UCB MATH 113 Notes

These are notes from MATH 113, at UC Berkeley. The course is taught by Aleksandra Utiralova.

Table of Contents

• Lecture 1
• Lecture 2
• Lecture 3
• Lecture 4
• Lecture 5
• Lecture 6
• Lecture 7
• Lecture 8
• Lecture 9
• Lecture 10
• Lecture 11
• Lecture 12
• Lecture 13
• Lecture 14
• Lecture 15
• Lecture 16
• Lecture 17
• Ring Constructions

Lecture 1

We start with natural numbers, with only addition and multiplication, integers, with addition, multiplication, and subtraction, rationals, with addition, subtraction, multiplication, and division, and of coures reals with all four operations. We can also consider complex numbers, which are of the form $a + bi$ where $a$ and $b$ are real numbers and we define addition, multiplication, division, and subtraction as usual.

$(\mathbb{N}, +)$ is what we call a monoid, $(\mathbb{Z}, +)$ is what we call a group, $(\mathbb{Z}, +, \times)$ is what we call a ring, and $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ are all examples of fields.

We discuss some basic set theory and logic but I'll omit that here since it's pretty commonly repeated in practically any math context.

We now discuss functions or maps.

Definition (Function). Let $A$ and $B$ be sets. A function $f$ from $A$ to $B$ denoted $f : A \rightarrow B$ is an assignment of a unique element of $B$ for each element in $A$. $A$ is called the domain of $f$ and $B$ is called the range.

We write $f : a \mapsto b$ if $b = f(a)$. An example of a function with $A = B = \mathbb{N}$ is $f:\mathbb{N} \rightarrow \mathbb{N}$, $n \mapsto n^2$.

Definition (Well-Defined). $f : A \rightarrow B$ is well-defined if $a_1 = a_2 \Rightarrow f(a_1) = f(a_2)$.

Given a function $f: A \rightarrow B$, we can define $f(A)$ to be a subset of $B$ defined as $\{b \in B \;|\; \exists a \in A, f(a) = b\}$. $f(A)$ is called the image of $f$. Similarly, if $C \subset B$, define $f^{-1}(C) = \{a \in A \;|\; f(a) \in C\} \subset A$ to be the preimage of $C$.

Definition (Injective). $f$ is injective if for any $a_1,a_2 \in A$, $f(a_1) = f(a_2) \Rightarrow a_1 = a_2$.

Definition (Surjective). $f$ is surjective if for any $b \in B$, $\exists a \in A$ such that $f(a) = b$. In other words, the image of $A$ is $B$.

Definition (Bijective). $f$ is called bijective or one-to-one if it is both surjective and injective.

Definition (Composition). If $f : A \rightarrow B$, $g : B \rightarrow C$, then we define $g \circ f : A \rightarrow C$ with $g \circ f (a) = g(f(a))$.

Definition (Identity). $Id_A : A \rightarrow A$, $a \mapsto a$

Definition (Left Inverse). If $f : A \rightarrow B$, $g : B \rightarrow A$, $g$ is called a left inverse of $f$ if $g \circ f = Id_A$.

Definition (Right Inverse). If $f : A \rightarrow B$, $g : B \rightarrow A$, $g$ is called a right inverse of $f$ if $f \circ g = Id_B$.

Definition (Inverse). $g$ is the inverse of $f$ if $g$ is both a left and right inverse of $f$ (we can show uniqueness).

Proof left for me later.

Our next topic is equivalence relations.

Definition (Binary Relation). $R \subset A \times A$ is called a binary relation on $A$, and we write $a\;R\;b$ if $(a,b) \in R$.

Definition (Equivalence Relation). A binary relation $\sim$ is called an equivalence relation if it satisfies three properties:

1. Reflexive Property
2. Symmetric Property
3. Transitive Property

Definition (Equivalence Class). Suppose we have an equivalence relation $\sim$ on $A$. For $a \in A$, we define $[a] = \{x \in A \;|\; x \sim a \} \subset A$ to be the equivalence class of $a$.

From this definition, we can see that any two equivalence classes are either disjoint or the same. Thus, the whole set is just a disjoint union of equivalence classes. In other words, $A = \sqcup [a]$. $A \backslash \sim = \{ [a] \;|\; a \in A\}$. We have $\pi : A \rightarrow A \backslash \sim$ is sujective, $a \mapsto [a]$.

Now we discuss properties of $\mathbb{Z}$.

Definition (Divides). We say that $d \neq 0$ divides if $a$ if $\exists$ some $b \in \mathbb{Z}$ such that $a = db$. We denote this relation by $d \;|\; a$.

Definition (Greatest Common Divisor). For all $a,b \in \mathbb{Z}_{\neq 0}$, there exists a unique $d \in \mathbb{Z}_{> 0}$ called the greatest common divisor (gcd) of $a$ and $b$ such that $d \;|\; a$ and $d \;|\; b$, and for all $c$ such that $c \;|\; a$ and $c \;|\; b$, $c \;|\; d$.

Definition (Least Common Multiple). For all $a, b \in \mathbb{Z}_{\neq 0}$, there exists a unique $l \in \mathbb{Z}_{> 0}$ called the least common multiple of $a$ and $b$ such that $a \;|\; l$ and $b \;|\; l$, and for all $m$ such that $a \;|\; m$ and $b \;|\; m$, $l \;|\; m$.

We can run this algorithm backwards to prove the following therorem:

Definition (Prime). $p \in \mathbb{Z}$ is called prime if tis divisors are $q$ and $p$ only.

Lecture 2

In other words, if you are able to express $n$ as a product of priems in an ordered fashion, any two such expressions must coincide.

Definition (Euler's $\varphi$ function). We define $\varphi : \mathbb{N} \rightarrow \mathbb{N}$, $\varphi(n) = \# \{a \in \mathbb{N} \;|\; \gcd(a,n) = 1, a \le n\}$.

so we can get the totient of any number from its prime factorization. The above theorem is a corollary of the Chinese Remainder Theorem.

We now consider integers modulo $n$. We say that $a \sim b$ if $n \;|\; a - b$. What we would like to consider is $\mathbb{N}/\sim = \mathbb{Z}/n\mathbb{Z}$.

We can then write $\mathbb{Z}/\sim = \{ [0], [1], [2], ..., [n+1]\}$.

We can then consider $\pi : \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$ which maps $a \mapsto [a]$. This is called "reduction mod $n$".

We can then define $[a] + [b]$ as taking representative elements, adding them together, and then reducing mod $n$ again. We need to check that this is well-defined. Given $a_1, a_2 \in [a]$ and $b_1, b_2 \in [b]$, $a_1 + b_1 - a_2 - b_2 = (a_1 - a_2) + (b_1 - b_2)$ which is divisible by $n$, so $[a_1 + b_1] = [a_2 + b_2]$ and our addition is indeed well-defined. We can similarly check multiplication.

Another thing we can check is that $0$ and $1$ behave as additive and multiplicative identites.

Definition (Units). Consider $(\mathbb{Z}/n\mathbb{Z})^\times = \{[a]\in\mathbb{Z}/n\mathbb{Z} \;|\; \exists b \in \mathbb{Z}, [a][b] = [1]\}$. This is the set of invertible elements, or units.

Proof.

Suppose $\gcd(a,b) = 1$. Then $\exists u, v \in \mathbb{Z}$ such that $au + nv = 1$. Then $[a][u] + [0] = [1]$, so $[u]$ is the inverse of $[a]$ as desired.

We now consider the reverse direction. Suppose $\gcd(a,b) = d \neq 1$. Define $m = \dfrac{n}{d} \in \mathbb{Z}$. Then $[m][a] = [0]$, so $n \;|\; ma$. Thus $[a]$ cannmot be invertible. This is because if $[a][b] = [1]$, then $[m][a][b] = [m]$, but then $[0] = [m]$ which is a contradiction.

$\blacksquare$

Proof by Binomial Theorem. Proof of Theorem.

We use induction. Simply consider base case $0^p \equiv 0 \pmod p$. Then $a^p \equiv a \pmod p \Rightarrow (a+1)^p \equiv a^p + 1^p \equiv a + 1 \pmod p$ as desired.

