Catchup 6

docs/math/abstract-algebra/16-rings.md

@@ -53,7 +53,7 @@ Note that some books impose the condition that $1 \neq 0$. If $1 = 0$, we can sh
 
 ## Section 16.3 - Ring Homomorphisms and Ideals
 
-**Definition** Given rins $R$ and $S$, and a mapping $\phi: R \leftarrow S$, we say that $\phi$ is a *ring homomorphism* if the following are satisfied for all elements of $R$:
+**Definition** Given rins $R$ and $S$, and a mapping $\phi: R \rightarrow S$, we say that $\phi$ is a *ring homomorphism* if the following are satisfied for all elements of $R$:
 
 $$
 \begin{align}
@@ -70,7 +70,7 @@ $$
 \ker \phi = \{ r \in R | \phi(r) = 0 \}
 $$
 
-**Definition**. Proposition 16.22: Let $\phi: R \leftarrow S$ be a ring homomorphism. Then,
+**Definition**. Proposition 16.22: Let $\phi: R \rightarrow S$ be a ring homomorphism. Then,
 
 1. If $R$ is a commutative ring, then $\phi(R) \subset S$ is a commutative ring.
 2. $\phi(0_R) = 0_S$
@@ -89,7 +89,7 @@ is an ideal in $R$. Specifically, $<a>$ is a *principal ideal*.
 
 **Example**. Theorem 16.25. Every ideal in $\mathbb{Z}$ is a principal ideal.
 
-**Examplee**. With $\phi: R \leftarrow S$, $\ker \phi$ is an ideal of $R$.
+**Examplee**. With $\phi: R \rightarrow S$, $\ker \phi$ is an ideal of $R$.
 
 **Remark**. 16.28: We are working with *two-sided ideals*. If rings are not commutative, we may deal with *left ideals* and *right ideals*.
 
@@ -99,9 +99,9 @@ $$
 (r + I)(s + I) = rs + I
 $$
 
-**Theorem**. 16.30: Let $I$ be an ideal of $R$. Then, the map $\phi: R \leftarrow R/I$ defined by $\phi(r) = r + I$ is a ring homomorphism of $R$ onto $R/I$ with $\ker \phi = I$.
+**Theorem**. 16.30: Let $I$ be an ideal of $R$. Then, the map $\phi: R \rightarrow R/I$ defined by $\phi(r) = r + I$ is a ring homomorphism of $R$ onto $R/I$ with $\ker \phi = I$.
 
-**Theorem**. 16.31, *First Isomorphism Theorem*. Let $\psi: R \leftarrow S$. Then, $\ker \psi$ is an ideal of $R$. Consider the isomorphism $\phi: R \leftarrow R/\ker \psi$. There exists an isomorphism $\eta: R / \ker \psi \leftarrow \psi(R)$ such that $\psi = \eta \phi$.
+**Theorem**. 16.31, *First Isomorphism Theorem*. Let $\psi: R \rightarrow S$. Then, $\ker \psi$ is an ideal of $R$. Consider the isomorphism $\phi: R \rightarrow R/\ker \psi$. There exists an isomorphism $\eta: R / \ker \psi \rightarrow \psi(R)$ such that $\psi = \eta \phi$.
 
 **Theorem**. 16.32, *Second Isomorphism Theorem*. Let $I$ be a subring of $R$ and $J$ be an ideal of $R$. Then, $I \cap J$ is an ideal of $I$ and
 
