- 4/3: Defined algebraic field extensions. Recalled that finite extensions are algberaic. Defined the algebraic closure of \(F\) in an extension \(E\), and proved
that it is always a subfield of \(E\). Proved that \(K/E\) and \(E/F\) are algebraic extensions if and only if \(K/F\) is an algebraic extensions. Defined algebraically closed and
an algebraic closure of a field \(F\). Mentioned that for a field \(\overline{F}\) the following statements are equivalent:
- \(\overline{F}\) is algebraically closed.
- Every irreducible polynomial in \(\overline{F}[x]\) is of degree 1.
- For every \(f(x)\in \overline{F}[x]\), there exists \(c,\alpha_1,\ldots,\alpha_d\in \overline{F}[x]\),
\(f(x)=c(x-\alpha_1)\cdots (x-\alpha_d)\).
Proved that for every field \(F\) there exists a field extension \(E\) and \(\alpha_f\in E\) for every monic irreducible \(f\in F[x]\)
such that (1) \(E\) is generated by \(F\) and \(\alpha_f\)'s as a ring, and (2) \(f(\alpha_f)=0\) for every monic irreducible \(f\in F[x]\).
- 4/5: Finished the proof of the existence of an algebraic closure of a field. Next we showed that if \(\theta:F\to F'\) is a field isomorphism, \(\overline{F}\) is an
algebraic closure of \(F\), and \(\overline{F}'\) is an
algebraic closure of \(F'\), then there exists \(\widehat{\theta}\in {\rm Isom}_\theta(\overline{F},\overline{F'})\); that means \(\widehat{\theta}\) is an extension of \(\theta\).
- 4/7: Proved that if \(E/F\) is an algebraic extension and \(\overline{F}\) is an algebraic closure of \(F\), then there exists an \(F\)-embedding of \(E\) into \(\overline{F}\).
Defined a splitting field of a family of polynomials and proved its existence. Proved that for every algebraic extension \(E/F\) the following statements are equivalent:
- \(E\) is a splitting field of a family of polynomials.
- Suppose \(\overline{E}\) is an algebraic closure of \(E\). Then for every \(\theta\in {\rm Aut}_F(\overline{E})\), \(\theta(E)=E\).
- For every algebraic extension \(K/E\) and every \(\theta\in {\rm Aut}_F(K)\), \(\theta(E)=E\).
- For every \(\alpha\in E\), \(m_{\alpha,F}(x)=\prod_{i=1}^d (x-\alpha_i)\) for some \(\alpha_i\in E\).
An algebraic extension is called a normal extension if it satisfies the above properties. We showed that if \(F\subseteq E\subseteq K\) is a tower of field extensions and
\(E/F\) is a normal extension, then the restriction map defines a well-define group homomorphism and gives us the following exact sequence:
\(1\to {\rm Aut}_E(K) \to {\rm Aut}_F(K) \to {\rm Aut}_F(E)\). In the next lecture we show that the restriction map is also surjective if \(K/F\) is a normal extension.
- 4/10: Prove that if \(F\subseteq E\subseteq K\), \(E/F\) and \(K/F\) are normal extensions, then the restriction map \(r_{KE}\) defines a group homomorphism form
\({\rm Aut}_F(K)\) and \({\rm Aut}_F(E)\), and the following is a SES
\(
1\to {\rm Aut}_E(K)\to {\rm Aut}_F(K)\xrightarrow{r_{KE}} {\rm Aut}_F(E)\to 1.
\)
Then we showed that for every normal extension \(E/F\), \(E\) is the union of \(K\)'s as \(K\) ranges in
\(I(E/F):=\{K\mid F\subseteq K\subseteq E, K/F \text{is a finite normal extension}\}\). Use this to deduce that
\(
{\rm Aut}_F(E)\simeq \varprojlim_{K\in I(E/F)} {\rm Aut}_F(K).
\)
Next, we started the proof of the following theorem: if \(\theta:F\to F'\) is an isomorphism, \(E\) is a splitting field of \(f\in F[x]\) over \(F\) and \(E'\) is a
splitting field of \(f^\theta\) over F', then \(|{\rm Isom}_{\theta}(E,E')|\leq [E:F]\), and equality holds if the irreducible factors of \(f\) do not have multiple zeros.
- 4/12: Proved that if \(\theta:F\to F'\) is an isomorphism, \(E\) is a splitting field of \(f\in F[x]\) over \(F\) and \(E'\) is a
splitting field of \(f^\theta\) over F', then \(|{\rm Isom}_{\theta}(E,E')|\leq [E:F]\), and equality holds if the irreducible factors of \(f\) do not have multiple zeros.
