 Lecture 1: Here is my note for the first lecture.
In this lecture, we recalled that \({\rm Max}(A)\) is the set of all the maximal ideals of \(A\) and \({\rm Spec}(A)\) is the set of all the primes
ideals of \(A\); \({\rm Max}(A)\subseteq {\rm Spec}(A)\); for any proper ideal \(\mathfrak{a}\) there is \(\mathfrak{m}\in {\rm Max}(A)\) such that
\(\mathfrak{a}\subseteq \mathfrak{m}\); for any multiplicatively closed subset \(S\) and \(\mathfrak{a}\unlhd A\) such that \(S\cap \mathfrak{a}=\varnothing\)
there is \(\mathfrak{p}\in {\rm Spec}(A)\) such that \(\mathfrak{a}\subseteq \mathfrak{p}\) and \(S\cap \mathfrak{p}=\varnothing\);
\({\rm Nil}(A)=\cap_{\mathfrak{p}\in {\rm Spec}(A)}\mathfrak{p}\); we defined the Jacobson radical \(J(A)\) of a unital commutative ring, and proved \(x\in J(A)\)
if and only if, for any \(y\in A\), \(1xy\in A^{\times}\); next we defined coprime ideals; pointed out that if \(\mathfrak{m}\) is a maximal ideal which is not a
divisor of \(\mathfrak{a}\), then \(\mathfrak{a}\) and \(\mathfrak{m}\) are coprime; proved that if \(\{\mathfrak{a}_i\}_i\) is a set of pairwise coprime ideals, then
\(\cap_{i=1}^n \mathfrak{a}_i=\prod_{i=1}^n \mathfrak{a}_i\); let
\(\phi:A\rightarrow \prod_{i=1}^n A/\mathfrak{a}_i, \hspace{1mm} \phi(a)=(a+\mathfrak{a}_1,\ldots,a+\mathfrak{a}_n)\). Then \(\phi\) is surjective if and only if
\(\mathfrak{a}_i\)'s are pairwise coprime, and \(\phi\) is injective if and only if \(\cap_{i=1}^n \mathfrak{a}_i=0\).
 Lecture 2: Here is my note for the second lecture.
We proved that, suppose \(\mathfrak{a}\unlhd A\), \(\mathfrak{p}_i\in {\rm Spec}(A)\), and \(\mathfrak{a}\subseteq \mathfrak{p}_1\cup \cdots \cup \mathfrak{p}_n\); then \(\mathfrak{a}\subseteq \mathfrak{p}_i\) for some \(i\); we mentioned McCoy's theorem which is the modified version of this statement after removing the primeness assumption; then we defined the set \(V(\mathfrak{a})\) of prime divisors of an ideal \(\mathfrak{a}\); we proved that (1) \(V((1))=\varnothing\), \(V(0)={\rm Spec}(A)\); (2) \(V(\sum_{i\in I} \mathfrak{a}_i)=\cap_{i\in I} V(\mathfrak{a}_i)\); and (3) \(V(\mathfrak{a}_1\cdots \mathfrak{a}_n)=\cup_{i=1}^n V(\mathfrak{a}_i)\). In particular, \(\{V(\mathfrak{a})\}_{\mathfrak{a}\unlhd A}\) can be considered as closed sets of a topology on \({\rm Spec}(A)\). This is called the Zariskitopology on \({\rm Spec}(A)\); next we defined the contraction map associated with a ring homomorphism \(f:A\rightarrow B\), and proved that we get a welldefined continuous map \(f^{\ast}:{\rm Spec}(B)\rightarrow {\rm Spec}(A)\); along the way we defined the extension of an ideal and proved \((f^{\ast})^{1}(V(\mathfrak{a}))=V(\mathfrak{a}^e)\); then we showed that \(\pi^{\ast}:{\rm Spec}(A/\mathfrak{a})\rightarrow {\rm Spec}(A)\) induces a bijection between \({\rm Spec}(A/\mathfrak{a})\) and \(V(\mathfrak{a})\) where \(\pi:A\rightarrow A/\mathfrak{a}\) is the natural quotient map; Finally we mentioned if \(S\) is a multiplicatively closed subset of \(A\) and \(f:A\rightarrow S^{1}A, f(a):=\frac{a}{1}\), then \(f^{\ast}\) induces a bijection from \({\rm Spec}(S^{1}A)\) to \(\{\mathfrak{p}\in {\rm Spec}(A) \mathfrak{p}\cap S=\varnothing\}\). We will prove this statement in the next lecture.
