Koszul Complexes and Regular Sequences — Examples

The one-element lemma gives the displayed complex. Multiplication by $u$ on $k[u,v]$ is injective by comparison of polynomial coefficients, so $H_1=0$ and $H_0=R\mathrm{/}uR\cong k[v]$; all other terms vanish.

For two elements the coordinate formula gives $d_2(c)=(-vc,uc)$ and $d_1(a,b)=ua+vb$, with $d_1{d}_2(c)=-uvc+vuc=0$.

If $ua+vb=0$, reduction modulo $u$ gives $v\bar{b}=0$ in $k[v]$. Multiplication by $v$ is injective, so $b=uc$ for some $c\in R$. Substitution gives $u(a+vc)=0$, hence $a=-vc$. Thus every degree-one cycle is $d_2(c)$, proving $H_1=0$.

If $d_2(c)=0$, then $uc=0$, so $c=0$ and $H_2=0$. Finally $H_0=R\mathrm{/}(u,v)\cong k$, and there are no terms outside degrees $0,1,2$. This proves both computations.

Successive quotients by the variables are polynomial rings, so each next variable is injective.

The regular-quotient resolution theorem gives a finite free resolution of $k$.

$x^2$ is regular in $k[x,y]$, and $y^3$ is regular modulo $(x^2)$.

The Koszul complex resolves the quotient with exterior ranks $1,2,1$, whose alternating sum is zero.

In $k[t]\mathrm{/}(t^2)$, multiplication by $t$ has kernel and image $(t)$.

The one-element formula gives $H_0\cong k$ and $H_1\cong (t)\cong k$.

Modulo $x$, the relation becomes $z=0$, leaving $k[y]$ where $(x-1)y$ becomes $-y$; also $x$ is a non-zero-divisor.

In reverse order $z\mathrm{\ne }0$ is killed by $x-1$, so the first regularity condition fails; the ring is not local.

After inverting $t$, the module $R\mathrm{/}(t)$ is zero, so every term of the complex localizes to zero.

Exact localization gives zero homology, agreeing with the localized Koszul complex.

The empty complex is $M$; for $0$ the two-term differential is zero, and for $1$ it is an isomorphism.

Thus $H_0=H_1=M$ in the zero case and the unit case is acyclic; the unit fails the proper-quotient regularity convention.

After one differential the three terms are $x_1{e}_2\wedge e_3-x_2{e}_1\wedge e_3+x_3{e}_1\wedge e_2$.

The second differential produces each $x_i{x}_j{e}_k$ twice with opposite signs, so the total is zero.

The annihilator of $s$ in $k[s,t]\mathrm{/}(st)$ is $(t)$.

Thus $H_0=R\mathrm{/}(s)$ and $H_1=(t)$; both are killed by $s$ and supported on $V(s)$.

The displayed upper-triangular matrix is invertible and expresses $(x+y,y)$ in terms of $(x,y)$.

Its exterior action intertwines differentials, and the inverse matrix supplies the inverse chain map.

$x,y$ are regular in the local polynomial ring because the successive quotients are domains.

Under the assumed AC, Positive Powers Of A Regular Sequence Remain Regular applied to $(x,y)$ gives $(x^2,y^3)$; Regular Sequences Permutable Local then swaps its two entries to give $(y^3,x^2)$. These two applications are the uses of Choice.

The squared variables are a regular sequence in the local polynomial ring and lie in its maximal ideal.

The minimal Koszul ranks are $\left(\genfrac{}{}{0px}{}{3}{0}\right),\left(\genfrac{}{}{0px}{}{3}{1}\right),\left(\genfrac{}{}{0px}{}{3}{2}\right),\left(\genfrac{}{}{0px}{}{3}{3}\right)=1,3,3,1$.