Said a polynomial \(f\in F[x]\) is separable if it does not have multiple roots in a splitting field of \(f\) over \(F\). Prove that \(f\in F[x]\) is separable if and only if \(\gcd(f,f')=1\) in
\(F[x]\). Showed that an irreducible polynomial \(f\in F[x]\) is separable if and only if \(f'\neq 0\); in particular, an irreducible polynomial \(f\in F[x]\) is not separable if and only if
the characteristic of \(F\) is \(p>0\) and \(f(x)=f_1(x^p)\) for some irreducible \(f_1\in F[x]\). Defined separable extensions. Mentioned that in chracteristic zero every algebraic extension is
separable. Formulated the following important theorem: suppose \(\theta:F\to F'\) is an isomorphism, \(E/F\) is a finite extension, and \(\overline{F}'\) is an algebraic extension of \(F'\).
Then the number of \(\theta\)-embedding of \(E\) into \(\overline{F}'\) is at most \([E:F]\) and equality holds if and only if \(E/F\) is separable.
- 4/14: Proved that: suppose \(\theta:F\to F'\) is an isomorphism, \(E/F\) is a finite extension, and \(\overline{F}'\) is an algebraic extension of \(F'\).
Then the number of \(\theta\)-embedding of \(E\) into \(\overline{F}'\) is at most \([E:F]\) and equality holds if and only if \(E/F\) is separable. Proved that for a finite extension
\(E/F\) the following statements are equivalent: (1) \(E\) is a splitting field of a polynomial \(f\in F[x]\) whose irreducible factors are separable. (2) \(|{\rm Aut}_F(E)|=[E:F]\).
(3) \(E/F\) is nomral and separable. An extension is called a Galois extension if it is normal and seprarable. If \(E/F\) is Galois, we write \({\rm Gal}(E/F)\) instead of \({\rm Aut}_F(E)\).
We noticed that (1) if \(\overline{F}\) is an algebraic closure of \(F\) and \(F\subseteq E\subseteq \overline{F}\), then for every \(\alpha\in E\) and every zero
\(\alpha'\in \overline{F}\) of \(m_{\alpha,F}\), there exists an \(F\)-embedding \(\theta:E\to \overline{F}\) such that \(\theta(\alpha)=\alpha'\). (2) \({\rm Gal}(E/F)\) acts transitively on the zeros of \(m_{\alpha,F}\) for every \(\alpha\in E\).
Next we want to show that \({\rm Fix}({\rm Gal}(E/F))=F\). We proved the following technical lemma: if \(G\) is a subgroup of \({\rm Aut}(E)\) and \(V\) is a \(G\)-invariant non-zero
subspace of \(E^n\), then the set \(V^G\) of \(G\)-fixed points is not zero.
- 4/17: Proved that if \(G\) is a finite subgroup of \({\rm Aut}(E)\), then \(E/{\rm Fix}(G)\) is a field extension of degree at most \(|G|\). Used this result to show that
\(E/{\rm Fix}(G)\) is a Galois extension, \({\rm Gal}(E/{\rm Fix}(G))=G\), and \([E:{\rm Fix}(G)]=|G|\). As a corollary, we showed that if \(E/F\) is a finite Galois extension, then
\({\rm Fix}({\rm Gal}(E/F))=F\). At this point, we were ready to prove the main theorem of Galois theory: suppose \(E/F\) is a finite Galois extension. Let \({\rm Sub}({\rm Gal}(E/F))\)
be the set of all the subgroups of \({\rm Gal}(E/F)\), and \({\rm Int}(E/F)\) be the set of all the intermidiate subfields. Let
\(
\Phi:{\rm Sub}({\rm Gal}(E/F))\to {\rm Int}(E/F),\quad \Phi(H):={\rm Fix}(H),
\)
and
\(
\Psi:{\rm Int}(E/F) \to {\rm Sub}({\rm Gal}(E/F)), \quad \Psi(K):={\rm Aut}_K(E).
\)
Then the following statements hold.
- \(\Phi\) and \(\Psi\) are inverse of each other; in particular, they are bijections.
- For every \(K\in {\rm Int}(E/F)\), \(E/K\) is Galois; and so \({\rm Aut}_K(E)={\rm Gal}(E/K)\).
- \(\Phi\) and \(\Psi\) induce bijections between normal subgroups of \({\rm Gal}(E/F)\) and intermediate subfields \(K\) such that \(K/F\) is a normal extension.
