- Lecture 1: Here is my note for the first lecture.
In this lecture, we mainly reviewed topics from ring theory; prime and irreducible elements; properties of the principal
ideals generated by primes and irreducible elements; Zorn's lemma, and why it is often useful in ring theory; Noetherian rings;
in a PID, \({\rm spec}(A)={\rm max}(A)\cup \{0\}\). Proved the following theorem: suppose \(D\) is a Noetherian integral domain. Then
irreducible elements are prime if and only if \(D\) is a UFD.
- Lecture 2: Here is my note for the second lecture.
We started this lecture by showing that \(\mathbb{Z}[\sqrt{-5}]\) is not a UFD. Then we proved \(A[x]\) is a PID if and only if \(A\) is a field.
Along the way we showed that \(\mathfrak{p}\in {\rm Spec}(A)\) if and only if \(\mathfrak{p}[x]\in {\rm Spec}(A[x])\). We defined a greatest common
divisor of non-zero elements \(a_1,\ldots,a_n\) of an integral domain; and pointed out that \(d\) is such an element if and only if \(\langle d \rangle\)
is the minimum principal ideal that contains \(\langle a_1,\ldots,a_n\rangle\).
- Lecture 3: Here is my note for the third lecture.
We defined the \(p\)-valuation \(v_p:D\setminus \{0\}\rightarrow \mathbb{Z}^{\ge 0} \), and mentioned the following basic properties of
\(v_p\): (1) \(v_p(ab)=v_p(a)+v_p(b)\), (2) \(v_p(a+b)\ge \min\{v_p(a),v_p(b)\}\); and equality holds if \(v_p(a)\neq v_p(b)\),
(3) a|b if and only if for any irreducible \(p\) in \(D\) we have \(v_p(a)\le v_p(b)\), (4) \(a\sim b\) if and only if for any irreducible \(p\) in \(D\)
we have \(v_p(a)=v_p(b)\). Used these to get basic properties of gcd: (1) \(v_p(\gcd(a_1,\ldots,a_n))=\min\{v_p(a_1),\ldots,v_p(a_n)\}\), (2)
\(\gcd(ca_1,\ldots,ca_n)=[c]\gcd(a_1,\ldots,a_n)\) and \(\gcd(a_1/d,\ldots,a_n/d)=[1]\) if \(d\) is a gcd of \(a_1,\ldots,a_n\). Then we defined the content of a polynomial with coefficients in a UFD; and showed that \(f(x)=c_f \overline{f}(x)\) where \(c(f)=[c_f]\) and \(\overline{f}(x)\) is primitive. Proved that product of two primitive polynomials is primitive; and deduced Gauss's lemma which asserts that \(c(fg)=c(f)c(g)\); Used Gauss's lemma to show the following:
Suppose \(D\) is a UFD and \(F\) is its field of fractions. Suppose \(f(x)\in D[x]\) and \(f_i(x)\in F[x]\) such that \(f(x)=\prod_{i=1}^n f_i(x)\); then there are \(c_i\in F\) such that (1) \(c_i f_i(x)\in D[x]\) (2) \(\prod_{i=1}^n c_i=1\).
- Lecture 4: Here is my note for the fourth lecture.
We proved \(D\) is a UFD if and only if \(D[x]\) is a UFD. Along the way we proved the following results:
Let \(D\) is a UFD and \(F\) be its field of fractions. Then \(d\in D\) is irreducible in \(D\) if and only if \(d\) is irreducible in \(D[x]\);
A primitive polynomial \(f(x)\in D[x]\) of positive degree is irreducible in \(D[x]\) if and only if it is irreducible in \(F[x]\); If \(f(x)\in D[x]\) is primitive and \(cf(x)\in D[x]\) for some \(c\in F\), then \(c\in D\). We also pointed out that the presented proofs in the previous lectures imply:
Let \(D\) be a Noetherian integral domain. Then any non-zero non-unit element can be written as a product of irreducibles. And If in an
integral domain any irreducible element is prime, then a decomposition to irreducibles is unique up to reordering and multiplying by units.
- Lecture 5: Here is my note for the fifth lecture.
We discussed various irreducibility criteria; discussed irreducible elements in \(\mathbb{Z}[i]\);
mentioned Hilbert's basis theorem; deduced any finitely generated \(k\)-algebra is Noetherian if \(k\) is a field; went through half of proof of
Hilbert's basis theorem.
- Lecture 6: Here is my note for the sixth lecture.
We finished proof of Hilbert's basis theorem; we defined the ring of fractions with respect to a multiplicative set, and proved its universal property.
- Lecture 7: Here is my note for the seventh lecture.
We started module theory; mentioned that as group actions are crucial in understanding groups, modules are fundamental in understanding rings. Along the way,
the opposite ring was defined; mentioned that modules can be viewed as a generalization of vector spaces; left ideals are modules; \(R^n\) is an \({\rm M}_n(R)\)-module;
if \(M\) is a \(S\)-module and \(f:R\rightarrow S\) is a unital ring homomorphism, then \(M\) can be viewed as an \(R\)-module; having a unital commutative ring \(R\) and \(T\in {\rm M}_n(R)\), we can view \(R^n\) as an \(R[x]\)-module via \((\sum_i c_i x^i)\cdot v:=\sum_i c_i T^iv\); defined submodule; module homomorphism; image and kernel of a module homomorphism; and prove the first isomorphism theorem for modules.
- Lecture 8: Here is my note for the eighth lecture.
We defined the ring of endomorphisms \({\rm End}(M)\) of an abelian group \(M\); proved that there is a bijection between the possible \(R\)-module structures on an abelian group \(M\) and \({\rm Hom}(R,{\rm End}(M))\). To be precise we show the following maps are well-defined and inverses of each other:
\[
\Phi:\{m:R\times M\rightarrow M|\hspace{1mm} m \text{ defines module structure}\}\rightarrow {\rm Hom}(R,{\rm End}(M)),
\]
\[
((\Phi(m))(r))(x):=m(r,x),
\]
\[
\Psi:{\rm Hom}(R,{\rm End}(M))\rightarrow \{m:R\times M\rightarrow M|\hspace{1mm} m \text{ defines module structure}\},
\]
\[
(\Psi(f))(r,x):=(f(r))(x).
