I am reading p. 20 of Introduction to Commutative Algebra by Atiyah and Macdonald. There there is a module decomposition
$$ A=\mathfrak{a}_1\oplus\cdots\oplus\mathfrak{a}_n $$
of a commutative ring $A$. Then the last sentence on the page says "The identity element $e_i$ of $\mathfrak{a}_i$ is an idempotent in $A$, and $\mathfrak{a}_i=(e_i)$." But do ideals have identity elements?
More context is important here. From the beginning of the paragraph, $A = \prod A_i$ is a product of rings, and the ideals $\mathfrak a_i$ are the image of the non-unit preserving maps $A_i \to A$ sending $a_i \mapsto (0,0,\ldots, 0,a_i,0,\ldots,0,0)$. This map identifies $\mathfrak a_i \simeq A_i$ as sets, so $e_i$ just means the element corresponding to the identity of $A_i$.
No ideals do not have identity elements, generally speaking they are non-unital rings. Even if they do have a unit, they are not considered subrings of the ring they are an ideal for (because they have a different unit).
You forgot the sentence before that quote:
Each $\mathfrak a_i$ is a ring (isomorphic to $A/\mathfrak b_i$).
(where $\mathfrak b_i = \bigoplus_{j \ne i} \mathfrak a_j$)
