Principal ideal
Dependencies:
Let $R$ be a ring and $a \in R$. Let $aR = \{ar: r \in R\}$. Then $aR$ is a right ideal. Such an ideal is called a principal ideal.
Similarly, $Ra$ is a left ideal.
If $R$ is commutative, $aR = Ra$ is a 2-sided ideal.
Proof
Let $ar_1, ar_2 \in aR$.
- $0 = a0 \in aR \Rightarrow aR \neq \{\}$.
- $ar_1 - ar_2 = a(r_1 - r_2) \in aR$ (by distributivity).
- $(ar_1)(ar_2) = a(r_1ar_2) \in aR$
Therefore, $aR$ is a subring of $R$.
$(ar)x = a(rx) \in aR$. Therefore, $aR$ is a right ideal.
Similarly it can be proved that $Ra$ is a left ideal.
When $R$ is commutative, \[ aR = \{ar: r \in R\} = \{ra: r \in R\} = Ra \] Therefore, $aR$ is a 2-sided ideal.
Dependency for:
- Every ideal of Z is a principal ideal
- F[x]/p(x) is isomorphic to F[x]/p(x)F[x]
- Every ideal in F[x] is principal
Info:
- Depth: 5
- Number of transitive dependencies: 9