$\blacksquare$

We finally get to groups.

Definition (Binary Operation). A binary operation $*$ on a set $G$ is a function $* : G \times G \rightarrow G$. We use notation $a * b = *(a,b)$.

Definition (Associativity). A binary operation $*$ on $G$ is associative if $(a*b)*c = a*(b*c)$ for all $a,b,c \in G$.

Definition (Commutativity). A binary operation $*$ on $G$ is commutative if $a * b = b * a$ for all $a, b \in G$.

Definition (Group). A group is a pair $(G, *)$ where $G$ is a set and $*$ is a binary operation on $G$ such that $*$ is associative, there exists $e \in G$ such that $ea = ae = a$ for all $a \in G$, and for each $a \in G$, there exists $a^{-1} \in G$ such that $aa^{-1} = a^{-1}a = e$.

Definition (Abelian Group). A group $(G, *)$ is called commutative or Abelian if $*$ is commutative.

Definition (Monoid). A monoid is a pair $(M,*)$ where $M$ is a set and $*$ is a binary operation on $M$ such that $*$ is associative. Sometimes, people also require that monoids have identity elements.

Lecture 3

We now discuss the free monoid and the free group (hehehe we discussed this in MUSA 174).

Definition (Free Group). The free group with generators $a_1, ..., a_n$ is the free monoid but also with the added symbols $a_1^{-1}, ..., a_n^{-1}$ under the condition that all words cannot have $a_i$ and $a_i^{-1}$ neighboring each other. The definitions of multiplication now are pretty clear.

Proof.

(1) Suppose $e_1$ and $e_2$ are identities. Then $e_1 = e_1e_2 = e_2$ as desired.

(2) Suppsoe $h_1$ and $h_2$ are inverses. Then $h_1 = h_1 e = h_1(gh_2) = (h_1g)h_2 = eh_2 = h_2$ as desired.

(3) $aa^{-1} = a^{-1}a = e$ so this follows from the uniqueness above and definition of inverse.

(4) $ab \cdot b^{-1} a^{-1} = aa^{-1} = e$, $b^{-1} a^{-1} a b = b^{-1} b = e$ as desired.

(5) can be shown via induction.

$\blacksquare$

Proof.

We first show that the solution is unique if it exists. Consider $ax = b$. We have $a^{-1} a x = a^{-1} b$ so $x = a^{-1} b$. We can then show that this solution actully is a solution: $a(a^{-1} b) = (aa^{-1})b = b$ as desired. Similarly, $ya = b$ implies that $yaa^{-1} = ba^{-1}$ so $y = ba^{-1}$, and $ba^{-1}a = b$ as desired. The particular case follows.

$\blacksquare$

Definition (Order). The order of an element $x \in G$ (denoted $|x|$) is the minimal natural number $n \in \mathbb{N} \cup \{\infty\}$ such that $x^n = e$.

Proof.

Let $x \in G$. Then $\{1, x, x^2, ...\} \subseteq G$, so not elements are distinct. Thus there exist $k < l$ such that $x^k = x^l$. Thus, $1 = x^{l- k}$, so the order is at most $l - k$. Thus $x$ has finite order.

$\blacksquare$

Definition (Symmetric Group). $\Omega$ a set, $S_{\Omega} = \{\text{bijections }\Omega \rightarrow \Omega\}$. The operation is composition.

Proof. $$(u \circ v) \circ w [\omega] = (u \circ v) (w(\omega))$$ $$ = u(v(w(\omega)))$$ $$= u \circ (v \circ w (\omega))$$ The identity sends each element to itself. Also, each bijection has an inverse.

$\blacksquare$

Note that when $\Omega = \{1, ..., n\}$, $S_{\Omega} = S_n$. Bijections from $\{1,...,n\}$ to itself are called permutations.

Definition. A cycle of length $k$ $(a_1, a_2,...,a_k) = w \in S_n$, $a_i \in \{1,...,n\}$, $a_i \neq a_j$ if $i \neq j$.

Definition (Transposition). A permutation which swaps two elements and keeps all other elements the same. (I give a loose definition here but obviously there is a better more rigorous definition).

$(a_1,...,a_{m_1})(a_{m_1 + 1},...,a_{m_2})...(a_{m_k+1},...,a_{m_{k+1}})$ is a product of disjoint cycles, where $a_j$ are pairwise distinct elements in $\{1,...,n\}$. This defines a permutation.

We will prove this next time.

Lecture 4

We now prove the theorem.

Rough Proof.

Smoothing up the proof is left to the reader. It's basically yeah induction = algorithm. Pick the smallest element such that it is not yet in a cycle. If there is no such element, we are done. Otherwise, call the element $a$. Let $b = w(a)$. If $b = a$, then close the bracket and go back to step 1. Otherwise, let $c = w(b)$. Let $b = c$ and repeat the process. We claim that this process must terminate. This is because otherwise the permutation is infinite. It is not precisely necessary, but we generally erase all cycles of length 1.

$\blacksquare$

Now suppose we have a cycle decomposition $(a_1,...,a_{m_1})(a_{m_1+1},...,a_{m_2})...(a_{m_{k-1}+1}...a_{m_k})$ then notice that we can actually compute squaring and other things by doing things individually on the cycles. For example, $(123)(45)(6)(789 \:10)$, the square is $(132)(45)(6)(79)(8\:10)$.

Now we have that the order of an element is the length of the cycle it is a part of, and $|w|$ is the $\text{lcm}$ of the orders of the individual elements, or $\text{lcm} (m_1,m_2,...,m_k)$.

A corollary from the remark is

Definition. Let $G$ be a group and let $S$ be a subset of $G$. Let $S = \{s_1,...,s_n\}$ be finite. We say that $G$ is generated by $S$ if every element $g \in G$ can be written as a product of elements in $S$ and their inverses.

If $g = w_1w_2...w_k$ such that $w_i \in S \cup S^{-1}$ (where $S^{-1}$ is th eset of the inverses), and $s_i$ does not neighbor $s_i^{-1}$, then we call the word a reduced expression for $g$.

Proof.

For (1), she gave an example and thus an algorithm. I propose instead a smaller lemma which does everything for us:

The proof is pretty much exactly the same, just slightly different. See the previous remark, proof left to reader since it's pretty simple. Basically if we have $(1234) = (14)(13)(12)$.

For (2), do the homework. It follows from (1).

Definition. A presentation of a group $G$ is the data of $\langle S \;|\; R_1 = ... = R_m = 1\rangle$, $S$ is the set generating $G$, and $R_i$ is a reduced expression in cuts of $S$.

We have $s_i^2 = 1$, $(s_is_j)^2 = 1$ if $|i-j|^2 > 1$, and $(s_is_{i+1})^3 = 1$. We can write: $$S_n = \langle s_1,...,s_{n-1} \;|\; s_i^2=1,(s_is_j)^2 = 1 \text{ if } |i - j| > 1, (s_is_{i+1})^3 = 1)\rangle$$

Definition (Dihedral Group). $D_{2n} = \{\text{symmetries of a regular }n-\text{gon}\}$. For example, you can rotate (for example by $\dfrac{2\pi}{n}$ radians). We can get all rotations by composing the smallest rotation. There are also reflections.

Notice that there are $n$ rotations and $n$ reflections (reflections needs a bit of casework on the parity of $n$). The reason we call it $D_{2n}$ and not $D_n$ is that $|D_{2n}| = 2n$ (1 identity, $n-1$ rotations, $n$ reflections).

We have an injective map from $D_{2n}$ to $S_n$, going from the symmetry to its action on the vertices. On the other hand, this map is not surjective since not every permutation is a bunch of symmetries on the $n$-gon.

Notice that each symmetry is determined by the image of two neighboring vertices.

Now if $r$ is a rotation by $\dfrac{2\pi}{n}$, then $r$ has order $n$. We have $D_{2n} = \{1,r,...,r^{n-1}, s, sr,sr^2,...,sr^{n-1}\}$ because it turns out that all of the rotations are $s$ (one of the rotations) composed with one of the rotations. In particulary, we can see that $r$ and $s$ generate $D_{2n}$. We will investigate these properties more thoroughly next time.

