# 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, $$R\mathrel{\vcenter{:}}={\mathbb{Z}} \left[ { { {S}^{-1}} } \right]$$ 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, $${\mathbb{Z}}_f = {\mathbb{Z}}{ \left[ { \frac{1}{p} } \right] } \subseteq {\mathbb{Q}}$$ 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
• $${\mathbb{Z}}_{\left\langle{p}\right\rangle} = {\mathbb{Z}}{ \left[ { \frac{1}{\ell} } \right] }_{\ell\neq p \text{ prime}}$$, 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 $${ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(C_3, {\mathbb{Z}}) \cong C_3$$.

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 To show $$R^n$$ is Noetherian, use induction since we know $$R^1$$ is Noetherian. Use the following diagram, using the snake lemma on $$s_1, s_3$$ to show $$s_2$$ is surjective: Show TFAE:

• $$M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}$$ is Noetherian.
• The ACC for submodules.

Hint: for $$1\implies 2$$, set $$M_\infty = \displaystyle\bigcup_i M_i \leq M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}$$. Write $$M_i = \left\langle{x_1,\cdots, x_n}\right\rangle$$ and use that each $$x_k \in M_{i_k}$$ to choose $$N\gg 1$$ with $$M_\infty = M_N$$. For $$2\implies 1$$, for $$M$$ non-Noetherian find $$S\leq M$$ infinitely generated as $$S = \left\langle{x_1,\cdots}\right\rangle$$ and take the chain $$\left\{{S_k}\right\}_{k\geq 0}$$ where $$S_k = \left\langle{x_1,\cdots, x_k}\right\rangle$$.

Show that the following are Noetherian:

• Fields
• PIDs
• $$R/I$$ for $$R$$ Noetherian and $$I$$ arbitrary
• For $$A$$ Noetherian, $$A[x]$$ (Hilbert’s basis theorem).1
• Similarly $$A{\left[\left[ x \right]\right] }$$.
• Localizations of Noetherian rings

If $$A$$ a Noetherian local ring and $$M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{{\mathrm{fg}}, \mathop{\mathrm{proj}}}$$, then $$M$$ is free.

Prove this!

Hints:

• Surjectivity:
• Choose a basis $$M/{\mathfrak{m}}M = \left\langle{\left\{{ \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu}\right\}_{k\leq n} }\right\rangle$$ and lifts $$x_k$$ to $$M$$.
• Take a surjective map $$A^n \to M$$ where $$e_i \mapsto x_i$$.
• Take the SES $$A^n \xrightarrow{f} M \twoheadrightarrow\operatorname{coker}f\to 0$$ and apply $$({-})\otimes_A A/{\mathfrak{m}}$$; use that $$(A/{\mathfrak{m}})^n { \, \xrightarrow{\sim}\, }M/{\mathfrak{m}}M$$ and apply Nakayama.
• Injectivity:
• Write $$0\to K = \ker f \to A^n \xrightarrow{f} M \to 0$$ and apply $$({-})\otimes_A A/{\mathfrak{m}}$$ to get \begin{align*} \cdots \to \operatorname{Tor}^0(M, A/{\mathfrak{m}}) \to K/{\mathfrak{m}}K \to (A/{\mathfrak{m}})^n { \, \xrightarrow{\sim}\, }M/{\mathfrak{m}}M\to 0 ,\end{align*} then $$K/{\mathfrak{m}}K = 0$$ and apply Nakayama to conclude $$K=0$$.

Try an example: $$k[x]\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$, which is free after reduction mod $${\mathfrak{m}}= \left\langle{x}\right\rangle$$ but not free before reduction.

Note that this works with projective replaced by flat.

Why care about Noetherian rings? Hilbert studied group actions $$G$$ on modules over (say) polynomial rings and wanted to find the submodules of $$G{\hbox{-}}$$invariants. There was an industry of writing down generating sets in order to show existence and finiteness, and the basis theorem (which is partially effective) showed that this is no longer necessary – finite generating sets always exist.