- 4/19: Used the main theorem of Galois theory and showed that \({\rm Int}(E/F)\) is finite if \(E/F\) is a finite separable extension. Then proved that for a finite extension
\(E/F\), \({\rm Int}(E/F)\) is finite if and only if \(E/F\) is a simple extension, i.e. \(E=F[\alpha]\) for some \(\alpha\in E\). Deduce that if \(E/F\) is a finite separable extension, then
\(E/F\) is a simple extension. Next, we showed that the following are equivalent:
- Every algebraic extension \(E/F\) is separable.
- \(\overline{F}/F\) is a separable extension where \(\overline{F}\) is an algebraic closure of \(F\).
- Either \(F\) is of characteristic zero or of characteristic \(p>0\) and \(F^p=F\).
A field which satisfies these conditions is called a perfect field. Then we started talking about the Galois group of \(\mathbb{F}_{p^n}/\mathbb{F}_p\).
- 4/21: Showed that \({\rm Gal}(\mathbb{F}_{p^n}/\mathbb{F})=\langle \sigma_0\rangle\) where \(\sigma_0:\mathbb{F}_{p^n}\to \mathbb{F}_{p^n}, \sigma_0(x):=x^p\). Next we showed
that if the characteristic of \(F\) does not divide \(n\), then a splitting field \(E\) of \(x^n-1\) over \(F\) is a Galois extension, it is of the form \(F[\zeta_n]\), and its Galois group
\({\rm Gal}(F[\zeta_n]/F)\) can be embedded into \((\mathbb{Z}/n\mathbb{Z})^\times\); in particular it is abelian. Then we considered a splitting field \(E\) of \(x^n-a\) over \(F\) where
\(a\in F^\times\). Assuming \(F\) contains \(n\) distinct zeros of \(x^n-1\), we showed that \(E=F[\sqrt[n]{a}]\) and it is a Galois extension of \(F\). Moreover
\(
{\rm Gal}(F[\sqrt[n]{a}]/F)\to \mu_n(F),\quad \sigma\mapsto \frac{\sigma(\sqrt[n]{a})}{\sqrt[n]{a}}
\)
is an injective group homomorphism; hence \({\rm Gal}(F[\sqrt[n]{a}]/F)\) is a cyclic group and its order divides \(n\). Next, we mentioned what it means for a polynomial to be solvable-by-radicals,
and started proof of Galois's theorem of solvability of polynomials.
- 4/24: Proved one side of Galois's theorem on solvability of polynomials: suppose \(F\) is a field of characteristic zero and \(f\in F[x]\). Suppose \(E\) is a splitting field
of \(f\) over \(F\). If \(f\) is solvable by radicals, then \({\rm Gal}(E/F)\) is a solvable. Then we proved Dirichlet's independence theorem.
- 4/26: Defined the norm \(N_{E/F}:E^\times \to F^\times\) assuming \(E/F\) is a finite Galois. Observed that \(N_{E/F}(\sigma(a)/a)=1\) for every \(\sigma\in {\rm Gal}(E/F)\)
and \(a\in E^\times\). Mentioned the connection with Pell's equation. Proved Hilbert's theorem 90: if \({\rm Gal}(E/F)=\langle \sigma_0\rangle\) and \(N_{E/F}(a)=1\), then \(a=\sigma_0(b)/b\) for some
\(b\in E^\times\). To show this, we considered the \(F\)-linear map \(T_a:E\to E, T_a(e):=a\sigma_0(e)\), and showed that its minimal polynomial is equal to its characteristic polynomial, and
it is \(x^n-N_{E/F}(a)\). Using Hilbert's theorem 90, we proved that if \(E/F\) is a cyclic extension of order \(n\) and \(F\) has \(n\) distinct \(n\)-th roots of unity,
then \(E=F[\sqrt[n]{a}]\) for some \(a\in F\).
- 4/28: Proved the other side of Galois's theorem: suppose \(F\) is a field of characteristic zero, \(f\in F[x]\), and \(E\) is a splitting field of \(f\) over \(F\).
Then \(f\) is solvable-by-radicals if and only if \({\rm Gal}(E/F)\) is solvable. Next, we defined the \(n\)-th cyclotomic polynomial \(\Phi_n(x)\). Proved that \(\Phi_n(x)\) is in
\(\mathbb{Z}[x]\) and it is irreducible in \(\mathbb{Q}[x]\). Used this to obtain that \({\rm Gal}(\mathbb{Q}[\zeta_n]/\mathbb{Q})\simeq (\mathbb{Z}/n\mathbb{Z})^\times\).
