A polynomial PID has a nonzero Kunneth Tor class

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.