Spellchecker

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

@@ -2,7 +2,7 @@
 
 ## Section 16.1 - Rings
 
-**Definition**. A nonempty set $S$ is a *ring* if, with two binary operations called addition and multipllication, the following are satisfied:
+**Definition**. A nonempty set $S$ is a *ring* if, with two binary operations called addition and multiplication, the following are satisfied:
 
 1. Addition is commutative. $a + b = b + a$ for $a, b \in R$
 2. Addition is associative. $(a + b) + c = a + (b + c)$ for $a, b, c \in R$
@@ -46,17 +46,17 @@ Note that some books impose the condition that $1 \neq 0$. If $1 = 0$, we can sh
 
 **Definition**. For any non-negative integer $n \in \mathbb{N}$ and $r \in R$, we say that $nr = r + \ldots + r \text{(n times)}$.
 
-**Definition**. The *charactaristic* of a ring is the leat possible $n \in \mathbb{N}$ such that $nr = 0$ for all $r \in R$.
+**Definition**. The *characteristic* of a ring is the least possible $n \in \mathbb{N}$ such that $nr = 0$ for all $r \in R$.
 
-**Example**. For every prime number $p$, $\mathbb{N}_p$ is a field of charactaristic $p$.
+**Example**. For every prime number $p$, $\mathbb{N}_p$ is a field of characteristic $p$.
 
-**Lemma**. 16.18: Given $R$ is a ring with identity, the charactaristic of $1$ is the charactartistic of the field.
+**Lemma**. 16.18: Given $R$ is a ring with identity, the characteristic of $1$ is the characteristic of the field.
 
-**Theorem**. 16.19: The charactaristic of an integral domain is prime or zero.
+**Theorem**. 16.19: The characteristic of an integral domain is prime or zero.
 
 ## Section 16.3 - Ring Homomorphisms and Ideals
 
-**Definition** Given rins $R$ and $S$, and a mapping $\varphi: R \rightarrow S$, we say that $\varphi$ is a *ring homomorphism* if the following are satisfied for all elements of $R$:
+**Definition** Given rings $R$ and $S$, and a mapping $\varphi: R \rightarrow S$, we say that $\varphi$ is a *ring homomorphism* if the following are satisfied for all elements of $R$:
 
 $$
 \begin{align}
@@ -92,7 +92,7 @@ is an ideal in $R$. Specifically, $\langle a \rangle$ is a *principal ideal*.
 
 **Example**. Theorem 16.25. Every ideal in $\mathbb{Z}$ is a principal ideal.
 
-**Examplee**. With $\varphi: R \rightarrow S$, $\ker \varphi$ is an ideal of $R$.
+**Example**. With $\varphi: R \rightarrow S$, $\ker \varphi$ 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*.
 
@@ -118,7 +118,7 @@ $$
 R/I \cong \frac{R/J}{I/J}
 $$
 
-**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 correspondence 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
 
@@ -140,6 +140,6 @@ $$
 (a + P)(b + P) = ab + P = 0 + P = P
 $$
 
-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.
+Thus, $ab \in P$. By symnetry, assume $a \notin P$. Thus, $b \in P$ by the definition 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

@@ -14,7 +14,7 @@ where $a_i \in R$ and $a_n \neq 0$ is called a *polynomial over $R$* with *indet
 
 **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**. The *degree* of $f$ is the largest nonnegative number such that $a_n \neq 0$, written as $\deg f(x) = n$. If no such number 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 $R[x]$.
 
@@ -46,7 +46,7 @@ where either $\deg r(x) < \deg g(x)$ or $r(x)$ is the zero polynomial.
 
 ## 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 $f(x)$.
+**Definition** A non-constant 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 $p(x) = \frac{r}{s} a(x)$.
 

docs/math/abstract-algebra/18-integral-domains.md

@@ -2,7 +2,7 @@
 
 ## Section 18.1 - Fields of Fractions
 
-**Definition**. Given an integral domain $D$, we can construct a field $F$ containing $D$ by stating that any $p/q \in F$, annd that any two elements $a/b = c/d$ if and only if $ad = bc$. We can consider this akin o a set of ordered pairs
+**Definition**. Given an integral domain $D$, we can construct a field $F$ containing $D$ by stating that any $p/q \in F$, and that any two elements $a/b = c/d$ if and only if $ad = bc$. We can consider this akin o a set of ordered pairs
 
 $$
 S = \{(a, b) : a, b \in D \text{ and } b \neq 0 \}
@@ -18,13 +18,13 @@ $$
 
 Additionally, $F_D$ is unique. That is, given field $E$ such that $E \supset D$, there exists a map $\psi: F_D \rightarrow D$ giving an isomorphism such that $\psi(a) = a$ for all $a \in D$.
 
