If $I$ is a right ideal of a ring $R$ and $e \in I$ is idempotent, then $I=eI$?

Question

Restatement of question: If $I$ is a right ideal of a ring $R$ and $e \in I$ is idempotent, then $I=eI$? I seen this result in a proof and am attempting to verify it.

To prove the equality, it is easy to see that $eI \subseteq I$, since $eI \subseteq eR \subseteq I$, but I cannot figure out why $I \subseteq eI$. It is probably really obvious but the solution is evading me. Any help is appreciated!

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Edit: As mentioned in the comments and answers, I made the mistake of not realizing that the idempotent in question has additional properties. i.e., if $R$ is a unital ring and a direct sum of right ideals $I$ and $J$, then $1_{R}=e+f$ for some $e \in I$ and $f \in J$. Then $\forall a \in I$, $a=1_{R}a=(e+f)a=ea+fa$. Then $a-ea=fa \in I \cap J=0$, so $a=ea$ and thus $I=eI$. $e$ is idempotent because $e^{2}=(1_{R}-f)e=e-fe=e$ similarly.

Answer 1

This is a reiteration of the edit I included in the question:

I made the mistake of not realizing that the idempotent in question has additional properties. i.e., if $R$ is a unital ring and a direct sum of right ideals $I$ and $J$, then $1_{R}=e+f$ for some $e \in I$ and $f \in J$. Then $\forall a \in I$, $a=1_{R}a=(e+f)a=ea+fa$. Then $a-ea=fa \in I \cap J=0$, so $a=ea$ and thus $I=eI$. $e$ is idempotent because $e^{2}=(1_{R}-f)e=e-fe=e$ similarly.

Answer 2

This is clearly not true in general: the element $0$ is idempotent in any ring, but $0I \neq I$ for any nonzero ideal $I$.

However, in your comment you say that in the course of the proof you're reading, it's shown that there is an idempotent element $e$ such that $I = eR$. In that case, $eI = I$. You've already observed that $eI\subseteq I$, and for the other direction, if $x\in I$, then $x\in eR$, so $x = ey$ for some $y\in R$, hence $ex = eey = ey = x$, and $x\in eI$.