- 5/1: Defined the set of divisiors, prime divisiors, and maximal prime divisors of an ideal. Defined the contraction and the extension maps associated with a ring
homomorphism. Proved that \(\mathfrak{a}^{ece}=\mathfrak{a}^e\) and \(\mathfrak{b}^{cec}=\mathfrak{b}^c\). Mentioned that the contraction defines a map \(f^\ast:{\rm Spec}(B)\to {\rm Spec}(A)\) for every ring homomorphism \(f:A\to B\). Proved that the set
of prime divisors of ideals as closed subsets of \({\rm Spec}(A)\) defines a topology on \({\rm Spec}(A)\) which is called the Zariski-topology. Defined the Jacobson radical of a ring.
- 5/3: Proved that \(x\in J(A)\) if and only if \(1-xy\in A^\times\) for all \(y\in A\). Showed that \(J(A[x])={\rm Nil}(A)[x]\). Showed that \(\pi^\ast\) induces a bijection
between \({\rm Spec}(A/\mathfrak{a})\) and \(V(\mathfrak{a})\). Proved that with respect to \(i:A\to S^{-1}A\), \(\mathfrak{a}^e=S^{-1}\mathfrak{a}\) and
\(\widetilde{\mathfrak{a}}^{ce}=\widetilde{\mathfrak{a}}\); in particular, \(i^\ast\) is injective. Proved that \(S^{-1}A/S^{-1}\mathfrak{p}\simeq \overline{S}^{-1}(A/\mathfrak{p})\) and
\(\overline{S}(A/\mathfrak{p})\) is an integral domain. Deduced that \(S^{-1}\mathfrak{p}\) is a prime ideal if \(p\) is.
- 5/5: Proved that \(i^\ast\) induces a bijection between \({\rm Spec}(S^{-1}A)\) and \(\mathcal{O}_S:=\{\mathfrak{p}\in {\rm Spec}(A)\mid \mathfrak{p}\cap S=\varnothing\}\) where
\(i:A\to S^{-1}A,\quad i(a):=\frac{a}{1}\). For a ring homomomorphism \(f:A\to B\) and \(\mathfrak{p}\in {\rm Spec}(A)\), proved that \(\theta^\ast\) gives us a bijection between
\({\rm Spec}(\frac{f(S_{\mathfrak{p}})^{-1}B}{f(S_{\mathfrak{p}})^{-1}\mathfrak{p}^e})\) and
\((f^\ast)^{-1}(\mathfrak{p})\) where \(S_{\mathfrak{p}}:=A\setminus \mathfrak{p}\) and \(\theta\) is the composite of \(i:B\to f(S_{\mathfrak{p}})^{-1}B\) and
\(\pi: f(S_{\mathfrak{p}})^{-1}B\to \frac{f(S_{\mathfrak{p}})^{-1}B}{f(S_{\mathfrak{p}})^{-1}\mathfrak{p}^e}\). Used this to show that
\(\mathfrak{p}\) is in the image of \(f^\ast\) if and only if \(\mathfrak{p}=\mathfrak{p}^{ec}\). Next, we proved a generalization of Nakayama's lemma (for commutative rings): suppose
\(M\) is a finitely generated \(A\)-module, \(\phi\in {\rm End}_A(M)\), \(\phi(M)\subseteq \mathfrak{a}M\) for some ideal \(\mathfrak{a}\). Then
\(
\phi^n+a_{n-1}\phi^{n-1}+\cdots+a_0=0
\)
for some \(a_i\)'s in \(\mathfrak{a}\).
- 5/8: Proved other versions of Nakayama's lemma: if \(M\) is a finitely generated \(A\)-module and \(M=\mathfrak{a}M\), then there exists \(x\in A\) such that \(xM=0\) and
\(x\equiv 1\pmod{\mathfrak{a}}\). Used this to show that if \(M\) is a finitely generated \(A\)-module and \(M=J(A)M\), then \(M=0\). Pointed out that this version is true for
non-commutative rings as well. Then we showed: if \(M\) is a finitely generated \(A\)-module and \(M/J(A)M\) is generated by \(x_1+J(A)M,\ldots,x_d+J(A)M\) as an \(A/J(A)\)-module,
then \(M\) is generated by \(x_i\)'s. This is particularly useful when \(A\) is a local ring. In this case, \(J(A)=\mathfrak{m}\) is the unique maximal ideal of \(A\);
and so \(A/J(A)\) is a field. Next, we defined a primary ideal. Proved that \(\mathfrak{q}\) is primary if and only if \(D_0(A/\mathfrak{q})={\rm Nil}(A/\mathfrak{q})\). We proved that the
radical of a primary ideal is prime. Gave an example where the converse does not hold. Then proved that if the radical of an ideal is maximal, then it is primary.
