(\sigma z)$, where $Rz$ is the span of $z$ in $R^n$, and these are isomorphic to $R/(\sigma)$. This is basically what is happening.
Formally, consider the map $R^n\to R^{n-q}\oplus\bigoplus_{j=s}^qR/(\sigma_j)$ as defined by $\sum_ir_iz_i\mapsto(\sum_{i>q}r_iz_i,\overline{r_s},\overline{r_{s+1}},\cdots,\overline{r_q})$. This is obviously a map of $R$-modules and is well defined since the $(z_\bullet)$ are a basis. It clearly also is surjective. Therefore, the only thing we need to do is prove the kernel of this map is $\mathrm{im}(\lambda)$.
By linear independence, $\sum_{i>q}r_iz_i$ is zero iff. $r_i=0$ for all $i>q$. The other terms are zero iff. $r_j=x_j\cdot\sigma_j$ for some $x_j\in R$, for all $s\le j\le q$. $r_i$ for $i<s$ can be absolutely anything. Therefore, the kernel is the span of $\{z_1,z_2,\cdots,z_{s-1},\sigma_sz_s,\cdots,\sigma_qz_q\}$. As $\sigma_i$ is a unit for $i<s$, this is equal to the span of $\{\sigma_1z_1,\cdots,\sigma_{s-1}z_{s-1},\sigma_sz_s,\cdots,\sigma_qz_q\}$. Oh, but that's exactly the image of $\lambda$!
Therefore $M\cong R^{n-q}\oplus\bigoplus_{j=s}^qR/(\sigma_j)$ and all terms of this direct sum, except possibly $R^{n-q}$, are nonzero modules.