# 1 Thursday, January 06

Topics: Localization and completion, Nakayama’s lemma, Dedekind domains, Hilbert’s basis theorem, Hilbert’s Nullstellensatz, Krull dimension, depth and Cohen-Macaulay rings, regular local rings.

References:

Some examples of module morphisms:

• $$\mathop{\mathrm{Hom}}_\mathsf{Ring}({\mathbb{Z}}, S) = \left\{{{\operatorname{pt}}}\right\}$$, since $$1\to 1$$ is necessary.

• $$\mathop{\mathrm{Hom}}_\mathsf{Ring}({\mathbb{Z}}[x], S) \cong S$$. Why? Since $$1\to 1$$ is forced, $$x\mapsto s$$ can be sent to any $$s\in S$$.

• $$\mathop{\mathrm{Hom}}_\mathsf{Ring}\qty{ { {\mathbb{Z}}[x, y]\over \left\langle{y^2-x^3-1}\right\rangle }, S} = \left\{{(a, b)\in S{ {}^{ \scriptscriptstyle\times^{2} } } {~\mathrel{\Big\vert}~}a^2-b^3=1}\right\}$$.

• $$\mathop{\mathrm{Hom}}_\mathsf{Ring}(R/I, S) = \left\{{f\in \mathop{\mathrm{Hom}}_\mathsf{Ring}(R, S) {~\mathrel{\Big\vert}~}f(I) = 0}\right\}$$

Show that $$\operatorname{Id}(R/I)\rightleftharpoons\left\{{J\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}J \supseteq I}\right\}$$ using \begin{align*} \Pi: R &\to R/I \\ x &\mapsto [x] \\ \pi^{-1}(J) &\mapsfrom J .\end{align*} Show that $$\pi^{-1}(J)$$ is in fact an ideal, construct a proposed inverse $$\Pi^{-1}$$, and show $$\Pi\circ \Pi^{-1}= \Pi^{-1}\circ \Pi = \operatorname{id}$$.

Show that $$R$$ is a field iff $$R$$ is a simple ring iff any ring morphism $$R\to S$$ is injective. For $$3\implies 1$$, directly shows that every nonzero element is a unit.

Chapter 1 of A&M:

• 1,8,10,13,15,16,19.

# 2 Thursday, January 13

Last time: fields are simple rings.

Let $$k\in \mathsf{Field}$$ and show that

• If $$f\in R\mathrel{\vcenter{:}}= k[x]$$ is irreducible then $$\left\langle{f}\right\rangle \in \operatorname{Spec}R$$.
• $$\left\langle{xy}\right\rangle \in \operatorname{Id}(k[x,y])$$ is not prime and not maximal.
• There exist nonzero non-prime maximal ideals.
• Show that $$I\in \operatorname{Spec}R \iff A/I$$ is an integral domain.
• Show that $$I\in \operatorname{mSpec}R \iff A/I\in \mathsf{Field}$$.

Prove that if $$f: R\to S$$ is a ring morphism then there is a well-defined map \begin{align*} f^*: \operatorname{Spec}S &\to \operatorname{Spec}R \\ {\mathfrak{p}}&\mapsto f^{-1}({\mathfrak{p}}) ,\end{align*} i.e. if $${\mathfrak{p}}$$ is prime in $$S$$ then $$f^{-1}({\mathfrak{p}})$$ is prime in $$R$$. Show that this doesn’t generally hold with $$\operatorname{Spec}$$ replaced by $$\operatorname{mSpec}$$.

Show that defining $$V(I) \mathrel{\vcenter{:}}=\left\{{{\mathfrak{p}}\in \operatorname{Spec}R {~\mathrel{\Big\vert}~}{\mathfrak{p}}\supseteq I}\right\}$$ as closed sets defines a topology on $$\operatorname{Spec}R$$. Show that $$\operatorname{Spec}R$$ is Hausdorff iff it’s discrete, and $$V$$ is functorial.

Describe

• $$\operatorname{Spec}k$$ for $$k\in \mathsf{Field}$$
• $$\operatorname{Spec}{\mathbb{Z}}$$, show it is not Hausdorff, and $$\left\langle{0}\right\rangle$$ is a generic point, and the subspace topology on its closed points is cofinite.
• $$\operatorname{Spec}R$$ for $$R$$ a local ring.
• $$\operatorname{Spec}R$$ for $$R$$ a DVR
• $$\operatorname{Spec}k[x,y]$$
• $$\operatorname{Spec}R$$ for $$R = {\mathbb{Z}}[i]$$.
• $$\operatorname{Spec}{\mathcal{O}}_K$$, the ring of integers of a number field $$K$$.
• $$\operatorname{Spec}R$$ for $$R$$ a Dedekind domain

Show that every $$I\in \operatorname{Id}(R)$$ is contained in some $${\mathfrak{m}}\in \operatorname{mSpec}R$$.

Show that the following rings are local:

• For $$p\in {\mathbb{Z}}$$ prime, $$R\mathrel{\vcenter{:}}={\mathbb{Z}} \left[ { \scriptstyle { {S}^{-1}} } \right]$$ for $$S\mathrel{\vcenter{:}}=\left\{{\ell \in {\mathbb{Z}}\text{ prime} {~\mathrel{\Big\vert}~}\ell \neq p}\right\}$$.
• $$k\in \mathsf{Field}$$
• $$k { \left[ {x} \right] }$$ with $${\mathfrak{m}}= \left\langle{x}\right\rangle$$
• $$k { \left[ {x, y} \right] }$$. What is the maximal ideal?

For $$(R, {\mathfrak{m}})$$ a local ring, show that $${\mathfrak{m}}= R\setminus(R^{\times})$$.

Recall that

• $$\sum I_j$$ is the smallest ideal containing all of the $$I_j$$.
• $$\displaystyle\bigcap I_j$$ is again an ideal
• $$IJ \mathrel{\vcenter{:}}=\left\langle{xy {~\mathrel{\Big\vert}~}x\in I, y\in J}\right\rangle$$ is an ideal
• $$IJ \subseteq I \cap J$$.
• $${\sqrt{0_{R}} }$$ is the set of nilpotent elements.

Show that $${\sqrt{0_{R}} }$$ is the intersection of all prime ideals.

Regarding elements $$f\in R$$ as functions on $$\operatorname{Spec}R$$, $$f$$ nilpotent is like being zero at every point of $$\operatorname{Spec}R$$.

