Kunneth Exactness and Splittings over Principal Ideal Domains — Examples

Since $k$ is a field, it is a domain. By [F1], $k[t]$ is a domain, $t\mathrm{\ne }0$, and $t$ is not a unit: $tf=1$ for nonzero $f$ would give $1+\mathrm{deg}f=0$. By [F2], degree on the nonzero polynomials satisfies the Euclidean division condition (the convention requires no multiplicative monotonicity). Thus [F3] applies and $R$ is a commutative PID. Multiplication by $t$ on $R$ is injective, so $H_{0}C=H_{0}D=R\mathrm{/}(t)$ and all other homology groups of the two input complexes vanish.

Write $u=e_1\otimes f_0$, $v=e_0\otimes f_1$, $w=e_0\otimes f_0$, and $h=e_1\otimes f_1$. The tensor total complex has terms $Rh$, $Ru\oplus Rv$, $Rw$ in degrees $2,1,0$. Its Koszul formula gives $d_1(u)=tw$, $d_1(v)=tw$, and $d_2(h)=tv-tu$. Thus in the ordered basis $(u,v)$, $d_1(a,b)=t(a+b)$ and $d_2(c)=(-tc,tc)$; $d_1{d}_2(c)=t(-tc+tc)=0$.

The image of $d_1$ is $tRw$, so $H_0=Rw\mathrm{/}tRw\cong R\mathrm{/}(t)$. Since $t$ is not a zero divisor, $d_1(a,b)=0$ exactly when $a+b=0$. Hence $\ker d_1=R(u-v)$ and $\mathrm{im}{d}_2=tR(u-v)$, giving $H_1\cong R\mathrm{/}(t)$ via $[a]\mapsto [a(u-v)]$. Also $d_2(c)=0$ forces $tc=0$ and hence $c=0$, so $H_2=0$. There are no chain terms in the other degrees.

First derive the DC hypothesis of [F6] from the assumed AC. Given an entire relation $E$ on a nonempty set $S$ and $a\in S$, AC chooses a successor function $f:S\to S$ with $xEf(x)$ for every $x$. By [F8], iterate $f$ from $a$; the resulting sequence satisfies [F9]. The length-one free resolution $R\stackrel{t}{\to }R\to R\mathrm{/}(t)$ of the first module, tensored with $R\mathrm{/}(t)$, has differential zero and no degree-two term. Thus [F6] gives ${\mathrm{Tor}}_1^R(R\mathrm{/}(t),R\mathrm{/}(t))=R\mathrm{/}(t)$, represented by $1\otimes [1]$. In the cycle-boundary presentation $B_{0}C=tR{e}_0\hookrightarrow Z_{0}C=R{e}_0$, identify its degree-one free module $R$ by $a\mapsto at{e}_0$. This is an isomorphism because $t$ is nonzero in a domain.

The map ${\rho }_1:C_1\to B_{0}C$ sends $e_1$ to $t{e}_0$, while ${\rho }_0$ is zero. Therefore $\rho \otimes 1$ sends the degree-one cycle $u-v$ to $t{e}_0\otimes f_0$, whose homology image is $t{e}_0\otimes [f_0]$. Under the resolution identification of step 2.2, this is exactly $1\otimes [1]$, the positive generator. Since $1\mathrm{\notin }(t)$, the image and the original class are nonzero. This calculation works also in characteristic two, where minus equals plus but $a+b=0$ still means $b=-a$.

In degree one the tensor term of [F4] is zero because only degree-zero input homology is nonzero. Hence the quotient is an isomorphism, with section explicitly $[a]\mapsto [a(u-v)]$, well defined by step 2.1 and inverse by step 3.1. It is a section of the kind guaranteed by [F5]. In degree zero, multiplication gives $R\mathrm{/}(t){\otimes }_{R}R\mathrm{/}(t)\cong R\mathrm{/}(t)$ with inverse $[a]\mapsto [a]\otimes [1]$: balancing gives $[a]\otimes [b]=[ab]\otimes [1]$. The cross product carries this tensor generator to $[w]$. These computations verify the zero and one degree endpoints and all higher vanishing.

A field is a PID: a nonzero ideal contains $c\mathrm{\ne }0$, hence also $c^{-1}c=1$ and so is the whole field; the remaining ideal is $(0)$. The displayed complexes are nonnegative and have free terms with their stated bases, so [F1] applies. Each factor has zero differential, and the Koszul formula gives $d(a\otimes u)=d(b\otimes u)=d(a\otimes v)=d(b\otimes v)=0$. The only tensor terms are exactly those displayed in degrees $0,1,2$.

First derive the DC hypothesis of [F3] from the assumed AC. Given an entire relation $E$ on a nonempty set $S$ and $a\in S$, AC chooses for each $x\in S$ an element $f(x)$ of the nonempty successor set $E_x=\{y:xEy\}$. By [F4], iterate $f$ from $a$ to obtain $x_0=a$ and $x_{m+1}=f(x_m)$, so $x_{m}E{x}_{m+1}$ for every $m$; this is [F5]. Now for any $k$-modules $U,V$, [F2] makes both projective under AC. Supply the resolutions having $U$, respectively $V$, only in degree zero and identity augmentation. The tensor resolution then has no positive-degree terms, so its homology in every degree $i>0$ is zero. Hence [F3] gives ${\mathrm{Tor}}_1^k(U,V)=0$, including zero $U$ or $V$. In particular every summand in the Kunneth Tor term vanishes for any complexes under discussion.

These tensors are bases: for example $(\lambda b)\otimes (\mu u)\mapsto \lambda \mu$ and $\nu \mapsto \nu (b\otimes u)$ are inverse linear maps $kb\otimes ku\leftrightarrow k$, since balancing identifies $(\lambda b)\otimes (\mu u)$ with $(\lambda \mu )(b\otimes u)$. The same formulas apply to the other three one-dimensional factors. In degree one the two bidegrees form a direct sum. As all differentials vanish, every element is a cycle and the only boundary is zero; this proves the listed homology bases and vanishing elsewhere.

Exactness in [F1] now makes the cross product ${\alpha }_n$ injective and surjective for every $n$. In this instance it sends $[a]\otimes [u]$, $[b]\otimes [u]$, $[a]\otimes [v]$, $[b]\otimes [v]$ to the four homology basis elements in that order. Its inverse sends those four elements back to the four tensors of classes, extended linearly. Both composites fix a basis and therefore every element. Degree zero has an empty Tor sum; degree one has the single zero group ${\mathrm{Tor}}_1(k,k)$; in higher degrees the same vanishing argument applies, including where every term is zero.

For arbitrary maps of such complexes write $h$ for the induced map on tensor-complex homology and $k_{\ast }$ for the induced map on the tensor of homologies. Naturality in [F1] gives $h{\alpha }_n={\alpha }_n^{\mathrm{\prime }}{k}_{\ast }$. Multiplying by the inverses just proved gives ${\alpha }_n^{\mathrm{\prime }-1}h=k_{\ast }{\alpha }_n^{-1}$. Hence the inverse is natural as well. There is no selected complement in this inverse: it is the unique inverse of the canonical cross product after the Tor term vanishes.