Fixes

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

@@ -73,12 +73,12 @@ $$
 
 **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.
+1. If $R$ is a commutative ring, then $\phi(R) \subseteq S$ is a commutative ring.
 2. $\phi(0_R) = 0_S$
 3. Let $1_R$ and $1_S$ be the identities in $R$ and $S$. If $\phi$ is onto, then $\phi(1_R) = 1_S$
-4. If $R$ is a field an $\phi(R) \neq \{0\}$, then $\phi(R) \subset S$ is a field.
+4. If $R$ is a field an $\phi(R) \neq \{0\}$, then $\phi(R) \subseteq S$ is a field.
 
-**Definition**. A subring $I \subset R$ is asn *ideal* of $R$ if, when given $a \in I, r \in R$, then $ar$ and $ra$ are both in $I$. That is, $rI \subset I$ and $Ir \subset I$.
+**Definition**. A subring $I \subseteq R$ is asn *ideal* of $R$ if, when given $a \in I, r \in R$, then $ar$ and $ra$ are both in $I$. That is, $rI \subseteq I$ and $Ir \subseteq I$.
 
 **Definition**. Given a commutative ring $R$ with identity, and $r \in R$, the set
 
@@ -120,13 +120,13 @@ $$
 
 ## 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 $M$, $I = R$.
+**Definition**. Consider ring $R$ and proper ideal $M \subseteq R$. 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 $b \in P$.
+**Definition**. Consider ring $R$ and proper ideal $P \subseteq R$. 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.
+**Theorem**. 16.38: Let $R$ be a commutative ring with identity $1$. Then, $P \subseteq R$ 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.
 

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

@@ -43,7 +43,7 @@ Additionally, $F_D$ is unique. That is, given field $E$ such that $E \supset D$,
 
 **Lemma**. 18.11: Let $D$ be an integral domain and $a, b \in D$. Then,
 
-1. $a | b$ if and only if $\langle b \rangle \subset \langle a \rangle$
+1. $a | b$ if and only if $\langle b \rangle \subseteq \langle a \rangle$
 2. $a$ and $b$ are associates if and only if $\langle b \rangle = \langle a \rangle$
 3. $a$ is a unit in $D$ if and only if $\langle a \rangle = D$.
 
@@ -51,7 +51,7 @@ Additionally, $F_D$ is unique. That is, given field $E$ such that $E \supset D$,
 
 **Collary**. 18.13: Let $D$ be a PID. For any $p \in D$, if $p$ is irreducible, then $p$ is prime.
 
-**Lemma**. 18.14: Let $D$ be a PID. Let $I_1 \subset I_2 \subset \ldots$. Then, there exists some integer $N$ such that $I_n = I_N$ for all $n > N$. That is, any chain of ideals converges.
+**Lemma**. 18.14: Let $D$ be a PID. Let $I_1 \subseteq I_2 \subseteq \ldots$. Then, there exists some integer $N$ such that $I_n = I_N$ for all $n > N$. That is, any chain of ideals converges.
 
 **Definition**. Any commutative ring that satisfies the above condition (the *ascending chain condition*), even if it's not a PID, is called a *Noetherien ring*.