Show that $$x\in {J ({R}) } \iff 1-xy\in R^{\times}$$ for all $$y\in R$$.

# 3 Tuesday, January 18

A reference for pictures: Mumford’s red book. Note the typo in A&M problem 10: it is false for the zero ring.

Recall some definitions:

• $$R{\hbox{-}}$$modules, what are some examples?
• Submodules and quotient modules.
• The submodule generated by a subset.
• A morphism of modules and the $$R{\hbox{-}}$$module structure on $$\mathop{\mathrm{Hom}}_R(M, N)$$, $$(rf)(x) \mathrel{\vcenter{:}}= r\cdot f(x)$$.
• This makes $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ into a category enriched over itself.
• $$\operatorname{im}f, \ker f, \operatorname{coker}f$$.
• $$\prod M_i, \bigoplus M_i$$

Show that if $$M\leq N \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$, then the following action makes $$M/N$$ into an $$R{\hbox{-}}$$module: \begin{align*} r\cdot [x] \mathrel{\vcenter{:}}=[rx] ,\end{align*} i.e. that if $$[x] = [y]$$ then $$r\cdot [x] = r\cdot [y]$$.

Show the universal properties of kernels and cokernels: given $$f: M\to N$$ and $$Q\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$, \begin{align*} \mathop{\mathrm{Hom}}_R(Q, \ker f) &= \left\{{ s\in \mathop{\mathrm{Hom}}_R(Q, M) {~\mathrel{\Big\vert}~}f\circ s = 0}\right\} \\ \mathop{\mathrm{Hom}}_R(\operatorname{coker}f, Q) &= \left\{{t\in \mathop{\mathrm{Hom}}_R(N, Q) {~\mathrel{\Big\vert}~}t\circ f = 0 }\right\} .\end{align*}

Show that if $$I$$ is a finite indexing set, there is an isomorphism \begin{align*} \bigoplus _{i\in I} M_i { { \, \xrightarrow{\sim}\, }}\prod_{i\in I} M_i , \end{align*} and that \begin{align*} \mathop{\mathrm{Hom}}_R\qty{ T, \prod M_i} &\cong \prod_{i\in I} \mathop{\mathrm{Hom}}_R\qty{T, M_i} \\ \mathop{\mathrm{Hom}}_R\qty{ \bigoplus M_i, T} &\cong \bigoplus \mathop{\mathrm{Hom}}_R\qty{ M_i, T} .\end{align*}

Show that by Yoneda, this satisfies the universal property.

Sketch:

• Define projections $$\pi_j: \prod_i M_i \to M_j$$.
• Send $$f\in \mathop{\mathrm{Hom}}(T, \prod_i M_i)$$ to $$\pi_j \circ f\in \prod_i \mathop{\mathrm{Hom}}(T, M_i)$$
• For the other direction, given $$(f_i) \in \prod_i \mathop{\mathrm{Hom}}(M_i, T)$$, send $$(f_i)$$ to $$\sum f_i$$.

Show that $$\prod$$ is a categorical product and $$\bigoplus$$ is a categorical coproduct. What are the product and coproduct in $${\mathsf{Top}}$$?

Given $$N \leq M \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$, define the colon ideal \begin{align*} (N: M) \mathrel{\vcenter{:}}=\left\{{a\in R{~\mathrel{\Big\vert}~}aM \subseteq N}\right\} \end{align*} and the annihilator of $$M$$ \begin{align*} \operatorname{Ann}(M) = (0: M) = \left\{{a\in R {~\mathrel{\Big\vert}~}aM = 0}\right\} .\end{align*}

Some annihilators:

• $$C_n \in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}$$, so $$\operatorname{Ann}(C_n) = n{\mathbb{Z}}= \left\langle{n}\right\rangle$$.
• Again in $${\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}$$, $$\operatorname{Ann}(C_n \oplus C_m) = n{\mathbb{Z}}\cap m{\mathbb{Z}}= \operatorname{lcm}(m, n) {\mathbb{Z}}$$.

An $$R{\hbox{-}}$$module is free iff $$R \cong \bigoplus _{i\in I} R$$, where $$I$$ can be infinite but (importantly) we need the direct sum instead of the direct product. Note that generally $$\prod_{i\in I} R$$ may not be free as an $$R{\hbox{-}}$$module.

A module is finitely generated if there exists a generating set $$X \mathrel{\vcenter{:}}=\left\{{x_i}\right\}_{i\leq n} \subseteq M$$ such that any submodule containing $$X$$ is all of $$M$$, or equivalently $$x\in M \implies x = \sum r_i x_i$$ for some $$r_i \in R$$.

Show that $$M$$ is finitely-generated iff there is a surjective morphism $$R^n \to M$$ for some $$n\in {\mathbb{Z}}_{\geq 0}$$.

Sketch:

• finitely-generated $$\implies$$ surjection:
• Take $$e_i = \left\{{0,\cdots, 1,\cdots, 0}\right\} \in R^n$$ and define $$f(e_i) \mathrel{\vcenter{:}}= x_i$$
• Use the universal property of the direct sum.
• $$\impliedby$$:
• Show that the $$f(e_i)$$ generate $$M$$: by surjectivity, $$m = f(x) = f(\sum r_i e_i) = \sum r_i f(e_i)$$.

Thus every $$M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ is the quotient of a free module: find a surjection $$f: F\to M$$, so $$\operatorname{im}f = M$$, then use that $$\operatorname{im}f \cong /\ker f$$. For example, one can take $$F \mathrel{\vcenter{:}}=\bigoplus _{m\in M} R \to M$$.

Recall

• The definition of exact sequences
• $$0\to A \xrightarrow{f} B$$ is exact iff $$f$$ is injective,
• $$A \xrightarrow{f} B \to 0$$ iff $$f$$ is surjective,
• $$0\to A \xrightarrow{f} B \to 0$$ iff $$f$$ is an isomorphism,
• $$0\to A\to B\to C\to 0$$ is called a short exact sequence.

Check that $$({-}) \oplus M$$ is exact.

Show that $$\mathop{\mathrm{Hom}}(N, {-})$$ and $$\mathop{\mathrm{Hom}}({-}, N)$$ are both left-exact for any $$N\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$, where $$0 \to A \xrightarrow{f} B \xrightarrow{g} C\to 0$$ is sent to $$0 \to \mathop{\mathrm{Hom}}(N, A) \xrightarrow{f\circ {-}} \cdots$$. Give an example of when right-exactness fails.