If $$A$$ is Noetherian then $$A[x]$$ is Noetherian.

Fix $$U \subseteq A[x]$$ an ideal, so $$I = \left\{{f\in I{~\mathrel{\Big\vert}~}f = \sum a_{f, i} x^i}\right\}$$. Write $$J = \left\langle{a_{f, \deg f}}\right\rangle$$ be the ideal generated by all leading coefficients of elements in $$I$$. Since $$A$$ is Noetherian, $$J$$ is finitely-generated, so write $$J = \left\langle{a_1,\cdots, a_n}\right\rangle$$ and choose $$\left\{{f_1,\cdots, f_n}\right\}$$ so that $$a_i$$ is the leading coefficient of $$f_i$$. Note that these exist since $$J = A\left\{{a_1,\cdots, a_n}\right\}$$ (i.e. these already form an ideal). Consider $$L\mathrel{\vcenter{:}}= I \cap A[x]^{\deg \leq d}$$: since $$A$$ is Noetherian, $$A[x]^{\deg \leq d} \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$, and $$L$$ also forms a finitely generated $$A{\hbox{-}}$$module. Write the generators as $$L = \left\{{g_1,\cdots, g_m}\right\}$$.

\begin{align*} I = I_{fg} \mathrel{\vcenter{:}}=\left\langle{f_1,\cdots, f_n, g_1,\cdots, g_m}\right\rangle .\end{align*}

If not, pick $$f\in I\setminus I_{fg}$$ of minimal degree. Then $$\deg f > d$$ by construction of $$L$$. Write $$f = \sum b_i x^i$$ where $$b_{\deg f} = \sum c_i a_i$$, and check $$f - \sum c_i f_i x^{\deg f - \deg f_i}\in I$$.

Idea: the $$f_k$$ take care of low degree elements, the $$g_k$$ knock things down in degree.

Any quotient $$k[x_1, \cdots, x_{n}]/I$$ is Noetherian, as is $${\mathbb{Z}}[x_1,\cdots, x_n]/I$$.

Some examples:

• $$k[x_1,\cdots, ]$$ is not Noetherian: take $$I = \left\langle{x_1,\cdots}\right\rangle$$.
• $$k[t. t^{1\over 2}, t^{1\over 3}, t^{1\over 4},\cdots]$$.
• $${\mathbb{Z}} { \left[ { \left\{{ 2^{1\over n}}\right\}_{n\geq 0} } \right] }$$. Note this is countable!

Next time: other finiteness conditions, integrality, the Nullstellensatz.

# 12 Thursday, February 17

Prove the correspondence theorem between $$\operatorname{Id}(A)$$ and $$\operatorname{Id}(S^{-1}A)$$. Use that

• $$\iota^{-1}I\in \operatorname{Id}(A)$$
• $$\iota(\iota^{-1}(I))$$ generates $$I$$.
• $${a\over s} = {1\over s}{a\over 1}$$ and $${a\over 1}$$ is in the image of the lower map.

Show that the localization of any Noetherian ring is again Noetherian.

Say $$M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{ \operatorname{fp} }$$ iff there is an exact sequence $$R^a\to R^b\to M\to 0$$. Show that if $$R$$ is Noetherian, $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{ \operatorname{fp} }= {\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{{\mathrm{fg}}}$$.

For $$f\in \mathsf{CRing}(A, B)$$ and $$b\in B$$ define \begin{align*} f_b: A[x] &\to B \\ x&\mapsto b \\ \sum a_i x^j &\mapsto \sum f(a_i) b^i .\end{align*} The element $$b$$ is integral over $$A$$ iff $$\operatorname{im}f_b \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$. The ring $$B$$ is integral over $$A$$ iff every $$b\in B$$ is integral over $$A$$ as above.

Show that $${\mathbb{Z}}{ \left[ { \frac{1}{2} } \right] } \subseteq {\mathbb{Q}}$$ is not integral.