-**Corollary**. 18.6: Let $F$ be a field of charactaristic $0$. Then, $F$ contains a subfield isomorphic to $\mathbb{Q}$.
+**Corollary**. 18.6: Let $F$ be a field of characteristic $0$. Then, $F$ contains a subfield isomorphic to $\mathbb{Q}$.
 
-**Corollary**. 18.6: Let $F$ be a field of charactaristic $p$. Then, $F$ contains a subfield isomorphic to $\mathbb{Z}_p$.
+**Corollary**. 18.6: Let $F$ be a field of characteristic $p$. Then, $F$ contains a subfield isomorphic to $\mathbb{Z}_p$.
 
 ## Section 18.2 - Factorization in Integral Domains
 
-**Definition**. Let $R$ be a commutative ring with identity, and $a, b \in R$. We say that $a$ *divideds* $b$, that is, $a | b$, if there exists some $c \in R$ such that $b = ac$.
+**Definition**. Let $R$ be a commutative ring with identity, and $a, b \in R$. We say that $a$ *divides* $b$, that is, $a | b$, if there exists some $c \in R$ such that $b = ac$.
 
 **Definition**. A *unit* element is any element that has a multiplicative inverse.
 
@@ -37,7 +37,7 @@ Additionally, $F_D$ is unique. That is, given field $E$ such that $E \supset D$,
 **Definition**. Given integral domain $D$, we say that $D$ is a *Unique Factorization Domain (UFD)* if it satisfies the following criteria:
 
 1. Given $a \in D, a \neq 0$, and $a$ is not a unit, $a$ can be written as a product of irreducible elements in $D$.
-2. Let $a = p_1 \ldots p_r = q_1 \ldots q_s$, where $p_i$ and $q_i$ are all irreducible. Then, $r = s$, and there exists some fuction $\pi \in S_r$ such that $p_i$ and $q_{\pi(j)}$ are associates for $j = 1, \ldots, r$.
+2. Let $a = p_1 \ldots p_r = q_1 \ldots q_s$, where $p_i$ and $q_i$ are all irreducible. Then, $r = s$, and there exists some function $\pi \in S_r$ such that $p_i$ and $q_{\pi(j)}$ are associates for $j = 1, \ldots, r$.
 
 **Definition**. A ring $R$ is a *principal ideal domain (PID)* if every ideal of $R$ is principal.
 
@@ -61,24 +61,24 @@ Additionally, $F_D$ is unique. That is, given field $E$ such that $E \supset D$,
 
 ---
 
-**Definition**. Any integral domain $D$ is a *euclidian domain* with a *euclidian function* $nu: D \\ \{0\} \rightarrow \mathbb{N}$ that satisfies the following:
+**Definition**. Any integral domain $D$ is a *Euclidean domain* with a *Euclidean function* $nu: D \\ \{0\} \rightarrow \mathbb{N}$ that satisfies the following:
 
 1. Given $a, b \neq 0$, then $\nu(a) \leq \nu(ab)$.
 2. Given, $a, b \in D$ and $b \neq 0$, there exists some $q, r \in D$ such that $a = bq + r$ and either $r = 0$ or $\nu(r) < \nu(b)$.
 
-**Example**. Absolute value on $\mathbb{Z}$ is a Euclidian validation.
+**Example**. Absolute value on $\mathbb{Z}$ is a Euclidean validation.
 
-**Example**. Degree on $F[x]$ is a Euclidian validation.
+**Example**. Degree on $F[x]$ is a Euclidean validation.
 
-**Example**. $\nu(a + bi) = a^2 + b^2$ is a Euclidian validation over $\mathbb{Z}[i]$.
+**Example**. $\nu(a + bi) = a^2 + b^2$ is a Euclidean validation over $\mathbb{Z}[i]$.
 
-**Theorem**. 18.21: Every Euclidian domain is a PID.
+**Theorem**. 18.21: Every Euclidean domain is a PID.
 