Hint: try $$0\to {\mathbb{Z}}\xrightarrow{\cdot 2} {\mathbb{Z}}\to C_2 \to 0$$ and apply $$\mathop{\mathrm{Hom}}_{\mathbb{Z}}({-}, C_2)$$.

Recall that $$\mathop{\mathrm{Hom}}_{{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}}({\mathbb{Z}}, {-}) \cong \operatorname{id}$$ and $$\mathop{\mathrm{Hom}}_{{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}}(C_n, {-}) \cong ({-})[n]$$ picks out the $$n{\hbox{-}}$$torsion.

# 4 Thursday, January 20

Last time: $$\mathop{\mathrm{Hom}}(N, {-})$$ and $$\mathop{\mathrm{Hom}}({-}, N)$$ are left exact. We can explicitly describe \begin{align*} \mathop{\mathrm{Hom}}_A(A/I, M) = \left\{{m\in M {~\mathrel{\Big\vert}~}im = 0 \forall i\in I}\right\} ,\end{align*} which is the $$I{\hbox{-}}$$torsion in $$M$$. Using that $$0\to I\to A \to A/I \to 0$$ is short exact, we’ll get a long exact sequence \begin{align*} 0 \to \mathop{\mathrm{Hom}}(I, M) \to \mathop{\mathrm{Hom}}(A, M) \xrightarrow{f} \left\{{I{\hbox{-}}\text{torsion in } M}\right\} \to \operatorname{Ext} ^1_A(I, M) \to \cdots ,\end{align*} where the $$\operatorname{Ext}$$ term measures failure of surjectivity of $$f$$.

Show that $$\mathop{\mathrm{Hom}}(N, {-})$$ and $$\mathop{\mathrm{Hom}}({-}, N)$$ are left exact.

Recall the snake lemma:

The recipe for the connecting morphism:

• Start with $$x\in \ker(C_1\to C_2)$$, choose a preimage in $$B_1$$
• Push to $$B_2$$
• Use exactness to pull to $$A_2$$
• Project along $$A_2\to \operatorname{coker}(A_1\to A_2)$$

An object $$M\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ is of finite presentation iff there is an exact sequence \begin{align*} R^m \to R^n\to M \to 0 ,\end{align*} i.e. there are finitely many generators and finitely many relations.

A nice application of the snake lemma: for finitely presented modules $$M, N$$, one can extend a morphism $$M\to N$$ to

Since $$R\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ is free, the extended maps can be represented by matrices, which is a significant simplification. In fact, $$f$$ can be recovered uniquely by knowing the map on generators.

Prove the snake lemma, and show exactness at all 6 places.

Recall the universal property of $$M\otimes_R N$$ in $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ in terms of bilinear morphisms.

Prove uniqueness of any object satisfying a universal property using the Yoneda lemma.

Prove using universal properties:

• $$R\otimes_R R \cong R$$, using universal properties. Why is this unique?
• $$R\otimes_R M \cong M$$.
• $$(M_1 \oplus M_2) \otimes_R N \cong (M_1 \otimes_R N) \oplus (M_2\otimes_R N)$$.

# 5 Tuesday, January 25

Do A&M:

• 2.12, 24, 25
• 3.1, 4, 12

See website: https://www.daniellitt.com/commutative-algebra

Last time:

• Defined $$M\otimes_R N$$, need to show it exists.
• Showed $$R\otimes_R M \cong M$$
• Showed sums commute with tensor products

Show that if $$A \xrightarrow{f} B \xrightarrow{g} C\to 0$$ is exact and $$A\otimes_R N, B\otimes_R N$$ exist, then $$C\otimes_R N$$ also exists.

Hint: Use the universal property to produce a map $$\psi: A\otimes N\to B\otimes N$$ and set $$C\otimes N \mathrel{\vcenter{:}}=\operatorname{coker}\psi$$.

Prove that $$M\otimes_R N$$ exists by constructing it.

Some hints: Construction 1: use $$0\to \ker f\to R{ {}^{ \scriptscriptstyle\oplus^{I} } } \xrightarrow{f} M\to 0$$, and find $$R{ {}^{ \scriptscriptstyle\oplus^{J} } }\to \ker f$$ to assemble an exact sequence \begin{align*} R{ {}^{ \scriptscriptstyle\oplus^{J} } }\to R{ {}^{ \scriptscriptstyle\oplus^{I} } }\to M\to 0 .\end{align*} Now apply $$({-})\otimes_R N$$ to exhibit $$M\otimes_R N$$ as a cokernel $$N{ {}^{ \scriptscriptstyle\oplus^{J} } }\to N{ {}^{ \scriptscriptstyle\oplus^{I} } }\to M\otimes_R N\to 0$$. Separately, use the hands-on construction and prove it satisfies the universal property.

Prove $$C_2 \otimes_{\mathbb{Z}}C_3 \cong 0$$ using construction 1 above.

Sketch:

Take $${\mathbb{Z}}\xrightarrow{\times 2} {\mathbb{Z}}\to C_2\to 0$$, apply $$({-})\otimes C_3$$, and check that $$\operatorname{coker}(C_3 \xrightarrow{\times 2} C_3) = 0$$ since multiplication by 2 is invertible. Alternatively, use bilinearity: \begin{align*} B(x,y) = 3B(x, y) - 2B(x, y) = B(x, 3y) - B(2x, y) = B(x, 0) - B(0, y) = 0 .\end{align*}