Lecture 5

Last time we discussed $D_{2n}$. We note that the identity and rotations are orientation preserving and the reflections change the orientation.

Proof.

We can already see that the rotations are all different, and we can see that all the reflections are different than the rotations. What we must show is that $sr^i \neq sr^j$ when $i,j < n$. We can see this by noting that $s(1) = 1$, $sr(n) = 1$, and so on until $sr^{n-1}(2) = 1$. Alternatively, suppose that $sr^i = sr^j$. Then $r^{-j} s^{-1} s r^i = 1 \Rightarrow r^{i-j} = 1$. But then $i = j$.

$\blacksquare$

Consider $rs$. Notice that $rs = sr^{-1}$. We can check this by noticing that $rs(1) = 2, rs(2) =1$, and $sr^{-1}(1) = 2, sr^{-1}(2) = 1$.

Proof by induction. Proof.

Notice that every element in $\langle r,s \;|\; r^n = s^2 = 1, rs = sr^{-1}\rangle$ can be expressed in terms of $r$ and $s$ (without inverses). This is because the first rule allows us to exchange an inverse with a larger power of $r$ or $s$. We can also use the first rule to bring the size of a string of consecutive $r$s or $s$s down to less than $n$ $r$s or less than 2 $s$s. In other words, every word can be written as $r^{m_1} s r^{m_2} s ... s r^{m_k}$, with $m_1,m_k$ allowed to be zero, and $m_i \le n-1$. The middle $m_i$s may not be zero. Finally, the last rule allows us to commute all $s$s in the above string to the left, resulting in $s^{k-1} r^{M}$. Finally, we can reduce powers using the first rule again to get $sr^{m}$, with $m < n$. Thus the element is equal to an element in $\{1,r,...,r^{n-1}, s, sr,sr^2,...,sr^{n-1}\}$ as desired.

$\blacksquare$

Definition (Homomorphism). A map that preserves the structure while changing the set. More precisely, it is a map $\phi : (G, \circ) \rightarrow (H, *)$, such that $\phi(a \circ b) = \phi(a) * \phi(b)$.

Definition.

1. A bijective homomorphism $\phi : G \rightarrow H$ is called an isomorphism.
2. If $\exists$ an isomorphism between $G$ and $H$, we say that $G$ is isomorphic to $H$. Denoted by $G \cong H$.
3. A homomorphism that maps $G$ to itself is called an endomorphism.
4. A bijective endomorphism is called an automorphism.

Notice that $G \cong G$ via the identity map, $G \cong H$ implies $H \cong G$, and the transitive property also holds. Thus, $\cong$ is an eqiuvalence relation. Essentially, this just means that we can mainly worry about groups up to this isomorphism.

Proof.

(1) follows directly from the isomorphism being a bijection. For (2), suppose $H$ is commutative, but $G$ is not. Then there must be $h_1, h_2$ such that $h_1h_2 \neq h_2h_1$. But then $\phi(h_1)\phi(h_2) = \phi(h_1h_2) \neq \phi(h_2h_1) = \phi(h_2) \phi(h_1)$ by injectivity which is a contradiction. For the other direction, use the fact that $\phi$ being an isomorphism implies that $\phi^{-1}$ is also an isomorphism. For (3), notice that $x^n = 1$ if and only if $\phi(x)^n = 1$, which completes the proof.

$\blacksquare$

Definition (Kernel). If $\phi : G \rightarrow H$ is a homomorphism, the kernel of $\phi$ $\text{Ker}(\phi) = \{g \in G \;|\; \phi(g) = e_H\}$.

Proof.

The first direction is obvious. Suppose $\ker (\phi) = \{e_G\}$ and suppose $\phi(a) = \phi(b)$. Then $\phi(ab^{-1}) = e_H$. Thus $ab \in \ker(\phi)$, so $ab^{-1} = e_G$, so $a = b$.

$\blacksquare$

Definition (Image). $\text{Im}(\phi) = \{\phi(g) \;|\; g \in G\}$.

Lecture 6

Consider homomorphism $\varphi: G \rightarrow H$. Let $G = \langle s_1,...,s_n \;|\; R_1 = ... = R_m = 1\rangle$ where $R_j = s_{i_1}^{\epsilon_1} ... s_{i_{k_j}}^{e_{k_j}}$ for $i_l \in \{1,...,n\}$ and $\epsilon_j \in \{-1,1\}$. We have $\varphi(s_i) = r_i \in H$, $\varphi(1) = 1$. $\varphi(s_{i_1}^{\epsilon_1} . . . s_{i_{k_j}}^{\epsilon_{k_j}} = r_{i_1}^{\epsilon_1} ... r_{k_j}^{\epsilon_{k_j}} \in H$. Additiionally, $\varphi(R_j) = R_j[r_1...r_n]$. We can verify $\varphi(1) = 1$.

$\varphi : D_{2n} \rightarrow S_n$, $r \mapsto (1,2,3...,n)$, $s \mapsto (2\; n)(3 \;n-1)...$

Definition (Subgroup). A $H \subset G$ is called a subgroup if it is closed under the binary operation of $G$ and also closed under inverses. Additionally, the identity is a part of $H$. We denote subgroups with $H \le G$ to distinguish from subset.

From the above claim, we could have written:

Proof.

Suppose $H$ satisfies (1) and (2). Take $x \in H$. $1 = xx^{-1} \in H$. For all $x \in H$, $1 \cdot x^{-1} = x^{-1} \in H$. For all $x, y \in H$, $y^{-1} \in H$. $x \cdot (y^{-1})^{-1} = xy \in H$. The other direction is trivial.

$\blacksquare$

Definition (Centralizer). $A \subset G$ any subset. The centralizer of $A$ in $G$ is $C_G(A) = \{g \in G\;|\; g a g^{-1} = a\; \forall a \in A\}$.

Proof.

Consider $x,y \in C_G(A)$. We have $yay^{-1} = a$ for all $a \in A$. Thus $a = y^{-1}yay^{-1}y = y^{-1}ay \Rightarrow y^{-1} \in C_G(A)$. Additionally, for all $a \in A$, $(xy^{-1}) a (xy^{-1})^{-1} = xy^{-1}ayx^{-1} = xax^{-1} = a$, so $xy^{-1} \in C_G(A)$. Thus, by the previous proposition, $C_G(A)$ is a subgroup.

$\blacksquare$

Definition (Normalizer). Consider $A \subset G$, so $g \in G$, $G \supset g A g^{-1} = \{gag^{-1} \;|\; a \in A\}$. The normalizer of $A$ in $G$ is $$N_G(A) = \{g \in G \;|\; gAg^{-1} = A\}$$

Proof.

Let $x,y \in N_G(A)$. Then $y Ay^{-1} = A$. Consider the map $f_y : A \rightarrow A$, $a \mapsto yay^{-1}$. Then $yay^{-1} = yby^{-1} \Rightarrow a = b$, so $f_y$ is injective and thus a bijection. Thus it must have an inverse. We claim that $f^{-1}_y : A \rightarrow A$, $a \mapsto y{-1}ay$. We have $f^{-1}_y f_y (a) = y^{-1}(yay^{-1})y = a$ as desired. Thus $y^{-1} \in N_G(A)$. Additionally, we have $xy^{-1} A yx^{-1} = xAx^{-1} = A$, so $xy^{-1} \in N_G(A)$. Thus the subgroup criterion implies that it is a subgroup.

$\blacksquare$

Definition (Center). The center of $G$ is $Z_G = \{g \in G \;|\; \forall h \in G, hg=gh\}$. Alternatively, $Z_G = C_G(G)$.

Definition (Conjugation). $gag^{-1}$ is conjugation of $a$ by $g$.

Consider $G = S_{n}$, and let $w$ and $u$ be two permutations.

Proof.

We have $wuw^{-1}(w(i)) = wu(i) = w(j)$ as desired.

$\blacksquare$

Proof.

If $w \in Z_{S_n}$. Suppose $w(i) = j$, $j \neq i$. Take $k$ such that $k \neq i$, $k \neq j$. $w(i\; k)w^{-1} = (ik) \neq (jw_k)$.