 Lecture 3: Here is my note for the third lecture.
In this lecture, first we proved that \(f^{\ast}:{\rm Spec}(S^{1}A)\rightarrow {\rm Spec}(A)\)
is injective and its image is \(\{\mathfrak{p}\in {\rm Spec}(A) \mathfrak{p}\cap S=\varnothing\}\). Then we deduced that image of
\(f^{\ast}:{\rm Spec}(A_{\mathfrak{p}})\rightarrow {\rm Spec}(A)\) is \(\{\mathfrak{q}\in {\rm Spec}(A) \mathfrak{q}\subseteq \mathfrak{p}\}\). Next we
considered a ring homomorphism \(f:A\rightarrow B\) and described fibers of \(f^{\ast}\). We started the proof of the following claim: there is a bijection
between \((f^{\ast})^{1}(\mathfrak{p})\) and
\({\rm Spec}(B\otimes_A k(\mathfrak{p}))\), where \(k(\mathfrak{p}):=A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}\) is the residue field of \(\mathfrak{p}\).
 Lecture 4: Here is my note for the fourth lecture.
We proved that there is a canonical bijection between \((f^{\ast})^{1}(\mathfrak{p})\) and
\({\rm Spec}(B\otimes_A k(\mathfrak{p}))\); proved that if \(\phi\in {\rm End}_A(M)\) where \(M\) is a finitely generated \(A\)module and
\(\phi(M)\subseteq \mathfrak{a} M\) for some ideal \(\mathfrak{a}\) of \(A\), then for some \(a_i\in \mathfrak{a}\) we have
\(\phi^n+a_{n1}\phi^{n1}+\cdots+a_0=0\) in \({\rm End}_A(M)\); then we deduced that, if \(M\) is a finitely generated \(A\)module and
\(\mathfrak{a}M=M\), then for some \(x\in A\) we have \(xM=0\) and \(x\equiv 1 \pmod{\mathfrak{a}}\); used this result to show Nakayama's lemma. In the lecture
note, I included an application of Nakayama's lemma related to the minimum number of generators of a module.
 Lecture 5: Here is my note for the fifth lecture.
We defined primary ideals; proved radical of a primary ideal \(\mathfrak{q}\) is prime and it is the minimum element of \(V(\mathfrak{q})\); proved that if
radical \(\mathfrak{m}\) of an ideal \(\mathfrak{q}\) is maximal, then \(V(\mathfrak{q})=\{\mathfrak{m}\}\); and so for any \(x\not\in \mathfrak{q}\) we have
\((\mathfrak{q}:x)\subseteq \mathfrak{m}\), which implies \(\mathfrak{q}\) is \(\mathfrak{m}\)primary; deduced that powers of a maximal ideal is always primary; used
this to show primary ideals of a PID are precisely ideals generated by irreducible elements; next we showed if \(\mathfrak{q}\) is \(\mathfrak{p}\)primary, then

\((\mathfrak{q}:x)=A\) if \(x\in \mathfrak{q}\);

\((\mathfrak{q}:x)=\mathfrak{q}\) if \(x\not \in \mathfrak{p}\);

\((\mathfrak{q}:x)\) is \(\mathfrak{p}\)primary if \(x\not\in \mathfrak{q}\).
Finally we defined a primary decomposition and a reduced primary decomposition, and a decomposable ideal.
 Lecture 6: Here is my note for the sixth lecture.
