Formatting Fix

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

@@ -1,4 +1,5 @@
 # Chapter 16 - Rings
+
 ## 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:
@@ -139,4 +140,4 @@ $$
 
 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.
+**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

@@ -1,4 +1,5 @@
 # Chapter 17 - Polynomial Rings
+
 ## Section 17.1 - Polynomial Rings
 
 Throughout this chapter, we will assume that $R$ is a commutative ring with identity.
@@ -63,4 +64,4 @@ Then, if $p | a_i$ for $0 \leq i < n$, but $p \nmid a_n$ and $p^2 \nmid a_0$, th
 
 **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.
+**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/math/abstract-algebra/18-integral-domains.md

@@ -98,4 +98,4 @@ As a direct consequence, we see the following.
 
 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.
-3. Given $D$ is a UFD, $D[x]$ is a UFD. Thus, $D[x_1, x_2]$ is a UFD, and by induction, $D[x_1, \ldots, x_n]$ is a UFD.
+3. Given $D$ is a UFD, $D[x]$ is a UFD. Thus, $D[x_1, x_2]$ is a UFD, and by induction, $D[x_1, \ldots, x_n]$ is a UFD.