Show that

• $$({-})\otimes_R N$$ is right exact
• Tensoring is associative, distributive, commutative
• There is a canonical isomorphism $$R\otimes_R M\to M$$ induced by $$r\otimes m \mapsto r.m$$.
• Morphisms $$f: A\to B\in \mathsf{CRing}$$ induce functors $$f^\sharp: {\mathsf{A}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{B}{\hbox{-}}\mathsf{Mod}}$$. Also show that $$M\otimes_A B$$ has a $$B{\hbox{-}}$$module structure given by $$b_1(m\otimes b_2) \mathrel{\vcenter{:}}= m\otimes(b_1 b_2)$$.
• For $$N\in{\mathsf{B}{\hbox{-}}\mathsf{Mod}}$$, there is an isomorphism $$\mathop{\mathrm{Hom}}_{A}(M, N) { { \, \xrightarrow{\sim}\, }}\mathop{\mathrm{Hom}}_{B}(M\otimes_A B, N)$$.
• Use $$f\mapsto (m\otimes b \mapsto bf(m)) \in \mathrm{Bil}_B(M\times B, N)$$, with inverse $$Q\mapsto Q({-}, 1)$$.
• Show that $$M\otimes_R R/I \cong M/IM$$, using $$R/I = \operatorname{coker}(I\hookrightarrow R)$$ and applying $$({-})\otimes M$$. Also show that $$I\otimes_R M{ { \, \xrightarrow{\sim}\, }}IM$$ canonically.
• Show that $$k[t]{ {}^{ \scriptscriptstyle\oplus^{2} } } \otimes_{k[t]} {k[t]\over\left\langle{t^2}\right\rangle} \cong \qty{k[t]\over \left\langle{t^2}\right\rangle}{ {}^{ \scriptscriptstyle\oplus^{2} } }$$.
• Show that $$R/I \otimes_R R/J \cong {R/I \over I \cdot R/J} \cong {R\over I+J}$$.
• Tensoring need not be left exact.
• $$C_p \not\in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}^\flat$$ .
• $$R\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}^\flat$$.
• $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}^\flat$$ is closed under $$\otimes_R$$ and $$\oplus$$.
• $${\mathbb{Q}}\in{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}^\flat$$ but not in $${\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}^{\mathrm{free}}$$

$${\mathsf{Alg}}_{/ {A}}$$ is the coslice category $$\mathsf{CRing}_{{A/}}$$: objects are rings $$B$$ equipped with ring morphisms $$A\to B$$, and morphisms are cones under $$A$$:

Examples of algebras:

• $$k[t] \in{\mathsf{Alg}}_{/ {k}}$$
• $$R\in \mathsf{CRing}\implies R\in{\mathsf{Alg}}_{/ {{\mathbb{Z}}}}$$
• Every $$B\in {\mathsf{Alg}}_{/ {A}}$$ is a quotient of some polynomial algebra $$A[t_1,\cdots]$$ on potentially infinitely many generators.

An object $$B\in{\mathsf{Alg}}_{/ {A}}$$ is finitely generated if it is a quotient of some $$A[t_1, \cdots, t_n]$$. Equivalently, there exist $$x_1,\cdots, x_n\in B$$ such that any subring containing the $$x_i$$ and the image of $$A$$ is all of $$B$$.

$$B$$ is finite if $$B\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$.

Examples:

• $$k[t] \in {\mathsf{Alg}_{/k} }$$ is finitely generated but not finite.
• $$k[t]/\left\langle{t^2}\right\rangle\in {\mathsf{Alg}_{/k} }$$ is finitely generated and finite.
• $$({-})\otimes_A ({-}) \in {\mathsf{Fun}}({\mathsf{Alg}}_{/ {A}} { {}^{ \scriptscriptstyle\times^{2} } }, {\mathsf{Alg}}_{/ {A}} )$$, defined by $$(b_1 \otimes c_1)(b_2\otimes c_2) \mathrel{\vcenter{:}}= b_1 b_2 \otimes c_1 c_2$$.
• $$\mathop{\mathrm{Hom}}_{{\mathsf{Alg}}_{/ {A}} }(B\otimes_A C, S) = \left\{{f: B\to S, g: C\to S {~\mathrel{\Big\vert}~}{ \left.{{f}} \right|_{{A}} } = { \left.{{g}} \right|_{{A}} }}\right\}$$.
• $$k[t_1] \otimes_k k[t_2] \cong k[t_1, t_2]$$ is not isomorphic to $$k[t_1]\times k[t_2]$$ via $$(f(t_1), g(t_2)) \mapsto f(t_1) \cdot g(t_2)$$, since e.g. $$h(t_1, t_2) \mathrel{\vcenter{:}}= t_1 + t_2$$ is not in the image of this map.

# 6 Thursday, January 27

Nakayama: called a lemma, but arguably the most important statement in commutative algebra! Recall that $${J ({A}) } = \cap_{{\mathfrak{m}}\in \operatorname{mSpec}A} {\mathfrak{m}}$$, and being zero mod every $${\mathfrak{m}}\in \operatorname{mSpec}A$$ means being zero mod $${J ({A}) }$$.

Let $$A \in \mathsf{CRing}$$ with $$I \subseteq {J ({I}) }$$ and let $$M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$. Then \begin{align*} M=IM \implies M=0 .\end{align*}

This reduces statements about local rings to statements about fields. Commonly used examples:

• $$I = {\sqrt{0_{R}} }$$.
• $$A\in \mathsf{Loc}\mathsf{Ring}$$ with $$I = {\mathfrak{m}}_A$$ its maximal ideal.

If $$A$$ is local and $$M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$, \begin{align*} M/{\mathfrak{m}}_A M = 0 \iff M = 0 .\end{align*}

$$M$$ yields a sheaf (or vector bundle) on $$\operatorname{Spec}A$$, so think of $$m\in M$$ as a function on $$\operatorname{Spec}A$$ in the following way: if $$m\in M$$, \begin{align*} m: {\mathfrak{p}}\mapsto m \operatorname{mod}{\mathfrak{p}}\in M/{\mathfrak{p}}M .\end{align*}

If $$n\in N\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$ and $$n\equiv 0 \operatorname{mod}{\mathfrak{m}}$$ for all $${\mathfrak{m}}\in \operatorname{mSpec}A$$, then $$M=0$$.

Show that if $$A \in \mathsf{Loc}\mathsf{Ring}$$ and $$M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$ with $$\left\{{x_i}\right\}_{i\leq n} \subseteq M$$, then \begin{align*} \left\langle{x_1,\cdots, x_n}\right\rangle = M \iff \left\langle{\mkern 1.5mu\overline{\mkern-1.5mux_1\mkern-1.5mu}\mkern 1.5mu, \cdots, \mkern 1.5mu\overline{\mkern-1.5mux_n\mkern-1.5mu}\mkern 1.5mu}\right\rangle = M/{\mathfrak{m}}M, \qquad \mkern 1.5mu\overline{\mkern-1.5mux_i\mkern-1.5mu}\mkern 1.5mu \mathrel{\vcenter{:}}= x_i \operatorname{mod}{\mathfrak{m}} .\end{align*}

$$\implies$$: If $$\mkern 1.5mu\overline{\mkern-1.5muy\mkern-1.5mu}\mkern 1.5mu\in M/{\mathfrak{m}}M$$, lift to $$y\in M$$ to write $$y = \sum c_i x_i$$ and thus $$\mkern 1.5mu\overline{\mkern-1.5muy\mkern-1.5mu}\mkern 1.5mu = \sum c_i \mkern 1.5mu\overline{\mkern-1.5mux_i\mkern-1.5mu}\mkern 1.5mu$$.

$$\impliedby$$: Present $$M$$ by $$A^n \xrightarrow{f} M \to C =\operatorname{coker}f \to 0$$ where $$f(e_i) = x_i$$ – we want to show $$C = 0$$. Reduce mod $${\mathfrak{m}}$$ by applying $$({-})\otimes_A A/{\mathfrak{m}}$$ to get \begin{align*} (A/{\mathfrak{m}})^n \xrightarrow{\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu} M/{\mathfrak{m}}M \to C/{\mathfrak{m}}C\to 0 .\end{align*} This is surjective so $$C/{\mathfrak{m}}C = 0$$. By Nakayama, $$C=0$$.

