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For $A=k[\epsilon ]/({\epsilon }^2)$ and $k=A/(\epsilon )$, the kernel and image of multiplication by $\epsilon$ both equal $(\epsilon )$. Thus the displayed sequence is a free resolution. Applying ${\mathrm{Hom}}_A(-,k)$ makes every differential zero, so ${\mathrm{Ext}}_A^i(k,k)\cong k$ for every $i\ge 0$. The module $k$ therefore has infinite projective dimension.

A finitely generated module over a Noetherian ring admits a possibly infinite free resolution whose every term is finitely generated. Surject a finite free module onto it; the kernel is finitely generated because a finite free module is Noetherian. Repeat this construction on the successive kernels. If the original module has finite projective dimension $n$, its $n$th syzygy is projective by dimension shifting of the Ext functor, giving a length-$n$ resolution with finite projective terms.

Given a free resolution $\cdots \to F_1\to F_0\to M\to 0$, its first syzygy module is $\ker (F_0\to M)$, the module of relations among the chosen generators of $M$. Higher syzygy modules are successive kernels. They depend on the choice of resolution; a minimal graded free resolution gives canonical ranks and degrees of generators over a standard graded ring structure on a polynomial ring.

The Tor functor is the homology of a tensor product of modules with a free resolution: ${\mathrm{Tor}}_i^R(M,N)=H_i(F_{\bullet }{\otimes }_{R}N)$ for a free resolution $F_{\bullet }$ of $M$. It measures the failure of tensoring to preserve exact sequences. For a commutative ring, one may resolve either argument; the double complex of the two resolutions proves that both computations agree.

Put $A=k[[t]]$ and $K=A[1/t]$. A free resolution has countably many generators $e_i$ and relations $e_i-t{e}_{i+1}$. Applying the Hom functor with coefficients in $A$ gives $(b_i)\mapsto (b_i-t{b}_{i+1})$ on sequence products. This is surjective: for prescribed $(c_i)$ use $b_i={\sum }_{j\ge 0}{t}^j{c}_{i+j}$, a well-defined formal power series. Therefore the degree-one Ext functor vanishes. But ${\mathrm{Hom}}_A(K,A)=0$, 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.