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Group theory

Definitions

Let $G$ be a group. The normalizer (in G) of a subset $A \subset G$ , denoted $N_{G} (A)$ , is defined as $N_{G} (A) = {g \in G | g A g^{- 1} = A} = {g \in G | g A = A g} .$ In other words, if $n \in N_{G} (A)$ , $a \in A$ , then $n a n^{- 1} \in A$ .

Preliminary theorems

Let $φ : G \to H$ be a homomorphism. Let $E \leq H$ . Then $φ^{- 1} [E] \leq G$ ; i.e., the homomorphic pullback of a subgroup is a subgroup.

Let $x, y \in φ^{- 1} [E]$ . Since $φ$ is a homomorphism, $\begin{matrix} e_{H} = φ (e_{G}) = φ (y y^{- 1}) = φ (y) φ (y^{- 1}), \\ e_{H} = φ (e_{G}) = φ (y^{- 1} y) = φ (y^{- 1}) φ (y); \end{matrix}$ consequently, $φ (y^{- 1}) = {φ (y)}^{- 1} .$ Therefore, $φ (x y^{- 1}) = φ (x) {φ (y)}^{- 1} \in E,$ so $x y^{- 1} \in φ^{- 1} [E] .$ So $φ^{- 1} [E] \leq G .$

The prior proposition holds if "subgroup" is replaced by "normal subgroup."

Suppose $E ⊴ H .$ Write $N = φ^{- 1} [E] .$ For any $g \in G,$ let $g n g^{- 1} \in g N g^{- 1} .$ Then since $φ (g) \in H$ and $E ⊴ H,$ $φ (g n g^{- 1}) = φ (g) φ (n) {φ (g)}^{- 1} \in E,$ so $g n g^{- 1} \in N .$ Consequently, $g N g^{- 1} \subseteq N,$ so $N ⊴ G .$

The kernel of a homomorphism is a normal subgroup.

Isomorphism theorems

First Isomorphism Theorem If $φ : G \to H$ is a homomorphism of groups, then $ker φ ⊴ G$ and $G / ker φ ≅ φ (G) .$

$ker φ ⊴ G$ : already proven.
$G / ker φ ≅ φ (G)$ : Let $ψ : G / ker φ \to φ (G)$ be defined by $ψ (g + ker φ) ≔ φ (g) .$ $ψ$ is a homomorphism since $ψ (g + ker φ) ψ (h + ker φ) = φ (g) φ (h) = φ (g h) = ψ (g h + ker φ) .$ To show that $ψ$ is injective, suppose that $φ (g) = φ (h) .$ Then $φ (g h^{- 1}) = φ (g h^{- 1}) = φ (g) {φ (h)}^{- 1} = e_{H},$ so $g h^{- 1} \in ker φ .$ Therefore $h + ker φ = (g h^{- 1}) h + ker φ = g + ker φ .$

Second Isomorphism Theorem If $A, B \leq G$ and $A \leq N_{G} (B)$ , then $A B \leq G, B ⊴ A B, A \cap B ⊴ A,$ and $A B / B ≅ A / (A \cap B) .$

$A B \leq G$ : $A B$ is nonempty as $e \in A B .$ Let $a_{1} b_{1}, a_{2} b_{2} \in A B$ . Write $b = b_{1} b_{2}^{- 1} \in B .$ Then $\begin{matrix} (a_{1} b_{1}) {(a_{2} b_{2})}^{- 1} & = a_{1} b_{1} b_{2}^{- 1} a_{2}^{- 1} \\ = a_{1} b a_{2}^{- 1} \\ = a_{1} (a_{2}^{- 1} a_{2}) b a_{2}^{- 1} \\ = a_{1} a_{2}^{- 1} (a_{2} b a_{2}^{- 1}) \in A B, \end{matrix}$ since $a_{2} \in N_{G} (B) .$
$B ⊴ A B$ : $B$ is nonempty by assumption. Let $x \in B, a b \in A B .$ Then $(a b) x {(a b)}^{- 1} = a (b x b^{- 1}) a^{- 1} \in B$ since $a \in N_{G} (B) .$
$A \cap B ⊴ A$ : $A \cap B$ is nonempty as $e \in A \cap B .$ Let $i \in A \cap B, a \in A .$ Then $a i a^{- 1} \in A$ since $i \in A$ and $a i a^{- 1} \in B$ since $i \in B$ and $a \in N_{G} (B) .$ Hence $a i a^{- 1} \in A \cap B .$
$A B / B ≅ A / (A \cap B)$ : We use the First Isomorphism Theorem. To this end, we construct a homomorphism $φ : A \to A B / B$ whose kernel is $A \cap B .$
Define $φ (a) = a B .$ $φ$ is a homomorphism since $φ (a_{1}) φ (a_{2}) = (a_{1} B) (a_{2} B) = a_{1} a_{2} B = φ (a_{1} a_{2}) .$ Further, $\begin{matrix} ker φ & = {a \in A | φ (a) = 1 B} \\ = {a \in A | a B = 1 B} \\ = {a \in A | a \in B} = A \cap B . \end{matrix}$

Ring theory

Let $R$ be a commutative ring with ideal $I$ . $I$ is a maximal ideal if and only if $R / I$ is a field.

Insert proof here

Field theory

Let $F$ be a field. Let $I$ be an ideal of $F .$ Then $I$ is $0$ or $F .$

Suppose $0 \neq x \in I$ . $F$ is a field, so $x^{- 1} \in F$ . $I$ is an ideal, so $1 = x x^{- 1} \in I$ , and consequently $y = y 1 \in I$ for any $y \in F$ , so $I = F$ .

Let $φ : F \to F^{'}$ be a homomorphism of fields. Then $φ$ is either identically $0$ or injective.

$ker (φ)$ is an ideal of $F$ .

The following theorem is due to Kronecker. It represents the nucleus of his attempt to set down the theory of irrational numbers on a solid foundation.

Fundamental Theorem of Field Theory Let $F$ be a field and let $p (x) \in F [x]$ be an irreducible polynomial. Then there exists an extension field of $F$ in which $p (x)$ has a root.

Let $K = F [x] / ⟨ p (x) ⟩ .$ We assert a few things about $K$ :

$K$ is a field. This is because $p (x)$ is irreducible, so $⟨ p (x) ⟩$ is maximal in $F [x]$ .
$K$ is an extension field of $F$ . Pithily, $F$ is the field of constants of $K$ .
To be picky about it, consider the map $π : F \to K$ given by $π (a) = a + ⟨ p (x) ⟩ .$ This is the restriction to $F$ of the canonical quotient map $φ : F [x] \to K$ . Like all canonical quotient maps, $φ$ is a homomorphism of rings, so $π$ , its restriction to the field $F$ , is a homomorphism of fields.
Because $p (x)$ is irreducible, and hence not $1$ , $π (1) = 1 + ⟨ p (x) ⟩ \neq 0$ . So $π$ is not identically zero. Therefore it is injective, and hence an iso-morphism, by the prior proposition. Hence $F$ is contained isomorphically in $K$ as its field of constants.
$K$ contains a root of $p (x)$ . Pithily, this is $π (x) = x + ⟨ p (x) ⟩ .$
More specifically, since $π$ is a homomorphism, we have This proof looks like a magic trick. I don't pick up where the mathematical subtlety is hidden. This is the cool thing and the irritating thing about textbook mathematics; it is done so elegantly that it looks fake. $\begin{matrix} p (π (x)) & = π (p (x)) \\ = p (x) + ⟨ p (x) ⟩ = 0 . \end{matrix}$

Abstract Algebra, Dummit & Foote