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This is a pretty basic question about principal ideals - on page 197 of Katznelson's A (Terse) Introduction to Linear Algebra, it says:

Assume that $\mathcal{R}$ has an identity element. For $g\in \mathcal{R}$, the set $I_g = \{ag:a\in\mathcal{R}\}$ is a left ideal in $\mathcal{R}$, and is clearly the smallest (left) ideal that contains $g$.

Ideals of the form $I_g$ are called principal left ideals...

(Note $\mathcal{R}$ is a ring).

Why is the assumption that $\mathcal{R}$ has an identity element important?

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Because if $\mathcal{R}$ has an identity, then $I_{g}$ is the smallest left ideal containing $g$. Without an identity, it might be that $g \notin I_{g}$.

For instance if $\mathcal{R} = 2 \mathbf{Z}$, then $I_{2} =\{a \cdot 2:a\in 2 \mathbf{Z} \} = 4 \mathbf{Z}$ does not contain $2$.

(Thanks Cocopuffs for pointing out an earlier mistake.)

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Rings without identity (sometimes called rngs) do not behave as rings with identity (for instance, they may not contain maximal ideals), and many theorems that hold for rings with identity don't hold for rngs.

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  • $\begingroup$ I thought a ring without i was a rong. $\endgroup$ – Stephen Mar 25 '13 at 1:58
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If $R$ is a rng, the left ideal generated by an element $g$ (i.e. the smallest left ideal containing $g$) is given by the $\mathbb{Z}$-span of $\{g\} \cup \{rg : r \in R\}$. Thus, elements of $\langle g \rangle$ have the form $z \cdot g + r \cdot g$ for $z \in \mathbb{Z}$ and $r \in R$.

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