Local conditions are quotient packets

Proof of Theorem 3.1

Write \(S=\bigotimes_{i=1}^{s}R_i\) for the ordinary tensor product. Its ideal of definition is

The scalar term retains the base adic topology when the family is empty. The ordinary quotient theorem gives a canonical algebra isomorphism

carrying the image of \({\mathfrak a}\) to the defining ideal on the right. The quotient topology on \(S/L\) is precisely this image ideal topology.

Here completion means the actual Hausdorff completion for the additive uniformity of the indicated adic topology. For every adic ring \(S\) and ideal \(L\), there is a canonical bicontinuous ring isomorphism

Indeed, the completed quotient map kills the closed extended ideal. In the other direction, the quotient map from \(S/L\) to \(\widehat S/\overline{L\widehat S}\) extends to its completion. The latter quotient is complete: the adic uniformity has a countable base, and the quotient of a complete countably based additive group is complete. Its closed defining subgroup makes it Hausdorff. The two composites agree with the identity on dense quotient representatives, so continuity proves both inverse laws.

Applying this comparison to the displayed ordinary tensor isomorphism proves the claim. The closed extended ideal is exactly \(J\), since extension commutes with the finite sum of the factor ideals. The maps preserve coefficients and each canonical factor map. The common residue field identifies \(S/{\mathfrak a}\) with \(k\); finite generation of \({\mathfrak a}\), supplied by the Noetherian factor hypotheses, identifies the completed topology with its maximal ideal topology. Proper quotients retain locality. If a condition is the unit ideal, the ring comparison remains valid with the zero ring, without treating it as a local ring with residue field \(k\).

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Proof of Corollary 3.2

Let \(f:R^{\mathrm{loc}}\to A\) be the point. Since \(A\) is Hausdorff and \(f\) is continuous, \(\ker f\) is closed. Thus it contains the closed ideal \(J\) if and only if it contains the algebraic ideal generated by the images of the \(I_i\). This is equivalent to every local restriction killing \(I_i\). The universal property of the actual ideal quotient gives the factors, and the quotient topology gives their continuity. Surjectivity of the quotient map gives uniqueness. If \(f\) reflects units, lift an element of the quotient to the original ring to see that its factor also reflects units.

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Proof of Proposition 3.3

Put \(A=R^{\mathrm{loc}}\), \(B=R^\square\), and let \(J\subseteq A\) be the closed local condition ideal. By Theorem 3.1, the independently constructed ring \(R^{\mathrm{loc},\tau}\) is canonically \(A/J\). The ordinary tensor identity

carries the defining ideal of the tensor to the image of \({\mathfrak m}_B\). Here the structural map is local by the stated global deformation datum, so it carries \({\mathfrak m}_A\) into \({\mathfrak m}_B\). Completing, and applying the completion and quotient comparison from the preceding proof, gives a bicontinuous isomorphism

The construction commutes with the canonical map from \(B\), which is therefore surjective. It is also unchanged by transporting the tensor factor from \(A/J\) to the actual completed tensor of the local quotients. The transported ideal is exactly its natural defining ideal.

For a continuous algebra point \(x:B\to C\) with complete Hausdorff target, the kernel of \(x\) is closed. Consequently \(x\) has a unique continuous factor through \(R^\tau\) if and only if it kills \(JB\), equivalently if its restriction to \(A\) kills \(J\). By Corollary 3.2, this is exactly the requirement that all local restrictions satisfy their quotient conditions. The factor is an algebra map because its composite with the canonical global map is \(x\). If \(x\) is local, surjectivity of that map shows that its factor reflects units and is local. Uniqueness also proves naturality under coefficient preserving maps of targets. This identifies the represented base change; it does not supply representability of a new arithmetic deformation problem.

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Proof of Proposition 3.4

Let \(e^2=e\). Since \(e+(1-e)=1\), at least one of \(e\) and \(1-e\) is a unit. If \(e\) is a unit, then \(e=1\). If \(1-e\) is a unit, then \(e=0\).

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