Example. A zero-radical pairing that is not perfect [fdip-001S]

✍️sourceAGENTDRAFTED

On the free rank-one \(\mathbb Z\)-module \(M=\mathbb Z\), the symmetric bilinear form \(b(x,y)=2xy\) separates vectors in both slots but is not a perfect pairing. Its currying map is \[\Phi :\mathbb Z\longrightarrow \operatorname {Hom}_{\mathbb Z} (\mathbb Z,\mathbb Z),\qquad \Phi (y)(x)=2xy.\]

If \(b(x,y)=0\) for every integer \(x\), testing against \(x=1\) gives \(2y=0\), hence \(y=0\). Symmetry gives the same conclusion in the other slot, so both radicals are zero and \(\Phi \) is injective. Every integer-linear functional \(\ell \) on \(\mathbb Z\) satisfies \(\ell (x)=x\ell (1)\), so it is determined by its value at \(1\). But \(\Phi (y)(1)=2y\) is even for every \(y\). Thus \(\Phi \) misses the linear functional \(x\mapsto x\), whose value at \(1\) is \(1\), and is not surjective. Freeness of \(M\) does not repair this failure of the chosen form.

The Hom convention is [stacks2017stacks, Section 10.10], Tag 0581; the calculation above proves the failure for this chosen form. For the perfect counting form and its explicit inverse, see Lemma [fdip-001F].