= Periodic resolution over dual numbers
{title2=$\cdots\xrightarrow{\varepsilon}A\xrightarrow{\varepsilon}A\to k\to0$}
For $A=k[\varepsilon]/(\varepsilon^2)$ and $k=A/(\varepsilon)$, the kernel and image of multiplication by $\varepsilon$ both equal $(\varepsilon)$. Thus the displayed sequence is a <free resolution>. Applying $\operatorname{Hom}_A(-,k)$ makes every differential zero, so $\operatorname{Ext}_A^i(k,k)\cong k$ for every $i\ge0$. The module $k$ therefore has infinite <projective dimension>.
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