A finitely localized polynomial ring in positive dimension is not a field

Choose an integer $N>1$ larger than every exponent occurring in $s$. Iterating [L1] defines $\phi :k[t_1,\dots ,t_r]\to k[u]$ by $t_i\mapsto u^{N^{i-1}}$. Distinct exponent tuples with entries below $N$ give distinct sums ${\sum }_i{e}_i{N}^{i-1}$: reduction modulo $N$ identifies the first entry, and subtraction and division by $N$ repeats the argument. Thus distinct monomials of $s$ retain distinct degrees, and $g:=\phi (s)\mathrm{\ne }0$. Also $\phi (t_1)=u$.

In $k[u][1\mathrm{/}g]$ the image of $s$ is the unit $g$, so [L2] gives a unital homomorphism $\mathrm{\Phi }:k[t_1,\dots ,t_r][1\mathrm{/}s]\to k[u][1\mathrm{/}g]$. It is surjective: $h(u)\mathrm{/}{g}^m$ is the image of $h(t_1)\mathrm{/}{s}^m$. The target is nonzero by [L3]. If the source were a field, every nonzero target element would be the image of a nonzero element and its inverse would be the image of the source inverse. Hence the target would be a field.

The target is not a field. If $g$ is constant, it is nonzero, so the target is $k[u]$; the polynomial $u$ has no inverse since a nonzero product $uh$ has positive degree. If $g$ is nonconstant, then $g+1$ is a nonzero nonconstant polynomial and remains nonzero after localization. An inverse $h\mathrm{/}{g}^m$ would imply $g^m=h(g+1)$ in $k[u]$ by [L3]. But $g^m-(-1{)}^m$ is divisible by $g+1$ (expand $(g+1-1{)}^m$), so this equality would make $g+1$ divide the nonzero constant $(-1{)}^m$, impossible by degrees. This also covers $m=0$.

By step 2.1 a field source would give a field target, which step 3.1 rules out. Therefore $k[t_1,\dots ,t_r][1\mathrm{/}s]$ is not a field.