Let us consider the subset
\[Z:=\{z\in R \mid zr=rz \text{ for any } r\in R\}.\]
(This is called the center of the ring $R$.)
This is a subgroup of the additive group $R$.
In fact, if $z, z’\in Z$, then we have for any $r\in R$,
\begin{align*}
(z-z’)r=zr-z’r=rz-rz’=r(z-z’).
\end{align*}
It follows that $z-z’\in Z$, and thus $Z$ is a subgroup of $R$.
Note that $0, 1 \in Z$, hence $Z$ is not a trivial subgroup.
Thus, we have either $|Z|=p, p^2$ since $R$ is a group of order $p^2$.
If $|Z|=p^2$, then we have $Z=R$.
By definition of $Z$, this implies that $R$ is commutative.
It remains to show that $|Z|\neq p$.
Assume that $|Z|=p$.
Then $R/Z$ is a cyclic group of order $p$.
Let $\alpha$ be a generator of $R/Z$.
Since $Z\neq R$, there exist $r, s\in R$ such that $rs\neq sr$.
Write
\[r=m\alpha+z \text{ and } s=n\alpha+z’\]
for some $m, n\in \Z$, $z, z’\in Z$.
Then we have
\begin{align*}
rs&=(m\alpha+z)(n\alpha+z’)\\
&=(m\alpha)(n\alpha)+m\alpha z’ + n z\alpha +z z’\\
&=(n\alpha)(m\alpha)+m z’ \alpha +n \alpha z +z’ z\\
&=(n\alpha+z’)(m\alpha+z)\\
&=sr.
\end{align*}
This contradicts $rs\neq sr$, and we conclude that $|Z|\neq p$.
Generators of the Augmentation Ideal in a Group Ring
Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by
\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\]
where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring […]
Primary Ideals, Prime Ideals, and Radical Ideals
Let $R$ be a commutative ring with unity. A proper ideal $I$ of $R$ is called primary if whenever $ab \in I$ for $a, b\in R$, then either $a\in I$ or $b^n\in I$ for some positive integer $n$.
(a) Prove that a prime ideal $P$ of $R$ is primary.
(b) If $P$ is a prime ideal and […]
There is Exactly One Ring Homomorphism From the Ring of Integers to Any Ring
Let $\Z$ be the ring of integers and let $R$ be a ring with unity.
Determine all the ring homomorphisms from $\Z$ to $R$.
Definition.
Recall that if $A, B$ are rings with unity then a ring homomorphism $f: A \to B$ is a map […]
Three Equivalent Conditions for a Ring to be a Field
Let $R$ be a ring with $1$. Prove that the following three statements are equivalent.
The ring $R$ is a field.
The only ideals of $R$ are $(0)$ and $R$.
Let $S$ be any ring with $1$. Then any ring homomorphism $f:R \to S$ is injective.
Proof. […]
Ideal Quotient (Colon Ideal) is an Ideal
Let $R$ be a commutative ring. Let $S$ be a subset of $R$ and let $I$ be an ideal of $I$.
We define the subset
\[(I:S):=\{ a \in R \mid aS\subset I\}.\]
Prove that $(I:S)$ is an ideal of $R$. This ideal is called the ideal quotient, or colon ideal.
Proof.
Let $a, […]
Every Integral Domain Artinian Ring is a Field
Let $R$ be a ring with $1$. Suppose that $R$ is an integral domain and an Artinian ring.
Prove that $R$ is a field.
Definition (Artinian ring).
A ring $R$ is called Artinian if it satisfies the defending chain condition on ideals.
That is, whenever we have […]
Equivalent Conditions For a Prime Ideal in a Commutative Ring
Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:
(a) The ideal $P$ is a prime ideal.
(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.
Proof. […]
Nilpotent Element a in a Ring and Unit Element $1-ab$
Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.
Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
We give two proofs.
Proof 1.
Since $a$ […]