Let the local ring $$R = k { \left[ {t} \right] }$$ with $${\mathfrak{m}}_R = \left\langle{t}\right\rangle$$, and let $$M = R{ {}^{ \scriptscriptstyle\oplus^{3} } }$$. Consider \begin{align*} {\left[ {1+t^3 + t^5, t+t^2, t^3} \right]}, {\left[ {t^{22}, 1+t^9, t^{10} + t^{10^{10}}} \right]}, {\left[ {1+ t^{10^{10^{10}}}, 1+t^5 + t^9, 1+t^7 + t^{11} } \right]} ,\end{align*} which reduced $$\operatorname{mod}t$$ yields \begin{align*} {\left[ {1,0,0} \right]}, {\left[ {0,1,0} \right]}, {\left[ {1,1,1} \right]} .\end{align*} So the original elements generate $$M$$.

Next goal: proving Nakayama. We’ll need a version of Cayley-Hamilton. Recall the definition of multiplicative subsets and localization, along with its universal property.

Examples:

• For any element $$f$$, $$S\mathrel{\vcenter{:}}=\left\{{1, f, f^2,\cdots}\right\}$$
• For $$A$$ an integral domain, $$S \mathrel{\vcenter{:}}= A\setminus\left\{{0}\right\}$$
• All nonzero zero divisors.

Prove $$S^{-1}A$$ exists and is unique for $$A\in \mathsf{CRing}$$. Give an example where $$s'a = sb$$ is not sufficient.

Remarks on localization:

• $${a\over s} = {b\over s'} \iff s'' s' a - s'' s b$$ for some $$s''\in S$$ – this is needed because it will hold in $$S^{-1}A$$, since $$s''\in (S^{-1}A)^{\times}$$ for any $$s''\in S$$ by construction.
• $${a\over s}=0$$ iff $$a$$ is annihilated by an element of $$S$$.
• Producing the actual map for the universal property: if $$f:A\to B$$ sends $$S$$ to invertible elements,

• $$A_f \mathrel{\vcenter{:}}= S^{-1}A$$ for $$S \mathrel{\vcenter{:}}=\left\{{1,f,f^2,\cdots}\right\}$$. For $$A = {\mathbb{Z}}$$ and $$f=p$$ a prime, $${\mathbb{Z}}_f = {\mathbb{Z}}{ \left[ { \scriptstyle \frac{1}{p} } \right] } \subseteq {\mathbb{Q}}$$ are fractions whose denominator is a power of $$p$$.
• For $$A$$ an integral domain and $$S=A\setminus\left\{{0}\right\}$$ yields $$S^{-1}A = \operatorname{ff}(A)$$ the fraction field.
• For $${\mathfrak{p}}\in \operatorname{Spec}A$$, $$S\mathrel{\vcenter{:}}= A\setminus{\mathfrak{p}}$$, then $$A_{\mathfrak{p}}\mathrel{\vcenter{:}}= S^{-1}A$$ is localization at a prime ideal
• $${\mathbb{Z}}_{\left\langle{p}\right\rangle} = {\mathbb{Z}}{ \left[ { \scriptstyle \frac{1}{\ell} } \right] }_{\ell\neq p \text{ prime}}$$, fractions in $${\mathbb{Q}}$$ with denominators not divisible by $$p$$.

Some exercises:

• Use the universal property to show $$({\mathbb{Z}}/15{\mathbb{Z}})_5 \cong {\mathbb{Z}}/3{\mathbb{Z}}$$.
• Show that for $$M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}$$, $$S^{-1}M$$ exists and is unique.
• Show that $$S^{-1}A \curvearrowright S^{-1}M$$.
• Show that $$S^{-1}M = M\otimes_A S^{-1}A$$ using the universal property. Use \begin{align*} M = M\otimes_A A \xrightarrow{\operatorname{id}\otimes S^{-1}({-})} M\otimes_A S^{-1}A \end{align*} where $$m\mapsto m\otimes 1$$ For $$f:M\to N$$ where $$s$$ acts invertibly on $$N$$, produce a map $$M\times S^{-1}A\to N$$ where $$(m, a/s)\mapsto as^{-1}f(m)$$ where $$s^{-1}$$ is the inverse of the action $$s: N\to N$$.
• Show that $$({-})\otimes_A S^{-1}A$$ is left exact and thus exact.
• Injectivity: use that $${m\over s}\mapsto 0 \iff s'f(m) = 0$$ for some $$s'\in S$$.
• Show that $$S^{-1}A \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^\flat$$.

# 7 Tuesday, February 01

For $$A\in \mathsf{CRing}, M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$, $${\mathfrak{a}}\in \operatorname{Id}(A)$$, and $$\phi: M\to {\mathfrak{a}}M \subseteq M$$, there exists $$\left\{{a_i}\right\}_{i\leq n} \subseteq {\mathfrak{a}}$$ such that \begin{align*} \phi^r + a_1 \phi^{r-1} + \cdots + a_r \operatorname{id}= 0 .\end{align*}

Reduce to showing this for $$M = A^r$$ a free module. Use the diagram:

Lift by sending $$e_i$$ to any element $$a_i \in {\mathfrak{a}}$$. Then: STS $$\tilde f$$ satisfies some polynomial, since $$\phi$$ will satisfy the same polynomial and map to zero by commutativity of the square above. $$\tilde f: A^r\to A^r$$ can be written as a matrix $$(a_{ij})$$. Now reduce to $${\mathbb{C}}$$: consider the map \begin{align*} {\mathbb{Z}}[ \left\{{ x_{ij}}\right\}_{i, j \leq r} ] &\to A \\ x_{ij} &\mapsto a_{ij} .\end{align*}