Show TFAE:

• $$b$$ is integral over $$A$$
• $$b$$ satisfies a monic polynomial over $$A$$, so $$b^n + \sum_{i = 1}^n a_i b^{n-i} = 0$$.

Hints:

• Write $$\operatorname{im}f_b = I\mathrel{\vcenter{:}}=\left\langle{1,b,\cdots, b^{n-1}}\right\rangle$$, suppose $$b^n\in I$$, and show $$b^{n+1} \in I$$ by expanding as a sum.
• Set $$M_i = \left\langle{1,b, \cdots, b^i}\right\rangle$$ and use that these stabilize to conclude $$b^n \in M_{n-1}$$ for some $$n$$.

$$B$$ Noetherian over $$A$$ does not imply $$B$$ is Noetherian! Consider $$\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu$$, the algebraic integers, which is integral over $${\mathbb{Z}}$$. Note that $$\left\langle{2^{1\over n}}\right\rangle$$ is not finitely-generated.

# 13 Toward the Nullstellensatz

Idea: there is a dictionary between $$k[x_1, \cdots, x_{n}]$$ and $${\mathbb{A}}^n_{/ {k}}$$ for $$k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }$$:

• Points in $${\mathbb{A}}^n$$ correspond to maps $$k[x_1, \cdots, x_{n}]\to k$$ given by $$x_i\mapsto c_i\in k$$.
• Rings $$R$$ correspond to their prime spectra $$\operatorname{Spec}R$$.
• Kernels correspond to maximal ideals
• Ideals $$I\in \operatorname{Id}(R)$$ correspond to $$V(I) =\left\{{{\mathfrak{m}}{~\mathrel{\Big\vert}~}I \subseteq {\mathfrak{m}}}\right\}$$
• Ideals $$I\in \operatorname{Id}(k[x_1, \cdots, x_{n}])$$ correspond to $$V(I) = \left\{{\mathbf{c}\in k{ {}^{ \scriptscriptstyle\times^{n} } } {~\mathrel{\Big\vert}~}f(\mathbf{c}) = 0 \, \forall f\in I}\right\}$$.

Any $${\mathfrak{m}}\in \operatorname{mSpec}k[x_1, \cdots, x_{n}]$$ is given by $$\ker \pi_p$$ for some morphism \begin{align*} \pi_p: k[x_1, \cdots, x_{n}]\to k \\ x_i &\mapsto c_i \end{align*} for some $$p = {\left[ {c_1,\cdots, c_n} \right]} \in k{ {}^{ \scriptscriptstyle\times^{n} } }$$. Equivalently, $${\mathfrak{m}}= \left\langle{x_1-c_1,\cdots, x_n - c_n}\right\rangle$$.

If $$f\in k[x_1, \cdots, x_{n}]$$ satisfies $${ \left.{{f}} \right|_{{V(I)}} } = 0$$ for a fixed $$I\in \operatorname{Id}(k[x_1, \cdots, x_{n}])$$, then $$f^n\in I$$ for some $$n>0$$. I.e. there is a bijective correspondence $$I\rightleftharpoons V(I)$$ for radical ideals $$\sqrt{I} = I$$.

Necessity of conditions:

• Why $$k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }$$ is needed: take $$\left\langle{x^-2}\right\rangle\in {\mathbb{Q}}[x]$$, this is a maximal ideal but $$V(I) = 0$$.

• Why the radical condition is needed: take $$I = \left\langle{x^{10} }\right\rangle$$ so $$V(I) = \left\{{0}\right\} \subseteq k$$, but $${ \left.{{x}} \right|_{{V(I)}} } = 0$$.

For $$k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }$$ and $$R\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}$$, \begin{align*} \operatorname{mSpec}R \rightleftharpoons\mathop{\mathrm{Hom}}_{{\mathsf{Alg}_{/k} }}(R, k) .\end{align*}

$$V(f) = k{ {}^{ \scriptscriptstyle\times^{n} } } = V(0)$$ iff $$f\in {\sqrt{0_{R}} }$$.