- 5/10: Showed that if \(\mathfrak{q}\) is \(\mathfrak{p}\)-primary and \(x\in A\), then (1) \((\mathfrak{q}:x)=A\) if \(x\in \mathfrak{q}\),
(2) \((\mathfrak{q}:x)=\mathfrak{q}\)
if \(x\not\in \mathfrak{p}\), and
(3) \((\mathfrak{q}:x)\) is \(\mathfrak{p}\)-primary if \(x\not\in \mathfrak{q}\). Proved that intersection of two \(\mathfrak{p}\)-primary
ideals is a \(\mathfrak{p}\)-primary ideal. Deduced that every decomposable ideal has a reduced primary decomposition. Next we showed that if \(\bigcap_{i=1}^n \mathfrak{q}_i\) is a
reduced primary decomposition of \(\mathfrak{a}\) and \(\mathfrak{q}_i\) is \(\mathfrak{p}_i\)-primary, then
\(\{\mathfrak{p}_1,\ldots,\mathfrak{p}_n\}={\rm Spec}(A)\cap \{\sqrt{(\mathfrak{a}:x)}\mid x\in A\}\); in particular, the set of \(\mathfrak{p}_i\) only depends on \(\mathfrak{a}\) and it
does not depend on the choice of a reduced primary decomposition. To show this, we argued why \(\sqrt{\bigcap_{i=1}^m \mathfrak{a}_i}=\bigcap_{i=1}^m \sqrt{\mathfrak{a}_i}\) and
\((\bigcap_i \mathfrak{a}_i:x)=\bigcap_i (\mathfrak{a}_i:x)\).
- 5/12: Proved that if \(\mathfrak{a}\) is a decomposable ideal, then for every \(\mathfrak{p}\in V(\mathfrak{a})\) there exists
\(\mathfrak{p}'\in {\rm Ass}(\mathfrak{a})\) such that \(\mathfrak{p}'\subseteq \mathfrak{p}\). Deduce that the sets of minimal elements of \(V(\mathfrak{a})\) and
\({\rm Ass}(\mathfrak{a})\)
are equal. Obtained that if \(0\) is decomposable, then \({\rm Nil}(A)=\bigcap_{\mathfrak{p}\in {\rm Ass}(0)} \mathfrak{p}\). Next, we showed that if \(\mathfrak{a}\) is decomposable, then
\( \bigcup_{\mathfrak{p}\in {\rm Ass}(\mathfrak{a})} \mathfrak{p}=\{x\in A\mid (\mathfrak{a}:x)\neq \mathfrak{a}\}.\) Used this to deduce that if \(0\) is a decomposable ideal, then the set of
zero-divisors is \(\bigcup_{\mathfrak{p}\in {\rm Ass}(0)} \mathfrak{p}\). Next, we defined \(S(\mathfrak{a}):=\mathfrak{a}^{ec}\) under the ring homomomorphism \(i:A\to S^{-1}A\).
Pointed out that \(S(\mathfrak{a})\) is the set of all the numerators in \(S^{-1}\mathfrak{a}\); equivalently, we showed \(S(\mathfrak{a})=\bigcup_{s\in S} (\mathfrak{a}:s)\). Recalled that
for every \(\mathfrak{p}\in \mathcal{O}_S\) (that means \(\mathfrak{p}\) is a prime ideal and \(\mathfrak{p}\cap S=\varnothing\)), we have \(S(\mathfrak{p})=\mathfrak{p}\). Next, we showed that if
\(\mathfrak{q}\) is \(\mathfrak{p}\)-primary and \(\mathfrak{p}\cap S\neq \varnothing\), then \(S^{-1}\mathfrak{q}=S^{-1}A\) and \(S(\mathfrak{q})=A\). Then, we showed that if \(\mathfrak{p}\in
\mathcal{O}_S\) and \(\mathfrak{q}\) is \(\mathfrak{p}\)-primary, then \(S(\mathfrak{q})=\mathfrak{q}\). Then, we proved that \(\mathfrak{q}\) is \(\mathfrak{p}\)-primary and \(\mathfrak{p}\in
\mathcal{O}_S\), then \(S^{-1}\mathfrak{q}\) is \(S^{-1}\mathfrak{p}\)-primary.
