20. Separate support from multiplicity
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Let \(R=\mathbb Q[t]\). Compare the modules \(M_1=R/(t)\) and \(M_2=R/(t^2)\). This is an elementary model for the support distinctions used in Paper XII.
Write the presentations
Each module has one generator and one relation:
\[R\xrightarrow{\ t\ }R\longrightarrow M_1\longrightarrow0,\qquad
R\xrightarrow{\ t^2\ }R\longrightarrow M_2\longrightarrow0.
\]
The presentation matrices are the one-entry matrices \([t]\) and \([t^2]\). Their zeroth Fitting ideals are generated by those entries: \((t)\) and \((t^2)\). These are also the annihilator ideals in these two cyclic examples.
Compare the fibers
Specializing the parameter to a rational value \(t=\lambda\) replaces the presentation matrices by \([\lambda]\) and \([\lambda^2]\) over \(\mathbb Q\). When \(\lambda\ne0\), both matrices are invertible and both cokernels are zero. At \(\lambda=0\), both are zero maps and both cokernels have dimension one.
Thus both modules have support at the point \(t=0\). More precisely, the only prime ideal in their support is \((t)\). Their support sets agree even though the ideals \((t)\) and \((t^2)\) retain different multiplicity information.
Inspect the action inside each module
As a rational vector space, \(M_1\) has basis \((1)\) and dimension one. The space \(M_2\) has basis \((1,t)\) and dimension two. Multiplication by \(t\) on \(M_2\) is
\[J=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad J\ne0,\qquad J^2=0.
\]
The modules cannot be isomorphic because their rational dimensions differ. Even the one-dimensional special fiber of \(M_2\) does not retain its full two-dimensional rational structure.
Check whether a vector-space projector respects the module
The matrix \(P=\mathrm{diag}(1,0)\) is an idempotent on the underlying two-dimensional space. However,
\[PJ=0,\qquad JP=J.
\]
It fails to commute with multiplication by \(t\), so it is not an \(R\)-linear projector. A decomposition of the vector space need not split the module.
In fact, an element of the quotient algebra has the form \(a+bt\) with \(t^2=0\). The idempotent equation gives
\[(a+bt)^2=a+bt\quad\Longrightarrow\quad a^2=a,\qquad 2ab=b.
\]
Over \(\mathbb Q\), either \(a=0\) or \(a=1\), and either choice forces \(b=0\). Thus this algebra has only the idempotents zero and one. Its single support point does not provide an extra packet splitting.
Try it
Replace the relation by \(t^3=0\). Find the rational dimension, Fitting ideal, special-fiber dimension and nilpotency index of multiplication by \(t\).
Answer: For \(M_3=R/(t^3)\) they are \(3\), \((t^3)\), \(1\) and \(3\). In the basis \((1,t,t^2)\), multiplication shifts \(1\) to \(t\) and \(t\) to \(t^2\), then sends \(t^2\) to zero. The support is still the point \(t=0\).
These calculations concern the supplied modules. Paper XII’s patching and automorphic conclusions also require its stated arithmetic inputs.
Papers, proofs, Lean, and code
Terms in this example: Commutative ring with identity · Module · Module presentation · Cokernel · Nilpotent operator · Module annihilator · Module support · Zeroth Fitting ideal · Multiplicity in the module example · Ideal · Prime ideal · Localization at a prime
Paper XII distinguishes module support from packet and arithmetic constructions. This example compares two explicitly supplied modules. Patching, classicality and the other arithmetic inputs remain separate.
Paper XII: Patching Support, Hecke Packets, and Conditional Automorphic Completion
Read · Written proofs · Lean coverage · Code and setup · Evidence
Source passages:
Related library guide and code examples. The guide covers related constructions and their software interfaces.
Python arithmetic check for this example: support_multiplicity(). This checks the displayed rational, modular or finite-set calculations exactly.
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