$\blacksquare$

Lecture 7

We consider conjugation in $S_n$.

We have that for $\Omega \subset \{1,2,...,n\}$, $S_{\Omega} \subseteq S_n$, and $S_\Omega$ is isomorphic to $S_m$ for some $m \le n$.

Consider $\Omega = \{3,4,...,n\}$ so $S_{\Omega} \le S_n$. Every element in $C_{S_n}((12))$ is either $w \in S_{\Omega}$ or $w(12)$ for $w \in S_{\Omega}$. Therefore $C_{S_n}((12)) \cong \mathbb{Z}/2\mathbb{Z} \times S_{n-2}$, $\varphi : \mathbb{Z} / 2\mathbb{Z} \times S_{\Omega} \rightarrow C_{S_n}((12))$. $\varphi : ([0], w) \mapsto w$, $([1],w) \mapsto (12)w$. $([1],u)\cdot([1],w) \mapsto (12)u(12)w = uw$. Then $\text{Ker} \varphi = \{([0], e)\}$, and $C_{S_n} ((12)) = N_{S_n} ((12))$. $\varphi(([0],u) \cdot ([1],w))) = \varphi([0], uw) = uw$.

Furthermore, we have that $C_{S_n}((12)) = N_{S_n}((12))$.

We now discuss subgroups generated by subsets.

Definition (Subgroup Generated by a Subset). Given $A \subset G$, $\langle A\rangle = \bigcap_{H \le G, A \subset H} H$ is the subgroup generated by $A$.

$G$ is generated by elements of $A \subseteq G$ if and only if $G = \langle A \rangle$.

Definition (Cyclic). A subgroup generated by a single element $x$ is called cyclic ($x$ is called a generator). Equivalently, $G$ is called cyclic if $G = \langle x \rangle$ for some $x \in G$.

Proof.

Pretty straightforward, I leave it to you, the reader (mwehehe).

Proof.

Suppose $G$ is finite. $G$ is cyclic if and only if $\exists x \in G$, $|x| = |G|$. Using a result form the homework, $|([1],[1])| = \text{lcm}(n,m) = nm$.

$\blacksquare$

Proof.

We start with the first part. If $x^a = 1$, then by the division algorithm $a = n q + r$, so $x^r = 1$ and $r$ is between 0 and $n-1$ so $r = 0$. Thus, $x^a = 1$ if and only if $a$ is divisible by $n$.

Now the second part.Consider $d = \gcd(n,a)$, $n = dc$, $a = db$, and $\gcd(c,b) = 1$. $(x^a)^c = x^{dbc} = (x^n)^b = 1$. If $(x^a)^l = 1$, then notice that $n \;|\; al$. Sine $n = dc$, and $a= db$, $dc \;|\; dbl$, so $c \;|\; bl$, so $c \;|\; l$ which is exactly what we wanted.

$\blacksquare$

If $\gcd(n,m) = d > 1$, then $|([a],[b])| = \text{lcm}(\dfrac{n}{\gcd(n,a)}, \dfrac{m}{\gcd(m,b)}) = \dfrac{n}{\gcd(n,a)} \cdot \dfrac{m}{\gcd(m,b)} < nm$ if $[1] = x$ and $[a] = x^a$.

Proof.

We start with the first part. $G= \{1,x,x^{-1}, x^2, x^{-2}\}$, $\varphi : \mathbb{Z} \rightarrow G$, $k \mapsto x^k$. For the second part, consider $\varphi : \mathbb{Z} / n\mathbb{Z} \rightarrow G$, with $[k] \mapsto x^k$.

$\blacksquare$

From this we get the other direction of the previous theorem:

Lecture 8

Proof.

We start with (1). If $K = \{1\}$, we are done. Suppose $K \neq \{1\}$. $$\{ n \in \mathbb{N} \;|\; x^n \in K \} \neq \varnothing$$ Let $l$ be a minimal such number. $K = \langle x^l \rangle$, because if $x^m \in K$, $x^m \cdot x^{-lq} = x^r \in K$. ($m = lq + r$, $0 \le r \le l$). Since $l$ is minimal, $r = 0$.

Now we consider (2). $K = \langle x^{n/a} \rangle$. $|x^{n/a} = a \Rightarrow |K| = a$. If $|K'| = a$, let $K = \langle x^l \rangle$. $|x^l| = \dfrac{n}{\gcd(n,l)} = a$, so $\gcd(n,l) = \dfrac{n}{a}$. Thus $\dfrac{n}{a} \;|\; l$. We now have $K=\langle x^{n/a} \rangle \ge \langle x^l \rangle = K'$. But $|K| = |K'|$ and are finite, so $K = K'$.

$\blacksquare$

Proof.

We can see that $C_G(\langle A \rangle) \le C_G(A)$ is fairly quick. Now suppose $g \in C_G(A)$. Then $gag^{-1} = a$ for all $a$. Now notice that $ga^{-1}g^{-1} = a^{-1}$ since the left and right hand sides of the second equation are the inverses of the left and right hand sides of the first equation. Then, notice that we have a multiplicative ness of the property. The rest is left for me later so that I can get a better handle of conjugation (and also because I feel like writing it down later and it's fairly simple to derive).

Definition (Group Action). A left action of a group $G$ on a set $A$ is a map $\rho : G \times A \rightarrow A$. $\rho(g,a) = g * a$. It must satisfy:

1. $g_1 * (g_2 * a) = g_1g_2 * a$ for all $g_1,g_2 \in G$ and $a \in A$.
2. $1 * a = a$ for all $a \in A$.

The above diagram commutes.

Apon experimentation that I didn't copy down right now, we can actually see that a group action is like a homomorphism from $G$ to the symmetric group on $A$. In particular, we can have $\tilde{\rho} : G \rightarrow S_A$, so $\rho(g,a) = \tilde{\rho}(g)(a)$.

Definition (Kernel). A kernel of a group action of $G$ on $A$ is $\{g \in G \;|\; ga = a \;\forall a \in A\} = \ker(\text{permutation rep})$. Essentially, we have $\ker(\tilde{\rho} = \{g\in G \;|\; \tilde{\rho}(g) = e\}$, where $\tilde{\rho} : G \rightarrow S_A$.

Definition (Faithful). An action of $G$ or $A$ is called faithful if $\ker(\tilde{\rho}) = \{e\}$, ($\tilde{\rho} : G \rightarrow S_A$ is injective).

Lecture 9

Consider $\tilde{\rho} : G \rightarrow S_A$ and $\ker(\tilde{\rho}) = \{e\}$. Then $\tilde{\rho}$ is injective and $\tilde{\rho} : G \rightarrow \text{Im}(\tilde{\rho}) \le S_A$ is bijective.

Definition. Let $G$ act on $A$ and let $a \in A$. The orbit of $a$ is a set $G \cdot a = \left\{ ga \;|\; g \in G\right\}$.

Definition. An action of $G$ on $A$ is called transitive if it has only one orbit.

Definition. Let $G$ act on the left on $A$. $G \backslash A = A / n = \{Ga \;|\; a \in A\}$ is the quotient of $A$ by the action of $G$. $$\pi : A \rightarrow G \backslash A, a \mapsto Ga$$

Definition. An automorphism of $G$ is called inner if it lies in the image of $\tilde{\rho}$. Essentially if it is equivalent to $\sigma g$ where $\sigma g$ is $ghg^{-1}$.

Definition. An orbit of $g \in G$ under the conjugation action is called a conjugacy class of $G$.

Lecture 10

We discovered last time that $G$ acts on itself by conjugation. We denote $\text{Cl}(x) = G \cdot x$ the conjugacy class of $x$.

Definition (Stabilizer). Definition. Let $G \acts X$, $x \in X$. $\text{Stab}(x) = \{g \in G \;|\; g \cdot x = x\}$.

Proof.

We start with the first part. Let $g,h \in G_x$. $hx = x$, $x = h^{-1} h x = h^{-1}x$, $gh^{-1}x = gx = x$, so $gh^{-1} \in G_x$, so it is a subgroup by the subgroup criterion.

