koszul euler characteristic redundant zero generator

In the ordered bases $e_1,e_2$ and $e_1\wedge e_2$, the complex is $0\to R\stackrel{d_2}{\to }{R}^2\stackrel{d_1}{\to }R\to 0$ with $d_1(a,b)=ta$ and $d_2(c)=(0,tc)$, because deletion gives $t{e}_2-0e_1$. Their composite is zero. Since $t$ is nonzero in a domain, $\ker d_2=0$ and $\ker d_1=0\oplus R$. Consequently $H_2=0$, $H_1=(0\oplus R)\mathrm{/}(0\oplus tR)\cong k$ and $H_0=R\mathrm{/}tR=k$.

The length formula gives ${\mathrm{\ell }}_R(k)=1$. Hence both nonzero homology modules have length one, and $\chi (K)=1-1+0=0$. The homology has total length $2$ by additivity, so its alternating cancellation is not acyclicity.

For $n\ge 0$ we have ${\mathrm{\ell }}_R(R\mathrm{/}{I}^{n+1})=n+1$, so $P(T)=T+1$. Thus $e_2=2![T^2]P=0$ and $e_1=1![T]P=1$. The DVR is Noetherian local, $R$ is finite over itself and $R\mathrm{/}I=k$ has finite length. The bridge theorem applies under AC with the actual sequence length $r=2$, agreeing with $\chi =0$.

To identify the dimension without a general Hilbert–Samuel dimension theorem, let $\mathfrak{p}$ be a nonzero prime ideal. It has the form $(t^a)$ with $a\ge 1$ because it is proper. Since $t^a\in \mathfrak{p}$, repeated primality gives $t\in \mathfrak{p}$. Thus $\mathfrak{p}=(t)$, as $(t)$ is maximal. Also $(0)$ is prime because $R$ is a domain, and $t\mathrm{\ne }0$ makes $(0)⊊(t)$. These are all primes, so the largest number of strict inclusions in a prime chain is one. Therefore the dimension is one, and the coefficient at that dimension is the $e_1=1$ already calculated.

The first-element identity provides another explicit check: removing $t$ gives $C=k$ and $T=0$ since multiplication by $t$ on $R$ is injective. The remaining sequence is $(0)$, whose complex on $k$ is $0\to k\stackrel{0}{\to }k\to 0$. It has one copy of $k$ in each homology degree, so $\chi (K(0;k))=1-1=0$, whereas $K(0;0)$ is zero. The reduction formula is therefore $0-0=0$, with $C\mathrm{/}0C=k$ finite length. This confirms that the redundant generator changes the coefficient index, not the generated ideal.