In this lecture, first we proved that the intersection of two \(\mathfrak{p}\)primary ideals is again \(\mathfrak{p}\)primary; and so any decomposable ideal
has a reduced primary decomposition. Then we proved that if \(\mathfrak{a}=\cap_{i=1}^n \mathfrak{q}_i\) is a reduced primary decomposition and \(\mathfrak{q}_i\)
is \(\mathfrak{p}_i\)primary, then \(\{\mathfrak{p}_1,\ldots,\mathfrak{p}_n\}={\rm Spec}(A)\cap \{\sqrt{(\mathfrak{a}:x)}x\in A\}\); and so this set just depends on
\(\mathfrak{a}\). We denote this set by \({\rm Ass}(\mathfrak{a})\) and we say these primes are associated with \(\mathfrak{a}\); next we showed that
\(\{x\in A (\mathfrak{a}:x)\neq \mathfrak{a}\}=\cup_{\mathfrak{p}\in {\rm Ass}(\mathfrak{a})}\mathfrak{p}\), and deduced that if 0 is decomposable, then the set
\(D(A)\) of zerodivisors of \(A\) is \(\cup_{\mathfrak{p}\in {\rm Ass}(0)}\mathfrak{p}\); then we showed that \({\rm Ass}(\mathfrak{a})\subseteq V(\mathfrak{a})\)
and for any \(\mathfrak{p}\in V(\mathfrak{a})\) there is \(\mathfrak{p}'\in {\rm Ass}(\mathfrak{a})\) such that \(\mathfrak{p}'\subseteq \mathfrak{p}\); and so the
the set of minimal elements of \(V(\mathfrak{a})\) is the same as the set of minimal elements of \({\rm Ass}(\mathfrak{a})\). Therefore \({\rm Nil}(A)\) is the
intersection of minimal elements of \({\rm Ass}(0)\) if 0 is decomposable; last we described primary ideals of \(S^{1}A\), and we will prove this result in the
next lecture.
 Lecture 7: Here is my note for the seventh lecture.
Let \(\bar{e}:\{\mathfrak{q}\unlhd A \mathfrak{q} \text{ is } \mathfrak{p}\text{primary }\}\rightarrow
\{\tilde{\mathfrak{q}}\unlhd S^{1}A \tilde{\mathfrak{q}} \text{ is } \tilde{\mathfrak{p}}\text{primary }\}\),
\(\bar e(\mathfrak{q}):=\mathfrak{q}^e\), and
\(\bar{c}:\{\tilde{\mathfrak{q}}\unlhd S^{1}A \tilde{\mathfrak{q}} \text{ is } \tilde{\mathfrak{p}}\text{primary}\}\rightarrow
\{\mathfrak{q}\unlhd A \mathfrak{q} \text{ is } \mathfrak{p}\text{primary }\}, \bar c(\tilde{\mathfrak{q}})=\tilde{\mathfrak{q}}^c\). Then \(\bar e\) and \(\bar c\) are inverse of each other. Moreover, if \(\mathfrak{q}\) is a \(\mathfrak{p}\)primary ideal of \(A\) and \(\mathfrak{p}\cap S\neq \varnothing\), then
\(S^{1}\mathfrak{q}=S^{1}A\). Then we defined \(S(\mathfrak{a})=(\mathfrak{a}^e)^c\); and so the above result implies for a \(\mathfrak{p}\)primary ideal \(\mathfrak{q}\) we have \(S(\mathfrak{q})=\mathfrak{q}\) if \(\mathfrak{p}\cap S=\varnothing\) and \(S(\mathfrak{q})=A\) if \(\mathfrak{p}\cap S\neq\varnothing\). Next we extended this result to the setting of a reduced primary decomposition: suppose \(\mathfrak{a}=\cap_{i=1}^n \mathfrak{q}_i\) is a reduced primary decomposition, \(\mathfrak{q}_i\) is \(\mathfrak{p}_i\)primary, \(S\cap \mathfrak{p}_j=\varnothing\) for \(1\le j\le m\) and \(S\cap \mathfrak{p}_j\neq \varnothing\) for \(m< j\le n\). Then \(S^{1}\mathfrak{a}=\cap_{i=1}^m S^{1}\mathfrak{q}_i\) and \(S(\mathfrak{a})=\cap_{i=1}^m \mathfrak{q}_i\). Based on this result, we proved the second uniqueness theorem: suppose \(\mathfrak{a}\) is decomposable, \(\Sigma\subseteq {\rm Ass}(\mathfrak{a})\) is an isolated subset, and \(\cap_{i=1}^n \mathfrak{q}_i\) is a reduced primary decomposition. Then \(\cap_{\sqrt{\mathfrak{q}_i}\in \Sigma} \mathfrak{q}_i\) just depends on \(\Sigma\) and it is independent of the choice of a reduced primary decomposition. In particular, if \(\mathfrak{p}\) is a minimal element of \({\rm Ass}(\mathfrak{a})\), then the \(\mathfrak{p}\)factor of all the reduced primary decompositions are the same.
 Lecture 8: Here is my note for the eighth lecture.
