Ring by Idempotent
Theorem
Let $\struct {A, +, \circ}$ be a commutative ring.
Let $e$ be an idempotent element of $A$.
Then the ideal $I := \ideal e$ generated by $e$ is a commutative ring with unity $\struct {I, +, \circ}$ with unity $e$.
Proof
Because $\struct {I, +}$ is an ideal of $\struct {A, +, \circ}$, it follows that $\struct {I, +, \circ}$ is a ring (not necessarily unital).
This article, or a section of it, needs explaining. In particular: Why is $\struct {I, +}$ an ideal? By definition it is the ideal generated by $e$. --Wandynsky (talk) 13:04, 30 July 2021 (UTC) I don't see this anywhere in the Definition:Generator of Ideal of Ring or Definition:Generated Ideal of Ring.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
By definition of generated ideal, $I$ is the intersection of all ideals of $A$ containing $e$.
Thus $e \mathop \in I$.
This article, or a section of it, needs explaining. In particular: No not "clearly", we need to link to a result. I'm sure it's there, somewhere under Cosets. Ah i produced some confusion with the notation. Here $eA$ is defined as $\ideal e$. I don't want to use any cosets. There should be some theorem like $(e) = eA$, where the RHS are the actual cosets. But I can't find it. --Wandynsky (talk) 13:26, 30 July 2021 (UTC) There is certainly a page which gives $e \in I$, for a start. As for a page containing $\ideal e = e A$, there is no such page. Thank you. Now the only things left are refactoring Definition:Generator of Ideal, proving it is an ideal (above) and proving the general form of an element of $I$ (below). --Wandynsky (talk) 15:59, 30 July 2021 (UTC)You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Let $e a \in I$ for some arbitrary $a \in A$.
This article, or a section of it, needs explaining. In particular: theorem on the form of a general element of $I$ --Wandynsky (talk) 15:50, 30 July 2021 (UTC) You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Then
\(\ds e \paren {e a}\) | \(=\) | \(\ds e^2 a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e a\) |
It follows that $e$ is a unity of $\struct {I, +, \circ}$.
Hence $\struct {I, +, \circ}$ is a commutative ring with unity with unity $e$.
$\blacksquare$