Is the ideal generated by $a$ simply $\{ au\mid u\in R, a\in I \}$ ($R$ is a ring)

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More specifically, $R$ is a commutative ring.

I'm trying to understand what the ideal "generated by $a$" is, where $a$ is an element of $R$. I believe this ideal is simply the set $\{a\cdot u\mid u\in R \}$

(I'd like to say the ideal is just all "multiples" of $a$ but am not sure whether that'd be correct).

Can anyone confirm or deny that what I'm thinking is right?

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Yes, that is correct, assuming your ring is commutative with unit. As a consequence, $aR$ is one of the common notations for the ideal generated by $a$, alongside $\langle a\rangle$ and $(a)$. "All multiples of $a$" isn't really wrong to say either, but no one says it because "the ideal generated by $a$" has become conventional and universal.

If $R$ doesn't have a unit, then $aR = \{a\cdot r\mid r\in R\}$ doesn't necessarily contain $a$, so it is not the ideal you're looking for. For instance, in the ring $R = 2\Bbb Z$ of even integers, "all multiples of $2$" will be the ring $4\Bbb Z$, which doesn't contain $2$.