\]
We pointed out the similarity with group actions; If \(M\) is an \(R\)-module and \(\theta:R\rightarrow {\rm End}(M)\) is the induced unital ring homomorphism, then \({\rm End}_R(M)=C_{{\rm End}(M)}(\theta(R))\); in particular, \({\rm End}_R(M)\) is a subring of \({\rm End}(M)\), and, when \(R\) is a unital commutative ring, \(\theta(R)\subseteq Z({\rm End}_R(M))\); we defined the submodule generated by a subset and summation of submodules.
- Lecture 9: Here is my note for the ninth lecture.
We defined external direct sum and direct product of a family of modules; proved the universal property of the external direct sum; compared it with free product of groups; explored the connection between the internal and the external direct sum; defined free module, and proved its universal property; mentioned that if \(A\) is a unital commutative ring and \(A^n\simeq A^m\) (as \(A\)-mod), then \(n=m\).
- Lecture 10: Here is my note for the tenth lecture.
We defined rank of a module: \({\rm rank}(M)\) is the maximum number of linearly independent elements of \(M\); proved that, if \(D\) is an integral domain, then
\({\rm rank}(D^n)=n\); (as part of your homework assignment you will prove the same result for any unital commutative ring); started proof of the following theorem:
Let \(D\) be a PID; and \(M\) be a submodule of \(D^n\). Then (a) \(M\) is a free module, and (b) there are \(x_1,\ldots,x_n\in D^n\) and \(a_1,\ldots,a_m\in D\setminus \{0\}\) such that (b-1) \(a_1| a_2 |\cdots |a_m\), (b-2) \(D^n=\bigoplus_{i=1}^n D x_i \), and (b-3) \(M=\bigoplus_{i=1}^m Da_ix_i\).
- Lecture 11: Here is my note for the eleventh lecture.
We finished the proof of characterizing submodules of a f.g. free module over a PID; proved the existence part of the Fundamental Theorem of Finitely Generated Modules Over a PID:
Let \(D\) be a PID and \(M\) be a finitely generated \(D\)-module. Then there are \(a_1,\ldots,a_m\in D\setminus \{0\}\) such that
\(M\simeq D^r\oplus \bigoplus_{i=1}^m D/Da_i\).
pointed out that for \(D=\mathbb{Z}\) we get the Fundamental Theorem of Finitely Generated Abelian Groups:
A finitely generated abelian group \(G\) is isomorphic to \(\mathbb{Z}^n \oplus \bigoplus_{i=1}^m \mathbb{Z}/a_i\mathbb{Z}\) for some non-zero integers \(a_i\) such that \(a_1|a_2|\cdots |a_m\).
Started proof of the uniqueness part; along the way proved: Let \(M:=D^n\oplus \bigoplus_{i=1}^m D/Da_i\). Then \( {\rm rank}(M)=n \); \({\rm Tor}(M)=\bigoplus_{i=1}^m D/Da_i\); in particular; \(M=F\oplus T\) where \(F\) is free and \(T\) is torsion, and so
A torsion free finitely generated module over a PID is free.
- Lecture 12: Here is my note for the twelfth lecture.
We finished proof of the uniqueness part of the Fundamental Theorem of Finitely Generated Modules Over a PID; so overall we get the following theorem:
Let \(D\) be a PID and \(M\) be a finitely generated \(D\)-module. Then there are \(a_1,\ldots,a_m\in D\setminus \{0\}\) such that
- \(M\simeq D^r\oplus \bigoplus_{i=1}^m D/Da_i\)
- \({\rm rank}(M)=r\)
- \({\rm Tor}(M)= \bigoplus_{i=1}^m D/Da_i\)
- For any irreducible element \(p\), the dual of the Young Tableau of the decreasing sequence
\(
\{\dim_{k(p)} p^i M_p/ p^{i+1} M_p\}_i
\)
is the Young Tableau of the increasing sequence \(\{v_p(a_i)\}_i\), where \(k(p):=D/\langle p\rangle\).
(We pointed out that for any \(\mathfrak{p}\in {\rm Spec}(A)\), \(k(\mathfrak{p}):=A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}\) is a field; and it is called the residue field of \(\mathfrak{p}\).)
Then we started the formulation and proof of Rational Canonical Form of a matrix over a field \(k\): let \(A\in {\rm M}_n(k)\). We view \(k^n\) as a
\(k[x]\)-module: \(x\cdot v:= Av\). We proved that \(k^n\) is a torsion \(k[x]\)-module; and so there are polynomials \(f_1,\ldots,f_m\in k[x]\) such that
\(f_1|f_2|\cdots |f_m\) and \(k^n\simeq k[x]/\langle f_1(x)\rangle \oplus \cdots \oplus k[x]/\langle f_m(x)\rangle\) as \(k[x]\)-modules; next we discussed that
\(\mathfrak{B}:=\{\bar{1},\bar{x},\ldots,\bar{x}^{d-1}\}\) is a \(k\)-basis of \(k[x]/\langle f(x)\rangle\) if \(f\) is a degree \(d\) polynomial, and the matrix associated with the multiplication by \(x\) in \(k[x]/\langle f(x)\rangle\) with respect to the basis \(\mathfrak{B}\) is given by the companion matrix of \(f\); we also discussed the converse of this statement.
- Lecture 13: Here is my note for the thirteenth lecture.
We finished proof of Rational Canonical Form theorem:
Let \(k\) be a field and \(A\in {\rm M}_n(k)\). Then there are unique monic polynomials \(f_1|f_2|\cdots|f_m \in k[x]\) such that \(A\) is similar to
\[
\begin{pmatrix}
c(f_1)&&\\
&\ddots&\\
&&c(f_m)\\
\end{pmatrix}
\]
where \(c(f)\) is the companion matrix of \(f\).
Polynomials \(f_i\) in the Rational Canonical Form Theorem are called the invariant factors of \(A\). So two matrices are similar if and only if they have the same invariant factors.
