I am reading through Atiyah and Macdonald's Dimension theory, chapter, but I can't understand a step in the proof. The relevant definitions are included below. $\lambda$ is hte length function.
Why do we get $g(t)$ above? Where does this come from? If we have an exact sequence $$ 0 \rightarrow K[k_s] \rightarrow M[k_s] \rightarrow M \rightarrow L \rightarrow 0 $$ of graded modules, then shouldn't we get $$ t^{k_s} \sum_{i = 0}^\infty \lambda(N_i)t^i - t^{k_s} \sum_{i = 0}^\infty \lambda(M_i)t^i + \sum_{i = 0}^\infty \lambda(M_i)t^i - \sum_{i = 0}^\infty \lambda(L_i)t^i = 0 $$ I get no polynomial $g(t)$ as far as I can understand, since It seems like $M_i \cong L_i$ for $0 \leq i < k_s$ anyways.
Multiplying $t^{n+k_s}$, $t^{n+k_s}λ(K_n)-t^{n+k_s}λ(M_n)+t^{n+k_s}λ(M_{n+k_s})-t^{n+k_s}λ(L_{n+k_s})=t^{k_s}t^{n}λ(K_n)-t^{k_s}t^{n}λ(M_n)+t^{n+k_s}λ(M_{n+k_s})-t^{n+k_s}λ(L_{n+k_s})$.
Summing with respect to $n$, $t^{k_s}\sum_{n=0}^{+\infty} t^{n}λ(K_n)-t^{k_s}\sum_{n=0}^{+\infty} t^{n}λ(M_n)+\sum_{n=0}^{+\infty} t^{n+k_s}λ(M_{n+k_s})-\sum_{n=0}^{+\infty} t^{n+k_s}λ(L_{n+k_s})$
$=t^{k_s}\sum_{n=0}^{+\infty} t^{n}λ(K_n)-t^{k_s}\sum_{n=0}^{+\infty} t^{n}λ(M_n)+(\sum_{n=0}^{+\infty} t^{n+k_s}λ(M_{n+k_s})+\sum_{n=0}^{k_{s}-1} t^{n}λ(M_{n})-\sum_{n=0}^{k_{s}-1} t^{n}λ(M_{n}))-(\sum_{n=0}^{+\infty} t^{n+k_s}λ(L_{n+k_s})+\sum_{n=0}^{k_{s}-1} t^{n}λ(L_{n})-\sum_{n=0}^{k_{s}-1} t^{n}λ(L_{n}))$
$=t^{k_s}\sum_{n=0}^{+\infty} t^{n}λ(K_n)-t^{k_s}\sum_{n=0}^{+\infty} t^{n}λ(M_n)+(\sum_{n=0}^{+\infty} t^{n}λ(M_n)-\sum_{n=0}^{k_{s}-1} t^{n}λ(M_{n}))-(\sum_{n=0}^{+\infty} t^{n}λ(L_n)-\sum_{n=0}^{k_{s}-1} t^{n}λ(L_{n}))=0$
From the last equality, $\sum_{n=0}^{+\infty} t^{n}λ(M_n)-t^{k_s}\sum_{n=0}^{+\infty} t^{n}λ(M_n)=\sum_{n=0}^{+\infty} t^{n}λ(L_n)-t^{k_s}\sum_{n=0}^{+\infty} t^{n}λ(K_n)+(\sum_{n=0}^{k_{s}-1} t^{n}λ(M_{n})-\sum_{n=0}^{k_{s}-1} t^{n}λ(L_{n})$.
By taking $g(t)=\sum_{n=0}^{k_{s}-1} t^{n}λ(M_{n})-\sum_{n=0}^{k_{s}-1} t^{n}λ(L_{n})$,
$(1-t^{k_s})P(M,t)=P(L,t)-t^{k_s}P(K,t)+g(t)$.