In this lecture, we defined the Krull dimension of a ring. Observed that the Krull dimension of a field is 0, and for an integral domain \(D\),
\(\dim D=1\) if and only if \(0\not\in {\rm Max}(D)\) and \({\rm Spec}(D)=\{0\}\cup {\rm Max}(D)\). And so \(\dim D=1\) if \(D\) is a PID and not a field. Then
we proved that if \(D\) is an integral domain of dimension 1, and \(\mathfrak{a}\) is a proper nonzero decomposable ideal, then \(\mathfrak{a}\)
can be uniquely written as a product of primary ideals with distinct radicals. Next we defined irreducible ideal, and proved in a Noetherian ring any irreducible
ideal is primary. Using this result, we proved in a Noetherian ring any proper ideal is decomposable. As a corollary we showed in a Noetherian ring \(A\), there
are finitely many minimal prime ideals \(\{\mathfrak{p}_1,\ldots,\mathfrak{p}_n\}\) such that \(\overline{\{\mathfrak{p}_1,\ldots,\mathfrak{p}_n\}}={\rm Spec}(A)\).
 Lecture 9: Here is my note for the ninth lecture.
We defined integral extension. And proved that for a ring extension \(A\subseteq B\) and \(b\in B\), the following are equivalent:
(1) \(b\) is integral over \(B\); (2) \(A[b]\) is a finitely generated \(A\)module; (3) There is a subring \(C\) of \(B\) such that \(A[b]\subseteq C\)
and \(C\) is a finitely generated \(A\)module; (4) There is a faithful \(A[b]\)module that is finitely generated as an \(A\)module. Then
we defined the integral closure of \(A\) in \(B\), and proved it is a subring of \(B\); and it is integrally closed in \(B\). To show this result, we showed
that if \(B/A\) and \(C/B\) are integral extensions, then \(C/A\) is an integral extension. Then we defined integrally closed integral domain, and recalled
that a UFD is integrally closed. We also proved that the ring of integers of a number field is integrally closed.
 Lecture 10: Here is my note for the tenth lecture.
We proved that if \(B/A\) is an integral extension, then \(B/\mathfrak{b}\) is an integral extension of \(A/\mathfrak{b}\cap A\) and \(S^{1}B\) is an
integral extension of \(S^{1}A\) where \(S\) is a multiplicatively closed subset of \(A\). Then we defined integrally closed ring and proved that an integral
domain \(A\) is integrally closed if and only if \(A_{\mathfrak{m}}\) is integrally closed for any \(\mathfrak{m}\in {\rm Max}(A)\). Then we showed if \(B/A\)
is an integral extension, then \(A\) is a field if and only if \(B\) is a field. Next we showed that if \(f:A \rightarrow B\) is an integral embedding, then
\(f^{\ast}:{\rm Spec}(B)\rightarrow {\rm Spec}(A)\) is onto.
 Lecture 11: Here is my note for the eleventh lecture.
First we proved that if \(f:A \rightarrow B\) is an integral embedding, then \(f^{\ast}:{\rm Spec}(B)\rightarrow {\rm Spec}(A)\) is a closed map. Then we proved
GoingUp Theorem: supppose \(B/A\) is an integral extension, \(\mathfrak{p}_0\subsetneq \cdots \subsetneq \mathfrak{p}_n\) is a chain in \({\rm Spec}(A)\), and
\(\mathfrak{q}_0\subsetneq \cdots \subsetneq \mathfrak{q}_m\) is a chain in \({\rm Spec}(B)\) such that \(f^{\ast}(\mathfrak{q}_i)=\mathfrak{p}_i\). Then there
are \(\mathfrak{q}_{m+1},\ldots,\mathfrak{q}_n\in {\rm Spec}(B)\) such that \(\mathfrak{q}_m\subsetneq \cdots \subsetneq \mathfrak{q}_n\) and
\(f^{\ast}(\mathfrak{q}_i)=\mathfrak{p}_i\) for any \(i\). Next we showed that if \(f:A \rightarrow B\) is an integral embedding, then each fiber
\((f^{\ast})^{1}(\mathfrak{p})\) has dimension zero. Using this result and the GoingUp Theorem, we proved that if \(f:A \rightarrow B\) is an integral embedding,
then \(\dim A=\dim B\). Then we defined integral over an ideal, and proved the following: suppose \(B/A\) is a ring extension, \(C\) is the integral closure
of \(A\) in \(B\), and \(\mathfrak{a}\unlhd A\). Then \(b\in B\) is integral over \(\mathfrak{a}\) if and only if \(b\in \sqrt{\mathfrak{a}^{e}}\) where
\(\mathfrak{a}^e\) is the extension of \(\mathfrak{a}\) in \(C\).