Along the way, we considered the following: for a matrix \(A\in {\rm M}_n(k)\), we defined the \(k\)-algebra homomorphism \(\phi_A:k[x]\rightarrow {\rm End}_k(k^n), \phi_A(f(x)):=f(A)\). This morphism makes \(k^n\) into a \(k[x]\)-module. We called this module \(V_A\).
We proved: \(V_A\simeq V_B\) if and only if \(A\) and \(B\) are similar; \(V_{c(f)}\) is isomorphic to \(k[x]/\langle f\rangle\). Then we proved
suppose \(f_1|f_2|\cdots|f_m \in k[x]\) are the invariant factors of \(A\). Then
- the minimal polynomial \(m_A(x)=f_m(x)\) and \(\{g(x)| g(A)=0\}=\langle m_A(x) \rangle\);
- the characteristic polynomial \(f_A(x):=\det(xI-A)\) is equal to \(\prod_{i=1}^m f_i(x)\);
- in particular, \(m_A(x)|f_A(x)\) and \(f_A(A)=0\) (this is called Cayley-Hamilton's theorem);
- if \(p(x)\) is an irreducible factor of \(f_A(x)\), then \(p(x)|m_A(x)\).
Then we started proof of Jordan Form Theorem.
- Lecture 14: Here is my note for the fourteenth lecture.
We prove the existence and uniqueness of the Jordan form of a matrix:
Let \(k\) be a field and \(A\in {\rm M}_n(k)\). Suppose the characteristic polynomial \(f_A\) of \(A\) is equal to \(\prod_{i=1}^s(x-\lambda_i)^{n_i}\). Then there are
unique positive integers \(m_{i1}\le m_{i2} \le \cdots \le m_{il_i} \le n_i\) such that \(A\) is similar to
\[
\begin{pmatrix}
J_{m_{11}}(\lambda_1)&&&\\
& J_{m_{12}}(\lambda_1)&&\\
&&\ddots &\\
&&& J_{m_{sl_s}}(\lambda_s)\\
\end{pmatrix},
\]
where \(J_m(\lambda)=
\begin{pmatrix}\lambda&&& \\
1&\lambda&&\\
&\ddots&\ddots&&\\
&&1&\lambda
\end{pmatrix}\) is a Jordan block.
Then we defined simple modules, and proved Schur's lemma. Next we defined exact and short exact sequences; and proved a short exact sequence
\(0 \rightarrow M_1 \xrightarrow{f_1} M_2 \rightarrow M_3 \rightarrow 0\) is isomorphic to the short exact sequence
\(0\rightarrow f_1(M_1)\rightarrow M_2 \rightarrow M_2/f_1(M_1)\rightarrow 0\).
- Lecture 15: Here is my note for the fifteenth lecture.
We defined a homomorphism of short exact sequences. Recalled that any short exact sequence \(0\rightarrow M_1 \rightarrow M_2 \xrightarrow{f_2} M_3 \rightarrow 0\) is isomorphic to \(0\rightarrow {\rm ker} f_2 \rightarrow M_2 \rightarrow M_2/ {\rm ker} f_2 \rightarrow 0\). Proved The Short Five Lemma:
Suppose \((\phi_1,\phi_2,\phi_3)\) is a homomorphism of short exact sequences. Then (a) if \(\phi_1,\phi_3\) are injective, then \(\phi_2\) is injective, (b)
if \(\phi_1,\phi_3\) are surjective, then \(\phi_2\) is surjective, and (c) if \(\phi_1,\phi_3\) are isomorphisms, then \(\phi_2\) is an isomorphism.
Then we defined a split short exact sequence and proved the following properties are equivalent for a short exact sequence \(0\rightarrow M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \rightarrow 0\):
- There is a submodule \(N_2\) of \(M_2\) such that \(M_2=N_2\oplus f_1(M_1)\).
- There is \(g_1\in {\rm Hom}_R(M_2,M_1)\) such that \(g_1\circ f_1={\rm id}_{M_1}\).
- There is an isomorphism of short exact sequences \(({\rm id}_{M_1},\phi, {\rm id}_{M_3})\) from \(0\rightarrow M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \rightarrow 0\) to \(0\rightarrow M_1 \rightarrow M_1\oplus M_3 \rightarrow M_3 \rightarrow 0\).
- There is \(g_2\in {\rm Hom}_R(M_3,M_2)\) such that \(f_2\circ g_2={\rm id}_{M_3}\).
At the end of lecture, we defined what a category and a functor are.
- Lecture 16: Here is my note for the sixteenth lecture.
We defined representable functor for a locally small category. Proved its analogue for the category of \(R\)-modules gives us a functor
\(F_M:R-{\rm mod} \rightarrow {\rm Ab}\) from \(R\)-mod to the category \({\rm Ab}\) of abelian groups. Proved if \(0\rightarrow N_1\xrightarrow{f_1}N_2\xrightarrow{f_2}N_3 \rightarrow 0\) is a short exact sequence, then
\(0\rightarrow F_M(N_1)\xrightarrow{F_M(f_1)}F_M(N_2)\xrightarrow{F_M(f_2)}F_M(N_3) \) is an exact sequence. Proved the following statements are equivalent:
-
\(F_M\) is an exact functor; that means if \(0\rightarrow N_1\xrightarrow{f_1}N_2\xrightarrow{f_2}N_3 \rightarrow 0\) is a short exact sequence, then \(0\rightarrow F_M(N_1)\xrightarrow{F_M(f_1)}F_M(N_2)\xrightarrow{F_M(f_2)}F_M(N_3)\rightarrow 0 \) is a short exact sequence.
-
If \(\phi\in {\rm Hom}_R(N,N')\) is surjective, then \(F_M(\phi)\) is surjective.
-
Suppose \(\phi \in {\rm Hom}_R(N,N')\) is surjective; then any \(\psi \in {\rm Hom}_R(M,N')\) has a lift to \(\overline{\psi} \in {\rm Hom}_R(M,N)\).
-
Any short exact sequence of the form \(0\rightarrow N_1\rightarrow N_2\rightarrow M \rightarrow 0\) splits.
-
\(M\) is a direct summand of a free module.