- 5/15: Proved that \(\mathfrak{q}\mapsto S^{-1}\mathfrak{q}\) and \(\widetilde{\mathfrak{q}}\mapsto \widetilde{\mathfrak{q}}^c\) give us bijections between
\(\mathfrak{p}\)-primary
and \(S^{-1}\mathfrak{p}\)-primary ideals if \(\mathfrak{p}\cap S=\varnothing\). Next, we showed that if \(\bigcup_{i=1}^n\mathfrak{q}_i\) is a reduced primary
decomposition of \(\mathfrak{a}\), \(\mathfrak{p}_i:=\sqrt{\mathfrak{q}}_i\), and \(\{1,\ldots,m\}=\{i\mid \mathfrak{p}_i\cap S=\varnothing\}\), then
\(S^{-1}\mathfrak{a}=\bigcup_{i=1}^m S^{-1}\mathfrak{q}_i\) and \(S(\mathfrak{a})=\bigcap_{i=1}^m \mathfrak{q}_i\). Next, we defined an isolated subset of \({\rm Ass}(\mathfrak{a})\),
and proved the 2nd uniqueness theorem for reduced primary decompositions. We proved that if \(\Sigma\) is an isolated subset of \({\rm Ass}(\mathfrak{a})\), then
\(\bigcap_{\mathfrak{p}\in \Sigma} \mathfrak{q}\) only depends on \(\mathfrak{a}\) and \(\Sigma\). In particular, if \(\mathfrak{p}\) is a minimal element of \({\rm Ass}(\mathfrak{a})\), then
the \(\mathfrak{p}\)-factor of a reduced primary decomposition is unique (i.e. it is independent of the choice of a reduced primary decomposition). We have almost proved this result; modulo the following
lemma: if \(\mathfrak{p}_i\)'s are prime ideals, \(\mathfrak{a}\) is an ideal, and \(\mathfrak{a}\subseteq \bigcup_{i=1}^n \mathfrak{p}_i\), then there exists \(i_0\) such that
\(\mathfrak{a}\subseteq \mathfrak{p}_{i_0}\).
- 5/17: Finished proof of the second uniqueness. Defined Krull dimension of a ring. Proved that an integral domain is of dimension 1 precisely when it is a field. Showed that an
integral domain has dimension 1 if and only if \({\rm Max}(A)={\rm Spec}\setminus\{0\}\). Used this to show that if \(D\) is an integral domain of dimension 1, then every decomposable ideal has a
unique reduced primary decomposition. Next, defined irreducible ideals. Showed that prime implies irreducible, but the converse does not hold. Then proved that in a Noetherian ring every
ideal can be written as an intersection of finitely many irreducible ideals. Next, we have almost proved why in a Noetherian ring every irreducible ideal is primary; and so in
a Noetherian ring every ideal has a primary decomposition.
- 5/19: Finished the proof of why in a Noetherian ring every irredicble ideal is primary. Then defined integral extension, and proved the following statements are equivalent:
- \(b\) is integral over \(A\).
- \(A[b]\) is a fintiely generated \(A\)-module.
- There exists an intermidiate subring \(A[b]\subseteq C\subseteq B\) which is a finitely generated \(A\)-module.
- There exists a faithful \(A[b]\)-module \(M\) which is a finitely generated \(A\)-module.
Next, we proved that if \(b_1,\ldots, b_n\) are integral over \(A\), then \(A[b_1,\ldots,b_n]\) is a finitely generated \(A\)-module. Used this to show that for a ring extension
\(B/A\), \(C:=\{b\in B\mid b \text{ integral over } A\}\) is a subring and \(C/A\) is an integral extension. This ring is called the integral closure of \(A\) in \(B\). We say \(A\)
is integrally closed in \(B\) if \(C=A\). After that we proved if \(B/A\) and \(C/B\) are integral extensions, then \(C/A\) is an integral extension. Used this result to show that the integrally
closure of \(A\) in \(B\) is integrally closed in \(B\).
5/22: Proved that if \(B/A\) is an integral extension, \(\mathfrak{b}\) is an ideal of \(B\), and \(S\subseteq A\) is multiplicatively closed, then
\(A/\mathfrak{b}^c\hookrightarrow B/\mathfrak{b}\) and \(S^{-1}A\hookrightarrow S^{-1}B\) are integral extensions. Next, we proved that if \(B/A\) is a ring extension and \(C\) is
the integral closure of \(A\) in \(B\), then \(S^{-1}C\) is the integral closure of \(S^{-1}A\) in \(S^{-1}B\). Using this result, we showed that for an integral domain \(D\)
the following are equivalent:
- \(D\) is integrally closed.
- \(D_{\mathfrak{p}}\) is integrally closed for every \(\mathfrak{p}\in {\rm Spec}(D)\).
- \(D_{\mathfrak{m}}\) is integrally closed for every \(\mathfrak{m}\in {\rm Max}(D)\).
Next, we said we are going to work towards proving the following important results:
Suppose \(B/A\) is an integral extension. Then
- \(f^\ast:{\rm Spec}(B)\to {\rm Spec}(A)\) is surjective.
- \(f^\ast({\rm Max}(B))={\rm Max}(A)\).
- \(f^\ast(V(\mathfrak{b}))=V(\mathfrak{b}^c)\).