 Lecture 12: Here is my note for the twelfth lecture.
First we proved the following result:
Let \(A\) be an integral domain that is integrally closed, and its field of fractions is \(F\). Suppose \(B\) is an integral extension of \(A\), and it is an
integral domain. Suppose \(\mathfrak{a}\unlhd A\) and \(b\in B\) is integral over \(\mathfrak{a}\). Suppose the minimal polynomial of \(b\) over \(F\) is
\(x^n+a_{n1}x^{n1}+\cdots+a_0\in F[x]\). Then \(a_i\in \sqrt{\mathfrak{a}}\) for any \(i\). Based on this result, we proved the GoingDown theorem:
Assumptions: \(A\): integral domain, integrally closed; \(B\): an integral extension over \(A\), integral domain;
\(\mathfrak{p}_0\subsetneq \cdots \subsetneq \mathfrak{p_n}\) is a chain in \({\rm Spec}(A)\);
\(\mathfrak{q}_{m+1}\subsetneq \cdots \subsetneq \mathfrak{q}_n\) is a chain in \({\rm Spec}(B)\);
Then Conclusion: there are \(\mathfrak{q}_0,\ldots,\mathfrak{q}_m\in {\rm Spec}(B)\) such that \(\mathfrak{q}_{0}\subsetneq \cdots \subsetneq \mathfrak{q}_n\) and \(\mathfrak{q}_i\cap A=\mathfrak{p}_i\).
To get a better understanding of this proof, we started an alternative approach.
 Lecture 13: Here is my note for the thirteenth lecture.
First we proved that: Let \(A\) be an integrally closed integral domain with field of fractions \(F\). Suppose \(E/F\) is a normal extension and \(B\) is the integral closure of \(A\) in \(E\). Let \(f:A\rightarrow B\) be the corresponding integral embedding. Then for any \(\mathfrak{p}\in {\rm Spec}(A)\) \({\rm Aut}(E/F)\) acts transitively on the fiber \((f^{\ast})^{1}(\mathfrak{p})\).
Along the way, we proved that if \({\rm Char}(F)=p\) and \(F':={\rm Fix}({\rm Aut}(E/F))\), then \(F'/F\) is a purely inseparable extension.
Based on this result, we presented a second proof for the GoingDown Theorem.
 Lecture 14: Here is my note for the fourteenth lecture.
First we proved:
Suppose \(A\) is an integrally closed integral domain, \(B\) is an integral domain, and \(f:A\rightarrow B\) is an integral embedding. Then
\(f^{\ast}:{\rm Spec}(B)\rightarrow {\rm Spec}(A)\) is an open function. Then in order to prove the integral closure of a Noetherian integrally closed integral
domain in a separable finite extension is Noetherian, we studied finite separable field extensions. We proved:
suppose \(E/F\) is a finite separable field extension. For \(e\in E\), let \(l_e:E\rightarrow E, l_e(e'):=ee'\). Then over an algebraic closure \(\overline{F}\) of \(F\), \(l_e\) can be represented by the diagonal matrix \({\rm diag}(\sigma_1(e),\ldots,\sigma_n(e))\) where \(\{\sigma_1,\ldots,\sigma_n\}={\rm Hom}_F(E,\overline{F})\). In particular, \(T_{E/F}(e)=\sum_{\sigma\in {\rm Hom}_F(E,\overline{F})} \sigma(e)\) and
\(N_{E/F}(e)=\prod_{\sigma\in {\rm Hom}_F(E,\overline{F})} \sigma(e)\). Next we proved, if \(E/F\) is a finite separable extension, then \(f(e_1,e_2):=T_{E/F}(e_1e_2)\) is a nondegenerate symmetric bilinear form.
 Lecture 15: Here is my note for the fifteenth lecture.
First we reviewed some of the results on nondegenerate bilinear forms; in particular, how one gets an isomorphism between a
finitedimensional vector space and its dual using a nondegenerate bilinear form; and for a symmetric nondegenerate bilinear form
\(f:V\times V\rightarrow k\) on a finitedimensional \(k\)vector space, \((W^{\perp})^{\perp}=W\) for any subspace \(W\). Next we used these results
for \(f(e_1,e_2):=T_{E/F}(e_1e_2)\) where \(E/F\) is a finite separable field extension; and proved the following:
Suppose \(A\) is an integrally closed, integral domain with field of fractions \(F\). Let \(E/F\) be a finite separable field extension. Suppose
\(B\) is the integral closure of \(A\) in \(E\). Then there are \(e_1,\ldots,e_n\in E\) such that \(B\subseteq Ae_1+\cdots+Ae_n\); in particular if \(A\)
is Noetherian, then \(B\) is Noetherian. We also proved that as an additive group the integral closure \(\mathcal{O}_k\) of \(\mathbb{Z}\) in a number field \(k\) is isomorphic to \(\mathbb{Z}^{[k:\mathbb{Q}]}\).