A module that satisfies the above properties is called a projective module. We showed that if \(D\) is an integral domain then
free module implies projective; projective implies torsion-free; and when \(M\) is a finitely generated \(D\)-modules and \(D\) is a PID, torsion-free implies free; and so projective implies free for f.g. modules over a PID. Then we showed the ideal \(\langle 2,\sqrt{-10}\rangle\) is not a free \(\mathbb{Z}[\sqrt{-10}]\)-module; but it is projective. We pointed out that later based on Kaplansky's thoerem we will see that a finitely presented module is projective if and only if it is locally free; and so for a ring that is locally PID we get that any ideal is projective.
- Lecture 17: Here is my note for the seventeenth lecture.
In this lecture, first we defined an \( (S,R)\)-bimodule; and proved that if \(M\) is an \((S,R)\)-bimodule, then the representable functor
\(F_M\) is a functor from the category of right \(R\)-modules to the category of right \(S\)-modules. For a right \(S\)-module \(N\) and an \((S,R)\)-bimodule, \(F_N\circ F_M\) gives us a functor from (right) \(R\)-mod to Ab. We tried to figure out if this is a representable functor or not. Along the way, we proved that the following is a bijection
\[
{\rm Hom}_S(N,{\rm Hom}(M,L))\rightarrow \mathcal{A},
\\
\phi \mapsto f_{\phi}, \text{ where } f_{\phi}(n,m):=(\phi(n))(m),
\]
where \(\mathcal{A}\) consists of functions \(f:N\times M\rightarrow L\) such that (1) \(f(ns,m)=f(n,sm)\) (we say \(S\)-balanced),
(2) \(f(n_1-n_2,m)=f(n_1,m)-f(n_2,m)\) (linear in the first factor), (3) \(f(n,m_1r_1+m_2r_2)=f(n,m_1)r_1+f(n,m_2)r_2\) (\(R\)-linear in the second factor).
Using an idea similar to proof of Yoneda's lemma (we did not say what this Lemma is), we saw how to define a right \(R\)-module \(\mathcal{F}(N,M)\) such that there would be a natural transformation between \(F_{\mathcal{F}(N,M)}\) and \(F_N\circ F_M\). This is how we defined and proved the following statements:
Let \(N\) be a right \(S\)-module and \(M\) be an \((S,R)\)-bimodule. Then there are a right \(R\)-module \(N\otimes_S M\) and \(f_0:N\times M\rightarrow N\otimes_S M, f_0(n,m):=n\otimes m\) such that \(f_0\) is \(S\)-balanced, linear in \(N\), and \(R\)-linear in \(M\), and we have the following universal property: if \(L\) is a right \(R\)-module and \(f:N\times M\rightarrow L\) is \(S\)-balance, linear in \(N\), and \(R\)-linear in \(M\), then there is a unique right \(R\)-module homomorphism \(\tilde{f}: N\otimes_S M \rightarrow L, \tilde{f}(n\otimes m)=f(n,m)\).
and we have the following result as well:
There is a natural transformation between \(F_{N\otimes_S M}\) and \(F_N\circ F_M\); in particular,
\(
{\rm Hom}_R(N\otimes_S M, L)\simeq {\rm Hom}_S(N,{\rm Hom}_R(M,L)).
\)
- Lecture 18: Here is my note for the eighteenth lecture.
First we recalled what we have proved on tensor products: its universal property and tensor-hom adjunction. We used it to prove:
Suppose \(M\) is an \( (S,R)\)-bimodule and \(N\) is a right \(S\)-module. Suppose \(M\) is a projective right \(R\)-module and \(N\) is a projective right \(S\)-module. Then \(N\otimes_S M\) is a projective right \(R\)-module.
In particular, if \(R\) is a commutative ring, then tensor product of any two \(R\)-modules is a projective \(R\)-module. (So for an integral domain \(D\) we can consider rank one projective modules and create a group out of them: this is called the Picard group of \(D\); this is a generalization of the class group of the ring of integers of a number field (as we will see in Math200C).) Then we showed the examples \(\mathbb{Z}/n\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq \mathbb{Z}/\gcd(n,m)\mathbb{Z} \) and \(\mathbb{Q}\otimes_{\mathbb{Z}} T=0\) where \(T\) is a torsion abelian group (we also discussed that \(\mathbb{Q}/\mathbb{Z}\) is a torsion abelian group). After these examples, we discussed the extension of scalars or base change: suppose \(\phi:S\rightarrow R\) is a ring homomorphism (think about the case where \(S\) is a subring of \(R\)). Then having a right \(S\)-module \(M\), we can extend the scalar multiplication to R (or change the base ring from \(S\) to \(R\)): \(R\) can be viewed as an \( (S,R)\)-bimodule via \(s\cdot r \cdot r':=\phi(s) r r'\), and we can consider the right \(R\)-module \(M_R:=M\otimes_S R\). Then we considered the base change induced by \(\pi:R\rightarrow R/I\) where \(I\) is a (two-sided) ideal of \(R\), and proved:
\(f: M\otimes_R R/I\rightarrow M/IM, f(x\otimes r+I):= xr+IM,\) and \(g:M/IM\rightarrow M\otimes_R R/I, g(x+IM):=x\otimes (1+I)\) are well-defined \(R/I\)-module isomorphisms that are inverses of each other.
Next we proved that if \(f\in {\rm Hom}_S(N_1,N_2)\) and \(g\in {\rm Hom}_{(S,R)}(M_1,M_2)\), then there is a unique \(f\otimes g\in {\rm Hom}_R(N_1\otimes_S M_1, N_2 \otimes_S M_1)\) such that \((f\otimes g)(n_1\otimes m_1)=f(n_1)\otimes g(m_1)\). Based on this we showed that if \(M\) is an \( (S,R)\)-bimodule, then \(\bullet\otimes_S M\) defines a functor from Right \(S\)-Mod to Right \(R\)-Mod.
- Lecture 19: Here is my note for the nineteenth lecture.
First we proved that the tensor functor \(\bullet\otimes_S M\) from Right \(S\)-Mod to Right \(R\)-Mod is right exact. Then we defined a flat module.
