<p>We show that <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="29_2025_1065_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="104" /> </InlineMediaObject> <EquationSource Format="TEX">\(\textrm{TR}_{2i-1}(S)=0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mtext>TR</mtext> <mrow> <mn>2</mn> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mrow> <mo stretchy="false">(</mo> <mi>S</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation> for all <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="29_2025_1065_Article_IEq2.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="40" /> </InlineMediaObject> <EquationSource Format="TEX">\(i\in \mathbb {N} \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>i</mi> <mo>∈</mo> <mi mathvariant="double-struck">N</mi> </mrow> </math></EquationSource> </InlineEquation> and all <i>S</i> quasiregular semiperfect. We show this by computing the algebraic <i>K</i>-theory of truncated polynomial algebras over certain quaisregular semiperfect rings and showing that for these rings the topological restriction homology is even via the curves on <i>K</i>-theory description of [<CitationRef CitationID="CR9">9</CitationRef>]. We then use prismatic cohomology to go from these cases to the general case.</p>

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TR of quasiregular semiperfect rings is even

  • Micah Darrell,
  • Noah Riggenbach

摘要

We show that \(\textrm{TR}_{2i-1}(S)=0\) TR 2 i - 1 ( S ) = 0 for all \(i\in \mathbb {N} \) i N and all S quasiregular semiperfect. We show this by computing the algebraic K-theory of truncated polynomial algebras over certain quaisregular semiperfect rings and showing that for these rings the topological restriction homology is even via the curves on K-theory description of [9]. We then use prismatic cohomology to go from these cases to the general case.