 Lecture 16: Here is my note for the sixteenth lecture.
Valuation rings are defined; we proved that a valuation ring is a local integrally closed ring. Then we proved
Let \(\Omega\) be an algebraically closed field, \(A_0\) be an integral domain, and \(\phi_0:A_0\rightarrow \Omega\) be a ring homomorphism. Suppose \(F\)
is the field of fractions of \(A_0\), and
\[
\Sigma:=\{(A, \phi)\hspace{1mm}A_0\subseteq A\subseteq F, \phi\in {\rm Hom}(A, \Omega), \phi_{A_0}=\phi_0\}.
\]
Then \(\Sigma\) has a maximal element with respect to the following ordering: \((A_1,\phi_1)\preceq (A_2,\phi_2)\) if \(A_1\subseteq A_2\) and \(\phi_2_{A_1}=\phi_1\). And if \((B,\phi)\) is a maximal element of \(\Sigma\), then \(B\) is a valuation ring with field fractions \(F\), and \(\ker \phi\) is its unique maximal ideal.
 Lecture 17: Here is my note for the seventeenth lecture.
Using the technical theorem proved in the previous lecture, we deduced that if \(A\) is an integral domain, its integral closure in its
field of fractions \(F\) is equal to the intersection of all the valuation subrings of \(F\) that contain \(A\) as a subring. Next we proved the following important result about lifts of ring homomorphisms:
Suppose \(\Omega\) is an algebraically closed field, \(A\) is an integral domain, \(B\) is a finitely generated \(A\)algebra that is an integral domain, and \(A\) is a subring of \(B\). Then for any
\(b_0\in B\setminus \{0\}\), there is \(a_0:=a_0(b_0)\in A\) such that for \(\phi\in {\rm Hom}(A,\Omega)\), if \(\phi(a_0)\neq 0\), then there is \(\widetilde{\phi}\in {\rm Hom}(B,\Omega)\) such that \(\widetilde{\phi}_A=\phi\) and \(\widetilde{\phi}(b_0)\neq 0\).
Using this result, we proved
Hilbert's Nullstellensatz (1st version) Suppose \(k\) is a field, and \(A\) is a finitely generated \(k\)algebra. If \(A\) is a field, then
\(A/k\) is a finite extension.
 Lecture 18: Here is my note for the eighteenth lecture.
We proved various versions of Hilbert's Nullstellensatz.
Hilbert's Nullstellensatz (2nd version) Suppose \(k\) is an algebraically closed field, \(\mathfrak{a}\unlhd k[x_1,\ldots,x_n]\), and
\(A:=k[x_1,\ldots,x_n]/\mathfrak{a}\). Then \({\rm Max}(A)=\{\mathfrak{m}_p/\mathfrak{a} p\in X(\mathfrak{a})\}\), where
\(X(\mathfrak{a}):=\{p\in k^n \forall f\in \mathfrak{a}, f(p)=0\}\) (common zeros of elements of \(\mathfrak{a}\)) and
\(\mathfrak{m}_p:=\langle x_1p_1,\ldots, x_np_n \rangle\) (here \(p_i\)'s are coordinates of \(p\).). (So there is a bijection between
\({\rm Max}(k[x_1,\ldots,x_n]/\mathfrak{a})\) and \(X(\mathfrak{a}\)).)
Hilbert's Nullstellensatz (3rd version) Suppose \(k\) is an algebraically closed field, and \(\mathfrak{a}\) is a proper ideal of \(k[x_1,\ldots,x_n]\); then \(X(\mathfrak{a})\neq \varnothing.\)
Hilbert's Nullstellensatz (4th version) Suppose \(k\) is an algebraically closed field, and \(\mathfrak{a}\unlhd k[x_1,\ldots,x_n]\); then
\(I(X(\mathfrak{a}))=\sqrt{\mathfrak{a}}\).
 Lecture 19: Here is my note for the nineteenth lecture.