Next we proved \(f:N\otimes_S (M\otimes_R L)\rightarrow (N\otimes_S M)\otimes_R L, f(n\otimes (m\otimes l)):=(n\otimes m)\otimes l\) is a well-defined abelian group isomorphism. We used this to prove
Suppose \(M\) is an \( (S,R)\)-bimodule and \(N\) is a right \(S\)-module. Suppose \(M\) is a flat right \(R\)-module and \(N\) is a flat right \(S\)-module. Then \(N\otimes_S M\) is a flat right \(R\)-module.
- Lecture 20: Here is my note for the twentieth lecture.
We proved that \(\oplus\) commutes with \(\otimes\); to be precise \((m_1,m_2)\otimes n \mapsto (m_1\otimes n, m_2\otimes n)\) gives us an isomorphism from \((M_1\oplus M_2)\otimes_R N\) to \((M_1\otimes_R N)\oplus (M_2\otimes_R N)\). Using this we proved that \((r_1,\ldots,r_n)\otimes m\mapsto (r_1m,\ldots,r_nm)\) gives us an isomorphism form \(R^n \otimes_R M\) to \(M^n\). Then we proved any free module is flat. Next we proved \(M_1\) and \(M_2\) are flat if and only if \(M_1\oplus M_2\) is flat. Then concluded that any projective module is flat. Then we defined an \(R\)-algebra; and mentioned that \(A\otimes_R B\) is an \(R\)-algebra if \(A\) and \(B\) are \(R\)-algebras; and its multiplication is given by \((a\otimes b)(a'\otimes b'):=aa'\otimes bb'\).
- Lecture 21: Here is my note for the twenty first lecture.
Let \(\phi:S\rightarrow R\) be a unital commutative ring homomorphism. Then we can view \(R\) as an \(S\)-algebra. Suppose \(I\) is an ideal of \(S[x]\). Then we pointed out that \(\phi\) can be extended to a ring homomorphism from \(S[x]\) to \(R[x]\) and \(R\phi(I)\) is an ideal of \(R[x]\); and we proved that \(R \otimes_S S[x]/I\simeq R[x]/R\phi(I)\) and an isomorphism is given by \(r\otimes p(x)+I \mapsto r\phi(p)+R\phi(I)\). Using this we showed \(\mathbb{Q}[i]\otimes_{\mathbb{Q}} \mathbb{Q}[i]\simeq \mathbb{Q}[i]\oplus \mathbb{Q}[i]\) as \(\mathbb{Q}\)-algebras. Then we analyzed \(\mathbb{Z}/p\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}[i]\); we proved that, for \(p=2\), it is isomorphic to \((\mathbb{Z}/2\mathbb{Z})[y]/\langle y^2\rangle\) in particular it has a nilpotent element; if \(x^2+1\) has a zero in \(\mathbb{Z}/p\mathbb{Z}\) and \(p\) is odd, then it is isomorphic to \(\mathbb{Z}/p\mathbb{Z}\oplus \mathbb{Z}/p\mathbb{Z}\); and if \(x^2+1\) has no zero in \(\mathbb{Z}/p\mathbb{Z}\), then it is isomorphic to a field of order \(p^2\). So overall, the only time \(\mathbb{Z}/p\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}[i]\) has a (non-zero) nilpotent element is when \(p=2\). Then we moved to field theory, and the following theorem was stated:
Let \(F\) be a field and \(E/F\) be a field extension. Suppose \(\alpha\in E\) is a zero of a non-zero polynomial \(p(x)\in F[x]\).
Then
- There is a unique monic polynomial \(m_{\alpha}(x)\in F[x]\) such that, for \(f(x)\in F[x]\), \(f(\alpha)=0\) if and only if \(m_{\alpha}(x)|f(x)\).
- \(m_{\alpha}(x)\) is irreducible in \(F[x]\).
- \(F[\alpha]:=\{f(\alpha)|\hspace{1mm} f(x)\in F[x]\}\simeq F[x]/\langle m_{\alpha}(x) \rangle\); and so \(F[\alpha]\) is a field.
- \(F[\alpha]:=\{\sum_{i=0}^{d-1} a_i \alpha^i|\hspace{1mm} a_i\in F\}\) where \(d:=\deg m_{\alpha}\); and \(\dim_F F[\alpha]=\deg m_{\alpha}\).
- Lecture 22: Here is my note for the twenty second lecture.
We proved the mentioned theorem about elements of \(E\) that are algebraic over \(F\). We said \(E\) is called a splitting field of \(f(x)\in F[x]\setminus F\) over F if there are \(\alpha_1,\ldots, \alpha_n\in E\) such that (1) \(f(x)=c(x-\alpha_1)\cdots (x-\alpha_n)\) for some \(c\in F\); and (2) \(E=F(\alpha_1,\ldots,\alpha_n)\). Then we proved existence and uniqueness of the splitting field of \(f(x)\in F[x]\setminus F\) over \(F\):
Suppose \(\phi:F\rightarrow F'\) is a field isomorphism. Then we get a natural isomorphism \(\phi:F[x]\rightarrow F'[x]\). Suppose \(f(x)\in F[x]\setminus F\),
\(E\) is a splitting field of \(f\) over \(F\), and \(E'\) is a splitting field of \(\phi(f)\) over \(F'\). Then there is an isomorphism \(\widetilde{\phi}:E\rightarrow E'\) such that \(\widetilde{\phi}|_F=\phi\).
- Lecture 23: Here is my note for the twenty third lecture.
We used the existence and uniqueness of splitting fields to show:
- For any prime \(p\) and positive integer \(d\), there is a unique (up to isomorphism) finite field of order \(q:=p^d\). (It is denoted by \(\mathbb{F}_q\) or \(\mathfrak{f}_q\).)
- We have \([\mathbb{F}_{p^d}:\mathbb{F}_p]=d\) and \(\prod_{\alpha\in \mathbb{F}_{q}}(x-\alpha)=x^q-x\).
- \(\mathbb{F}_q\) is the splitting field of \(x^q-x\) over \(\mathbb{F}_p\).
Then we proved
if \(E/F\) and \(K/E\) are finite field extensions, then \(K/F\) is a finite field extension and \([K:F]=[K:E][E:F]\).