The second part is fairly obvious.

The third part follows from $\bigcap_{x \in X} G_x = \{g \in G\;|\; \forall x g\in G\} = \{ g \in G \;|\; \forall x\in X. gx = x\} = \text{Ker}(\rho)$ as desired.

Proof.

Let $h \in G_y$ the stabilizer of $y$. Then $hy = hgx = gx = y$, so $g^{-1}hgx = x$, so $g^{-1}hg \in G_x$. If $a \in G$, $a\cdot x = x$, $gag^{-1}y = gag^{-1} gx = gax = gx = y$. Thus $gag^{-1} \in G_y$.

We now consider right actions. We have defined left actions, but the difference is that now wherease we had $h \cdot g \cdot x = (hg)\cdot x$, we now have $x \cdot g \cdot h = x \cdot (gh)$ so the order has been swapped.

Definition (Anti-Homomorphism). A map $\varphi : G \rightarrow H$ is called an anti-homomorphism if $\varphi(ab) = \varphi(b)\cdot \varphi(a)$.

Definition. A right action of $G$ on $X$ is a map $\rho : X \times G \rightarrow X$, $\rho(x,g) = x \cdot g$, such that

1. $x \cdot \cdot h = x \cdot (gh)$
2. $x \cdot 1 = x$

How does this relate to homomorphisms? Consider $\sigma_g:X \rightarrow X$, $x \mapsto x \cdot g$. Then $\sigma_h \circ \sigma_g(x) = (x \cdot g)\cdot h = x \cdot(gh) = \sigma_{gh}(x)$ so $\tilde{\rho}: G \rightarrow S_X$ is an antihomomorphism.

Proof. $$\tilde{\tau}(g) = \tilde{\rho}(g^{-1})$$ $$\tilde{\tau} = \tilde{\rho} \cdot \text{inv}$$

Definition (Coset).

1. $g \cdot H$ is called a left coset of $H$ in $G$
2. $H \cdot g$ is called a right coset of $H$ in $G$

Now we can see that $H \backslash G$ is the set of right cosets, and $G / H$ is the set of left cosets.

Consider $f^{-1}(x) = \{g \in G \;|\; f(g) = x\}$. This is just $G_x$. Now consider $y = h \cdot x$. $f^{-1}(y) = \{g \in G \;|\; h^{-1}gx = x\}$ which is $h G_x$.

We essentially have a bunch of fibers that get glued together. In our case, the fibers are all the same size.

We obtain: $\# G = \sum_{y \in Gx} \#f^{-1}(y)$, but $\#f^{-1}(y)$ is the same for every $y$, and it is $\# G_x$. Thus $\# G = \# G_x \cdot \# G\cdot x$.

$\blacksquare$

Lecture 11

Consider $G$ acting on $X$. The map $f : G \rightarrow G x$ surjective, $g \mapsto g\cdot x$. $f^{-1}(gx) = g\cdot \text{Stab}(x)$.

If $G$ acts transitively on $X$, fix $x \in X$, then $f : G \rightarrow X$, $g \mapsto g\cdot x$ is surjective.

Now $f(g_1) = f(g_2) = gx$ if and only if $g_1, g_2 \in g \text{Stab}(x)$. Notice that $G / \text{Stab}(x) = \{g \text{Stab}(x) \;|\; g \in G\}$. $G \rightarrow G/\text{Stab}(x)$, $g \mapsto g \text{Stab}(x)$. $g_1 \sim g_2$ if they live in the same equivalence class $\sim^2$, so if they live in the same coset. $g_1 \text{Stab}(x) = g_2 \text{Stab}(x)$.

Basically we can come up with two equivalence relations on $G$ and they end up commuting somehow? Might need to rewatch this part later.

We get $G / \text{Stab}(x) \rightarrow X$ with $g \text{Stab}(x) \mapsto g \cdot x$.

Definition. If $G$ acts on $X$ and $G$ acts on $Y$, we will say that $X$ is isomorphic to $Y$ as a set with $G$-action if there exists a bijection $f : X \rightarrow Y$ such that $f(gx) = gf(x)$.

$G$ acts on $G/H$ by left multiplication, $g_1 \cdot (g_2H) = g_1g_2 H$. $G/\text{Stab}(x) \rightarrow X$, $g \text{Stab}(x) \mapsto gx$. $\tilde{f}(\text{Stab}(x)) = x$, $\tilde{f}(g\cdot \text{Stab}(x)) = g\cdot x$, $\tilde{f}(g_1\cdot (g_2\text{Stab}(x)) = g_1 \tilde{f}(g_2 \text{Stab}(c))$.

Transitive action: $$g_1 \cdot (g_2H) = g_1g_2 H$$ Wait are we sure? Rewatch.

$t \in \mathbb{C}$ is a complex point.

Consider $G / H = \{g \cdot H \;|\; g \in G\}$. We ask how many elements there are in this set (we know it is at most $|G|$).

Definition (Index). $\#G/H = [G : H]$ is called the index of $H$ in $G$.

Definition. $g_1,...,g_k \in G$ with $k = [G :H]$ are called coset representatives if $g_i H \neq g_k H$ if $i \neq k$ (or $\{g_1H, g_2H, ..., g_kH\} = G/H$).

Lecture 12

Cyclic Groups

A cyclic group is a group of the form $G = \langle x \rangle$, $x \in G$. Then $G = \{x^k \;|\; k \in \mathbb{Z}$. If $G$ is infinite, then $x^k \neq x^l$ for any $k$ and $l$. If it is finite, then $|G| = |x| = n$, and $G = \{1,x,x^2,...,x^{n-1}\} \cong \mathbb{Z}/n\mathbb{Z}$. Then $|x^a| = \dfrac{n}{\gcd(n,a)}$, and all subgroups are cyclic with $|\langle x^a \rangle| = \dfrac{n}{\gcd(n,a)}$. We also have some facts: $G$ is cyclic if adn only if $\exists x \in G$ such that $|x| = |G|$, and $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$ is cyclic if and only if $\gcd(m,n) = 1$. In particular, if $n$ and $m$ are coprime, $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z}$.

I'll make a blog post about centralizers of elements in $S_n$ later so I didn't take notes on that too much.

Lecture 14

$G/H$ is a group under the operation $g_1 H \cdot g_2 H = g_1g_2 H$ if and only if $H \le G$ (i.e. $N_G(H) = G$). $\pi : G \rightarrow G/H$ with $g \mapsto gH$ is surjective group homomorphism, $\varphi : G \rightarrow \overline{G}$ surjective, then $\overline{G} \cong G/\text{Ker}(\varphi)$, and $\text{Ker}(\varphi) \trianglelefteq G$.

Definition (Simple). A group $G$ is called simple if $\{1\}$ and $G$ are the only normal subgroups in $G$.

Let's categorize Subgroups and Quotients: A subgroup is a group along with an injective group homomorphism. We have a sort of universal property:

with $\text{Im}(\varphi) \le i(H)$. Consider quotients. We have $\pi : G \rightarrow \overline{G}$ surjective, ($N = \text{Ker}(\pi)$) with $N \le \text{Ker}(\tilde{\varphi})$. $\text{Hom}(\overline{G},H) = \{\tilde{\varphi} \in \text{Hom}(G,H), \text{Ker}(\tilde{\varphi})\ge N\}$. We can define the quotient like this by $N$. And we can show that this universal property determines the group up to isomorphism.

We define the sign map $\text{sign} : S_n \rightarrow \{1, -1\} \le \mathbb{R}_{\neq 0}$. Define $w \in S_n$.

Definition (Set of Inversions). The set of inversions $\text{Inv}(w) = \{(a,b) \;|\; a,b \in \{1,...,n\}, a < b, w(a) > w(b) \}$.

Definition. $\text{inv}(w) = |\text{Inv}(w)|$

Notice that $\text{inv}(w)$ is just the number of intersections, 4.

Definition. $\text{sign}(w) = (-1)^{\text{inv}(w)}$.

Notice that $\{w \in S_n \;|\; \text{inv}(w) = 1\} = \{s_i \;|\; s_i = (i\; i +1)\}$.

