Finding elements of the principal ideal $(10)$.

Question

Let $R=2\mathbb{Z}$ be a commutative ring that does not contain $1$. What elements of $(10)$ are not in $10R$?

$R=2\mathbb{Z}=\{0,\pm 2,\pm4,\dots\}$

So, $10R=\{0,\pm 20,\pm40,\dots\}$

So am I looking for an element in the principal ideal $10$ that does not have an identity? Sorry, need some help clarifying what the question is asking. Been stumped on this problem.

Answer 1

In $\Bbb Z$,

$(10) = 10\Bbb Z = \{\ldots, -30, -20, -10, 0, 10, 20, 30, \ldots \} = \{10z, \; z \in \Bbb Z \}; \tag 1$

that is, $(10) \subset \Bbb Z$ is the set of all integer multiples of $10$; in particular, it contains the odd multiples of $10$, which are absent from $10(2\Bbb Z) = 10R \subset R$.

Answer 2

$(10)$ is the smallest ideal or $R$ which contains $10$, therefore (recall that ideals are additive subgroups) $$(10)=\{0,\pm10,\pm20,\pm30\cdots\}$$

Assuming that the definition is $aR=\{ak\,:\, k\in R\}$, then $$10R=\{0,\pm20,\pm40,\pm60,\cdots\}$$