@@ -115,4 +115,28 @@ $$
 R/I \cong \frac{R/J}{I/J}
 $$
 
-**Theorem**. 16.34, *Correspondence Theorem*. Let $I$ be an ideal of $R$. Then, $S \mapsto S/I$ is a one-to-one correspeondence between the set of subrings $S$ containing $I$ (that is, $I \in S$) and the set of subrings of $R/I$. Furthermore, the ideals of $R$ containing $I$ correspond to the ideals of $R/I$.
+**Theorem**. 16.34, *Correspondence Theorem*. Let $I$ be an ideal of $R$. Then, $S \mapsto S/I$ is a one-to-one correspeondence between the set of subrings $S$ containing $I$ (that is, $I \in S$) and the set of subrings of $R/I$. Furthermore, the ideals of $R$ containing $I$ correspond to the ideals of $R/I$.
+
+## Section 16.4 - Maximal and Prime Ideals
+
+**Definition**. Consider ring $R$ and proper ideal $M \subset R$. Then, $M$ is a *maximal ideal* of $R$ if the ideal $M$ is not a subset of any ideal except $R$ itself. That is, given any ideal $I$ properly containing $M$, $I = R$.
+
+**Theorem**. 16.35: Given a commutative ring with identity $R$, $M$ is a maximal ideal if and only if $R/M$ is a field.
+
+**Definition**. Consider ring $R$ and proper ideal $P \subset R$. Then, $P$ is a *prime ideal* if given $ab \in P$, either $a \in P$ or $b \in P$.
+
+**Theorem**. 16.38: Let $R$ be a commutative ring with identity $1$. Then, $P \subset R$ is a prime ideal of $R$ if and only if $R/P$ is a field.
+
+Let us assume that $P$ is an ideal in $R$ and $R/P$ is an integral domain. Take two elements $ab \in P$. Now, consider $a + P$ and $b + P$ in $R/P$ such that $(a+P)(b+P) = 0+P = P$. As $R/P$ is a field, either $a + P = 0 + P = P$ or $b + P = 0 + P = P$, meaning either $a \in P$ or $b \in P$. Thus, $P$ is as prime ideal.
+
+Now, assume the opposite. Let $P$ be prime. Now, we want to show that $R/P$ is an integral domain.
+
+Consider two elements $a + P$, $b + P$ in $R/P$. We know that
+
+$$
+(a + P)(b + P) = ab + P = 0 + P = P
+$$
+
+Thus, $ab \in P$. By symnetry, assume $a \notin P$. Thus, $b \in P$ by the  devinition of a prime ideal, so $b + P = 0 + P$, meaning $R/P$ is an integral domain.
+
+**Theorem**. 16.40: In a commutative ring with identity, every maximal ideal is also a prime ideal.