Proof omitted here.

But an idea from the proof of (1) is that $(\text{Inv}(s_iw) \cup \text{Inv}(w))\backslash (\text{Inv}(s_iw) \cap \text{Inv}(w))$ contains one element.

We won't show the proof here, but yeah.

Definition (Alternating Subgroup). We define the alternating subgroup $A_n = \text{Ker}(\text{sign})\trianglelefteq S_n$.

Since $\text{sign} : S_n \rightarrow \mathbb{Z}/2\mathbb{Z}$, so $|A_n| = \dfrac{|S_n|}{|\mathbb{Z}/2\mathbb{Z}|} = \dfrac{n!}{2}$. Also, it can be shown that $A_n$ is simple.

Definition. $H, K \le G$. $H \cdot K = \{ hk \;|\; h \in H, k \in K\}$.

Proof left to me for later.

Lecture 15

We continue with direct and semidirect products of groups.

Definition. If $H,K$ are groups and $G$ is such that $H \trianglelefteq G$, $G/H \cong K$ then $G$ is called an extension between $K$ and $H$.

Proof.

Consider $\varphi : H \times K \rightarrow G$, $(h,k) \mapsto hk$. $$\varphi((h_1,k_1) \cdot (h_2,k_2)) = \varphi((h_1h_2,k_1k_2)) = h_1h_2k_1k_2 = (h_1k_1)(h_2k_2)$$

Recall from homework that if $H,K \le G$, $H \cap K = \{1\}$, then for all $h \in H$ and $k \in K$, $hk = kh$.

$\varphi$ is surjective because of (3) and injective because of (2) (simple to check).

We wish to relax the condition that $K$ must be a normal subgroup of $G$ to $K$ being just a subgroup if $G$, $H \cap K = \{1\}$, $HK = G = KH$, $\forall g \in G$, $g = hk$ for a unique $h \in H$ and a unique $k \in K$. Define $\pi |_k : K \rightarrow G/H$, $G = \bigcup_{k\in K} Hk$. If $Hk_1 = Hk_2$, then $k_1k_2^{-1} \in H \cap K$, so $k_1 = k_2$.

$\pi|_k : k \mapsto Hk$, so $\pi |_k$ is an isomorphism, so $G$ is an extension between $K$ and $H$.

We wish to generalize things with the next definition:

Definition (Semidirect Product). Let $H$ and $K$ be groups and let $K$ act on $H$ by group automorphisms $(\sigma : K \rightarrow \text{Aut}(H) \le S_H$). Let $\sigma(k)(h) = k * h$. Define $H \rtimes_\sigma K$ the semidirect product is a group with underlying set $H \times K$ and group operation $(h_1,k_1) \cdot (h_2, k_2) = (h_1 (k_1 * h_2), k_1k_2)$.

We can check that $(1,1)$ is the identity of this group and the inverse of $(h,k)$ is $(k^{-1} * h^{-1}, k^{-1})$.

Interesting point from proof.

An interesting and important idea is $$(1,k)(h,1) (1,k^{-1}) = (k * h, 1)$$ which shows that this recovers $k$!

The idea of the previous proposition is that we can rewrite the old proposition as

Proof is the same but now with $\varphi : H \rtimes K \rightarrow G$, $(h,k) \mapsto hk$.

Lecture 16

Proof Idea. Since $G = \langle a_1, ..., a_n\rangle$, $\varphi : \mathbb{Z}^n \rightarrow G$ is surjective, with $\varphi((k_1,...,k_n)) \mapsto \sum k_ia_i$. Then $G = \mathbb{Z}^n/\text{Ker}(\varphi)$. $L = \text{Ker}(\varphi) \le \mathbb{Z}^n$. We take as a black box that $L$ is finitely generated. We later need to use another fact, that $A = \text{Mat}_n(\mathbb{Z})$ is invertibel if and only if $\text{det}(A)$ is invertible in $\mathbb{Z}$. $\text{det}(A) = \pm 1$. Another acct is Smith's normal form: there exists a basis $y_1,...,y_n$ of $\mathbb{Z}^n$ and $n_1,...,n_m \in \mathbb{Z}$ such that $n_{i+1} \;|\; n_i$ such that $L = \langle n_1y_1,...,n_my_m\rangle$. Fun fact to prove some day: One abelian group that we encounter that we don't completely understand is the invertible elements in $\mathbb{Z}/n\mathbb{Z}$. But for prime $p$, $(\mathbb{Z}/p\mathbb{Z})^\times \cong \mathbb{Z}/(p-1)\mathbb{Z}$. In particular, it turns out that $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic.

Lecture 17

We start by defining rings. I won't include it here since I understand the definition pretty well. For this class, we work with unital rings as the normal rings and nonunital rings are special.

Definition. A ring $R$ is commutative if $ab = ba$ for all $a, b \in R$.

Definition. A ring $R$ is a division ring if for all $a \neq 0 \in R$, there exists $b \neq 0 \in R$ such that $ab = ba = 1$.

Definition. A commutative division ring is a field if $0 \neq 1$. There is debate over whether $0 \neq 1$ should be a condition.

An interesting ring is quaternions, $\mathbb{H} = \{a + bi + cj + dk \;|\; a,b,c \in \mathbb{R}, i^2 = j^2 = k^2 = -1$, and $ijk = -1\}$. It can be checked that $\mathbb{H}$ is a division ring.

Definition (Zero Divisor). $a \neq 0 \in R$ is a zero divisor if there exists $b \neq 0 \in R$ such that $ab = 0$ or $ba = 0$.

Definition (Unit). $a \neq 0 \in R$ is a unit if there exists $b \neq 0 \in R$ such that $ab = ba = 1$. $R^\times$ is the group of units in $R$.

Before we prove this, we need to be able to use a few things quite freely. We check some things:

Proof (Normally I omit this but this felt juicy for some reason).

1. $0 \cdot a = (0 + 0)a = 0 \cdot a + 0 \cdot a \Rightarrow 0 = 0 \cdot a$
2. $(-a) \cdot b + ab = (- a + a)b = 0b = 0$ and $a \cdot (-b) + ab = a(-b + b) = a0 = 0$
3. $(-a)(-b) = -(-a)b = -(-ab) = ab$
4. $1_a = 1_a1_b = 1_b$, and the last part is simply from part 2.

$\square$

Now we can prove the lemma. Suppose $ab = 0$ and $a^{-1}a = 1$. Then $0 = a^{-1}0 = a^{-1}ab = b$ contradiction.

$\square$

Definition (Integral Domain). A commutative ring with no zero divisors is called an integral domain.

Proof. $$ab - ac = a(b-c) = 0 \Rightarrow a = 0 \text{ or } b = c$$

$\blacksquare$

Proof.

Consider $a \neq 0 \in R$, and $m_a : R \rightarrow R$ which is injective, $b \mapsto ab$. Then since $R$ is finite, $m_a$ is a bijection, so there exists $c$ such that $ac = 1$.

$\blacksquare$

Definition (Subring). $S \subseteq R$ is a subring if $(S, +) \le (R,+)$ and $1 \in S$, $S$ is closed under multiplication.

Non-unital Rings

As a temporary tangent, we might want to consider nonunital rings.

Ok... that's enough heh. Back to other stuff.

Ring Constructions

$X$ nonempty set, $R$ is a ring. $\text{Fun}(X,R)$ inherits properties of $R$. $\text{Fun}(X,R)$ is commutative if and only if $R$ is commutative. For all $r \in R$, $C_r \in \text{Fun}(X,R)$. $Cr : x \mapsto r$, $R \subset \text{Fun}(X,R)$

$\begin{bmatrix} r & & 0 \\ & ... & \\ 0 & & r\end{bmatrix} \in \text{Mat}_n(R)$, $R \subset \text{Mat}_n(\mathbb{R})$

$R$ a ring, $G$ a finite group. $RG$ is called a group ring with coefficients in $R$. This is $\{ \sum_{g \in G} c_g \cdot g \;|\; c_g \in R\}$. It will make our lives easier if we make $R$ a commutative ring, so we actually do that. Then $(c\cdot g) (d \cdot h) = cd \cdot (gh)$.