Let $$R\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}$$, so $$R = k[x_1, \cdots, x_{n}]/I$$ for some $$n$$, and let $$V(I) \subseteq k{ {}^{ \scriptscriptstyle\times^{n} } }$$. Given $$J \subseteq R$$ radical, $$V(J) \subseteq V(I)$$, and $$f\in R$$ vanishes on $$V(J)$$ iff $$f^n\in J$$ for some $$n>0$$.

For $$k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }$$ and $$A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}$$ with $${\mathfrak{m}}\in \operatorname{mSpec}A$$, \begin{align*} A/{\mathfrak{m}}\cong k .\end{align*}

For $$k$$ an arbitrary field and $$A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}$$ with $${\mathfrak{m}}\in \operatorname{mSpec}A$$, \begin{align*} A/{\mathfrak{m}} \text{ is a finite extension of } k .\end{align*}

If $$k \subseteq F$$ fields with $$F\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}$$, then $$F_{/ {k}}$$ is a finite extension of fields.

If $$R\in {\mathsf{Alg}_{/{\mathbb{Z}}}}^{\mathrm{fg}}$$ and $${\mathfrak{m}}\in \operatorname{mSpec}R$$, then $$R/{\mathfrak{m}}$$ is finite.

Check $${\mathfrak{m}}\cap{\mathbb{Z}}= {\mathfrak{p}}\in \operatorname{Spec}{\mathbb{Z}}$$ and $$R/{\mathfrak{m}}$$ is a finitely generated extension of $${\mathbb{F}}_p$$ and hence finite.

If $$A\in \mathsf{CRing}^{ \mathrm{Noeth} }$$ and $$A \subseteq B \subseteq C$$ with

• $$C\in {{A}{\hbox{-}}\mathsf{Alg}}^{\mathrm{fg}}$$ and
• $$C\in {\mathsf{B}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}$$,

then $$B\in {{A}{\hbox{-}}\mathsf{Alg}}^{\mathrm{fg}}$$.

Sketch:

• Choose $$x_1,\cdots, x_n$$ generating $$C$$ as an $$A{\hbox{-}}$$algebra
• Choose $$y_1,\cdots, y_m$$ generating $$C$$ as a $$B{\hbox{-}}$$module.
• Write $$x_i = \sum_j b_{ij} y_j$$ with $$b_{ij}\in B$$
• Write $$x_i x_j = \sum_k b_{ijk} y_k$$ with $$b_{ijk}\in B$$
• Let $$B_0 \subseteq B$$ be the $$A{\hbox{-}}$$algebra generated by the $$b_{ij}$$ and $$b_{ijk}$$.
• Observe that $$B_0$$ is Noetherian by the Hilbert basis theorem since it’s finitely generated as an $$A{\hbox{-}}$$algebra.
• For any $$c\in C$$, write $$c = \sum_i b_i y_i$$ since the $$y_i$$ generate $$C$$ as a $$B_0$$ module.
• Idea: can rewrite in terms of lower degree monomials?
• Since each $$b_i = \sum a_{I}^i x^I$$, we have $$c = \sum_i \sum_I a_I^i x^I y_i$$, which is a polynomial in $$\sum b_{ij} y_j$$.
• Then $$y_{i} \cdots y_{i_k} = \sum p_s y_s$$ where the $$p_s$$ are polynomials in the $$b_{ijk}$$?
• Since $$B_0$$ is Noetherian and $$B \subseteq C$$, $$B$$ is finitely generated as a $$B_0$$ module and thus as a $$B_0$$ algebra.
• Since $$A \subseteq B_0 \subseteq B$$ and $$B_0$$ is finitely-generated as an $$A$$ algebra and $$B$$ is finitely-generated as a $$B_0$$ algebra, we have that $$B$$ is finitely-generated as an $$A$$ algebra.

