Set of Ring Elements forming Zero Product with given Element is Ideal

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Theorem

Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.

Let $a \in R$ be an arbitrary element of $R$.

Let $A$ be the subset of $R$ defined as:

$A = \set {x \in R: x \circ a = 0_R}$


Then $A$ is an ideal of $A$.


Proof

By definition of ring zero:

$\forall x \in R: x \circ 0_R = 0_R$

Hence $0_R \in A$ and so $A \ne \O$.


Let $a, b \in A$.

\(\ds \forall x \in R: \, \) \(\ds x \circ b\) \(=\) \(\ds 0_R\)
\(\ds \leadsto \ \ \) \(\ds -\paren {x \circ b}\) \(=\) \(\ds 0_R\)
\(\ds \leadsto \ \ \) \(\ds x \circ \paren {-b}\) \(=\) \(\ds 0_R\)


Thus:

\(\ds \forall x \in R: \, \) \(\ds x \circ a\) \(=\) \(\ds 0_R\)
\(\, \ds \land \, \) \(\ds x \circ \paren {-b}\) \(=\) \(\ds 0_R\)
\(\ds \leadsto \ \ \) \(\ds \paren {x \circ a} + \paren {x \circ \paren {-b} }\) \(=\) \(\ds 0_R\) as $0_R$ is identity of $\struct {R, +}$
\(\ds \leadsto \ \ \) \(\ds x \circ \paren {a + \paren {-b} }\) \(=\) \(\ds 0_R\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \leadsto \ \ \) \(\ds a + \paren {-b}\) \(\in\) \(\ds A\) Definition of $A$


Then:

\(\ds \forall x \in R: \, \) \(\ds x \circ a\) \(=\) \(\ds 0_R\)
\(\, \ds \land \, \) \(\ds x \circ b\) \(=\) \(\ds 0_R\)
\(\ds \leadsto \ \ \) \(\ds \paren {x \circ a} \circ b\) \(=\) \(\ds 0_R\) Definition of Ring Zero
\(\ds \leadsto \ \ \) \(\ds x \circ \paren {a \circ b}\) \(=\) \(\ds 0_R\) Ring Axiom $\text M1$: Associativity of Product
\(\ds \leadsto \ \ \) \(\ds a \circ b\) \(\in\) \(\ds A\) Definition of $A$

Hence the result, from Test for Ideal:

$\blacksquare$


Sources

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