Forming the matrix $$M = (x_{ij})$$ yields a commutative diagram:

Since $${\mathbb{Z}}[\left\{{ x_{ij} }\right\}]$$ is an integral domain, we can pass to the fraction field $${\mathbb{Q}}(\left\{{x_{ij}}\right\})$$, where by linear algebra $$M$$ has a characteristic polynomial in the $$x_{ij}$$. So for some homogeneous polynomials $$p_i$$ in the $$x_{ij}$$, we have \begin{align*} \tilde f^r + p_1(x_{ij}) \tilde f^{r-1} + \cdots + p_r(x_{ij}) = 0 .\end{align*} Now choose an arbitrary embedding $${\mathbb{Q}}(x_{ij}) \hookrightarrow{\mathbb{C}}$$, and prove Cayley-Hamilton here using (e.g.) Jordan normal form.

Write a careful proof of Cayley-Hamilton for arbitrary fields.

Useful strategy: reduce to the “universal matrix.”

Recall that there exists an adjugate of any square matrix satisfying $$M \operatorname{adj}(M) = \operatorname{adj}(M) M = \operatorname{det}(M) \operatorname{id}$$. Apply this to $$M \mathrel{\vcenter{:}}= tI - \tilde f \in \mathop{\mathrm{End}}_{A[t]}(A(t){ {}^{ \scriptscriptstyle\times^{r} } })$$. Write $$\operatorname{adj}(tI-\tilde f) = \sum B_i t^i$$ with $$B_i \in \operatorname{Mat}_{n\times n}(A)$$. We have \begin{align*} (tI-\tilde f) \operatorname{adj}(tI - \tilde f) = \operatorname{det}(\tilde f)I ,\end{align*} but we can’t plug in $$\tilde f$$ here because we don’t know if this lands in a commutative ring, so e.g. $$t m t^r \neq mt^{r+1}$$ doesn’t necessarily hold.

Note that $$B_i$$ commutes with $$tI - \tilde f \iff B_i$$ commutes with $$\tilde f$$, e.g. by equating coefficients. Write $$R = Z(\tilde f)$$ for the centralizer, those matrices commuting with $$\tilde f$$. Then $$(tI - \tilde f) \in R[t]$$, which reduces us to the world of commutative rings. Then $${ \left.{{ \operatorname{det}(tI - \tilde f) }} \right|_{{\tilde f}} } = (\tilde f - \tilde f)\cdot g = 0$$.

Show that if $$M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$ and $${\mathfrak{a}}\in \operatorname{Id}(A)$$, \begin{align*} M = {\mathfrak{a}}M \implies \exists x\cong 1{\mathsf{{\mathfrak{a}}}{\hbox{-}}\mathsf{Mod}}\text{ such that } xM = 0 .\end{align*}

Hint: apply Cayley-Hamilton to $$\operatorname{id}_M$$.

For $$M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}, I \in \operatorname{Id}(A)$$ with $$I \in {J ({A}) }$$, \begin{align*} M = IM \implies M = 0 .\end{align*}

Apply the corollary to get $$x\equiv 1 \operatorname{mod}I$$ with $$xM = 0$$. But then $$x \equiv 1 \operatorname{mod}{J ({A}) }$$, so $$x$$ is a unique and $$xM = M$$.

Alternatively, pick a minimal set of generators $$\left\{{x_i}\right\}$$ of $$M$$, so $$m = \sum a_i x_i$$ with $$a_i\in I$$ since $$M=IM$$. Since $$1-a_n\in {J ({A}) }$$ and is a unit, so \begin{align*} (1-a_n)x_n = \sum_{i\leq n-1} a_i x_i \implies x_n = (1-a_n)^{-1}\sum_{i\leq n-1}a_i x_i .\end{align*} $$\contradiction$$

Notes:

• Proved last time: $$A\in \mathsf{Loc}\mathsf{Ring}, M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$, if $$X\mathrel{\vcenter{:}}=\left\{{x_i}\right\} \subseteq M$$ with $$\left\{{\mkern 1.5mu\overline{\mkern-1.5mux_i\mkern-1.5mu}\mkern 1.5mu}\right\}$$ generating $$M/{\mathfrak{m}}M$$, then $$X$$ generates $$M$$.
• Suppose $$f\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}(M, N)$$ with $$\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu: M/{\mathfrak{m}}_A M \to N/{\mathfrak{m}}_A N$$ an isomorphism – $$f$$ is not necessarily an isomorphism.
• Counterexample: $$k[[x]]\to k$$ is not an isomorphism in $${\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$ but reduces mod $$x$$ to $$k \xrightarrow{\operatorname{id}} k$$.
• Show that $$f$$ need not be injective, but is always surjective.
• For surjectivity: use $$M \xrightarrow{f} N \to C \to 0$$, use that $$C/{\mathfrak{m}}C =0$$ to conclude $$C=0$$ by Nakayama.
• For injectivity: use $$0\to K\to M \xrightarrow{f} N \to 0$$ and try to show $$K=0$$. Not true, take $$0\to \left\langle{x}\right\rangle \to k[[x]] \to k\to 0$$ and apply $$({-})\otimes_{k[[x]]} k$$ to get $$k \xrightarrow{0} k \xrightarrow{\operatorname{id}} k\to 0$$.
• The special SES $$0\to M \to M \oplus N \to N \to 0$$ has a left-section $$s: M \oplus N\to M$$; applying $$({-})\otimes_A S$$ or $$\mathop{\mathrm{Hom}}_A(S, {-})$$ actually produces a SES, since this induces left sections on the resulting sequences.
• Prove that free modules are projective.
• Prove that divisible abelian groups are injective in $${\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}$$.
• Prove that $${\mathbb{Q}}$$ and $${\mathbb{Q}}/{\mathbb{Z}}$$ are injective.
• Show that a SES splits iff it admits a right section or a left section.
• Show that $$0\to I\to M\to P\to 0$$ splits if either $$P$$ is projective or $$I$$ is injective.
• Show that $$({-})\otimes_A S, \mathop{\mathrm{Hom}}_A({-}, S), \mathop{\mathrm{Hom}}_A(S, {-})$$ are exact on SES’s $$0\to A\to B \to P\to 0$$ with $$P$$ projective.