-**Corollary**. Every Euclidian domain is a UFD.
+**Corollary**. Every Euclidean domain is a UFD.
 
 ---
 
-**Definition**. Given a polynomial $p(x) \in D$, with $D$ bein an integer domain, we say that the *content* of $p(x)$ is the greatest common divisor of its coefficients. Additionally, if the content is $1$, we say that $p(x)$ is *primitive*.
+**Definition**. Given a polynomial $p(x) \in D$, with $D$ being an integer domain, we say that the *content* of $p(x)$ is the greatest common divisor of its coefficients. Additionally, if the content is $1$, we say that $p(x)$ is *primitive*.
 
 **Theorem**. 18.24: Let $D$ be a UFD, and $f(x), g(x) \in D[x]$ be primitive. Then, $f(x)g(x)$ is primitive.
 
@@ -94,7 +94,7 @@ As a direct consequence, we see the following.
 
 **Theorem**. If $D$ is as UFD, then $D[x]$ is a UFD.
 
-**Corollary**. This theorem has several collaries:
+**Corollary**. This theorem has several corollaries:
 
 1. Given a field $F$, since $F$ is a PID, it is also a UFD. Thus, $F[x]$ is a UFD.
 2. The ring of polynomials over integers, $\mathbb{Z}[x]$ is a UFD.

docs/math/abstract-algebra/DF-10-modules.md

@@ -15,13 +15,13 @@
 
 **Remark**. Modules over a field $F$ and vector spaces over $F$ are identical.
 
-**Definition** An *R-submodule* is a subset$N \subseteq M$ which is closed under the action taken forall $r \in R$. That is, given $r \in R, n \in N$, tthen $rn \in N$. Every module has at least two submodules: itself and the trivial (empty) submodule.
+**Definition** An *R-submodule* is a subset$N \subseteq M$ which is closed under the action taken forall $r \in R$. That is, given $r \in R, n \in N$, then $rn \in N$. Every module has at least two submodules: itself and the trivial (empty) submodule.
 
-**Remark**. If $F$ is a field, submodules are equivilent to subspaces.
+**Remark**. If $F$ is a field, submodules are equivalent to subspaces.
 
 ---
 
-**Example**. Let $F$ be a field and $F[x]$ a polynomial ring. Then, let $V$ be a vector space of $F$, and $T$ be a linear transformaion from $V$ to itself. That is, $V: T \rightarrow T$. We know that $V$ is an $F$-module. We will want to show that $V$ can be written as an $F[x]$-module for some choice of $T$. That is, we want an action $F[x] \cross V \rightarrow V$.
+**Example**. Let $F$ be a field and $F[x]$ a polynomial ring. Then, let $V$ be a vector space of $F$, and $T$ be a linear transformation from $V$ to itself. That is, $V: T \rightarrow T$. We know that $V$ is an $F$-module. We will want to show that $V$ can be written as an $F[x]$-module for some choice of $T$. That is, we want an action $F[x] \cross V \rightarrow V$.
 
 Now, for a given linear transformation $T$, consider some polynomial $p(x) = a_n x^n + \ldots + a_0$ and some $v \in V$. We define $p(x) \cross v$ by$
 
@@ -42,17 +42,17 @@ with $T^n$ being defined as applying $T$ a total of $n$ times.
 
 **Recall**. The center of a ring $A$ is the subring $A'$ such that for all $x, y \in R'$, then $xy = yx$. In other words, it is the commutative subring of $A$.
 
-**Definition**. Given two $R$-algebras $A, B$, an *$R$-algebra homomorphhism$ is a ring homomorphism $\varphi: A \rightarrow B$ that maps $1_A \rightarrow 1_B$ such that $\varphi(ra) = r\varphi(a)$.
+**Definition**. Given two $R$-algebras $A, B$, an *$R$-algebra homomorphism$ is a ring homomorphism $\varphi: A \rightarrow B$ that maps $1_A \rightarrow 1_B$ such that $\varphi(ra) = r\varphi(a)$.
 
 ## Section 10.2 - Quotient Modules and Module Homomorphisms
 
-**Definition**. Let $R$ be a ring and $M, N$ be $R$-modules. then a ring homomorphhism $\varphi: M \rightarrow N$ is an *$R$-module homomorphism* if for all $r \in R$, $\varphi(rx) = r\varphi(x)$.
+**Definition**. Let $R$ be a ring and $M, N$ be $R$-modules. then a ring homomorphism $\varphi: M \rightarrow N$ is an *$R$-module homomorphism* if for all $r \in R$, $\varphi(rx) = r\varphi(x)$.
 
