Koszul Euler Characteristics and Hilbert–Samuel Multiplicity — Examples

The empty Koszul complex has $K_0=M$, all other terms zero and all differentials zero. Thus $H_0=M$ and $H_i=0$ for $i\mathrm{\ne }0$, giving $\chi (K)={\mathrm{\ell }}_R(M)$.

For every $n\ge 0$, $0^{n+1}M=0$, so $M\mathrm{/}{0}^{n+1}M=M$ and $P(T)={\mathrm{\ell }}_R(M)$. Hence $e_0=0![T^0]P={\mathrm{\ell }}_R(M)$. A composition series also makes $M$ finitely generated: take a lift of one nonzero generator of each simple factor; induction through the finite series shows these finitely many lifts generate $M$. The module-relative hypothesis is exactly finite length of $M$, so the bridge theorem with $r=0$ applies under AC and agrees with this direct calculation.

In particular take $M=k=R\mathrm{/}\mathfrak{m}$. Its only submodules are zero and $k$, since any nonzero vector spans this one-dimensional $k$-space; its length is $1$. We obtain $H_0=k$, $P=1$ and $\chi =e_0=1$. With $M=0$ the empty series has length zero and the same calculation gives $P=\chi =e_0=0$.

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 ${\mathrm{\ell }}_R(C)=2$.

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.