Then we proved that a finite extension is an algebraic extension. Based on these two we proved:
Suppose \(E/F\) is a field extension. Let \(L:=\{a\in E|\hspace{1mm} a\) is algebraic over \(F\}\). Then \(L/F\) is an algebraic field extension and \(L\) is called the algebraic closure of \(F\) in \(E\).
Then algebraically closed fields were defined, and we started proof of the following theorem:
For any field \(F\), there is a field extension \(E/F\) such that \(E\) is algebraically closed.
- Lecture 24: Here is my note for the twenty fourth lecture.
We proved the existence of a field extension \(E/F\) where \(E\) is algebraically closed. Next we proved that if \(E/F\) and \(K/E\) are algebraic extensions, then
\(K/F\) is algebraic. Next we defined an algebraic closure of a field \(F\). We proved that, if \(\Omega/F\) is a field extension and \(\Omega\) is algebraically closed, then the algebraic closure of \(F\) in \(\Omega\) is an algebraic closure of \(F\). Next we started proof of uniqueness of algebraic closure of a field up to an isomorphism:
- Let \(\Omega\) be an algebraically closed field, \(\sigma:F\rightarrow \Omega\) be an embedding of fields,
\(f(x)\in F[x]\setminus F\) be a monic polynomial, and \(E\) be a splitting field of \(f(x)\) over \(F\). Then there is an embedding
\(\widetilde{\sigma}:E\rightarrow \Omega\) such that \(\widetilde{\sigma}|_F=\sigma\).
-
Let \(F\) be a field, and \(E,E'\) be two algebraic closures of \(F\). Let \(\sigma:F\rightarrow E'\) be an embedding. Then there is an isomorphism \(\widetilde{\sigma}:E\rightarrow E'\) such that \(\widetilde{\sigma}|_F=\sigma\).
- Lecture 25: Here is my note for the twenty fifth lecture.
We finished proof of the theorem mentioned in the previous lecture. As corollaries, we got that any two algebraic closures of a given field are isomorphic, and \(\{\sigma|_F:F\rightarrow \overline{F}|\hspace{1mm} \sigma\in {\rm Aut}(\overline{F})\}=\{\theta:F\rightarrow \overline{F}|\hspace{1mm}\) embedding \(\}\). Then we defined a splitting field of a family of non-constant polynomials, and proved the following:
Let \(F\) be a field and \(\overline{F}\) be an algebraic closure of \(F\). Let \(F\subseteq E\subseteq \overline{F}\). Then the following statements are equivalent:
- \(E/F\) is a field extension and, for any \(\sigma\in {\rm Aut}(\overline{F}/F), \sigma(E)=E\).
-
\(E/F\) is a field extension, and for any \(\alpha\in E\), there are \(\alpha_i\in E\) such that \(m_{\alpha,F}(x)=\prod_i (x-\alpha_i)\).
-
\(E\) is a splitting field of a non-empty family \(\mathcal{F}\subseteq F[x]\setminus F\) of monic polynomials.
-
There are a family \(\{E_i\}_{i\in I}\) of fields such that
- \(E_i\subseteq \overline{F}\) and \(E_i\) is a splitting field of a polynomial \(f_i(x)\in F[x]\) over \(F\).
- For any \(i,j\in I\), there is \(k\in I\), such that \(E_i\cup E_j\subseteq E_k\).
- \(E=\bigcup_{i\in I} E_i\).
An extension \(E/F\) is called a normal extension if the above properties hold.
- Lecture 26: Here is my note for the twenty sixth lecture.
Suppose \(\overline{F}\) is an algebraic closure of \(F\), and \(F\subseteq E_1\subseteq E_2\subseteq\overline{F}\) are intermediate subfields. Suppose \(E_i/F\) are normal extensions. Then we proved that the restriction function \(r_{E_1E_2}:{\rm Aut}(E_2/F)\rightarrow {\rm Aut}(E_1/F), r_{E_1E_2}(\sigma):=\sigma|_{E_1}\) is a well-defined onto group homomorphism; and deduced that \({\rm Aut}(E_2/E_1)\) is a normal subgroup of \({\rm Aut}(E_2/F)\) and \( {\rm Aut}(E_2/F)/{\rm Aut}(E_2/E_1)\simeq {\rm Aut}(E_1/F)\).
Next we proved the following:
Let \(L/F\) be a normal extension, and \(I:=\{E|\hspace{1mm} F\subseteq E\subseteq F, E/F\) is a normal extension \(\}\); then
\[
r:{\rm Aut}(L/F)\rightarrow \{(\phi_E)_{E\in I}\in \prod_{E\in I} {\rm Aut}(E/F)|\hspace{1mm} \phi_{E_2}|_{E_1}=\phi_{E_1} \text{ for any } E_1\subseteq E_2, E_i\in I\},
r(\sigma):=(\sigma|_E)_{E\in I}.
\]
is a group isomorphism. The RHS is called the inverse limit of \({\rm Aut}(E/F)\)'s; and so \({\rm Aut}(L/F)\simeq \varprojlim_{E\in I} {\rm Aut}(E/F)\). We remark that by Tychonoff's theorem, we can show that with respect to the topology induced by the product topology \(\varprojlim_{E\in I} {\rm Aut}(E/F)\) is compact; and the restriction group homomorphisms \(r_{E_1E_2}\) are continuous; and so \({\rm Aut}(E_2/E_1)\) is a closed normal subgroup of \({\rm Aut}(E_2/F)\).
Next we proved the following:
Let \(\sigma_0:F\rightarrow F'\) be an isomorphism of fields, and \(f(x)\in F[x]\setminus F\) is a square-free monic polynomial. Suppose \(E\) is a splitting field of \(f(x)\) over \(F\) and \(E'\) is a splitting field of \(\sigma_0(f)\) over \(F'\). Then
\[
|\{\sigma:E\xrightarrow{\sim} E'|\hspace{1mm} \sigma|_F=\sigma_0\}|\le [E:F],
\]
and equality holds exactly when \(f(x)\) has no multiple zeros in \(E\).