# 8 Thursday, February 03

Exercises:

• Find a way to remember which $$\hom$$ is covariant vs contravariant.
• Show $$P$$ is projective $$\iff \mathop{\mathrm{Hom}}(P, {-})$$ is exact, $$I$$ is injective $$\iff \mathop{\mathrm{Hom}}({-}, I)$$ is exact (sends injections to surjections).
• Find a non-free projective module.
• Show that $$P$$ is projective iff $$P$$ is a direct summand of a free module. Use that $$0\to K\to A{ {}^{ \scriptscriptstyle\oplus^{I} } } \to P\to 0$$ and lift $$P \xrightarrow{\operatorname{id}_P} P$$ to get $$A{ {}^{ \scriptscriptstyle\oplus^{I} } } = K \oplus P$$. For the other direction:

• Show that in $${\mathbb{Z}{\hbox{-}}\mathsf{Mod}}$$, $$I$$ is injective iff divisible.
• For one direction, show $$ni' = i$$ using the following diagram:

• For the other direction, produce a map:

• Use Zorn’s lemma on pairs $$(Y, f)$$ where $$(Y, f) \leq (Y', g) \iff Y \subseteq Y'$$ and $${ \left.{{g}} \right|_{{Y}} } = f$$. Show every chain has an upper bound by setting $$Y_\infty = \displaystyle\bigcup Y_i$$ and $$f_\infty = \cup f_i$$, thus producing $$Y_{\mathrm{max}}, f_{\mathrm{max}}$$ Take a SES $$0 \to \left\langle{y}\right\rangle \to \left\langle{y, Y_{\mathrm{max}}}\right\rangle \to \left\langle{y, Y_{\mathrm{max}}}\right\rangle/\left\langle{y}\right\rangle\to 0$$ and map the last term to $$I$$.

• Recall that projective resolutions are complexes $${ {P}_{\scriptscriptstyle \bullet}} = \cdots P_1\to P_0\to 0$$ with $$H_{i>0}{ {P}_{\scriptscriptstyle \bullet}} = 0$$ and $$H_0{ {P}_{\scriptscriptstyle \bullet}} = M$$, equivalently an exact complex $$\cdots\to P_1\to P_0\to M\to 0$$.

• Find a projective resolution of $$C_2\in{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}$$ and $$k[t]/\left\langle{t^2}\right\rangle\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$. Why must the latter necessarily be infinite length?

• Compute $$\operatorname{Tor}_*^{\mathbb{Z}}(C_2, C_2) \in {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}$$ and $$\operatorname{Tor}_*^{\mathbb{Z}}({\mathbb{Z}}{ {}^{ \scriptscriptstyle\oplus^{2} } }, C_2)$$.

• Show $$\operatorname{Tor}_*^{R}(k, k) = \bigoplus_{i\geq 0} k$$ for $$R = k[t]/\left\langle{t^2}\right\rangle$$? Use the resolution $$P_i = k[t]/\left\langle{t^2}\right\rangle$$ with maps $$({-})\times t$$, using that $$t\curvearrowright k$$ by zero.

• Show that if $$f\simeq g$$, then $$f-g$$ induces the zero map in homology. Use $$d_{i+1} s_i + s_{i-1} d_i: H_i(C)\to H_i(D)$$, pick $$\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu\in C_i$$ with $$d_i \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu = 0$$ and check $$(d_{i+1} s_i + s_{i-1} d_{i})\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu\in \operatorname{im}d_{i+1}$$.

# 9 Tuesday, February 08

Show that chain-homotopic maps induce the same map in homology.

$${ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N)$$ for $$M, N\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ is well-defined, since the identity $$M \xrightarrow{\operatorname{id}_M} M$$ induces an isomorphism on $${ {P}_{\scriptscriptstyle \bullet}} \otimes N \to { {P}_{\scriptscriptstyle \bullet}} '\otimes N$$ for any two projective resolution $${ {P}_{\scriptscriptstyle \bullet}} , { {P}_{\scriptscriptstyle \bullet}} ' \rightrightarrows M$$.

If $$f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(M, N)$$, then there is an induced morphism $$\tilde f\in \mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}({ {P}_{\scriptscriptstyle \bullet}} ^M, { {P}_{\scriptscriptstyle \bullet}} ^N)$$ between resolutions $${ {P}_{\scriptscriptstyle \bullet}} ^M\rightrightarrows M, { {P}_{\scriptscriptstyle \bullet}} ^N\rightrightarrows N$$, where $$\tilde f$$ is unique up to homotopy.

For existence, use projectivity to lift through surjections onto kernels. For uniqueness, create $$s_0$$ such that $$d_1 s_0 = d_0-g_0$$ and check that $$\operatorname{im}f_0 - g_0 \subseteq \ker d_0$$ after lifting through $$P_1^N\to \ker d_0$$:

Show that a SES of modules induces a SES of chain complexes between their projective resolutions.

Hint: use the following diagram.

Show that a SES of chain complexes induces a LES in their homology. There are about eight conditions one needs to check here (as in the snake lemma, of which this is a special case for two-term complexes).

Show that given $$0\to M_1 \to M_2\to M_3\to 0$$, taking projective resolutions and applying $$({-})\otimes N$$ yields a SES \begin{align*} 0\to { {P}_{\scriptscriptstyle \bullet}} ^1\otimes N \to { {P}_{\scriptscriptstyle \bullet}} ^2\otimes N \to { {P}_{\scriptscriptstyle \bullet}} ^3\otimes N\to 0 ,\end{align*} so there is an induced LES in $${ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}$$.

Show

• $$\operatorname{Tor}_0^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) = M\otimes_R N$$
• $$\tau_{\geq 1} { {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) = 0$$ if either $$M$$ or $$N$$ is projective.
• $${ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^R(M, N)$$ is uniquely determined by these properties.

Hint: \begin{align*} \operatorname{Tor}_0^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) &= H_0({ {P}_{\scriptscriptstyle \bullet}} ^M \otimes_R N) \\ &= \operatorname{coker}\qty{P_1\otimes N \to P_0\otimes_R N}\\ &= \operatorname{coker}(P_1\to P_0) \otimes_R N \\ &= M\otimes_R N .\end{align*}

For vanishing, use that projective implies flat and exact complexes have zero higher homology. Note that if $$M$$ is projective, it is its own projective resolution.