Consider $R$ a commutative ring. $R[x]$ is the polynomials with coefficients in $R$. $R[x] = \{a_n x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0 \;|\; a_i \in R\}$. $x^n x^m = x^{n + m}$. For all $r \in R$, $r$ is a constant polynomial. Let $p = a_nx^n + ... + a_1 x + a_0$, with $a_n \neq 0$. We say that the degree of $p$ is $n$ ($\text{deg} (p) = n$). We define the degree of $0 \in R[x]$ to be $-\infty$.

Midterm 2 Revieww

Lecture $N$

Definition (Principal Ideal Domain). A ring $R$ is a principal ideal domain (PID) if it is an integral domain and all ideals in $R$ are principal.

Proof.

Let $(p)$ be a prime ideal and $(p) \subset (m) \Rightarrow p = mr$ for some $r$. $mr \in (p)$ implies that $m = os \Rightarrow p =psr \Rightarrow 1 = sr$, and $r = ps \Rightarrow p = pms \Rightarrow 1 = ms$. Etc.

Definition (Noetherian). A ring is called Noetherian if every left ascending chain of ideals stabilizes, i.e. $$\forall I_1 \subseteq I_2 \subseteq ...$$ then there exists $N$ such that for all $n,m \ge N$, $I_n = I_m$.

Definition (Irreducible). Let $R$ be a commutative ring, we say $r \in R$ is irreducible if $r = pq \Rightarrow p \in R^\times$ or $q \in R^\times$.

Definition (Distinct). $p$ and $q$ are called distinct primes if $(p) \neq (q)$.

Field of Fractions

Consider a general comutative ring $R$. We can introduce an equivalence relation on ordered pairs as follows: $$F(R) = \{\dfrac{m}{n} \;|\; m \in R, \; n \in R_{\neq 0} \}/ \sim$$ where $\dfrac{m}{n} \sim \dfrac{a}{b}$ if $bm = an$. We can define addition as $$\dfrac{m}{n} + \dfrac{p}{q} = \dfrac{qm + pn}{nq}$$ $$-\dfrac{m}{n} = \dfrac{-m}{n}$$ $$0 = \dfrac{0}{1}$$ $$\dfrac{m}{n} \cdot \dfrac{p}{q} = \dfrac{mp}{nq}$$ $$(\dfrac{m}{n})^{-1} = \dfrac{n}{m}$$ $$1 = \dfrac{1}{1}$$ We can show all of this forms a field by easily verifying all of the conditions. The only problem is, multiplication is not well defined if we allow zero divisors, since $nq$ could be zero. Thus, we can force $R$ to be an integral domain, and then

Definition (Field of Fractions). If $R$ is an integral domain then $F(R)$ is a field called the field of fractions of $R$.

Lecture $N + 2$

Proof by isomorphism theorem.

Definition (Cyclic). $M$ is cyclic if $M = Rx$ for some $x \in M$.

Definition (Free). A module $M$ is free on the set $A \subseteq M$ if for all $m \in M_{\neq 0}$ there exists unique scalars $r_1,...,r_n \in R$ and $a_1,...,a_n \in A$ such that $m = \sum_{i=1}^n r_i a_i$. Definition (Basis). $A$ os a basis of $M$.

Definition (Tortion). $M \in R$-mod $$\text{Tor}(M) = \{m \in M \;|\; \exists r \in R_\neq 0, rm = 0\}$$

Definition (Rank). The rank $\text{rk }M$ is the maximal number of $R$-lineraly independent elements in $M$.

Proof.

Since $R$ is an integral domain, we can embed $R$ into its field of fractions $F(R)$. We can also embed $M$ into $F^n$, $$\sum r_i x_i$$ if $a_0,...,a_n\in M$ there exists $y_0,...,y_n \in F$, $\sum_{i=0}^n y_i a_i = 0$, $y_i = \dfrac{b_i}{c_i}, c = \prod_{i=0}^n c_i$, $cy_i \in R$, $\sum_{i=0}^r (cy_i)a_i = 0$.

Consider $\varphi, \psi$ in $\text{Hom}_R(M,N)$. Notice that $(\varphi + \psi)$ is also a homomorphism and $r \varphi$ is also a homomorphism.

Definition. If $N_1,...,N_k$ are submodules of $M$, then $N_1 + ... + N_k := \{n_1 + ... + n_k \;|\; n_i \in N\}$

Definition. Let $N,M \in R$-mod. $N \times M \in R$-mod, $r (n,m) = (rn,rm)$.

Proof I leave for myself later.

Defintiion. If $N_1,...N_k$ satisfy these conditions then write $N_1 \oplus ... \oplus N_k = N_1 + ... + N_k$.

Proof.

We start by trying to define $a_1$. Consider the $\varphi \in \text{Hom}_R(M,R)$. Then $\varphi(N) \subseteq R$ is a submodule, so it is an ideal. It is also a principle ideal, so we define $\varphi(N) = (a_\varphi)$. We then define $\sum = \{(a_\varphi) \;|\; \varphi \in \text{Hom}_R(M,R)$.

Proof.

Consider any $I_1 \in \sum$. If it is maximal, we are done. Otherwise, there exists another ideal $I_2$ so that $I_1 \subset I_2$. Thus, we can either find $I_2$ to be maximal or we continue to find a chain of inclusions. We claim that we must stop at some point because our ring must be Noetherian, so any chain must stabilize. We can then call this maximal ideal $(a_\nu)$, and define $a_1 = a_\nu$ ($a_1$ from the satement of the theorem.)

Now there exists some $y \in N$ such that $\nu(y) = a_1$. Let $M = Rx_1 \oplus ... \oplus Rx_n$. Consider $\pi_i : M \rightarrow R$ projections to the ith coordinates $(r_1,...,r_n) \mapsto r_i$. If $a_1 = 0$, then $\sum = \{(0)\}$. Therefore, $\varphi(N) = 0$ for all $\varphi$, so for all $n = (r_1,...,r_n)$, $\pi_i(n) = 0$, so $n = 0$, and $N = 0$. Then the result is trivial.

Assume $N \neq 0$, so $a_1 \neq 0$.

Proof.

Consider the gcd: $(a_1,\varphi(y)) = (d)$. $d = ra_1 + s\varphi(y)$. $\psi = r\nu + s\varphi \in \text{Hom}_R(M,R)$. Consider $$\psi(y) = r\nu(y) + s\varphi(y) = d$$ because $\nu(y) = a_1$. Consider $d \in \psi(N)$. $(d) \subseteq \psi(N) = a(\psi)$, $(a_1 \subseteq (d)$. Since $(a_1) \subseteq (a_\psi)$ immplies that $(a_1) = (a_\psi)$, which implies $(a_1) = (d)$, $\varphi(y) \in (a_1)$.

Bcak to the theorem. $\pi_i(y)= a_1b_i$, $y = (a_1b_1,a_1b_2,...,a_1b_n)$, $y_1 = (b_1,...,b_n)$, $y = a_1y_1$.

Proof. For the first part, $x = \nu(x)y_1 + (x-\nu(x)y_1)$ $\nu(x - \nu(x)y_1) = \nu(x) - \nu(x) \nu(y_1) = 0$. $ry_1 \in \text{Ker}(\nu)$, $\nu(ry_1) = r\nu(y_1) = r = 0$ which implies $Ry_1 \cap \text{Ker}(0) = 0$. For the second part, $x \in N$, $x = \nu(x) y_1 + (x-\nu(x) y_1)$.

Now that we have gotten all of this preliminary stuff out of the way, we can show part 1 of the theorem by induction on $m$. We have that $M = Ry_1 \oplus \tex{Ker }\nu$ and $N = Ra_1y_1 \oplus (\text{Ker }\nu \cap N)$. If the rank of $N$ is 0, then $\text{Tor}(N) = N = 0$, and $\text{Tor}(N) \subseteq \text{Tor}(M) = 0$. If the rank of $(\text{Ker }\nu \cap N)$ is $m-1$, then $?$ free implies that $\simeq R^{m-1}$ which implies $N \simeq R \times R^{m-1}$ is also free.