To prove these results, we started with a lemma: suppose \(B/A\) is an integral extension and \(A\) and \(B\) are integral domains. Then \(A\) is a field if and only if \(B\) is a field.
Used this to show that if \(B/A\) is an integral extension, then for every \(\mathfrak{q}\in {\rm Spec}(B)\),
\(\mathfrak{q}\) is maximal if and only if \(f^\ast(\mathfrak{q})\) is maximal.
- 5/24: We finshed the proof of
Suppose \(B/A\) is an integral extension. Then
- \(f^\ast:{\rm Spec}(B)\to {\rm Spec}(A)\) is surjective.
- \(f^\ast({\rm Max}(B))={\rm Max}(A)\).
- \(f^\ast(V(\mathfrak{b}))=V(\mathfrak{b}^c)\).
Next, we proved the Going-Up theorem. After that we showed that if \(B/A\) is an integral extension, \(\mathfrak{q}_0\subseteq\mathfrak{q}_1\in {\rm Spec}(B)\), and \(\mathfrak{q}_0^c=\mathfrak{q}_1^c\),
then \(\mathfrak{q}_0=\mathfrak{q}_1\). We concluded that if \(B/A\) is an integral extension, then \(\dim A=\dim B\). We metioned some of the consequences for the integral closure
\(\mathcal{O}_k\) of
\(\mathbb{Z}\) in a finite extension \(k\) of \(\mathbb{Q}\): every non-zero ideal can be written as a product of primary ideals in a unique way. We also discussed why
every non-zero prime ideal of \(\mathcal{O}_k\) is maximal. Finally, we said what it means to be integral over an ideal. We stated a relavent result which will be proved in the
next lecture.
- 5/26: Proved that for a ring extension \(B/A\) and an ideal \(\mathfrak{a}\) of \(A\), \(b\in B\) is integral over \(\mathfrak{a}\) if and only if \(b\) is in
\(\sqrt{\mathfrak{a}^e}\), where \(\mathfrak{a}^e\) is the extension of \(\mathfrak{a}\) in the integral closure of \(A\) in \(B\). Next, we considered a ring extension of
integral domains \(B/A\), assumed that \(A\) is integrally closed, \(F\) is a field of fractions of \(A\), and \(\mathfrak{a}\) is an ideal of \(A\). Assumed that \(b\in B\) is integral
over \(\mathfrak{a}\). Proved that the minimal polynomial of \(b\) over \(F\) is of the form \(x^n+a_{n-1}x^{n-1}+\cdots+a_0\) for some \(a_i\in \sqrt{\mathfrak{a}}\). Next, we stated the
Going-Down theorem, and reduced its proof to the following statement: suppose \(A\) is integrally closed, \(B/A\) is an integral extension, and \(B\) is an integral domain. Suppose
\({\rm Max}(A)=\{\mathfrak{p}\}\) and \(\mathfrak{q}\in {\rm Spec}(B)\) such that \(\mathfrak{q}^c=\mathfrak{p}\). Then we want to prove that
\({\rm Spec}(B_{\mathfrak{q}})\to {\rm Spec}(A)\) is surjective. We discussed that to show this it suffices to prove \(\mathfrak{p}'^{ec}=\mathfrak{p}'\) for every
\(\mathfrak{p}'\in {\rm Spec}(A)\). We also pointed out that \(a\in \mathfrak{p}'^{ec}\) implies that for some \(s\in B\setminus \mathfrak{q}\), \(sa\) is integral over
\(\mathfrak{p}'\). We will exploit this property in the next lecture to finish the proof of the Going-Down theorem.
- 5/31: Finished proof of Going-Down Theorem. Proved that if \(B/A\) is an integral extension, \(B\) is an integral domain, \(A\) is integrally closed, then
\(f^\ast:{\rm Spec}(B)\to {\rm Spec}(A)\) is an open map. Mentioned that \(\{\mathcal{O}_b\mid b\in B\}\) is a basis for the open sets of \({\rm Spec}(B)\) where
\(\mathcal{O}_b:=\{\mathfrak{q}\in {\rm Spec}(B)\mid b\not\in \mathfrak{q}\}\). Then proved that if the minimal polynomial of \(b\) over the field of fractions of \(A\) is
\(x^n+a_{n-1}x^{n-1}+\cdots+a_0\), then \(f^\ast(\mathcal{O}_b)=\bigcup_{i=0}^{n-1} \mathcal{O}_{a_i}\). Next, we defined a valuation integral domain, and proved that if \(A\)
is a valuation ring, then it is a local ring and it is integrally closed.
