From cde9c5fb67451721966e5e163d7f626e5f355775 Mon Sep 17 00:00:00 2001 From: Deyao Chen Date: Wed, 3 Dec 2025 13:31:37 +0000 Subject: [PATCH 1/2] add helper function and one example --- ha/3-tp.typ | 2 +- ha/4-enough.typ | 2 +- libs/template.typ | 11 +++++++++++ 3 files changed, 13 insertions(+), 2 deletions(-) diff --git a/ha/3-tp.typ b/ha/3-tp.typ index eb73700..f1f236b 100644 --- a/ha/3-tp.typ +++ b/ha/3-tp.typ @@ -415,7 +415,7 @@ $ Then this makes $hom_S (B, -)$ a functor from $ModS$ to $ModR$. -#theorem[ +#theorem("Tensor-hom Adjunction")[ Let $R$, $S$ be rings. Let $A$ be a #rrm, $B$ be an $R$-$S$-bimodule, and $C$ be a right $S$-module. Then we have a canonical isomorphism $ tau: hom_S (A tpr B, C) bij hom_R (A, hom_S (B, C)), $ where for $f : A tpr B -> C$, $a in A$, and $b in B$, diff --git a/ha/4-enough.typ b/ha/4-enough.typ index 2d140a8..5855559 100644 --- a/ha/4-enough.typ +++ b/ha/4-enough.typ @@ -366,6 +366,6 @@ With this proposition, we can prove that an abelian category has enough projecti // Exercise: $e_M$ is one-to-one (mono). (like what we did before.) [TODO] We would like to show that $e_M$ is an injective function. We only need to show that for any $0 != m in M$, there exists $phi : M -> hom_Ab (R, QQ over ZZ)$ such that $phi(m) != 0$. Then since $ phi in homr(M, hom_Ab (R, QQ over ZZ)) iso hom_Ab (M, QQ over ZZ) $ - by @tensor-hom, + by #thmref(), we only need to find some $phi : M -> QQ over ZZ$ in $Ab$ so that $phi(m) != 0$, which is given by @map-to-q-over-z. ] diff --git a/libs/template.typ b/libs/template.typ index 4124c3e..d73e07b 100644 --- a/libs/template.typ +++ b/libs/template.typ @@ -117,6 +117,17 @@ breakable: true, )[_#title._ #term #h(1fr) $qed$] +// Helper to reference theorems/props with their optional name. +#let thmref(id) = { + context { + if query(id).at(0).caption != none { + [#ref(id) (#query(id).at(0).caption.body)] + } else { + [#ref(id)] + } + } +} + #let project(title: "", authors: (), date: none, body) = { // Set the document's basic properties. set document(author: authors, title: title) From 471555281e4fa366c233c6ec46f13580d3d6be3d Mon Sep 17 00:00:00 2001 From: Deyao Chen Date: Wed, 3 Dec 2025 14:11:13 +0000 Subject: [PATCH 2/2] change Baer's Criterion --- ha/4-enough.typ | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/ha/4-enough.typ b/ha/4-enough.typ index 5855559..cf00ba5 100644 --- a/ha/4-enough.typ +++ b/ha/4-enough.typ @@ -239,7 +239,7 @@ For most of our homological algebra to work, an abelian category needs to have e #proof[@rotman[Corollary 3.35] and @notes[Corollary 5.9]. Let $M$ be an injective $R$-module, and let $m in M$ and $r in R without brace.l 0 brace.r$. Set $J eq r R$ (which is an ideal of $R$) and define - $f colon J arrow.r M$ by $f lr((r)) eq m$. By Baer’s Criterion, we may + $f colon J arrow.r M$ by $f lr((r)) eq m$. By #thmref(), we may extend $f$ to a homomorphism $tilde(f) colon R arrow.r M$. Then $ m eq f lr((r)) = tilde(f)(r)eq tilde(f) lr((r dot.op 1)) eq r dot.op tilde(f) lr((1)). @@ -254,7 +254,7 @@ For most of our homological algebra to work, an abelian category needs to have e $m eq f lr((r))$. Then since $M$ is divisible, there is some $m' in M$ such that $m eq r m'$. Define $tilde(f) colon R arrow.r M$ by $tilde(f) lr((1)) eq m'$. Clearly $tilde(f)$ is an extension of $f$, so - $M$ is injective by Baer’s Criterion. + $M$ is injective by #thmref(). ] #corollary[