# 14 Thursday, February 24

## 14.1 Going up, going down

Let $$A \subseteq B\in \mathsf{CRing}$$ with $$B$$ integral over $$A$$. Then for every $$p\in \operatorname{Spec}A$$ there is a $$q\in \operatorname{Spec}B$$ with $$p = A \cap q$$.

Idea: $$\operatorname{Spec}B\to \operatorname{Spec}A$$ has finite fibers.

If $$A \hookrightarrow B$$ is an integral extension, then $$A\in \mathsf{Field}\iff B\in \mathsf{Field}$$.

$$\implies$$: Last time.

$$\impliedby$$: Given $$x\in A$$ we have $$x^{-1}\in A$$. Then if $$f(x) = x^{-n} + \sum_{0\leq k \leq n-1} a_k x^{-k}$$, with $$f(x) = 0$$, then $$x^{n-1}f(x) = 0 \implies x^{-1}= -\sum a_k x^k \in A$$.

Note $$A \left[ { { {p}^{-1}} } \right] \hookrightarrow B \left[ { { {p}^{-1}} } \right]$$ is still integral, and let $$q'\in \operatorname{mSpec}B \left[ { { {p}^{-1}} } \right] \neq \emptyset$$. Write $$p$$ for the maximal ideal of $$A$$, then (claim) $$A \left[ { { {p}^{-1}} } \right] \cap q' = p$$. STS $$A \left[ { { {p}^{-1}} } \right] \cap q'$$ is maximal, since these are local rings, and so it’s ETS $$A \left[ { { {p}^{-1}} } \right]/(A \left[ { { {p}^{-1}} } \right] \cap q') \in \mathsf{Field}$$. Since this is integral iff $$B \left[ { { {p}^{-1}} } \right]/q'$$ is integral, which it is, by the lemma this is a field too.

Show tat taking the preimage of $$q'$$ in $$B$$ works.

Given $$A \hookrightarrow B$$ integral and \begin{align*} p_1 &\subseteq p_2 \subseteq \cdots \subseteq p_n \in \operatorname{Spec}A \\ q_1 &\subseteq q_2 \subseteq \cdots \subseteq q_m \in \operatorname{Spec}B \\ ,\end{align*} where $$p_i = q_i \cap A$$ and e.g. with $$m\leq n$$, then there exist $$q_{m+1}, \cdots, q_n$$ with $$q_j \subseteq q_{j+1}$$ and $$q_i \cap A = p_i$$.

Note that it’s enough to lift one stage and induct. So given $$q_1 \subseteq \cdots q_n$$, it’s ETF $$q_{m+1} \supseteq q_m$$ with $$q_{m+1} \cap A = p_{m+1}$$. Strategy:

• Replace $$A$$ with $$A/p_n$$ and $$B$$ with $$B/q_n$$.
• Find $$\mkern 1.5mu\overline{\mkern-1.5muq\mkern-1.5mu}\mkern 1.5mu \in B/q_m$$ with $$\mkern 1.5mu\overline{\mkern-1.5muq\mkern-1.5mu}\mkern 1.5mu \cap A/p_n$$ the image of $$p_{n+1}$$
• Use that $$\mkern 1.5mu\overline{\mkern-1.5muq\mkern-1.5mu}\mkern 1.5mu$$ exists by Going Up I.

The geometry: $$A\to B\leadsto \operatorname{Spec}B \xrightarrow{\pi} \operatorname{Spec}A$$. Increasing chains $$p_i$$ means $$p_{i+1}\in { \operatorname{cl}} _{\operatorname{Spec}A} p_i$$, and “going up” means sequences can be completed with points in closures in $$\operatorname{Spec}B$$ I.e. $$\pi$$ is a closed map, i.e. closed under specialization (passing to a point in the closure). Idea: covering map, possibly with ramification or splitting.