To prove the second part, by induction on $n$, By 1, $\text{Ker }\nu$ is free of rank $n-1$. $\text{Ker } \nu \cap N \subseteq \text{Ker }\nu$. There exists $a_2,...,a_m,y_2,...,y_m$, $(\text{Ker }\nu \cap N) = Ra_2 y_2 \oplus ... \oplus Ra_m y_m$, $a_2\;|\; a_3 \;|\; ... \;|\; a_m$. $\text{Ker }\nu = Ry_2 \oplus ... \oplus Ry_m$.

Lecture $N + 3$

Consider $R$ a PID, and $M$ an $R$-module. $M$ is free of rank $n$, $N \subseteq M$.

To check that $a_1$ divides $a_2$, consider $\psi : M \rightarrow R$, $\psi(y_1) = \psi(y_2) = 1$, $\psi(y_k) = 0$ for all $k > 2$. $\psi(a,y_1) = a_1 \in \psi(N) \in \Sigma$, $\psi(a_2y_2) = a_2 \in \psi(N)$, $(a_1) \subseteq \psi(N) \Rightarrow \psi(N) = (a_1)$, $a_2 \in (a_1)$, so $a_1 \;|\; a_2$.

Proof.

Let $m_1$, ..., $m_n$ be generators of $M$, and $\varphi: R^n \rightarrow M$, $(r_1,...,r_n) \mapsto \sum_{i=1}^n r_i m_i$, surjective, $R$-linear, $M \simeq R^n/\text{Ker }\varphi$. Mptoce $R^n = Ry_1 \oplus ... \oplus Ry_n$, $\text{Ker }\varphi simeq R\Tilde{a}_1 y_1 \oplus ... \oplus R\Tilde{a}_m y_m$. $R^n/\text{Ker }\varphi \simeq R/\Tilde{a}_1 \times ... \times R/\Tilde{a}_m \times R^{n-m}$. etc etc then we get existence.

To get uniqueness, we need to introduce the following definition:

Definition. Consider $p \in R$, $M \in R-\mathsf{mod}$. $pM = \{pm \;|\; m \in M\}$.

Assume $M = \text{Tor}(M)$. $M = R/(a_1) \times ... \times R/(a_k) = R/(b_1) \times ... \times R/(b_m)$.

Modules over $F[x]$. Let $X$ be a linear operator on a finite dimensional vector space $V$. $V \simeq F[x]/(a_1) \times ... \times F[x]/(a_k)$, $a_1,..., a_k \in F[x]$. If $V \simeq F[x]/(a)$, assume $a$ is monic, so the coefficient of its highest degree term is 1. We can always divide by the first coefficient to make this true. So we have WLOG. $F[x]/(a)$ has basis $\overline{1}, $\overline{x}$ (overline means image of), $\overline{x}^2$, ..., $\overline{x}^{n-1}$ over $F$.

In this basis, we can write the matrix $X$: $$X = \begin{bmatrix} 0 & 0 & 0 & ... & 0 & -a_0 \\ 1 & 0 & 0 & ... & 0 & -a_1 \\ ... & ... & ... & ... & ... & ... \\ 0 & 0 & 0 & ... & 1 & -a_n\end{bmatrix}$$

We can also get Jordan normal form as a corollary of this classification theorem along with the Chinese Remainder Theorem. $$F[x]/(a(x)) \simeq F[x]/(p_1)^{\alpha_1} \times ... \times F[x]/(p_k)^{\alpha_k}$$ $$a = p_1^{\alpha_1}...p_k^{\alpha_k}$$

We first cover the fundamental theorem of algebra. Irreducible polynomials over $\mathbb{C}$ are only linear ones.

We now consider Field Extensions.

Definition. $K$ is a field extension of $F$ (denoted $K / F$) if $F$ is a subfield of $K$.

Definition. The degree of $K$ over $F$, denoted $[K : F]$, is the dimension of $K$ as a vector space over $F$. $[K : F] = \dim_F K$.

Definition. $\text{ch } F$ is $p$ such that $1 + ... + 1 = 0$ ($p$ additions of 1). If $p = \infty$, $p = 0$.

Definition. A prime subfield $F$ of $K$ is a smallest subfield (equivalently a subfield generated by 1).

Definition. An extension $K/F$ is finite if $[K : F]$ is finite.

Proof.

Consider $F[x]/p(x) = K$, $[K : F] = \deg p$.

Definition. Let $K/F$, $\alpha \in K$. $F(\alpha)$ is the minimal subfield of $K$ containing $F,\alpha$. Also, $F(\alpha)/F$ is the simple extension.

Definition. $\alpha_1, ..., \alpha_k \in K/F$, $F(\alpha_1,...,\alpha_k)$ is the minimal self contaiing $F$, $\alpha_1,...,\alpha_k$.

Definition. $K/F$ is finitely generated if $\exists \alpha_1,...,\alpha_k$ such that $F(\alpha_1,..,\alpha_k) = K$.

Definition. $K/F$ is algebraic if for all $\alpha \in K$, there exists $f(x) \in F[x]$ such that $f(\alpha) = 0$.

Definition. If $\alpha \in K/F$ is an algebraic element, there exists a monic polynomial of minimum degree $m_\alpha(x) \in F[x]$ such that $m_\alpha(\alpha) = 0$. This is called the minimal polynomial of $\alpha$.

Proof. $L = K\langle \alpha_1,...,\alpha_n\rangle$ and $K = F\langle \beta_1,...,\beta_k\rangle$, so $L = F\langle \alpha_i \beta_j\rangle_{i \le n, j \le k}$, $x \in L$, $x = a_1\alpha-1 + ... + a_n \alpha_n$, $a_i \in K$. Proof.

$K = F(\alpha_1,...,\alpha_k)$, $F_j = F(\alpha_1,...,\alpha_j)$. $[K : F] = [F_K : F_0] = [F_k : F_{k-1}] [F_{k-1} : F_{k-2}] ... [F_1 : F_0]$.

Consider the extension $F_j/F_{j-1}$, and $F_j = F_{j-1}(\alpha_j)$. Since $\alpha_j$ is algebraic over $F_{j-1}$, $[F_j : F_{j-1}] < \infty$.

Definition (Algebraic Closure). The algebraic closure of $F$ denoted $\overline{F}$ is the minimal field containing roots of all polynomials with coefficients in $F$.

Definition (Algebraically Closed). $F$ is called algebraically closed if $\overline{F} = F$.

Proof.

Let $p(x) = a_0 + ... + a_{n-1} x^{n-1} + x^n$ be a polynomial in $\overline{F}[x]$. Consider $K := F(a_0,...,a_{n-1}) \subseteq \overline{F}$, and consider $K[x]/(p(x)) = L$. We have $F \subseteq K \subseteq L$. We have $[L :F] < \infty$, so $L \subseteq \overline{F}$.

Straight-Edge and Compass Constructions

Three old problems we solve:

1. Doubling the cube
2. Trisecting an angle
3. Squaring the circle
4. For which $n$ is it possible to construct a regular $n$-gon

First of all, $p(x,y)$ (a point at $(x,y)$) is constructable if and only if segments of lengths $x$ and $y$ are constructable.

$F = \{x \in \mathbb{R} \;|\; \text{segment of length $x$ or $-x$ is constructable}\} \subseteq \mathbb{R}$.

Proposition. $\alpha \in F$ if and only if $\mathbb{Q}(\alpha)$ is obtained by a sequence of degree 2 extensions.

One direction was shown in class.

$\{ \varphi \;|\; \varphi(a) = a$ for all $a \in F\} = \text{Aut}(K/F) \subseteq \text{Aut}(K)$. $\sigma \in \text{Aut}(K/F)$ if $\alpha$ is a root $f(x)$ in $F[x]$, $\sigma(\alpha)$ is alsoa root of $f(x)$.

There is a one to one corerspondence between $F \subseteq L \subseteq K$ and subgroups of $\text{Aut}(K/F)$. $L = K^H$ you get a subfield. The set of subgroups of the automorphisms of $K/F$ is called the Galois group of $K/F$.