For uniqueness, induct on $$i$$: write $$0\to K \to R{ {}^{ \scriptscriptstyle\oplus^{J} } }\to M\to 0$$, use that free implies projective, and consider the LES:

Now induct up using the isomorphisms in the LES.

Show that \begin{align*} { {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) \cong {{ {H}_{\scriptscriptstyle \bullet}} }({ {P}_{\scriptscriptstyle \bullet}} ^M \otimes_R N) \cong {{ {H}_{\scriptscriptstyle \bullet}} }(M\otimes_R { {P}_{\scriptscriptstyle \bullet}} ^N )\cong { { \operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(N, M) .\end{align*}

# 10 Thursday, February 10

Let $$R= k[x,y]$$ and $$M = k \cong k[x,y]/\left\langle{x, y}\right\rangle$$, and compute $${ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(k, k)$$.

Hint: use the following resolution

Compute $${ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(k, k)$$ for $$R=k[x_1, \cdots, x_{n}]$$ and $$M = k = k[x_1, \cdots, x_{n}]/\left\langle{x_1,\cdots, x_n}\right\rangle$$.

$$\operatorname{Tor}$$ measures failure of injectivity of tensoring against a module $$M$$, $${ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}}$$ measures failure of surjectivity when mapping against $$M$$.

Show that for $$R\in\mathsf{CRing}$$, every $$M\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ admits an injective resolution.

Hint: it suffices to show any $$M$$ injects into an injective object. Use the following diagram:

First show this for $$R={\mathbb{Z}}$$ using that $${\mathbb{Q}}{ {}^{ \scriptscriptstyle\oplus^{J} } }/K$$ is divisible and thus injective:

Now reduce to $$R={\mathbb{Z}}$$ using that $$M\hookrightarrow D$$ for $$D$$ a divisible abelian group, and show $$M\hookrightarrow I \mathrel{\vcenter{:}}=\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(R, D)\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$. Form the map as the composition:

Then show that $$I\mathrel{\vcenter{:}}=\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(R, D)$$ is injective using the universal property:

Show $${\mathbb{Z}}\in {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}$$ has an injective resolution.

\begin{align*} 0 \to {\mathbb{Z}}\hookrightarrow{\mathbb{Q}}\twoheadrightarrow{\mathbb{Q}}/{\mathbb{Z}}\to 0 .\end{align*}

Show $${ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(C_3, {\mathbb{Z}}) \cong C_3[1]$$.

Hint: apply $$\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(C_3, {-})$$ to the above resolution and use $$\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(C_3, {\mathbb{Q}}/{\mathbb{Z}}) \cong C_3$$

Show that if $$f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(M, N)$$ then there is an induced chain map $$\tilde f\in \mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}( { {I}^{\scriptscriptstyle \bullet}} _M, { {I}^{\scriptscriptstyle \bullet}} _N)$$ which is unique up to homotopy. Conclude that $${ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N)$$ is independent of injective resolution.

Hint: take $$f=\operatorname{id}_M$$.

Show that if $$0\to A\to B\to C\to 0$$ is a SES in $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ with $${ {I}^{\scriptscriptstyle \bullet}} _A \leftleftarrows A, { {I}^{\scriptscriptstyle \bullet}} _C\leftleftarrows B$$, then there is a complex $${ {I}^{\scriptscriptstyle \bullet}} _B\leftleftarrows B$$ making $$0\to { {I}^{\scriptscriptstyle \bullet}} _A \to { {I}^{\scriptscriptstyle \bullet}} _B \to { {I}^{\scriptscriptstyle \bullet}} _C \to 0$$ a SES in $$\mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$.

Show that a SES in $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ induces a LES in $${ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}$$. Do this for both homs: start with $$0\to A\to B\to C\to 0$$, and produce a LES for $${ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}} (M, {-})$$ and $${ { \operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}({-}, M)$$.

Hint: for the first case, apply $$\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, {-})$$ to the SES of chain complexes of injective resolutions, and use that if $$I_1$$ is injective then the SES splits. For the second, use that $$\mathop{\mathrm{Hom}}({-}, I)$$ is exact iff $$I$$ is injective and take an injective resolution of $$M$$.

Show that $${ { \operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N)$$ is uniquely characterized by

1. $$\operatorname{Ext} ^0_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) \cong \mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N)$$
2. $$\tau_{\geq 1} { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) = 0$$ if $$M$$ is projective or $$N$$ is injective.
3. The two LESs above exist.

Hints:

• Resolve $${ {I}^{\scriptscriptstyle \bullet}} _N \leftleftarrows N$$, apply $$\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M,{-})$$, and identify $$\operatorname{Ext} ^0_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}$$ as a kernel.
• $$N$$ is its own injective resolution when $$N$$ is injective.
• $$\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M,{-})$$ is exact when $$M$$ is projective.
• For uniqueness, use that if $$0\to N\to I \to C\to 0$$ with $$I$$ injective, then the middle terms in the LES vanish to get isomorphisms.

# 11 Tuesday, February 15

Check that $$\operatorname{Ext} _R^i(M, {-})$$ is independent of injective resolutions, and $$\operatorname{Ext} _R^i({-}, N)$$ is independent of projective resolutions.

Check that $${ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}}$$ is determined by

• $$\operatorname{Ext} ^0 = \mathop{\mathrm{Hom}}$$
• $$\operatorname{Ext} ^{i>0}(P, I) = 0$$ if $$P$$ is projective or $$I$$ is injective.
• It extends SESs to LESs

Show \begin{align*} { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{k[t]/\left\langle{t^2}\right\rangle } (k, k) = \bigoplus_{i\geq 0} k[i] .\end{align*} Use the projective resolution with entries $$k[t]/\left\langle{t^2}\right\rangle$$ with differential $${\partial}= \cdot t$$

Compute $${ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{k[x_1, \cdots, x_{n}]}(k, k)$$ using the Koszul resolution $${ {\bigwedge\nolimits}^{\scriptscriptstyle \bullet}} k[x_1, \cdots, x_{n}]\rightrightarrows k$$.

Defining Noetherian rings and modules:

• $$A\in \mathsf{CRing}$$ is Noetherian iff every $$\operatorname{Id}(A) \subseteq {\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$.
• $$M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}$$ is Noetherian iff every $$N\leq M$$ satisfies $$N\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$

Thank you Emmy Noether!!

Show that TFAE:

• $$R$$ is Noetherian
• Every $$M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$ is Noetherian.

For $$1\implies 2$$, it STS that $$R^n$$ is Noetherian. To reduce, use the diagram