Updates

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

@@ -5,7 +5,7 @@
 **Definition**. Let $R$ be a ring. A *left $R$-module* or a *left module over $R$* is a nonempty set $M$ together with
 
 1. A binary operation $+$ on $M$ under which $M$ is an abelian group
-2. An action $\cross$ of $R$ on $M$, that is, a map or function $R \cross M \leftarrow M$, denoted $rm$, that for all $r, s \in R, m, n \in M$ satisfies
+2. An action $\cross$ of $R$ on $M$, that is, a map or function $R \cross M \rightarrow M$, denoted $rm$, that for all $r, s \in R, m, n \in M$ satisfies
     - $(r + s)m = rm + sm$
     - $(rs)m = r(sm)$
     - $r(m + n) = rm + rn$
@@ -21,7 +21,7 @@
 
 ---
 
-**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, $V: T \leftarrow T$. 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 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 $F[x] \cross V \rightarrow V$.
 
 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$
 
@@ -98,11 +98,11 @@ This is the smallest submodule that contains both $A$ and $B$.
 **Definition**. Let $M$ be an $R$-module and $N_1, \ldots, N_n$ be submodules of $M$.
 
 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 a ny 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 $N$.
+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.
 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 *drirect product* is defined as
+**Definition**. Let $M_1, \ldots, M_k$ be a collection of $R$-modules. Then, the *direct product* is defined as
 
 $$
 M_1 \otimes \ldots M_k = (m_1, \ldots, m_k), m_i \in M_i
@@ -112,8 +112,8 @@ This direct product is in itself an $R$-module.
 
 **Proposition**. Let $N_1, \ldots, N_n$ be submodules of the $R$-module $M$. Then, the following are equivalent:
 
-1. The map $\pi: N_1 \otimes \ldots \otimes N_k \leftarrow N_1 + \ldots + N_k$ defined by $\pi(a_1, \ldots, a_n) = a_1 + \ldots + a_n$ is an isomorphism
+1. The map $\pi: N_1 \otimes \ldots \otimes N_k \rightarrow N_1 + \ldots + N_k$ defined by $\pi(a_1, \ldots, a_n) = a_1 + \ldots + a_n$ is an isomorphism
-2. $N_j \cup (N+1 + \ldots + N_{j-1} + N{j+1} + \ldots + N_n) = 0 $ for all $j \in \{1, 2, \ldots, k\}$
+2. $N_j \cup (N+1 + \ldots + N_{j-1} + N{j+1} + \ldots + N_n) = 0$ for all $j \in \{1, 2, \ldots, k\}$
 3. Every $x \in N_1 + \ldots + N_n$ can be written uniquely in the form $a_1 + \ldots + a_n$, with $a_i \in N_i$
 
 **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 $F$.

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

@@ -8,7 +8,7 @@ $$
 M_1 \subseteq M_2 \subseteq \ldots
 $$
 
-there xists 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 thaht 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.
 
@@ -36,7 +36,7 @@ $$
 \Ann_R(N) = \{r \in R | rn = 0 \text{ for all } n \in N \}
 $$
 
-This ideal of $R$ is the **annihilator of $N$*. That is, $\Ann(N)$ is the set of elements of $R$ such that $(r)N = \{ 0 \}$.
+This ideal of $R$ is the *annihilator of $N$*. That is, $\Ann(N)$ is the set of elements of $R$ such that $(r)N = \{ 0 \}$.
 
 Note that if $N$ is not a torsion submodule of $M$, then $\Ann(N) = (0)R$. Additionally, given $N, L$ are submodules of $M$ with $N \subseteq L$, then $\Ann(N) \subseteq \Ann(L)$.
 
@@ -49,4 +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 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 $r_1 | r_2 | \ldots | r_n$
+