Then we defined a separable polynomial and separable extension, and proved the following:
Let \(E/F\) be a finite extension. Then the following are equivalent:
- \(E\) is a splitting field of a separable polynomial \(f(x)\in F[x]\setminus F\) over \(F\).
- \(|{\rm Aut}(E/F)|=[E:F]\)
- \(E/F\) is a normal separable extension.
We called a field extension \(E/F\) Galois if it is normal and separable. Finally we proved the following technical lemma, which we need later.
Let \(E\) be a field, and \(G\) be a subgroup of \({\rm Aut}(E)\). Let \(G\) act on the vector space \(E^n\). Suppose \(V\) is a non-zero subspace of \(E^n\) that is \(G\)-invariant. Then \(V^G:=\{v\in V| \hspace{1mm}\forall \sigma\in G, \sigma(v)=v \}\) is non-zero.
- Lecture 27: Here is my note for the twenty seventh lecture.
First we used the mentioned technical lemma in the previous lecture to prove, for any finite group \(G\) of \({\rm Aut}(E)\), \({\rm Fix}(G)\) is a field and \([E:{\rm Fix}(G)]\le |G|\). Then this is used to prove the important parts of Fundamental Theorem of Galois Theory:
Let \(E/F\) be a finite Galois extension. Then
- For any intermediate field \(F\subseteq K\subseteq E\), \(E/K\) is a Galois extension, and \(K/F\) is a separable extension.
- Let \(\mathcal{F}:=\{K|\hspace{1mm} F\subseteq K\subseteq E \text{ subfield}\}\) and \(\mathcal{G}:=\{H|\hspace{1mm} H\leq {\rm Gal}(E/F)\}\). Let \(\Phi:\mathcal{F}\rightarrow \mathcal{G},\Phi(K):={\rm Gal}(E/K)\) and \(\Psi:\mathcal{G}\rightarrow \mathcal{F}, \Psi(H):={\rm Fix}(H)\). Then \(\Phi\) and \(\Psi\) are well-defined maps and inverse of each other.
- Let \(\mathcal{F}_N:=\{K\in \mathcal{F}|\hspace{1mm} K/F \text{ is normal}\}\) and \(\mathcal{G}_N:=\{N|\hspace{1mm} N\unlhd {\rm Gal}(E/F)\}\). Then \(\Phi(\mathcal{F}_N)=\mathcal{G}_N\) and \(\Psi(\mathcal{G}_N)=\mathcal{F}_N\). Moreover if \(K\in \mathcal{F}_N\), then \(K/F\) is a Galois extension, and
\({\rm Gal}(K/F)\simeq {\rm Gal}(E/F)/{\rm Gal}(E/K)\).
- Lecture 28: Here is my note for the twenty eighth lecture.
First we proved that if \(E/F\) is a finite separable extension, then there is \(L/E\) such that \(L/F\) is Galois. And so there are only finitely many intermediate fields
\(F\subseteq K\subseteq E\). Next we proved that there are only finitely many intermediate fields \(F\subseteq K\subseteq E\) if and only if \(E/F\) is a simple extension;
that means \(E=F[\theta]\) for some \(\theta\in E\). Then we proved if \(F\) is a field of characteristic zero, then any non-constant polynomial in \(F[x]\) is separable; and
if \({\rm char}(F)=p>0\), then for any irreducible \(f(x)\in F[x]\) there is an irreducible separable \(g(x)\in F[x]\) such that \(f(x)=g(x^{p^k})\). Using this result we proved that the following
statements are equivalent:
- Either char\((F)=0\), or char\((F)=p> 0\) and \(F^p=F\).
- \(\overline{F}/F\) is a Galois extension where \(\overline{F}\) is an algebraic closure of \(F\).
- Any finite extension \(E/F\) is separable.
We say \(F\) is perfect if the above statements hold. Hence if \(F\) is perfect, then any finite extension \(E/F\) is simple.
- Lecture 29: Here is my note for the twenty ninth lecture.
We proved that \({\rm Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p})=\langle \sigma_{p,n}\rangle\) where \(\sigma_{p,n}:\mathbb{F}_{p^n}\rightarrow \mathbb{F}_{p^n}, \sigma_{p,n}(\alpha)=\alpha^p\). Using this we deduced that \({\rm Gal}(\overline{\mathbb{F}_{p}}/\mathbb{F}_{p})\simeq \varprojlim \mathbb{Z}/n\mathbb{Z}\); this is called the profinite closure of \(\mathbb{Z}\) and it is denoted by \(\widehat{\mathbb{Z}}\). Then we studied a splitting field
\(E\subseteq \mathbb{C}\) of \(x^n-1\) over \(\mathbb{Q}\). We proved that \(E=\mathbb{Q}[\zeta_n]\) where \(\zeta_n:=e^{2\pi i/n}\), and \(\theta:{\rm Gal}(\mathbb{Q}[\zeta_n]/\mathbb{Q})\rightarrow (\mathbb{Z}/n\mathbb{Z})^{\times}, \theta(\sigma):=i_{\sigma}\) where \(\sigma(\zeta_n)=\zeta_n^{i_{\sigma}}\) is an injective group homomorphism. We observed that to show \(\theta\) is an isomorphism, it is enough to show the degree of the minimal polynomial of \(\zeta_n\) over \(\mathbb{Q}\) is \(\phi(n):=|(\mathbb{Z}/n\mathbb{Z})^{\times}|\). To find the minimal polynomial of \(\zeta_n\) over \(\mathbb{Q}\) we proved the following: suppose \(F[\alpha]/F\) is a Galois extension; then \(m_{\alpha,F}(x)=\prod_{\sigma\in {\rm Gal}(F[\alpha]/F)} (x-\sigma(\alpha))\). This motivated us to define the \(n\)-th cyclotomic polynomial \(\Phi_n(x):=\prod_{1\le j< n, \gcd(j,n)=1} (x-\zeta_n^j)\). And we pointed out that to prove \(\theta\) is an isomorphism it is enough to prove: \(\Phi_n(x)\in \mathbb{Q}[x]\) and \(\Phi_n(x)\) is irreducible in \(\mathbb{Q}[x]\). And we proved \(\prod_{d|n} \Phi_d(x)=x^n-1\).
