@ -285,7 +285,7 @@ Quotient group construction is useful as a way of actually manufacturing all the
\end{theorem}
\section{Rings}
\section{Rings and Fields}
\begin{definition}[Ring]
A set $A$ with operations called \emph{addition} and \emph{multiplication} which satisfy the following axioms:
@ -312,6 +312,19 @@ Quotient group construction is useful as a way of actually manufacturing all the
If $A$ is a commutative ring with unity in which every nonzero element is invertible, $A$ is called a \emph{field}.
\end{definition}
\begin{theorem}[Finite Field must be over p prime ($\mathbb{F}_p$)]
Proof from \href{https://github.com/aragonzkresearch/blog/blob/main/pdf/Aragon_Math_Seminar.pdf}{Matan Prasma seminars}:\\
One of the axioms of a field is $\exists$ multiplicative inverse.\\
If $\mathbb{Z}_n$ with $n$ no prime, then $n= k \cdot l$ for some $1\leq k,~l \leq n-1$.\\
Then in $\mathbb{Z}_n$, $k \cdot l =0$, but if $k \cdot l=0$ means that either $k=0$ or $l=0$ (otherwise, we could multiply by (eg) $k^{-1}$ and get $k^{-1}\cdot k \cdot l = k^{-1}\cdot0$, which leads to $1\cdot l =0$).\\
which is a contradiction here (since $1\leq k,~l \leq n-1$).\\
Thus $\mathbb{Z}_n$ with $n$ not prime can not be a field.\\
Conversely, if $n = p$ prime,\\
for $0\neq x \in\mathbb{Z}_p$, $gcd(x, p)=1$, so Extended Euclidean Algorithm gives $u, v \in\mathbb{Z}$ such that $u x + v p =1$.\\
Then, $ux=1\pmod p$, so $u=x^{-1}$, so inverses exist.\\
Thus $\mathbb{Z}_p$ is a field.
\end{theorem}
\begin{definition}[Divisor of zero]
In any ring, a nonzero element a is called a \emph{divisor of zero} if there is a
nonzero element b in the ring such that the product ab or ba is equal to
@ -347,7 +360,7 @@ Every field is an integral domain, but the converse is not true (eg. $\mathbb{Z}
\begin{definition}[Characteristic n]
Let $A$ be a ring with unity, the \emph{characteristic} of $A$ is the least positive integer $n$ such that
$$1+1+\cdots+1=0$$
$$\underbrace{1+1+\cdots+1}_{n-times}=0$$
If there is no such positive integer $n$, $A$ has characteristic $0$.