 **Theorem**. An $R$-module homomorphism is an *isomorphism* if it is 1-1 and onto, and said modules are *isomorphic*.
 
 **Definition**. Let $M, N$ be $R$-modules. The set $\Hom_R(M, N)$ is the set of all homomorphisms from $M$ to $N$.
 
-**Promposition**. Let $M$, $N$, and $L$ be $R$-modules. Then,
+**Proposition**. Let $M$, $N$, and $L$ be $R$-modules. Then,
 
 1. A function $\varphi: M \rightarrow N$ is an $R$-module homomorphism if and only if $\varphi(rx + y) = r\varphi(x) + \varphi(y)$ for all $x, y \in M$ and $r \in R$.
 2. Let $\varphi, \psi \in \Hom_R(M, N)$. Then, define $\varphi + \psi$ as
@@ -85,13 +85,13 @@ $$
 
 This is the smallest submodule that contains both $A$ and $B$.
 
-**Theorem**. First Isomorphism Theorem. Let $M, N$ be $R$-modules, and $\varphi: M \rightarrow N$ be an $R$-module homomorphhiism. Then, $\ker \varphi$ is a submodule of $M$, and $M / \ker \varphi \cong \varphi(M)$.
+**Theorem**. First Isomorphism Theorem. Let $M, N$ be $R$-modules, and $\varphi: M \rightarrow N$ be an $R$-module homomorphism. Then, $\ker \varphi$ is a submodule of $M$, and $M / \ker \varphi \cong \varphi(M)$.
 
 **Theorem**. Second Isomorphism Theorem. Let $A, B$ be submodules of the $R$-module $M$. Then, $(A + B)/B \cong A/(A \cap B)$.
 
 **Theorem**. Third Isomorphism Theorem. Let $M$ be an $R$-module, and $A \subseteq B$ be submodules of $M$. Then, $\frac{M/A}{B/A} \cong M/B$.
 
-**Theorem**. Lattice Isomorphism Theorem. Let $N$ be a submodule of the $R$-module $M$. Then, there is a bijection between submoudles of $M$ containing $N$ and submodules of $M/N$. This is given by $A \leftrightarrow A/N$, for $A \supseteq N$.
+**Theorem**. Lattice Isomorphism Theorem. Let $N$ be a submodule of the $R$-module $M$. Then, there is a bijection between submodules of $M$ containing $N$ and submodules of $M/N$. This is given by $A \leftrightarrow A/N$, for $A \supseteq N$.
 
 ## Section 10.3 - Generation of Modules, Direct Sums, and Free Modules
 
@@ -99,7 +99,7 @@ This is the smallest submodule that contains both $A$ and $B$.
 
 1. The *sum* of $N_1, \ldots, N_n$ is the set of all finite sums of elements from the sets $N_i$. That is, $N_1, \ldots, N_n := \{a_1 + a_2 + \ldots + a_n | a_i \in N_i\}$
 2. For any subset $A$ of $M$, let $RA = \{r_1 a_1 + r_2 a_2 + \ldots + r_m a_m | r_i \in R, a_i \in A\}$. If $N$ is a submodule of $M$ such that $N = RA$, then $A$ is called the *generating set* for $N$.
-3. A submodule $N$ of $M$ is *finitely generaated* if there is some finite subset $A$ of $M$ such that $N = RA$. That is, $N$ is generated by some finite subset.
+3. A submodule $N$ of $M$ is *finitely generated* if there is some finite subset $A$ of $M$ such that $N = RA$. That is, $N$ is generated by some finite subset.
 4. A submodule of $M$ (up to equality) is $cyclic$ if there exists some element $a \in M$ such that $N = Ra = \{ra | r \in R\}$.
 
 **Definition**. Let $M_1, \ldots, M_k$ be a collection of $R$-modules. Then, the *direct product* is defined as
@@ -118,13 +118,13 @@ This direct product is in itself an $R$-module.
 