We finished proof of Hilbert's Nullstellensatz (4th version); proved that a finitely generated \(k\)algebra is a Jacobson ring if \(k\) is algebraically closed; deduced that the Jacobson radical of such algebra is nilpotent. We started proof of Noether's normalization lemma:
Noether's normalization lemma Suppose \(k\) is a field, and \(A\) is a finitely generated \(k\)algebra. Then there are \(x_1,\ldots x_n\in A\) such that (1) \(x_i\)'s are algebraically independent over \(k\); (2) \(A\) is integral over \(k[x_1,\ldots,x_n]\).
 Lecture 20: Here is my note for the twentieth lecture.
We finished proof of Noether's normalization lemma, and then deduced that for any finitely generated \(k\)algebra \(A\), there are \(x_i\) in \(A\)
that are algebraically independent over \(k\), and \(\dim A=\dim k[x_1,\ldots,x_n]\) and \({\rm GKdim}(A)={\rm GKdim}(k[x_1,\ldots,x_n])=n\). Later we will prove that \(\dim k[x_1,\ldots,x_n]=n\); and so \(\dim A={\rm GKdim} A\). Then we studied Krull dimension zero Noetherian rings.
 Lecture 21: Here is my note for the twenty first lecture.
We studied modules with a composition series and length of a module. Then we proved a module is of finite length if and only if it is both Noetherian and Artinian. Next we proved the following lemma: suppose \(\mathfrak{m}_1,\ldots,\mathfrak{m}_n\in {\rm Max}(A)\) and \(\mathfrak{m}_1\cdots \mathfrak{m}_n=0\); then \(A\) is Noetherian if and only if it is Artinian.
 Lecture 22: Here is my note for the twenty second lecture.
In this lecture, we mainly studied Artinian rings. We proved \(A\) is Artinian if and only if \(A\) is Noetherian and \(\dim A=0\). If \(A\) is Artinian, then \({\rm Spec}(A)\) is finite and discrete. If \(A\) is Artinian, its Jacobson radical is nilpotent. Any Artinian ring is a unique (up to isomorphism) product of local Artinian rings.
 Lecture 23: Here is my note for the twenty third lecture.
First we studied structure of a local Artinian ring: suppose \(A\) is a local Artinian ring with the unique maximal ideal \(\mathfrak{m}\); then TFAE
(1) any ideal of \(A\) is principal, (2) \(\mathfrak{m}\) is principal, (3)\(\dim_{A/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2 \le 1\). Next we focused on
local integral domains of dimension 1:
Suppose \(D\) is a Noetherian, local, integral domain, \(\dim D=1\), and \( {\rm Max}(D)=\{\mathfrak{m}\} \); then TFAE (1) \(D\) is integrally closed, (2) \(\mathfrak{m}\) is principal, (3) \(\dim_{A/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2 = 1\), (4) any proper nonzero ideal is a power of \(\mathfrak{m}\), (5) there is \(\pi\in D\) such that any nonzero proper ideal is generated by a power of \(\pi\), (6) there is a valuation
\(v:F \rightarrow \mathbb{Z}\cup \{\infty\}\) such that \(D=\{a\in F v(a)\ge 0\}\) (here \(F\) is the field of fractions of \(D\)).
Such a ring is called a Discrete Valuation Ring (DVR).
 Lecture 24: Here is my note for the twenty fourth lecture.
We proved the global analogue of what we proved for DVRs:
Suppose \(A\) is Noetherian, integral domain, and \(\dim A=1\). Then TFAE (1) \(A\) is integrally closed, (2) for any \(\mathfrak{m}\in {\rm Max}(A)\), \(A_{\mathfrak{m}}\) is a DVR, (3) \(\mathfrak{q}\) is primary if and only if \(\mathfrak{q}=\mathfrak{p}^n\) for some \(\mathfrak{p}\in {\rm Spec}(A)\) and positive integer \(n\).
Such a ring is called a Dedekind domain. As a corollary we deduced:
Suppose \(k\) is a number field, and \(\mathcal{O}_k\) is the integral closure of \(\mathbb{Z}\) in \(k\). Then any proper nonzero ideal of \(\mathcal{O}_k\) can be uniquely written as a product of prime ideals.
Next we defined fractional ideals and proved for an integral domain \(A\) TFAE (1) a fractional ideal \(M\) is invertible, (2) \(M\) is finitely generated,
and for any maximal ideal \(\mathfrak{m}\), \(M_{\mathfrak{m}}\) is invertible in \({\rm Frac}(A_{\mathfrak{m}})\). As a corollary we deduced that the set of fractional ideals of a Dedekind domain forms a group; and defined the class group of a Dedekind domain which measures how far it is from being a PID.