docs/math/abstract-algebra/17-polynomial-rings.md

@@ -0,0 +1,66 @@
+# Chapter 17 - Polynomial Rings
+## Section 17.1 - Polynomial Rings
+
+Throughout this chapter, we will assume that $R$ is a commutative ring with identity.
+
+**Definition**. Any expression of the form
+
+$$
+f(x) = \sum_{i=0}^n a_i x^i = a_0 + a_1x + a_2x^2 + \ldots + a_n x^n
+$$
+
+where $a_i \in R$ and $a_n \neq 0$ is called a *polynomial over $R$* with *indeterminate* $x$. The elements $a_0, a_1, \ldots, a_n$ are the *coefficients* of $f$. The coefficient $a_n$ is the *leading coefficient*.
+
+**Definition**. A polynomial is known as *monic* if the leading coefficient is equal to $1$.
+
+**Definition**. The *degree* of $f$ is the largest nonnegative number such that $a_n \neq 0$, written as $\deg f(x) = n$. If no such mumber exists, that is, $f(x) = 0$, we say the degree of $f$ is $-\infty%.
+
+**Definition**. We denote the set of all polynomials with coefficients in $R$ as $R[x]$.
+
+Two polynomials are equal if and only if their corresponding coefficients are equal. When combined with standard addition and multiplication, $R[x]$ forms a ring.
+
+**Theorem**. If $R$ is commutative and has identity, so does $R[x]$.
+
+**Definition**. The *ring of polynomials with $n$ indeterminates and coefficients in $R$* is defined as $R[x_1][x_2][\ldots][x_n] = R[x_1, x_2, \ldots, x_n]$.
+
+**Definition**. The *evaluation homomorphism* is the homomorphism $\phi: R[x] \rightarrow R$ defined as $\phi(p(x)) = p(\alpha)$ for some $\alpha \in R$.
+
+## Section 17.2 - The Division Algorithm
+
+**Theorem**. Given $f(x), g(x) \in F[x]$, where $F$ is a field and $g(x) \neq 0$, there exist unique polynomials $q(x), r(x) \in F[x]$ such that
+
+$$
+f(x) = g(x)q(x) + r(x)
+$$
+
+where either $\deg r(x) < \deg g(x)$ or $r(x)$ is the zero polynomial.
+
+**Collary**. Let $F$ be a field. Then, an element $\alpha \in F$ is a zero of $p(x) \ in F[x]$ if and only if $(x-\alpha)$ is a factor of $p(x)$.
+
+**Collary**. Let $F$ be a field. Then, a nonzero polynomial $p(x) \in F[x]$ with degree $n$ can have at most $n$ distinct zeros in $F$.
+
+**Definition**. A monic polynomial $d(x)$ is the *greatest common divisor* of polynomials $p(x), q(x) \in F[x]$ if $d(x)$ evenly divides both $p(x)$ and $q(x)$. We write $\gcd(p(x), q(x)) = d(x)$. This polynomial is unique.
+
+**Definition**. Two polynomials are *relatively prime* if their greatest common divisor is $1$.
+
+## Section 17.3 Irreducible Polynomials
+
+**Definition** A nonconstant polynomial $f(x) \ in F[x]$ is *irreducible* over a field $F$ if it cannot be expressed as the product of two non-identity polynomials $g(x)$ and $h(x)$ in $F[x]$, with the degree of both polynomials strictly less than the  degree of $f(x)$.
+
+**Lemma**. Let $p(x) \in \mathbb{Q}[x]$. Then, with $r, s \in \mathbb{Z}, a(x) \in \mathbb{N}[x]$, we can write $p(x) = \frac{r}{s} a(x)$.
+
+**Lemma**. Gauss's Lemma. Let $p(x) \in \mathbb{Z}[x]$ be monic such that $p(x)$ factors into two polynomials $\alpha(x), \beta{x} \in \mathbb{Q}[x]$, with the degrees of both strictly less than the degree of $p(x)$. Then, ther exist two polynomials $a(x), b(x) \in \mathbb{Z}[x]$ such that $p(x) = a(x)b(x)$, and $\deg \alpha(x) = \deg a(x)$ and $\deg \beta(x) = \deg b(x)$.
+
+**Collary**. Let $p(x) \in \mathbb{Z}[x]$ be monic with constant term $a_0$. Then, if $p(x)$ has a zero in $\mathbb{Q}$, then it also has a zero $\alpha$ in $\mathbb[Z]$. Furthermore, $\alpha$ divides $a_0$.
+
+**Theorem**. Eisenstein's Criterion. Let $p$ be prime, and suppose that
+
+$$
+f(x) = a_n x^n + \ldots + a_0 \in \mathbb{Z}[x]
+$$
+
+Then, if $p | a_i$ for $0 \leq i < n$, but $p \nmid a_n$ and $p^2 \nmid a_0$, then $f(x)$ is irreducible over $\mathbb{Q}[x]$.
+
+**Theorem**. If $F$ is a field, then every ideal in $F[x]$ is a principal ideal.
+
+**Theorem**. Let $F$ be a field, and suppose $p(x) \in F[x]$. Then, the ideal $<p(x)>$ is maximal if and only if $p(x)$ is irreducible.

docs/physics/electrostatics/3-electro-magnetic-potentials.md

@@ -38,7 +38,7 @@ $$
 W = -q_m \frac{Q_m}{4 \pi \mu_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_m \frac{Q_m}{4 \pi \mu_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_m \frac{Q_m}{4 \pi \mu_0} \frac{1}{r}
 $$
 
-Letting the potential as $\vb{r} \leftarrow \infty$ equal $0$ be our reference and dividing out $q$, we find that the voltage for arrangement is the following:
+Letting the potential as $\vb{r} \rightarrow \infty$ equal $0$ be our reference and dividing out $q$, we find that the voltage for arrangement is the following:
 
 $$
 V_e(\vb{r}) = \frac{Q_e}{4 \pi \epsilon_0 r} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 r}