- Lecture 30: Here is my note for the thirtieth lecture.
First we proved that \(\Phi_n(x)\in \mathbb{Z}[x]\); next we showed \(\Phi_n(x)\) is irreducible in \(\mathbb{Q}[x]\); and we deduced that \(m_{\zeta_n,\mathbb{Q}}(x)=\Phi_n(x)\in \mathbb{Z}[x]\) and \({\rm Gal}(\mathbb{Q}[\zeta_n]/\mathbb{Q})\simeq (\mathbb{Z}/n\mathbb{Z})^{\times}\) and the isomorphism is given by \(\sigma \mapsto i_{\sigma}\) where \(\sigma(\zeta_n)=\zeta_n^{i_{\sigma}}\). Then we studied a splitting field \(E\subseteq \overline{F}\) of \(x^n-a\) over \(F\) when \({\rm char}(F)\nmid n\) and \(\mu_n:=\{\zeta\in \overline{F}| \hspace{1mm} \zeta^n=1\}\subseteq F\). We proved that \(E=F[\sqrt[n]{a}]\) where \(\sqrt[n]{a}\) is a zero of \(x^n-a\). Then we said \(\mu_n\) is cyclic, say \(\mu_n=\langle \zeta\rangle\). Then we showed the map \({\rm Gal}(F[\sqrt[n]{a}]/F)\xrightarrow{\theta} \mathbb{Z}/n\mathbb{Z}\) given by \(\sigma\mapsto j_{\sigma}\) where \(\sigma(\sqrt[n]{a})=\zeta^{j_{\sigma}}\sqrt[n]{a}\) is an injective group homomorphism; and so \(F[\sqrt[n]{a}]/F\) is a cyclic extension. Using these results, we proved:
(Galois) Suppose \(F\) is a field of characteristic zero, \(f(x)\in F[x]\), and \(E\) is a splitting field of \(f(x)\) over \(F\). Then if \(f(x)\) is solvable in radicals over \(F\), then \({\rm Gal}(E/F)\) is solvable.
To prove the converse of the above statement, we proved a technical lemma due to Dirichlet that has applications in other places as well: let \(G\) be a group and \(E\) be a field. Suppose \(\chi_1,\ldots,\chi_m:G \rightarrow E^{\times}\) are distinct group homomorphisms. Then \(\chi_i\)'s are \(E\)-linearly independent. We will be using Dirichlet's independence of characters to prove Hilbert's Theorem 90. We stated Hilbert's Theorem 90, and it will be proved in the next lecture.
- Lecture 31: Here is my note for the thirty first lecture.
First we proved Hilbert's Theorem 90: suppose \(E/F\) is a cyclic extension and \({\rm Gal}(E/F)=\langle \sigma \rangle\). Then \(N_{E/F}(\alpha)=1\) if and only if there is \(\beta\in E^{\times}\) such that \(\alpha=\sigma(\beta)/\beta\). Using this we studied cyclic extensions with enough roots of unity in the base field: suppose \({\rm char}(F)\nmid n\), \(\mu_n\subseteq F\), and \({\rm Gal}(E/F)=\langle \sigma\rangle\) has order \(n\). Then \(E=F[\sqrt[n]{a}]\) for some \(a\in F^{\times}\). Based on this result, we proved
(Galois) Suppose \(F\) is a field of characteristic zero, \(f(x)\in F[x]\), and \(E\) is a splitting field of \(f(x)\) over \(F\). Then if \({\rm Gal}(E/F)\) is solvable, then \(f(x)\) is solvable in radicals over \(F\).
(We had proved the converse of this statement in the previous lecture.)
- Lecture 32: Here is my note for the thirty second lecture.
In the final lecture, we talked about Kummer theory, and we covered the cyclic case. We proved the following statements:
Let \(F\) be a field. Suppose \({\rm char}(F)\nmid n\), \(\mu_n\subseteq F\), and \(a_1,a_2\in F^{\times}\). Then \(F[\sqrt[n]{a_1}]=F[\sqrt[n]{a_2}]\) if and only if \(\langle a_1(F^{\times})^n\rangle=\langle a_2(F^{\times})^n\rangle\) as subgroups of \(F^{\times}/(F^{\times})^n\).
Using this we proved the cyclic case of Kummer theory:
(Kummer theory: the cyclic case) Let \(F\) be a field. Suppose \({\rm char}(F)\nmid n\) and \(\mu_n\subseteq F\). Then the following is bijection
\[
\{H|\hspace{1mm} H\leq F^{\times}/(F^{\times})^n, \text{ cyclic}\}\rightarrow \{E|\hspace{1mm} E/F\text{ is a cyclic extension of exponent } n\},
\langle a(F^{\times})^n\rangle \mapsto F[\sqrt[n]{a}].
\]
Moreover \({\rm Gal}(F[\sqrt[n]{a}]/F)\simeq \langle a(F^{\times})^n\rangle\).
Then we explained that Kummer theory extends this bijection to all abelian extensions with exponent \(n\), and the following maps are inverse of each other:
(Kummer Theory: the general case) Let \(F\) be a field. Suppose \({\rm char}(F)\nmid n\) and \(\mu_n\subseteq F\). Let \(\mathcal{A}:=\{H|\hspace{1mm} H\leq F^{\times}/(F^{\times})^n\}\) and \(\mathcal{B}:=\{E|\hspace{1mm} E/F \text{ is an abelian extension of exponent } n\}\). Then the following maps are inverse of each other:
\[
\mathcal{A}\rightarrow \mathcal{B}, H\mapsto F[H^{\frac{1}{n}}],
\]
\[
\mathcal{B}\rightarrow \mathcal{A}, E\mapsto ((E^{\times})^n\cap F^{\times})/(F^{\times})^n.
\]
Moreover \({\rm Gal}(F[H^{\frac{1}{n}}]/F)\) is isomorphic to \({\rm Hom}(H, \mu_n)\) for any \(H\leq F^{\times}/(F^{\times})^n\).
We defined Kummer pairing, but did not go through the details of proof of the general case. But I have included its proof in my note.
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