- 6/2: Proved that if \(F\) is a field, \(A\) is a subring of \(F\), and \(\phi:A\to \Omega\) is a ring homomomorphism where \(\Omega\) is an algebraically closed field,
then there exists a valuation ring \(\mathcal{O}\subseteq F\) and \(\widehat{\phi}:\mathcal{O}\to \Omega\) such that the field of fractions of \(\mathcal{O}\) is \(F\),
\(\widehat{\phi}\) is an extension of \(\phi\), and \(\ker(\widehat{\phi})\) is the unique maximal ideal of \(\mathcal{O}\). Used this to proved that for every integral domain \(A\)
the integral closure of \(A\) in its field of fractions is equal to \(\bigcap \mathcal{O}\) as \(\mathcal{O}\) ranges over the valuation subrings of \(F\) which contain \(A\).
- 6/5: Proved that if \(B\) is a finitely generated \(A\)-algebra, \(A\) and \(B\) are integral domains, and \(b_0\in B\setminus \{0\}\), then there exists
\(a_0\in A\setminus\{0\}\) such that the following statement holds. For all ring homomorphisms \(\phi:A\to \Omega\) where \(\Omega\) is an algebraically closed field, if \(\phi(a_0)\neq 0\),
then there exists an extension \(\widehat{\phi}:B\to \Omega\) of \(\phi\) such that \(\widehat{\phi}(b_0)\neq 0\). We mentioned that one can think about it in terms of the image of
\(h_B(\Omega)\to h_A(\Omega)\) where \(h_B\) and \(h_A\) are the representable functors and \(h_B(\Omega)\) can be viewed as the set of common zeros of the defining relations of
\(B\).
We are proving that those points in \(h_A(\Omega)\) which are not in the "set of zeros" of \(a_0\) are in the image of points that are not "zeros" of \(b_0\). This is used to prove Chevalley's
theorem that the image of a constructible set is constructible. We used the mentioned result to prove the first version of Nullstellensatz:
if \(B\) is a finitely generated \(k\)-algebra and \(k\) and \(B\) are fields, then \(B/k\) is a finite extension. Next, we proved the second version of Nullstellensatz:
suppose \(k\) is algebraically closed and \(\mathfrak{a}\) is an ideal of \(k[x_1,\ldots,x_n]\). Let \(A:=k[x_1,\ldots,x_n]/\mathfrak{a}\). Then
\(
{\rm Max}(A)=\{\mathfrak{m}_p/\mathfrak{a}\mid p\in X(\mathfrak{a})\},
\)
where \(X(\mathfrak{a})=\{p\in k^n\mid f(p)=0 \text{ for all } f\in \mathfrak{a}\}\) and \(\mathfrak{m}_p=\langle x_1-p_1,\ldots,x_n-p_n \rangle\). We also mentioned that
\(I(X(\mathfrak{a}))\supseteq \sqrt{\mathfrak{a}}\) where \(I(X):=\{f\in k[x_1,\ldots,x_n]\mid f|_X=0\}\).
- 6/7: Proved that if \(\mathfrak{a}\) is a proper ideal of \(k[x_1,\ldots,x_n]\) and \(k\) is algebraically closed, then \(X(\mathfrak{a})\neq \varnothing\); that means
elements of \(\mathfrak{a}\) have a common zero. Next we proved the final version of Nullstellensatz: for every ideal \(\mathfrak{a}\) of \(k[x_1,\ldots,x_n]\) we have
\(I(X(\mathfrak{a}))=\sqrt{\mathfrak{a}}\) if \(k\) is algebraically closed. Then we defined what a Jacobson ring is
and proved that every finitely generated \(k\)-algebra is Jacobson. Finally, we almost proved Noether's normalization lemma: if \(A\) is a finitely generated \(k\)-algebra,
then there exist \(\xi_1,\ldots,\xi_n\in A\) such that (1) \(\xi_i\)'s are transcendental over \(k\) and (2) \(A/k[\xi_1,\ldots,\xi_n]\) is integral. We mentioned a geometric
interpretation of this statement.
- 6/9: We finished the proof of Noether's normalization lemma; along the way, we mentioned the analogue of the upper-triangular matrices in the group of automorphisms
of the ring of polynomials. We ended the class by reviewing what we can say about Noetherian rings of dimension zero. We gave an overview of why a zero-dimensional ring is Noetherian
if and only if it is Artinian. Mentioned that an Artinian ring can be written as a direct product of finitely many local Artinian rings in a unique way. We finished by saying that
Artinian rings can be used to study \(D/\mathfrak{a}\) where \(D\) is a one-dimensional integral domain and \(\mathfrak{a}\) is a non-zero ideal.
|
ucsd 
edu)