koszul euler characteristic annihilator correction

The complex is $0\to R\oplus k\stackrel{t}{\to }R\oplus k\to 0$, in degrees $1,0$, and the map is $(a,b)\mapsto (ta,0)$. Since a DVR is a domain and $t\mathrm{\ne }0$, its kernel is $0\oplus k$ and its image is $tR\oplus 0$. Thus $H_1\cong k$, $H_0\cong (R\mathrm{/}tR)\oplus k=k\oplus k$, and all remaining homology is zero.

The DVR length formula at $a=1$ gives ${\mathrm{\ell }}_R(k)=1$, and the split sequence $0\to k\to k\oplus k\to k\to 0$ gives length $2$. Therefore $M\mathrm{/}tM$ has finite length and $\chi (K)=2-1=1$.

For every $n\ge 0$, $t^{n+1}M=t^{n+1}R\oplus 0$. Hence $M\mathrm{/}{t}^{n+1}M\cong R\mathrm{/}(t^{n+1})\oplus k$ has length $(n+1)+1=n+2$. Thus $P(T)=T+2$ and $e_1=1![T]P=1$. The DVR is Noetherian local, $M$ is finite, and the finite-colength hypothesis was verified above, so the bridge theorem under AC gives the same value $1$.

Removing $t$ leaves the empty sequence with coefficients $C=M\mathrm{/}tM=k\oplus k$ and $T=0{:}_{M}t=0\oplus k$. For an empty sequence its Euler characteristic is the coefficient module's length. Thus the first-element identity reads $\chi (K(t;M))={\mathrm{\ell }}_R(C)-{\mathrm{\ell }}_R(T)=2-1=1$. Both lengths are finite, so all its hypotheses hold. The nonzero annihilator term is exactly the discrepancy with ${\mathrm{\ell }}_R(C)=2$.