 **Definition**. An $R$-module $F$ is said to be *free* on the subset $A$ of $F$ if for every nonzero $x \in F$, there exists nonzero elements $r_1, \ldots, r_n$ of $R$ and unique $a_1, \ldots, a_n$ such that $x = r_1 a_1 + \ldots + r_n a_n$ for some $n \in \mathbb{Z}^+$. That is, $A$ is a *basis* or *set of free generators* of $F$.
 
-**Theorem**. For any set $A$, there is a free $R$-module $F(A)$ on $A$ such that $F(A)$ satisfies the universal property: if $M$ is any $R$-module, and $\varphi: A \rightarrow M$ is a map of sets, there eixts a unique $R$-module homomorphism: $\Phi: F(A) \rightarrow M$ such that $\Phi(a) = \varphi(a)$ for all $a \in A$.
+**Theorem**. For any set $A$, there is a free $R$-module $F(A)$ on $A$ such that $F(A)$ satisfies the universal property: if $M$ is any $R$-module, and $\varphi: A \rightarrow M$ is a map of sets, there exists a unique $R$-module homomorphism: $\Phi: F(A) \rightarrow M$ such that $\Phi(a) = \varphi(a)$ for all $a \in A$.
 
 **Corollary**. If $F_1$ and $F_2$ are free modules on $A$, then there is a unique isomorphism between $F_1$ and $F_2$, which is the identity map on A.
 
 **Corollary**. If $F$ is a free $R$-module with basis $A$, then $F \cong F(A)$.
 
-**Definition** For a free module $F$ with basis $A$, if $R$ is commutative, tthen the *rank* of $F$ is the cardinality of $A$.
+**Definition** For a free module $F$ with basis $A$, if $R$ is commutative, then the *rank* of $F$ is the cardinality of $A$.
 
 ## Section 10.4 - Tensor Products of Modules
 

docs/math/abstract-algebra/DF-12-modules-pids.md

@@ -1,4 +1,4 @@
-# Chapter 12 - Modules over Principal Ideal Domains
+# Dummit & Foote Chapter 10 Chapter 12 - Modules over Principal Ideal Domains
 
 ## Section 12.1 The Basic Theory
 
@@ -8,7 +8,7 @@ $$
 M_1 \subseteq M_2 \subseteq \ldots
 $$
 
-there exists some $k \in \mathbb{N}$ such thaht given any $n \in \mathbb{N}$ with $n \geq k$, then $M_n = M_k$.
+there exists some $k \in \mathbb{N}$ such that given any $n \in \mathbb{N}$ with $n \geq k$, then $M_n = M_k$.
 
 **Definition**. A ring $R$ is *Noetherian* if it is Noetherian when viewed as a left $R$-module over itself.
 
@@ -20,7 +20,7 @@ there exists some $k \in \mathbb{N}$ such thaht given any $n \in \mathbb{N}$ wit
 
 **Corollary**. If $R$ is a principal ideal domain (PID), then all nonempty set of ideals of $R$ has a maximal element. Additionally, $R$ is as Noetherian ring.
 
-**Proposition**. Let $R$ be an integral doman, and $M$ be a free $R$-module of rank $n < \infty$. Then, given $S$ is subset $M$ with $|S| > n$, the elements of $S$ are $R$-linearly dependent.
+**Proposition**. Let $R$ be an integral domain, and $M$ be a free $R$-module of rank $n < \infty$. Then, given $S$ is subset $M$ with $|S| > n$, the elements of $S$ are $R$-linearly dependent.
 
 **Definition**. Given $R$ an integral domain and $M$ an $R$-module,
 
@@ -49,5 +49,5 @@ Additionally, if $R$ is a PID, as $\Ann_R(N)$ is an ideal, $\Ann(N) = (n)R$ and
 **Theorem**. Let $R$ be a principal ideal domain, and $M$ be a free $R$-module of finite rank $m$, and $N$ be a submodule of $M$. Then,
 
 1. $N$ is a free submodule with rank $n \leq m$.
-2. There exiss a basis $y_1, y_2, \ldots, y_m$ of $M$ so that $r_1 y_1, r_2 y_2, \ldots, r_m y_n$ is a basis of $N$ for some $r_i \in R$ and $r_1 | r_2 | \ldots | r_n$
+2. There exists a basis $y_1, y_2, \ldots, y_m$ of $M$ so that $r_1 y_1, r_2 y_2, \ldots, r_m y_n$ is a basis of $N$ for some $r_i \in R$ and $r_1 | r_2 | \ldots | r_n$