 Lecture 25: Here is my note for the twenty fifth lecture.
First we improved the 1st uniques theorem of primary decompositions: suppose \(A\) is Noetherian and \(\mathfrak{a}\) is a proper ideal. Then \({\rm Ass}(\mathfrak{a})={\rm Spec}(A)\cap \{(\mathfrak{a}:x) x\in A\}.\) We deduced that if \(A\) is Noetherian and \({\rm ht}(\langle a\rangle)=0\), then \(a\) is a zerodivisor. We pointed out that the converse is not correct. We stated Krull's Principal Ideal Theorem and its generalization Krull's Height Theorem. We started proof of Krull's HT based on Krull's PIT.
 Lecture 26: Here is my note for the twenty sixth lecture.
Krull's Principal Ideal Theorem Suppose \(A\) is Noetherian, \(a\not\in A^{\times}\), and \(\mathfrak{p}\) is a minimal prime divisor of \(\langle a\rangle\). Then \({\rm ht}(\mathfrak{p})\le 1\).
This is a special case of
Krull's Height Theorem Suppose \(A\) is Noetherian, \(\mathfrak{a}:=\langle a_1,\ldots, a_n\rangle\) is a proper ideal, and \(\mathfrak{p}\) is a minimal prime divisor of \(\mathfrak{a}\). Then \({\rm ht}(\mathfrak{p})\le n\).
We proved Krull's HT by induction on dimension using Krull's PIT. Using Krull's HT, we proved that height of any maximal ideal of \(k[x_1,\ldots,x_n]\) is n; and so \(\dim k[x_1,\ldots,x_n]=n\). Then we started proof of Krull's PIT.
 Lecture 27: Here is my note for the twenty seventh lecture.
We finished proof of Krull's PIT. Next we proved a converse of Krull's HT: suppose \(A\) is Noetherian, \(\mathfrak{p}\in {\rm Spec}(A)\), and \({\rm ht}\mathfrak{p}=d\). Then there are \(a_1,\ldots,a_d\in A\) such that \(\mathfrak{p}\) is a minimal prime divisor of \(\langle a_1,\ldots,a_d\rangle\). Using this result we showed: if \(A\) is a local Noetherian ring with the unique maximal ideal \(\mathfrak{m}\), then \(\dim A=\min\{d(\mathfrak{q})\mathfrak{q}\) is
\(\mathfrak{m}\)primary \(\}\), where \(d(\mathfrak{q})\) is the minimum number of generators of \(\mathfrak{q}\). And so by Nakayama's lemma
\(\dim A\le \dim_{A/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2 \).
 Lecture 28: Here is my note for the twenty eighth lecture.
We proved that if \(A\) is a Noetherian ring and \(a\not \in A^{\times}\cup D(A)\), then \(\dim A/\langle a\rangle=\dim A1\). Then we defined an
\(A\)regular sequence, and proved if \((a_1,\ldots,a_m)\) is an \(A\)regular sequence, then for any \(l\le m\) \(\dim A/\langle a_1,\ldots,a_l\rangle=\dim Al\); in particular length of any \(A\)regular sequence is at most \(\dim A\). Defined the depth of \(A\) to be the largest length
of an \(A\)regular sequence; and so \({\rm depth}(A)\le \dim A\). We pointed out that in general equality does not necessarily holds and if it does hold
we say \(A\) is a CohenMacaulay ring. Then we proved that
Suppose \(A\) is a local Noetherian ring and any proper ideal \(\mathfrak{a}\) that can be generated by at most \(\dim(A/\mathfrak{a})\)many elements is unmixed. Then \(A\) is a CM ring.
Next we recalled that for any \(\mathfrak{p}\in {\rm Spec}(A)\), \({\rm ht} \mathfrak{p}+\dim(A/\mathfrak{p})\le \dim A\); and we gave an example where the equality does not hold.
 Lecture 29: Here is my note for the twenty ninth lecture.
We proved that: if \(k\) is a field and \(A\) is a finitely generated \(k\)algebra that is an integral domain, then for any \(\mathfrak{p}\in {\rm Spec}(A)\), \[{\rm ht} \mathfrak{p}+\dim(A/\mathfrak{p})= \dim A.\] This is done based on a better documented version of Noether's normalization lemma.
