diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf
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diff --git a/commutative-algebra-notes.tex b/commutative-algebra-notes.tex
index 4fd9100..5c88d5c 100644
--- a/commutative-algebra-notes.tex
+++ b/commutative-algebra-notes.tex
@@ -92,7 +92,91 @@
 \section{Ideals}
 
 \subsection{Definitions}
-% WIP
+
+\begin{defn}{ideal}
+  $I \subset R$ ($R$ ring) such that $0 \in I$ and $\forall x \in I,~ r \in R,~ xr, rx \in I$.\\
+  \hspace*{2em} ie. $I$ absorbs products in $R$.
+\end{defn}
+
+\begin{defn}{prime ideal}
+  if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a in P$ or $b \in P$.
+\end{defn}
+
+\begin{defn}{principal ideal}
+  generated by a single element, $(a)$.
+
+  $(a)$: principal ideal, the set of all multiples $xa$ with $x \in R$.
+\end{defn}
+
+\begin{defn}{maximal ideal}
+  $\mM \subset A$ ($A$ ring) with $m \neq A$ and there is no ideal $I$ strictly between $\mM$ and $A$. ie. if $\mM$ maximal and $\mM \subseteq I \subseteq A$, either $\mM=I$ or $I=A$.
+\end{defn}
+
+
+\begin{defn}{unit}
+  $x \in A$ such that $xy=1$ for some $y \in A$. ie. element \emph{which divides 1}.
+\end{defn}
+
+\begin{defn}{zerodivisor}
+  $x \in A$ such that $\exists 0 \neq y \in A$ such that $xy=0 \in A$. ie. $x$ \emph{divides 0}..
+
+
+  If a ring does not have zerodivisors is an integral domain.
+\end{defn}
+
+\begin{defn}{prime spectrum - $Spec(A)$}
+  set of prime ideals of $A$. ie.
+
+  $$Spec(A) = \{ P ~|~ P \subset A~ \text{is a prime ideal} \}$$
+\end{defn}
+
+\begin{defn}{integral domain}
+  Ring in which the product of any two nonzero elements is nonzero.
+
+  ie. no zerodivisors.
+
+  ie. $\forall~ 0 \neq a,~ 0 \neq b \in A,~ ab \neq 0 \in A$.
+
+  Every field is an integral domain, not the converse.
+\end{defn}
+
+\begin{defn}{principal ideal domain - PID}
+  integral domain in which every ideal is principal. ie.
+  ie. $\forall I \subset R,~ \exists~ a \in I$ such that $I = (a) = \{ ra ~|~ r \in R \}$.
+\end{defn}
+
+\begin{defn}{nilpotent}
+  $a \in A$ such that $a^n=0$ for some $n>0$.
+\end{defn}
+\begin{defn}{nilrad A}
+  set of all nilpotent elements of $A$; is an ideal of $A$.
+
+  if $nilrad A = 0 ~\Longrightarrow$ $A$ has no nonzero nilpotents.
+
+
+  $$nilrad A = \bigcap_{P \in Spec(A)} P$$
+\end{defn}
+
+\begin{defn}{idempotent}
+  $e \in A$ such that $e^2=e$.
+\end{defn}
+
+\begin{defn}{radical of an ideal}
+  $$rad I = \{ f \in A | f^n \in I~ \text{for some} n \}$$
+
+  $rad I$ is an ideal.
+
+  $nilrad A = rad 0$
+
+  $rad I = \bigcap_{\substack{P \in \operatorname{Spec}(A)\\ P \supset I}} P$
+\end{defn}
+
+\begin{defn}{local ring}
+  A \emph{local ring} has a unique maximal ideal.
+
+  Notation: locall ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$:
+  $$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$
+\end{defn}
 
 \subsection{Lemmas, propositions and corollaries}
 \begin{thm}{AM.1.X}{Zorn's lemma} \label{zorn}
@@ -339,6 +423,7 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi
 \begin{proof}
   By \ref{2.5}: since $\aA M = M$, we have $x M =0$ for some $x \equiv 1 \pmod {Jac(A)}$. (notice that at \ref{2.5} is $\pmod \aA$ but here we use $\pmod {Jac(A)}$, since we have $\aA \subseteq Jac(A)$).
 
+  (recall \ref{1.9}: $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$).\\
   By \ref{1.9}, $x$ is a unit in $A$ (thus $x^{-1}\cdot x=1$).
 
   Hence $M = x^{-1} \cdot \underbrace{x~ \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$.
@@ -381,7 +466,7 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi
   $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$
 
   thus every $y \in \aA M +N$ can be written as
-  $$y=x+n,~~ \text{with} x \in \aA M,~ n\in M$$
+  $$y=x+n,~~ \text{with} x \in \aA M,~ n\in N$$
   which comes from \eqref{eq:2.7.1}.
 
   Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient.
@@ -500,9 +585,8 @@ As in with rings, it is equivalent to say that
 \end{proof}
 
 \begin{lemma}{R.3.4.L}
-  for submodules $M_1 \subset M_2 \subset M$,\\
-
-  $L \cap M_1 = L \cap M_2$ and $\beta(M_1) = \beta(M_2) ~\Longrightarrow~ M_1 = M_2$.
+  for submodules $M_1 \subset M_2 \subset M$,
+  $$L \cap M_1 = L \cap M_2 ~\text{and}~ \beta(M_1) = \beta(M_2) ~\Longrightarrow~ M_1 = M_2$$
 \end{lemma}
 \begin{proof}
   if $m\in M_2$, then $\beta(m) \in \beta(M_1) = \beta(M_2)$, so that there is an $n \in M_1$ such that $\beta(m)=\beta(m)$.