629: Principal Ideal of Ring

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of principal ideal of ring

---

Topics

About: ring

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Note

---

Starting Context

• The reader knows a definition of ideal of ring.

---

Target Context

• The reader will have a definition of principal ideal of ring.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the rings }\}$
$p$: $\in R$
$\ast I_l(p)$: $=Rp$, $\in \{\text{ the left ideals of }R\}$
$\ast I_r(p)$: $=pR$, $\in \{\text{ the right ideals of }R\}$
$\ast I_b(p)$: $=\{\sum _{j\in \{1,...,k\}}(p_{l,j}p{p}_{r,j})|k\in \mathbb{N},p_{l,j}\in R\wedge p_{r,j}\in R\}$, $\in \{\text{ the both-sided ideals of }R\}$
//

Conditions:
//

$I_l(p)$ is called "left principal ideal by $p$"; $I_r(p)$ is called "right principal ideal by $p$"; $I_b(p)$ is called "both-sided principal ideal by $p$" or "principal ideal by $p$".

---

2: Natural Language Description

For any ring, $R$, and any element, $p\in R$, $I_l(p):=Rp$ is called "left principal ideal by $p$"; $I_r(p):=pR$ is called "right principal ideal by $p$"; $I_b(p):=\{\sum _{j\in \{1,...,k\}}(p_{l,j}p{p}_{r,j})|k\in \mathbb{N},p_{l,j}\in R\wedge p_{r,j}\in R\}$ is called "both-sided principal ideal by $p$" or "principal ideal by $p$"

---

3: Note

$I_l(p)$ is indeed a left ideal: $p_{1}p+p_{2}p=(p_1+p_2)p\in I_l(p)$; $0=0p\in I_l(p)$; $-(p_{1}p)=(-p_1)p\in I_l(p)$; $RRp=Rp$.

$I_r(p)$ is indeed a right ideal: $p{p}_1+p{p}_2=p(p_1+p_2)\in I_r(p)$; $0=p0\in I_r(p)$; $-(p{p}_1)=p(-p_1)\in I_r(p)$; $pRR=pR$.

$I_b(p)$ is indeed a both-sided ideal: $\sum _{j\in \{1,...,k\}}(p_{l,j}p{p}_{r,j})+\sum _{j\in \{1,...,k^{\prime }\}}(p_{l,j}^{\prime }p{p}_{r,j}^{\prime })\in I_b(p)$; $0=0p0\in I_b(p)$; $-(\sum _{j\in \{1,...,k\}}(p_{l,j}p{p}_{r,j}))=\sum _{j\in \{1,...,k\}}((-p_{l,j})p{p}_{r,j})\in I_r(p)$; $R{I}_b(p)=I_b(p)R=I_b(p)$.

We cannot define $I_b(p)$ as $RpR$, because $p_{l,1}p{p}_{r,1}+p_{l,1}^{\prime }p{p}_{r,1}^{\prime }$ is not necessarily in $RpR$.

When $R$ is commutative, each left principal ideal or right principal ideal is a both-sided principal ideal, because for example, while $I_l(p)\subseteq I_b(p)$ is obvious, for each $\sum _{j\in \{1,...,k\}}(p_{l,j}p{p}_{r,j})$, $=\sum _{j\in \{1,...,k\}}(p_{l,j}{p}_{r,j}p)=\sum _{j\in \{1,...,k\}}(p_{l,j}{p}_{r,j})p\in Rp$, so, $I_b(p)\subseteq I_l(p)$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>