= Vanishing top Ext without finite generation
{title2=$\operatorname{pd}_{k[[t]]}k((t))=1,\quad\operatorname{Ext}_{k[[t]]}^1(k((t)),k[[t]])=0$}
Put $A=k[[t]]$ and $K=A[1/t]$. A <free resolution> has countably many generators $e_i$ and relations $e_i-te_{i+1}$. Applying the <Hom functor> with coefficients in $A$ gives $(b_i)\mapsto(b_i-tb_{i+1})$ on sequence products. This is surjective: for prescribed $(c_i)$ use $b_i=\sum_{j\ge0}t^jc_{i+j}$, a well-defined <formal power series>. Therefore the degree-one <Ext functor> vanishes. But $\operatorname{Hom}_A(K,A)=0$, since every image is infinitely divisible by $t$. A nonzero projective module has a nonzero coordinate map into $A$, so $K$ is not projective. Its displayed resolution proves that its <projective dimension> is exactly one. The finite-generation hypothesis in <top Ext detects finite projective dimension> is essential.
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